Joseph Rotman
GADS THEORY SECOND EDITION
Springer
Universitext Editorial Board (North America):
S. Axler F.W. Gehr...
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Joseph Rotman
GADS THEORY SECOND EDITION
Springer
Universitext Editorial Board (North America):
S. Axler F.W. Gehring K.A. Ribet
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Universitext Editors (North America): S. Axler, F.W. Gehring, and K.A. Ribet Aksoy/Khamsi: Nonstandard Methods in Fixed Point Theory Andersson: Topics in Complex Analysis Aupetit: A Primer on Spectral Theory Berberian: Fundamentals of Real Analysis Booss/Bleecker: Topology and Analysis Borkar: Probability Theory  An Advanced Course Carleson/Gamelin: Complex Dynamics Cecil: Lie Sphere Geometry: With Applications to Submanifolds Chae: Lebesgue Integration (2nd ed.) Charlap: Bieberbach Groups and Flat Manifolds Chern: Complex Manifolds Without Potential Theory Cohn: A Classical Invitation to Algebraic Numbers and Class Fields Curtis: Abstract Linear Algebra Curtis: Matrix Groups DiBenedetto: Degenerate Parabolic Equations Dimca: Singularities and Topology of Hypersurfaces Edwards: A Formal Background to Mathematics I a/b Edwards: A Formal Background to Mathematics H afb FouIds: Graph Theory Applications Friedman: Algebraic Surfaces and Holomorphic Vector Bundles Fuhrmann: A Polynomial Approach to Linear Algebra Gardiner: A First Course in Group Theory Garding/Tambour: Algebra for Computer Science Goldblatt: Orthogonality and Spacetime Geometry Gustafson/Rao: Numerical Range  The Field of Values of Linear Operatol and Matrices Hahn: Quadratic Algebras, Clifford Algebras, and Arithmetic Witt Groups Holmgren: A First Course in Discrete Dynamical Systems Howe/Tan: NonAbelian Harmonic Analysis: Applications of SL(2, R) Howes: Modern Analysis and Topology Humi/Miller: Second Course in Ordinary Differential Equations Hurwitz/Kritikos: Lectures on Number Theory Jennings: Modern Geometry with Applications Jones/Morris/Pearson: Abstract Algebra and Famous Impossibilities Kannan/Krueger: Advanced Analysis Kelly/Matthews: The NonEuclidean Hyperbolic Plane Kostrikin: Introduction to Algebra Luecking/Rubel: Complex Analysis. A Functional Analysis Approach MacLane/Moerdijk: Sheaves in Geometry and Logic Marcus: Number Fields McCarthy: Introduction to Arithmetical Functions Meyer: Essential Mathematics for Applied Fields Mines/Richman/Ruitenburg: A Course in Constructive Algebra Moise: Introductory Problems Course in Analysis and Topology Morris: Introduction to Game Theory Polster: A Geometrical Picture Book Porter/Woods: Extensions and Absolutes of Hausdorff Spaces Ramsay/Richtmyer: Introduction to Hyperbolic Geometry Reisel: Elementary Theory of Metric Spaces Rickart: Natural Function Algebras
Joseph Rotman
Galois Theory Second Edition
Springer
Joseph Rotman Department of Mathematics University of Illinois at UrbanaChampaign Urbana, IL 61801 USA Editorial Board (North America):
S. Axler Mathematics Department San Francisco State University San Francisco, CA 94132 USA
F.W. Gehring Mathematics Department East Hall University of Michigan Ann Arbor, MI 48109 USA
K.A. Ribet Department of Mathematics University of California at Berkeley Berkeley, CA 947203840 USA Mathematics Subject Classification (1991): 1201, 12F10 With 9 Figures Library of Congress CataloginginPublication Data Rotman, Joseph J., 1934– Galois theory / Joseph Rotman. — 2nd ed. cm. — (Universitext) p. Includes bibliographical references (p. – ISBN 0387985417 (softcover : alk. paper) 1. Galois theory. I. Title. QA2I4.R685 1998 512'.3—dc21
) and index.
983967
Printed on acidfree paper. CD 1998, 1990 SpringerVerlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (SpringerVerlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names. trademarks, etc.. in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Jenny Wolkowicki; manufacturing supervised by Jacqui Ashri. Cameraready copy prepared from the author's AmS 1X files. Printed and bound by BraunBrumfield, Inc., Ann Arbor, Ml. Printed in the United States of America. 
987654321 ISBN 0387985417 SpringerVerlag New York Berlin Heidelberg SPIN 10679225
To my teacher Irving Kaplansky
Preface to the Second Edition
There are too many errors in the first edition, and so a "corrected nth printing" would have been appropriate. However, given the opportunity to make changes, I felt that a second edition would give me the flexibility to change any portion of the text that I felt I could improve. The first edition aimed to give a geodesic path to the Fundamental Theorem of Galois Theory, and I still think its brevity is valuable. Alas, the book is now a bit longer, but I feel that the changes are worthwhile. I began by rewriting almost all the text, trying to make proofs clearer, and often giving more details than before. Since many students find the road to the Fundamental Theorem an intricate one, the book now begins with a short section on symmetry groups of polygons in the plane; an analogy of polygons and their symmetry groups with polynomials and their Galois groups can serve as a guide by helping readers organize the various definitions and constructions. The exposition has been reorganized so that the discussion of solvability by radicals now appears later; this makes the proof of the AbelRuffini theorem easier to digest. I have also included several theorems not in the first edition. For example, the Casus Irreducibilis is now proved, in keeping with a historical interest lurking in these pages. I am indebted to Gareth Jones at the University of Southampton who, after having taught a course with the first edition as text, sent me a detailed list of errata along with perspicacious comments and suggestions. I also thank Evan Houston, Adam Lewenberg, and Jack Shamash who made valuable comments as well. This new edition owes much to the generosity of these readers, and I am grateful to them. Joseph Rotman Urbana, Illinois, 1998
Preface to the First Edition
This little book is designed to teach the basic results of Galois theory fundamental theorem; insolvability of the quintic; characterization of polynomials solvable by radicals; applications; Galois groups of polynomials of low degree—efficiently and lucidly. It is assumed that the reader has had introductory courses in linear algebra (the idea of the dimension of a vector space over an arbitrary field of scalars should be familiar) and "abstract algebra" (that is, a first course which mentions rings, groups, and homomorphisms). In spite of this, a discussion of commutative rings, starting from the definition, begins the text. This account is written in the spirit of a review of things past, and so, even though it is complete, it may be too rapid for one who has not seen any of it before. The high number of exercises accompanying this material permits a quicker exposition of it. When I teach this course, I usually begin with a leisurely account of group theory, also from the definition, which includes some theorems and examples that are not needed for this text. Here I have decided to relegate needed results of group theory to appendices: a glossary of terms; proofs of theorems. I have chosen this organization of the text to emphasize the fact that polynomials and fields are the natural setting, and that groups are called in to help. A thorough discussion of field theory would have delayed the journey to Galois's Great Theorem. Therefore, some important topics receive only a passing nod (separability, cyclotomic polynomials, norms, infinite extensions, symmetric functions) and some are snubbed altogether (algebraic closure, transcendence degree, resultants, traces, normal bases, Kummer theory). My belief is that these subjects should be pursued only after the reader has digested the basics. My favorite expositions of Galois theory are those of E. Artin, Kaplansky, and van der Waerden, and I owe much to them. For the appendix on
X
PREFACE TO THE FIRST EDITION
"oldfashioned Galois theory," I relied on recent accounts, especially [Edwards], [Gaal], [Tignol], and [van der Waerden, 1985], and older books, especially [Dehn] (and [Burnside and Panton], [Dickson], and [Netto]). I thank my colleagues at the University of Illinois, Urbana, who, over the years, have clarified obscurities; I also thank Peter Braunfeld for suggestions that improved Appendix C and Peter M. Neumann for his learned comments on Appendix D. I hope that this monograph will make both the learning and the teaching of Galois theory enjoyable, and that others will be as taken by its beauty as I am. Joseph Rotman Urbana, Illinois, 1990
To the Reader
Regard the exercises as part of the text; read their statements and do attempt to solve them all. A result labeled Theorem 1 is the first theorem in the text; Theorem G1 is the first theorem in the appendix on group theory; Theorem R1 is the first theorem in the appendix on rulercompass constructions; Theorem H1 is the first theorem in the appendix on history.
Contents
Preface to the Second Edition Preface to the First Edition To the Reader
vii ix xi
Symmetry
1
Rings
7
Domains and Fields
13
Homomorphisms and Ideals
17
Quotient Rings
21
Polynomial Rings over Fields
24
Prime Ideals and Maximal Ideals
31
Irreducible Polynomials
38
Classical Formulas
44
Splitting Fields
50
The Galois Group
59
Roots of Unity
63
Solvability by Radicals
71
Independence of Characters
76
Galois Extensions
79
The Fundamental Theorem of Galois Theory
83
XiV
CONTENTS
Applications
85
Galois's Great Theorem
90
Discriminants
95
Galois Groups of Quadratics. Cubics, and Quartics
100
Epilogue
107
Appendix A: Group Theory Dictionary
109
Appendix B: Group Theory Used in the Text
112
Appendix C: RulerCompass Constructions
129
Appendix D: Oldfashioned Galois Theory
138
References
151
Index
153
Galois Theory
Galois theory is the interplay between polynomials, fields, and groups. The quadratic formula giving the roots of a quadratic polynomial was essentially known by the Babylonians. By the middle of the sixteenth century, the cubic and quartic formulas were known. Almost three hundred years later, Abel (1824) proved, using ideas of Lagrange and Cauchy, that there is no analogous formula (involving only algebraic operations on the coefficients of the polynomial) giving the roots of a quintic polynomial (actually Ruffini (1799) outlined a proof of the same result, but his proof had gaps and it was not accepted by his contemporaries). In 1829, Abel gave a sufficient condition that a polynomial (of any degree) have such a formula for its roots (this theorem is the reason that, nowadays, commutative groups are called abelian). Shortly thereafter, Galois (1831) invented groups, associated a group to each polynomial, and used properties of this group to give, for any polynomial, a necessary and sufficient condition that there be a formula of the desired kind for its roots, thereby completely settling the problem. We prove these theorems here.
Symmetry Although Galois invented groups because he needed them to describe the behavior of polynomials, we realize today that groups are the precise way to describe symmetry. The Greek roots of the word symmetry mean, roughly, measuring at the same time. In ordinary parlance, there are at least two meanings of the word, both involving an arrangement of parts somehow balanced with respect to the whole and to each other. One of these meanings attributes an aesthetic quality to the arrangement, implying that sym
2
GALOIS THEORY
metry is harmonious and wellproportioned. This usage is common in many discussions of art, and one sees it in some mathematics books as well (e.g., Weyl's Symmetry). Here, however, we focus on arrangements without considering, for example, whether a square is more pleasing to the eye than a rectangle. Before giving a formal definition of symmetry, we first consider mirror images.
Figure 1 Let F denote the figure pictured in Figure 1. If one regards the line AB as a mirror, then the left half of F is the reflection of the right half. This figure is an example of bilateral symmetry: each point P on one side of AB corresponds to a point P' (its mirror image) on the other side of AB; for example, C' corresponds to C and D' corresponds to D. We can describe this symmetry in another way. Regard the plane 1R 2 as a flat transparent surface in space, having F (without the letters) drawn on it. Imagine turning over this surface by flipping it around the axis AB. If one's eyes were closed before the flip and then reopened after it, one could not know, merely by looking at F in its new position, whether the flip had occurred. Indeed, if F lies in the plane so that AB lies on the yaxis and CC' lies on the xaxis, then the linear transformation r : 1R 2 4. R2 , defined by (x, y) i— (— x , y) and called a reflection, carries the figure into itself; that is,
r(F) = F. On the other hand, if T is some scalene triangle in the plane (say, with its center at the origin), then it is easy to see that there are points P in T whose mirror images P' = r(P) do not lie in T; that is, r(T) 0 T. Another type of symmetry is rotational symmetry. Picture an equilateral triangle A in the plane with its center at the origin. A (counterclockwise)
SYMMETRY
3
rotation p by 120 0 carries A into itself; if one's eyes were closed before p takes place and then reopened, one could not detect that a motion had occurred. B
C
A
AA Before
A B
After
C
Figure 2
If we identify the plane with the complex numbers C, then the rotation p : C > C can be described by p : re ie 1> rei (9 +2713) , and p(A) = A.
Definition. A linear transformation a : 1R 2 * R 2 is called orthogonal if it is distance preserving; that is, if I U  VI denotes the distance between points U and V, then la(U)  o (V)I = IU  V. There are distance preserving functions that are not linear transformations; for example, a translation is defined by (x, y) i  . (x ± a, y ± b) for fixed numbers a and b; geometrically, this translation sends any vector (x, y) into (x, y) + (a, b). (It is a theorem that every distance preserving function is a composite of reflections, rotations, and translations and, if it fixes the origin, then it is a composite of reflections and rotations alone.) It can be shown that every orthogonal transformation a is a bijection, 1 1 exists; moreover, one can prove that a'sothainverfuc is also orthogonal. The set 0 (2, IR) of all orthogonal transformations is a group under composition, called the real orthogonal group. injection (one also says that f is onetoone) if distinct points have distinct images; that is, if x 0 x', then f (x) 0 f (x'); the contrapositive, f (x) = f (x') implies x = x', is often the more useful statement. A function f is a surjection (one also says f is onto) if, for each y E Y, there exists x E X with f (x) = y. A function f is a Injection (one also says f is a onetoone correspondence) if it is both an injection and a surjection. Finally, a function f : X > Y is a bijection if and only if it has an inverse; that is, there is a function g : Y > X with both composites gf and f g identity functions. 1 A function f : X > Y is an
4
GALOIS THEORY
Lemma 1. Every orthogonal transformation a preserves angles: if A, V and B are points, then Z AV 8 = Z A' V' B', where A' = a (A), V' = and B' = a (B).
Proof. We begin by proving the special case when V is the origin 0. First, identify a point X with the vector starting at 0 and ending at X. Recall the formula relating lengths and dot product: I X1 2 = (X, X), so that IA — BI 2 = (A — B, A — B) = IAI 2 — 2(A, B) ± IBI 2 There is a similar equation for A' and B'. Since, by hypothesis, IA' — B'l = IA — BI, !NI = IAI, and IB'I = IBI, it follows that (A', B') = (A, B). But (A, B) = IAIIBI cos 0, where 0 = LA08. Therefore, LAO B = LA' 0 B' . But 0' = a(0) = 0, because a is a linear transformation, and so LA' 0 B' = LA' 0' 8', as desired. Now consider LAVB, where V need not be the origin 0. If r : W iW — V is the translation taking V to the origin, and if r' : W 1 W + a (V) is the translation taking the origin to a (V) = V', then the composite ea r takes W 1—> W — I 1 1—> cr(W — V) = a (W) — cr(V ) 1—> a(W) — a (V) + a (V) = a (W).
Thus, a (W) = ear(W) for all W, so that a = r'ar. Since the translations r and e preserve all angles, not merely those with vertex at the origin, the composite preserves LA VB. • The following definition, a common generalization of reflections and rotations, should now seem natural. Definition. Given a figure F in the plane, 2 its symmetry group E(F) is the family of all orthogonal transformations a : R 2 —> R2 for which o (F) = F.
The elements of E(F) are called symmetries. 2 It is clear that these definitions can be generalized: for every n > 1, there is an
n —dimensional real orthogonal group 0(n, R) consisting of all the distance preserving linear transformations of Rn , and symmetry groups of figures in higher dimensional euclidean space are defined as for planar figures.
SYMMETRY
5
It is easy to prove that the symmetry group is a subgroup of the orthogonal group, and so it is a group in its own right. The wonderful idea of Galois was to associate to each polynomial f (x) a group, nowadays called its Galois group, whose properties reflect the behavior of f(x). Our aim in this section is to set up an analogy between the symmetry group of a polygon and the Galois group of a polynomial. Since our major interest is the Galois group, we merely state the fact that if a is orthogonal and if U and V are points, then the image of the line segment U V is also a line segment, namely, U' V', where U' = a (U) and V' = a (V). (The basic idea of the proof is a sharp form of the triangle inequality: if W is a point on the line segment UV, then 1U WI +1W VI = 1UVI, while if W V UV, then 1UW1+1WVI > 1UVI.) Lemma 2. If P is a polygon, then every orthogonal transformation a E (P) permutes Vert(P), the set of vertices of P.
E
Proof. Let V be a vertex of P; if M, V, and N are consecutive vertices, then LM V N 0 180 0 . If V' = a (V), then either V' lies on the perimeter of P or it lies in the interior of P.
Figure 3 In the first case, Lemma 1 gives LMVN = LM'V'N'. But if V' is not a vertex, then LM'V'N' = 180°, a contradiction. Therefore, V' must be a vertex in this case. In the second case, V' lies inside of P, and so there is a (2dimensional) disk D with center V' lying wholly inside of P. Since a (P) = P. every point in D lies in the image of a. Now a 1 is also an orthogonal transformation, and a 1 (V') = V. In the disk D, every angle between 0° and 360° arises as LJV'K for some points J and K in D. Now a 1 (LJV'K) = ZPV K' for some points J' and K' in P. But the only such angles satisfy
0 < LIT K' < ZMV K.
6
GALOIS THEORY
Therefore, there are angles that the orthogonal transformation a 1 does not preserve, and this is a contradiction. We conclude, for every vertex V, that a (V) is also a vertex; that is, the restriction a l of a maps Vert(P) to itself. Since a is an injection, so is its restriction ai ; since Vert(P) is finite, a l must also be a bijection. Thus, if Vert(P) = I VI, ... , VO, then (VI , ... , Vn I = fa(Vi), ... , a(V)1 = tai(Vi), ... ,ai(Vn)}, and so al is a permutation of Vert(P). • Theorem 3. If P is a polygon with n vertices Vert(P) = (VI, ... , Vn 1, then E(P) is isomorphic to a subgroup of the symmetric group S n . A
B
A
A
AA/ CB CB C Figure 4
Proof. If a c E(P), denote its restriction to Vert(P) by a 1 . By the lemma, al is a permutation of Vert( P); that is, a l E Svert(P). It follows that the assignment a 1› a l is a well defined function f: E(P) —> Sver(p). To see that f is a homomorphism, suppose that a, r c E(P). It is easy to see that if V E Vert(P), then (a r) 1 and a1 r1 both have the same value on V, namely, a (r (V)). Therefore, (0 r) 1 = a1 r1, and so f is a homomorphism: f(ar) = f (a)i(r).
Finally, f is an injection, i.e., ker f = 1, for if f (a) = a l = 1, then a fixes every vertex V E Vert(P). But regarding the vertices as vectors in R 2 , there are two such that are linearly independent (neither is a scalar multiple of the other), and so these two vectors comprise a basis of R 2 . Since a is a linear transformation fixing a basis of 1ft 2, it must be the identity. Therefore, f is an isomorphism between E(P) and a subgroup of SveroP)''' g Spa . • Corollary 4. Let A be a triangle with vertices A, B, and C. If A is equi2 S3; ifA is only isosceles, then E(A)"' Z2; if A is lateral, then E(A) z' scalene, then E(A) has order 1.
RINGS
7
Proof. By the Theorem, E (A) is isomorphic to a subgroup of S3. If A is equilateral, then we can exhibit 6 symmetries of it: the reflections about any of the 3 altitudes and the rotations of 0 0 , 120° and 240 0 . Since I S3 I = 6, it follows that E (A)"L" S3. If A is isosceles, say, IA CI = I AB I, then the reflection about the altitude through A is in E (A). This is the only nonidentity symmetry, for every symmetry a must fix A because the angle at A is different than the angles at B and C (lest A be equilateral). Thus, Z2. Finally, if A is scalene, then any symmetry fixes all the vertices, for no two angles are the same, and hence it is the identity. • We shall see later that the Galois group of a polynomial having n distinct roots is also isomorphic to a subgroup of S. Moreover, there may be permutations of the roots that do not arise from the Galois group, just as there may be permutations of the vertices that do not arise from symmetries; for example, in Corollary 4 we saw that only two of the six permutations of the vertices of an isosceles triangle arise from symmetries. Exercises
1.
(i) If F is a square, prove that E(F) . D8, the dihedral group of order 8. (ii) If F is a rectangle that is not a square, prove that E(F) a' V. where V denotes the 4group (V " Z2 X Z2).
(iii) Give an example of quadrilaterals Q and Q' with E(Q) ''= Z2 and 1 (V) = I.
2. A polygon is regular if all the angles at its vertices are equal. Prove that a polygon P is regular if and only if E (P) acts transitively on Vert(P).
3. Prove that if Pn is a regular polygon with n vertices, then E(P) where D2n is the dihedral group of order 2n.
4. Prove that if F is a circular disk, then E(F) is infinite.
Rings The algebraic system encompassing fields and polynomials is a commutative ring with 1. We assume that the reader has, at some time, heard the
8
GALOIS THEORY
words group, ring, and homomorphism; our discussion is, therefore, not leisurely, but it is complete.
Definition. A commutative ring with 1 is a set R equipped with two binary operations, addition: (r, r') 1> r + r' and multiplication: (r, r') i> rr 1, such that: (i) R is an abelian group under addition; (ii) multiplication is commutative and associative; (iii) there is an element 1 E R with 1 0 0 and 1r = r
for all r E R;
(iv) the distributive law holds: r(s + t) = rs + rt
for all r, s,t
E
R.
The additive group of R is the abelian group obtained from it by forgetting its multiplication. From now on, we will write ring instead of "commutative ring with 1."
Example 1. The most familiar rings are Z (the integers), Q (the rational numbers), R (the real numbers), and C (the complex numbers); each of them is equipped with the usual addition and multiplication. Example 2. For a fixed positive integer n, define the ring Z n of integers modulo n as follows. Its elements are the subsets of Z [a] = [nzeZ:m=amodn} = {mEZ:m=a+kn for some k E Z}, where a E Z ([a] is called the congruence class of a mod n). Addition and multiplication are given by [a] ± [b] = [a + b]
and
[a][b] = [ab],
and [1] is "one." It is routine to check that addition and multiplication are well defined (that is, if a: _—_ a' mod n and b _ b' mod n, then a + b ma' + b' mod n and ab 7,7_ a' b' mod n; i.e., [a] + [b] = [al + [b] and [a][b] = [al[b1), and that 74 is a ring under these operations.
RINGS
9
Recall that Z n has exactly n elements, namely Zn = 1 [0], [1], .. . , [n  1] }, for if a E Z, the division algorithm provides a quotient q and a remainder r with a = qn ± r, where 0 < r < n; it follows that a r mod n, and so [a] = [r]. One can also prove that the congruence classes [r] for r in the indicated range are all distinct. It is a common practice, when working within Z n , to eliminate the brackets from the notation. In Z3, for example, it is correct to write 2 + 2 = 1. Example 3. If R is a ring, define a polynomial f (x) with coefficients in R (briefly, a polynomial over R) to be a sequence
f (x) = (co, ci, ... , cn , 0, 0, ... ) with ci E R for all i and ci = 0 for all i > n. If g(x) = (do , d i , . ..) is another polynomial over R, it follows that f (x) = g(x) if and only if ci = di for all i. Denote the set of all such polynomials by R[..x], and define addition and multiplication on R[x] as follows: (co, ci, • • • ) + (do, di, ) = (co + do, ci + d1, . . . ) and (C o , cl
, • • • )(do, d1, • • • ) = (eo, e 1 , • • • ), where eo = codo, el = codi ± cid°, and, in general, ek = E cid], the summation being over all i, j with i + j = k. Define the zero polynomial to be (0, 0, . . . ), and denote it by 0; similarly, denote (1, 0, 0, . . . ) by 1 (there are now two meanings for these symbols). It is routine but tedious to verify that R[x] is a ring, the polynomial ring over R. What is the significance of the letter x in the notation 1(x)? Let x denote the specific element of R[x]:
x= (0, 1, 0, 0, . .. ). It is easy to prove that x 2 = (0, 0, 1, 0, 0, .. . ) and, by induction, that xi is the sequence having 0 everywhere except for 1 in the ith spot. We now recapture the usual notation: f (x) = (co , ci, . . . , cn , 0, 0, . . .) = (co, 0, 0,  ) + (0, ci , 0, 0,  ) ± (0,0, c2, 0, 0,  ) +   = co (1, 0, 0,  ) ± ci(0, 1,0,0, ... ) ± c2(0, 0, 1,0, . • • ) + • • • = co + cix + • •  ± cnxn
10
GALOIS THEORY
We have written co = co1 after identifying co with (co, 0, 0, ... ) in R[x]. Notice that x is an honest element of a ring and not a variable; its role as a variable, however, will be given when we discuss polynomial functions. We remind the reader of the usual vocabulary associated with f (x) = co + c ix + • • • + cn xn . If f (x) is not the zero polynomial, its leading coefficient is cn , where n is the largest integer with cn 0 0; one calls n the degree and denotes it by a(f) [n is the highest exponent of x occurring in f (x)]. A monic polynomial is one whose leading coefficient is 1. The zero polynomial 0 = (0, 0, . . . ) does not have a degree, for it has no leading coefficient. The constant term of f (x) is co ; a constant (polynomial) is either the zero polynomial 0 or a polynomial of degree 0; linear, quadratic, cubic, quartic (or biquadratic), and quintic polynomials have degrees, respectively, 1, 2, 3, 4, and 5. Definition. Let f (x) = E ci xi be a polynomial over a ring R. A root of f (x) in R is an element a E R such that Co ± c la + • • • + cn an = O.
Remark. The polynomial f (x) = x 2 — 2 is a polynomial over Q, but we usually say that ..,h is a root of f (x) even though Nh is irrational. We will soon modify the definition of root of a polynomial f (x) over R to allow roots to lie in some ring larger than R. Recall from linear algebra that a linear homogeneous system over a field with r equations in n unknowns has a nontrivial solution if r < n; if r = n, one must examine a determinant. If f (x) = (x —al) . . . (x —an ) = E cixi, then it is easy to see, by induction on n, that Cn1 Cn2 Cn _3
=
—Eai i = Eceia, i,J = — E aia jak i<j 5.
Theorem 5. Let R be a ring. (i) The "one" in R is unique. (ii) 0 • r = 0 for every r E R; (iii) If — r is the additive inverse of r
E
R, that is, —r I r = 0, then
—r = (1)r; (iv) (1)(—r) = r for every r E R [in particular, (1)(1) = 1].
Proof. (i) Suppose that e E R satisfies er = r for all r E R. In particular, when r = 1, we have el = 1. But the defining property of 1 gives el = e, and so e = 1. (ii) The distributive law gives 0•r=(0+0)r=0•rd0•r, and subtracting 0 • r from both sides gives 0 • r = 0. (iii) 0 = 0. r = (1 I 1)r = (1)r ± r; now add —r to both sides of the equation. (iv)
0 = 0 • (—r)
Now add r to both sides.
=
(1 + 1)(—r)
=
(1)(—r) — r.
•
Suppose we do not insist, in the definition of ring, that 1 0 0. If R is a "ring" in which 1 = 0 and if r E R, then
r = lr = 0  r = 0; hence R consists of exactly one element, namely, 0. This algebraic system is not very interesting, and so we do not consider it as a bona fide ring. We can now see, in any ring R, why "dividing by zero" is forbidden. If a, b E R, then alb, should it exist, is an element of R such that
b(alb) = a; after all, dividing by b is the operation inverse to multiplying by b. In particular, if 1/0 exists, then it is an element of R with 0 • (1/0) = 1. But 0 • (1/0) = 0, by Theorem 5(i), and this forces 1 = 0, contrary to the inequality in the definition of ring.
12
GALOIS THEORY
Definition. A subring of a ring R is a subset S of R which contains 1 and which is closed under subtraction and multiplication.
For example, Z is a subring of Q which, in turn, is a subring of I', which is a subring of C. If R is a ring, then R' = ((r, 0, 0, ...):r E RI is easily seen to be a subring of Mx]. One usually identifies R' with R; once this is done, the string of subrings above can be lengthened: C is a subring of C[x].
Exercises
5. Show that the intersection of any family of subrings of R is a subring. 6. Prove that the binomial theorem holds in any ring R: if n? 1, then
(a + 13)n = where ( that
E ( nz. )ai
7 ) denotes the binomial coefficient n! 1 i!(n—i)!.
(Hint: First prove
(n. — 1 ± (n —1\ = (n) .) i—1 ) i ) W 7. If p is a prime, prove that p is a divisor of ( that 4 is not a divisor of ( 42 ) = 6.)
r) for i 0 0 and i 0 p. (Note
8. If R is any ring and f (x) c R[x j, say, f (x) = ro + rix +. . . + rn xn , define its derivative 3 by
t(x) = ri I 2r2x + . . . + nrnx" . Prove that
11(x) + g(x)r = f' (x) + g' (x) and
if (x)g(x)r = f (x)g' (x) + f' (x)g(x). 3 There is no notion of limit in most rings, and so we are taking the usual formula from
calculus and using it to define derivative over arbitrary rings.
DOMAINS AND FIELDS
13
9. If R is a ring and S is a set, let RS denote the set of all functions S —> R. Equip Rs with the operations of pointwise addition and multiplication; that is, if f, g : S+ R, then f + g : s 1> f (s) + g(s),
and fg : s 1> f (s)g(s).
Prove that RS is a ring. (Hint. "Zero" is the constant function Z with z (s) = 0 for all S E S, and "one" is the constant function e with e(s) = 1 for all S E S.)
Domains and Fields Two types of ring are especially important: domains and fields. Definition. A ring R is a domain (or integral domain) if the product of any two nonzero elements in R is itself nonzero. Example 4. Note that 4 is not a domain because [2] 0 0 and [3] 0 0, but [2][3] = [6] = 0. Theorem 6. A ring R is a domain if and only if it satisfies the cancellation law: if ra = rb and r 0 0, then a = b. Proof. Assume that R is a domain, that r 0 0, and that ra = rb. Then r (a — b) = 0. Since R is a domain, the inequality a — b 0 0 is untenable, and hence a — b = 0 and a = b. Conversely, assume that the cancellation law holds in R. Suppose there are nonzero elements r and a in R with ra = 0; then ra = 0 = r0 implies a = 0, a contradiction. • Example 4 can be generalized. Theorem 7. Zn is a domain if and only if n is prime. Proof. If n is not a prime, then it is composite, and so there is a factorization n = ab with 1 1, then v(n) = Ilk EZ:1 R', defined by q) : r i> (r, 0, 0 , . . .) , is an isomorphism from R to R'. Example 7. Let R be a domain with field of fractions F = Frac(R). It is easy to see that R"=frIlEF:r ER) is a subring of F, and that : r i 711 is an isomorphism from R to R". Example 8. The map it : Z —> Zn , defined by it : a i— [al, is a surjective ring map. Definition. If : R —> S is a ring map, then its kernel is ker = fr E R : co(r) = 0), and its image is im (p, = {s E S : s = q(r) for some r E R). It is easy to check that if q : R —> S is a ring homomorphism, then ker cp is an additive subgroup of R that is closed under multiplication (it is not a subring because it does not contain 1), and im p is a subring of S. In group theory, the kernel of a homomorphism is not merely a subgroup; it is a normal subgroup. Similarly, in ring theory, kernels have added structure. Definition. An ideal in a ring R is a subset I containing 0 such that: (i) a, b E i imply a — b E l; (ii) a E I and r E R imply ra E I. An ideal ! in a ring R is a proper ideal if I 0 R.
HOMOMORPHISMS AND IDEALS
19
Every ring R contains the ideals R itself and {0}. Theorem 10. If w : R —> S is a ring homomorphism, then ker w is a proper ideal in R. Moreover, w is an injection if and only if ker w = {0}. Proof. If one forgets the multiplications in the rings R and S and remembers only that they are additive abelian groups, then w is a homomorphism of groups, and so ker w is an additive subgroup of the additive group R. If a E i and r E R, then Ora) = w(r)w(a) = w(r) • 0 = O.
Therefore, ra E ker w, and so ker w is an ideal in R. Since coo (1) = 1 0 0, we see that ker w 0 R, and so ker cp is a proper ideal. If w is an injection, then distinct points have distinct images. In particular, if r 0 0, then p(r) # cp (0) = 0, so that r it ker w, and ker w = {0}. Conversely, suppose that ker w = {0}. If w(r) = w(r'), then 0 = w(r) — w(r) = w(r — r'); hence, r — r'
E
ker w = {O}, and so r = r'; therefore, w is an injection. •
Exercises
23. If R is a field, prove that the map R  Frac(R), given by a 1> a/1, is an isomorphism. Conversely, prove that if R is a domain and the map a 1> a/1 is an isomorphism, then R is a field. 24. If v : R > S is an isomorphism between domains, prove that there is an isomorphism Frac(R) > Frac(S), namely, alb i25. Let R be a subring of a field F, and let K be the intersection of all the subfields of F that contain R. Prove that K:',' Frac(R). 26.
(i) If w : R > S is an isomorphism, then v 1 : S > R is also an isomorphism. (ii) If p : R + S and * : S > T are ring homomorphisms, then so is their composite Op : R > T.
27. If a E R is a unit in R and if v : R > S is a ring map, then v(a) is a unit in S.
20 28.
GALOIS THEORY
(i) If R is a ring, prove that v : R[x] —> R, where 6,9 : f (x) 1—> co, the constant term of f (x), is a ring map. (ii) What is ker v?
29. (i) If a : R —> S is a ring map, prove that a* : R[x] —> S[x], defined by rix i 1—> Ea froxi,
E
is also a ring map. (ii) If r : S —> T is a ring map, prove that (ro )* : R[x] —> T[x] is equal to r*o * . (iii) Prove that if a is an isomorphism, then so is a*. 30.
(i) The intersection of any family of ideals in R is an ideal in R. Conclude that if X is any subset of a ring R, there is a smallest ideal, denoted by (X), containing X. One calls (X) the ideal generated by X, namely, the intersection of all the ideals in R that contain X. (ii) Prove that (X) is the "smallest" ideal containing X in the following sense: (X) is an ideal containing X and, if J is any ideal in R containing X, then (X) C J.
31.
(i) If a c R, prove that tra : r E R} is the ideal generated by a; it is called the principal ideal generated by a, and it is denoted by (a). (ii) If al , .. • , a n are elements in a ring R, prove that the set of all linear combinations, I = trial I  •  ± rn an : ri c R, i = 1, . . . , n},  
is equal to (ai, ... , an ), the ideal generated by fal , . . • , an ). 32. Let u be a unit in a ring R. (i) Prove that if an ideal I contains u, then I = R. (ii) If r c R, then (ur) = (r). In particular, every nonzero principal ideal (f (x)) in R = F[x], where F is a field, can be generated by a monic polynomial. (iii) If R is a domain and r, s c R, then (r) = (s) if and only if s = ur for some unit u in R. 33. Prove that a ring R is a field if and only if it has only one proper ideal,
namely, (0).
QUOTIENT
RINGS
21
(i) The set / of all f(x) c Z[x] having even constant term is an ideal in Z[x]; it consists of all the linear combinations of x and 2; that is, / = (x, 2).
34.
(ii) Prove that (x, 2) is not a principal ideal in Z[x]. 35. Prove that if F is a field and S is a ring, then a ring map q) : F —> S must be an injection and im w is a subfield of S isomorphic to F.
Quotient Rings Let I be an ideal in R. Forgetting the multiplication for a moment, I is a subgroup of the additive group of R; moreover, R abelian implies that I is a normal subgroup, and so the quotient group RII exists. The elements of RII are the cosets r + I, where r E R, and addition is given by (r + I) + (r' + I) = (r + r') + I ;
in particular, the zero element is 0 + I = I. Recall that r + I = r' + I if and only if r — r' E I. Finally, the natural map ir : R > R I I is the surjective (group) homomorphism defined by r 1> r + I. Theorem 11. Let I be a proper ideal in a ring R. Then the additive abelian group RII can be equipped with a multiplication which makes it a ring and which makes the natural map n  : R —> RII a surjective ring homomorphism.
Proof. Define multiplication on RII by (r + l)(r' + I) = rr' + I .
To see that this is well defined, suppose that r + / =s+Iand that r'+I = s' + I; we must show that rr' + I = ss' + I; that is, rr' — ss' E I. But rr' — ss' = (rr' — rs') + (rs' — ss') = r(r' — s') + (r — s)s' .
Now r' — s' c I and r — s E I, by hypothesis; hence r(r' — s') E I and (r — s)s' E I, because I is an ideal. Finally, the sum of two elements of I is again in I, so that rr' — ss' E I and rr' + I = ss' + I, as desired. It is routine to see that the abelian group RII equipped with this multiplication is a ring; in particular, "one" is 1 + I. Since I is a proper ideal, 1 /, and so 1 + I 0 0 + I = 0. The formula (r + I)(r' + I) = rr' + I says that n (r)7r(r') = n (rr'), where n :ri r + / is the natural map. It follows that g is a surjective ring homomorphism. •
22
GALOIS THEORY
Definition. If I is an ideal in a ring R, then R I I is called the quotient ring of R modulo I. In Exercise 36, one sees that 4 is equal to the quotient ring RII, where R = Z and ! = (n), the principal ideal generated by n.
Example 9. Let R = F[x], the polynomial ring over a field F; let I be the (principal) ideal generated by some particular polynomial p(x) of degree n, so that I = f(x)p(x) : f (x) E F[x]}. If g(x) E F[x], then the division algorithm gives q(x) and r(x) in fix] with g(x) = q(x)p(x) + r(x), where r = 0 or No < n. Note that g(x) + I = r(x) + 1, so we may assume that every coset (except ! itself) has a representative of degree < n. Indeed, each such coset has a unique representative r(x) of degree < n: if there were a second such, say, r'(x), then r —r' E l = (p), so that p I r —r' and r — r' = pf for some f (x) E F[x]. But r — r' has degree < n = 8 (p), while a (pf) > 0 (p), and this is a contradiction. The multiplication in R/! can be simplified: (f (x) + I)(g(x) +1) = f (x)g(x) +1 = r(x) +1, where r(x) is the remainder after dividing f (x)g(x) by p(x).
Example 10. Consider the special case of the preceding example in which F = IR and p(x) = x2 + 1. In R[x]// , where ! = (x 2 + 1), every element has the form a + + I ,where a, b
E
R, for x 2 + 1 has degree 2. Moreover,
(a + bx + I)(c + dx + I) =
+ bx)(c + dx) + I
= ac + (bc + ad)x + bdx2 +1. Now x2 —1 mod (p(x)), so that x2
+ I = —1 + I .
It follows that (a + bx + l)(c + dx +1) = ac — bd + (bc + ad)x +1. Now R[.x]/ I is actually a field, for it is easy to exhibit the multiplicative inverse of a + bx +1 (where a 0 0 or b 0), namely, c + dx + I, where c = a / (a2 + b2 ) and d = —b/(a2 + b2). We let the reader prove that the map 01 : R[x]// C, defined bya+b+11—>a+bi,is an isomorphism of fields. In particular, the "imaginary" number i with i 2 = —1 is equal to the coset x + I.
QUOTIENT RINGS
23
Theorem 12 (First Isomorphism Theorem). Ifv:R> Sisaring homommphism with ker co = I, then there is an isomorphism R I I .—> im co given by r ± I 1> v(r).
Proof. If we forget the multiplication in R and in S, then the First Isomorphism Theorem for groups says that the function (13 : R I I > imp, defined by (I) : r + I 1> yo(r), is an isomorphism of additive groups. Since (I) (1 ± I) = co(1) = 1, the proof will be complete if we prove that (I) preserves multiplication. Now if r, r' c R, then (I) ((r ± I)(r' + I)) = 0(r r' ± I) = co(rr') = because
v is a ring map. But (I) (r + 1)0(r' ± I) = v(r)co(r')
as well, and
SO
OW ± l)(r' ± I)) = (1)(r ± I)(1)(r' ± I),
as desired. • As in group theory, there is a correspondence theorem (see Exercise 38). There are also second and third isomorphism theorems for rings, but they are less interesting than their group theoretic analogues.
Exercises
36. Let n be a positive integer and let I = (n) be the principal ideal in Z generated by n. Show that the quotient ring Z// is equal to Z n , the ring of integers modulo n. (Hint. These rings have the same elements (La] = a ± I) and the same addition and multiplication.) 37. Prove that if R is a ring and I = (x) is the principal ideal in R[x] generated by x, then R[x]/l 1' R. 38. Prove the Correspondence Theorem for Rings. If I is a proper ideal in a ring R, then there is a bijection from the family of all intermediate ideals J, where / c J c R, to the family of all ideals in RII, given by J 1> rr(J)=J11={a+I:a€.1},
where n : R > RII is the natural map. Moreover, if J c J' are intermediate ideals, then n(J) C r (f). (Compare with Theorem G.9.) 39. Let / be an ideal in a ring R, let J be an ideal in a ring S, and let q) : R —> S be a ring isomorphism with v(/) = J. Prove that the function 0 : r + I F> w(r) ± J is a (well defined) isomorphism RII > SIJ.
24
GALOIS THEORY
Polynomial Rings over Fields Theorem 13. If F is afield, then every ideal in F[x] is a principal ideal. Proof. Let I be an ideal in Mx]. If I = {O}, then ! = (0), the principal ideal generated by 0. If! 0 {0}, choose a polynomial m(x) in I having smallest degree; we claim that I = (m(x)). Clearly, (m(x)) c I. For the reverse inclusion, take f (x) in I. By the division algorithm, there are polynomials q(x) and r(x) with f (x) = q(x)m(x) + r(x),
where either r(x) = 0 or a(r) < a(m). Now r (x) = f (x) — q (x)m (x) E I; if r(x) 0 0, then we have contradicted m(x) having the smallest degree of all polynomials in I. Therefore r(x) = 0 and f (x) = q(x)m(x) E
(WO. • By Exercise 32(ii), one may choose m(x) to be monic (since F is a field). Definition. A ring R is called a principal ideal domain (PID) if it is a domain in which every ideal is a principal ideal.
Of course, the reader knows that Z is a PID. Theorem 13 shows that F[x] is a PID when F is a field; on the other hand, Z[x] is not a PID (in Exercise 34, it is shown that the ideal I in Z[x] consisting of all the polynomials having even constant term is not a principal ideal). Definition. Let R be a ring; if r, s E R, then r divides s (or s is a multiple of r) if there exists r' E R with rr' = s; one writes r I s in this case.
Note that r I s if and only if s E (r), the principal ideal generated by r. It is easy to see that r I 0 for every r E R, but that 0 I r if and only if r = 0; also, r 1 r for every r G R, and r is a unit if and only if r I 1. Definition. Let R be a domain, and let f (x), g(x) E R[x]. The greatest common divisor (gcd) of f (x) and g(x) is a polynomial d(x) E R[x] such that: (i) d(x) is a common divisor of f (x) and g(x); that is, d I f and d I g;
(ii) if c(x) is any common divisor of f (x) and g(x), then c(x) I d(x); (iii) d(x) is monic.
POLYNOMIAL RINGS OVER FIELDS
25
One often denotes d(x) by (f, g). If (f, g) = 1, then f (x) and g(x) are called relatively prime. Note that the gcd d of f and g, if it exists, is unique. If d' is another gcd, then regard it only as a common divisor and use (ii) to obtain d' I d; similarly, d I d' if one regards d merely as a common divisor. By Exercise 4, d' = ud for some unit u c F[x]; that is, d' = ud for some nonzero constant u (Exercise 32). Since d and d' are both monic, however, u = 1 and d' = d.
If a linear combination of polynomials f and g is 1, say, there are polynomials a(x) and b(x) with 1 = a(x) f (x) + b(x)g(x), then f and g must be relatively prime. After all, any common divisor c(x) of f and g must also divide 1, and hence c is a unit. The next result shows that the gcd in F[x I, when F is a field, is always a linear combination.
Theorem 14. Let F be afield and let f (x), g(x) E F[x] with g(x) 0. Then the gcd (f (x), g(x)) = d(x) exists, and it is a linear combination of f (x) and g(x); that is, there are polynomials a(x) and b(x) with d(x) = a(x) f (x) + b(x)g(x).
Proof. By Exercise 31, I = {a(x) f (x) + b(x)g(x) : a(x), b(x)
E F[x])
is an ideal in F[x] containing both f (x) and g(x). Since F is a field, F[x] is a PID and I is a principal ideal. By Exercise 32, we may choose a monic polynomial d(x) with I = (d(x)); as is every element of I, the generator d is a linear combination of f and g. Now d is a common divisor of f and g because f, g E I = (d). Finally, if c is a common divisor, then c I f and c I g; that is, f = cc' and g = cc". Hence, d = af + bg = acc' + bcc" = c(ac' + bc"), and so c I d. Therefore, d(x) is the gcd. •
Example 11. Let R = F[x]l I , where F is a field and I is the principal ideal generated by some polynomial p(x). If f (x) and p(x) are relatively prime, then there are polynomials s(x), t (x) E F[x] with s(x) f (x) + t(x)p(x) = 1; in R this equation becomes s(x) f (x) + I = 1 + I.
26
GALOIS THEORY
Thus f (x) + I is a unit in R with inverse s(x) I I. The converse is also true. If f (x) +1 is a unit in R, then there is g(x) E R = F[x] with 1 + I = (f (x) + I)(g(x) + I) = f (x)g(x) + I; that is, f (x)g(x) — 1 = h(x)p(x)
for some h(x)
E F[x].
Therefore, f (x) and g(x) are relatively prime.
Corollary 15 (Euclid's Lemma). Let F be afield. If p(x)
E F[x] is irre
ducible and p(x) divides a product qi(x) • • • q 1 (x), then p(x) divides qi(x) for some j.
Proof. By induction on n > 2, it suffices to prove that if (f (x), g(x)) = 1 and f (x) divides g(x)h(x), then f (x) divides h(x). There are polynomials a(x) and b(x) with 1 = af + bg. Hence h = afh + bgh. By hypothesis, gh = f k for some polynomial k, so that h = af h + bf k = f (ah + bk) and f divides h. • The proof of Theorem 14 yields the following fact; it also explains why the gcd of f and g is denoted by (f, g). Corollary 16. Let F be afield, let f (x), g(x)
E F[x], and let the ideal
generated by f (x) and g(x) be I = (f (x), g (x)). Then I = (d(x)), where d(x) is the gcd of f (x) and g(x).
The proof of Euclid's lemma is just an adaptation of the usual proof of Euclid's lemma in Z; the same is true for the euclidean algorithm to be proved next. If one is given explicit polynomials f (x) and g(x), how can one compute their gcd? How can one express the gcd as a linear combination? Theorem 17 (Euclidean Algorithm). There are algorithms to compute the gcd and to express it as a linear combination. Proof. The idea is just to iterate the division algorithm. Consider the list of equations: = qig ±ri g = q2ri ± r2 r1 = q3r2 ± r3
f
afro 8 (r2) 8 (r3)
F by f. If F is infinite, then the function co : F[x] > P(F), given by (p: f (x) 1> f, is an isomorphism.
Proof. It suffices to prove that ker . v = in Suppose that f (x) c ken') is not the zero polynomial, and let n = 8( f). Since f (a) = 0 for all a E F, each of the infinitely many elements a c F is a root of f (x), and this contradicts Theorem 22. • Exercises
40. Prove that there are domains R containing a pair of elements having no gcd. (Hint. Let F be a field and let R be the subring of F[x] consisting of all polynomials having no linear term; i.e., f (x) c R if and only if f (x) = ao + a2x 2 + a3x 3 ± • • • . Show that x 5 and x 6 have no gcd by noting that their monic divisors are 1, x 2 , and x 3 , none of which is divisible in R by the other two.)
41.
(i) Define the gcd of integers al, . . . , a n to be a positive integer d which is a common divisor, i.e., d I ai for all i, that is divisible by every common divisor. Prove that the gcd d of al , . .. , an exists, and that d is a linear combination of ai, ... , an . (Hint. Let d be the positive generator of the ideal in Z generated by al, .  • , an.) (ii) Define the gcd of polynomials ft , . . . , fn E F[x], where F is a field, to be a monic polynomial d which is a common divisor, i.e., d I fi for all i, that is divisible by every common divisor. Prove the generalization of Corollary 16 that the gcd d of fit ... , fn exists, and that d is a linear combination of ft , — , fn.
42. Prove that if al , at, ... , a n are distinct elements in a field F, then for all i, the polynomials x — ai + i and (x — ai)(x — a2) • •  (x — ai) are relatively prime.
PRIME IDEALS AND MAXIMAL IDEALS
31
43. In the ring R = Z[x], show that x and 2 are relatively prime, but there are no polynomials f (x) and g(x) E Z[x] with 1 = xf (x) ± 2g (x). 41. Let f (x) = il(x —ai) E F[x], where F is a field and ai E F for all i. Show that f (x) has no repeated roots [i.e., f (x) is not a multiple of (x — a) 2 for any a E F] if and only if (f (x), f' (x)) = 1, where f' (x) is the derivative of f (x). 45. Find the gcd of x 3 — 2x 2 ± 1 and x 2 — x — 3 in Q[x] and express it as a linear combination. 46. Prove that Z2[x]// is a field, where p(x) = x 3  F x + 1 E Z2[x] and I = (p(x)). 47. If R is a ring and a E R, let ea : R[x] + R be evaluation at a. Prove that ker ea consists of all the polynomials over R having a as a root, and so ker ea = (x — a), the principal ideal generated by x — a. 48. Let F be a field, and let f (x), g(x) E F[x]. Prove that if a (f) _< a (g) = n and if f (a) = g (a) for n + 1 elements a c F, then f (x) = g(x).
Prime Ideals and Maximal Ideals The notion of prime number can be generalized to polynomials.
Definition. Let F be a field. A nonzero polynomial p(x) E F[x] is irreducible over F if a (p) > 1 and there is no factorization p(x) = f (x)g(x) in F[x] with a( f) < a(p) and a (g) < a (p). Notice that irreducibility does depend on the coefficient field F. For example, x2 + 1 is irreducible over R, but it factors over C. It is easy to see that linear polynomials (degree 1) are irreducible over any field F for which they are defined. It follows from Corollary 21 that irreducible polynomials of degree > 2 over a field F have no roots in F. The converse is false, 2 fr actors over R, but it has however, for f (x) = x4 ± 2x2 ± 1 = (x21. 1)no real roots. 4 This notion can be generalized to any ring R. A nonzero element r c R is called ir
reducible if r is not a unit and, in every factorization r = st in R, either s or t is a unit. If F is a field and R = F[x], then this notion coincides with our definition of irreducible polynomial. In Z[x ], however, 2x + 2 = 2(x + 1) is not irreducible, yet it does not factor into polynomials each of which has smaller degree.
32
GALOIS THEORY
Let p(x), f (x) E F[x], where F is a field. If p(x) is a monic irreducible polynomial, then its only monic divisors are 1 and p(x); hence, the gcd (p, f) is either 1 or p(x). It follows that if p(x) does not divide f (x), then p(x) and f (x) are relatively prime. Definition. An ideal I in a ring R is called a prime ideal if it is a proper ideal and ab E I implies a E I or b E I. Example 14. We claim that if p > 2, then the ideal (p) in Z is a prime ideal if and only if p is a prime. If p is prime and ab E (p), then p I ab. By Euclid's lemma, either p I a or p I b; that is, either a E (p) or b E (p). Therefore, (p) is a prime ideal. Conversely, if p is not a prime, then it has a factorization p = ab with a < p and b < p. It follows that neither a nor b lies in (p), and so (p) is not a prime ideal. Theorem 24. If F is afield, then a nonzero polynomial p(x)
E
F[x] is
irreducible if and only if (p(x)) is a prime ideal.
Proof. Suppose that p(x) is irreducible. If ab E (p), then p I ab, and so Euclid's lemma gives either p I a or p I b. Thus, a E (p) or b E (p). Finally, (p) is a proper ideal; otherwise, 1 E R = (p), and so there is a polynomial f (x) with 1 = p(x) f (x). But the constant 1 has degree 0, whereas a(pf) = a(p) + a(f) a(p) > 1. This contradiction shows that (p) is a proper ideal, and hence it is a prime ideal. Conversely, suppose that p(x) is not irreducible; there is thus a factorization p(x) = a(x)b(x)
with 8(a) < a(p) and 8 (b) < a(p). As every nonzero polynomial in (p) has degree > 8(p), it follows that neither a nor b lies in (p), and so (p) is not a prime ideal. • Theorem 25. A proper ideal I in R is a prime ideal if and only if RI I is a domain.
Proof. Let ! be a prime ideal. If 0 = (a+ I)(b+ I) = ab+ I , then ab e I. Since ! is a prime ideal, either a E I or b E I; that is, either a ± I = 0 or b +1 = 0. Hence, R 1 I is a domain. The converse is just as easy. •
PRIME IDEALS AND MAXIMAL IDEALS
33
Definition. An ideal I in a ring R is a maximal ideal if it is a proper ideal and there is no ideal J with I J R. Theorem 26. A proper ideal I in a ring R is a maximal ideal if and only if R/1 is a field. Proof. The Correspondence Theorem (Exercise 38) shows that I is a maximal ideal if and only if RII has no ideals other than {0} and RII itself; Exercise 33 shows that this property holds if and only if RII is a field. • Corollary 27. Every maximal ideal I in a ring R is a prime ideal. Proof. If I is a maximal ideal, then RII is a field. Since every field is a domain, RII is a domain, and so I is a prime ideal. • The converse of the last corollary is false. For example, the principal ideal (x) in ZIA is prime but not maximal; by Exercise 37, we have Z[x]/(x) and Z is a domain but not a field.
Theorem 28. If R is a principal ideal domain, then every nonzero prime ideal I is a maximal ideal. Proof. Assume there is an ideal J 0 I with I cJc R. Since R is a PID, I = (a) and J = (b) for some a, b E R. Now a E J implies that a = r b for some r E R, and so rb c I. Since I is prime, either r c I or b c I. If b c I, then JC I,a contradiction. If r E I, then r = sa for some s E R, and so a = rb = sab; hence l = sb and J = (b) = R, by Exercise 32(i). Therefore, I is maximal. • Corollary 29. If F is a field and p(x) E F[x] is irreducible, then the quotient ring Fix11(p(x)) is afield containing (an isomorphic copy of) F and a root of p(x). Proof. Since p(x) is irreducible, the principal ideal I = (p(x)) is a nonzero prime ideal; since F[x] is a PID, I is a maximal ideal, and so E = F[x]/I is a field. Now the map a i> a + I is an isomorphism from F to F' = fa I I : a E F) c E (one usually identifies F with this subfield F' of E).
34
GALOIS THEORY
Let 0 = x + I e E; we claim that 0 is a root of p(x). Write p(x) = ao ± aix ± • • • ± anxn , where ai E F. Then, in E:
P( 0 )
= (ao + 0 + (ai + 00 + • • • + (a n + I )0n
=
(
ao + n + (ai ± 1)(x ± I) +  •  ± (an ± .0(x + O n nxn ± I)
=(ao±I)+ix•(a
= ao ± aix I • • • + a nxn ± I = p(x) ± I = I , because I = (p(x)). But I = 0 + I is the zero element of F[x]l I , and hence 0 is a root of p(x). • For example, x 2 + 1 is an irreducible polynomial in r [x], and the quotient ring R[x]/(x 2 + 1) is a field containing R and an element i with i 2 = —1, namely, i = x ± I. We have seen, in Example 10, that r [x]/(x 2 + 1) is isomorphic to the field of complex numbers C. Definition. A polynomial f (x) linear factors in F[x].
E
F[x] splits over F if it is a product of
Of course, f (x) splits over F if and only if F contains all the roots of f (x). Theorem 30 (Kronecker). Let f (x) E F[x], where F is afield. There exists afield E containing F over which f (x) splits. Proof. The proof is by induction on a(f). If a(f) = 1, then f (x) is linear and we can choose E = F. If a(f) > 1, write f (x) = p(x)g(x), where p(x) is irreducible. The corollary provides a field B containing F and a root 0 of p(x). Hence p(x) = (x — 0)h(x) in B[x]. By induction, there is a field E containing B over which h(x)g(x), hence f (x), splits. • We now modify, for a polynomial f (x) E F[x], the definition of repeated roots appearing in Exercise 44 so that roots may belong to some larger field of coefficients than F. Definition. If F is a field and f (x) E F[x], then f (x) has repeated roots if there is a field E containing F and a factorization in E[x] of the form
f (x) = (x — a) 2h(x).
PRIME IDEALS AND MAXIMAL IDEALS
35
Using Exercise 44 and Corollary 18, one sees that 1(x) has no repeated roots if and only if (f, f') = 1, where f (x) is the derivative of (x). Notice that the criterion (f, f') = 1 can be checked over the original field F; there is no need to examine factorizations over E. Kronecker's theorem can be used to construct finite fields other than the fields Z p . Before giving the construction, we introduce an important property of fields. Definition. The prime field of a field F is the intersection of all the subfields of F. By Exercise 19, the prime field is a subfield. Theorem 31. If F is a field, then its prime field is isomorphic to either
Q
or Zp for some prime p.
F by n 1> n1 (where 1 is the "one" in F); it is easy Proof. Define x : to see that x is a ring map. If I = ker x, then Z// is a domain (because it is isomorphic to a subring of the field F). Therefore, I is a prime ideal, and hence I = (0) or I = (p) for some prime p. If! = (0), then x imbeds Z in F. By Exercise 25, the prime field is isomorphic to Q in this case. If I = (p), the first isomorphism theorem gives im x Z/(p) = Zp , which is a field; hence im x is the prime field of F. • Definition. A field has characteristic 0 if its prime field is isomorphic to Q; it has characteristic p if its prime field is isomorphic to Zp. Lemma 32. Let F be afield of characteristic p >
0.
(i) For all a E F, we have pa = O. (ii) (a + b)P = aP + bP for all a, b G F. (iii) (a
b)1
k = ap
k by for all a, b E F and all k > 1.
Proof. (i) For the moment, let us denote "one" in F by e. Now pa means the sum of p terms each equal to a: pa = a +   + a = (e + • • • + e)a.
In Zp , however, the sum of p terms each equal to [1] is 0. Since F has characteristic p, we have e +... + e = 0, and so pa = 0 in F.
36
GALOIS THEORY
(ii) The binomial theorem gives (a + b)P = aP
(P )a i
+ b".
i=1
By Exercise 7, ( = 0 in Zp for all 1 < i < p — 1. (iii) The proof is by induction on k> 1, the base step being part (ii). For the inductive step, ± by k÷i
= [(a + b)P k r
= Eap k
bp k i p
= ap k+I
bp k+I
•
It follows from this lemma that if F is a field of characteristic p and if q = plc, then the function a 1> aq is a ring homomorphism from F to itself. The following elementary remark is very useful. If F is a subfield of a field E, then the additive group of E may be viewed as a vector space over F. Define scalar multiplication by letting ca, for c c F and a E E, be the product of the two elements c and a under the given multiplication on E. Viewing the appropriate axioms in the definition of a field in this light, one can see that they are also the axioms of a vector space over F. In particular, a finite field E must have characteristic p for some prime p, for its prime field cannot be 0, and so it is a vector space over Z p . If fah . . . an} is an ordered basis of E, then each a E E has coordinates (c1, . . . , c n ) for ci in Z. Therefore, every finite field has if' elements, for some prime p and some positive integer n. Theorem 33 (Galois). For every prime p and every positive integer n, there exists a field having exactly pn elements.
Proof. If there were a field K with I K I = pn = q, then K 4 = K — {0} would be a multiplicative group of order q — 1; by Lagrange's theorem, aq 1 = 1 for all a e K 4 . It follows that every element of K would be a root of the polynomial g(x) =x' — x. We now begin the construction. By Kronecker's theorem, there is a field E containing Zp over which g(x) splits. Define F = [a E E: g(a) = 01; that is, F is the set of all the roots of g(x). Since the derivative g'(x) = qxq 1 — 1 = —1 (because q = pn and E has characteristic p), Lemma 32 shows that the gcd (g, g') = 1, and so g(x) has no repeated roots; that is,
IFI = q =
PRIME IDEALS AND MAXIMAL IDEALS
37
We claim that F is a field, which will complete the proof. If a, b E F, then aq = a and bq = b. Therefore, (ab) = aqbq = ab, and ab E F. By Lemma 32(iii), replacing b by —b, we have (a — b) = aq — bq = a — b, so that a—b c F. Finally, if a 0 0, then aq 1 = 1 so that a 1 = aq 2 E F (because F is closed under multiplication). • In Corollary 53 we shall see that any two fields of order I)" are isomorphic. It will follow that there are no finite fields other than those just constructed.
Exercises
49. A polynomial p(x) E F[x] of degree 2 or 3 is irreducible over F if and only if F contains no root of p(x). (This is false for degree 4: the polynomial (x2 + 1) 2 factors in Ek[x], but it has no real roots.) 50. Let p(x) E F[x] be irreducible. If g(x) c F[x] is not constant, then either (p(x), g(x)) = 1 or p(x) I g(x). 51.
(i) Every nonzero polynomial f (x) in F[x] has a factorization of the form f (x) = aPi(x) •  • where a is a nonzero constant and the p1 (x) are (not necessarily distinct) monic irreducible polynomials;
(ii) the factors and their multiplicities in this factorization are uniquely determined. (This analogue of the fundamental theorem of arithmetic has the same proof as that theorem: if also f (x) = bqi(x) .. . q s (x), where b is constant and the qi(x) are monic and irreducible, then uniqueness is proved by Euclid's lemma and induction on maxit , sl. One calls F[x] a unique factorization domain when one wishes to call attention to this property of it.) 52. Let f(x) = api(x) k i • • • p 1 (x) kz and g(x) = bpi(x)n' • • • p t (x)nt , where ki > 0, ni > 0, a, b are nonzero constants, and the pi (x) are distinct monic irreducible polynomials (zero exponents allow one to have the same p 1 (x) in both factorizations). Prove that gcd(f, g) = pi(x) m '    p 1 (x) m ' and lcm(f, g) = p1 (x )Ml   • p t (x) mi ,
where mi = min{ki,ni} and Mi = maxiki,ni).
38 53.
GALOIS THEORY (i) Prove that the zero ideal in a ring R is a prime ideal if and only if R is a domain. (ii) Prove that the zero ideal in a ring R is a maximal ideal if and only if R is a field.
54. The ideal 1 in 7Z[x] consisting of all polynomials having even constant term is a maximal ideal. 55. Let f (x), g(x) E F[x I. Then (f, g) 0 1 if and only if there is a field E containing both F and a common root of f (x) and g (x). 56.
(i) Prove that if f (x) E Zp[X1, then (f (x))P = f (xP). (Hint: Use Fermat's theorem: aP 1—__ a mod p.) (ii) Show that the first part of this exercise may be false if Z p is replaced by an infinite field of characteristic p.
57. Exhibit an infinite field of characteristic p. (Hint: Exercise 20.) 58. If F is a field, prove that the kernel of any evaluation map F[x]  F is a maximal ideal. 59. If F is a field of characteristic 0 and p(x) E F[x] is irreducible, then p(x) has no repeated roots. (Hint: Consider (p(x), p'(x)).) 60. Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of x 4 — x to Z2. 61. Give the addition and multiplication tables of a field having eight elements. (Hint: Factor x 8 — x over Z2.) 62. Show that a field with four elements is not (isomorphic to) a subfield of a field with eight elements.
Irreducible Polynomials Our next project is to find some criteria for irreducibility of polynomials; this is usually difficult, and it is unsolved in general. We begin with an elementary result, using Exercise 29: Ifa:R—>Sis a ring map, then a* : R[x] —> S[.x], defined by a* :
is also a map of rings.
E rix i
1—> Ea(ri)xi,
IRREDUCIBLE POLYNOMIALS
39
Theorem 34. Let R be a domain and F be a field, let a : R —* F be a ring map, and let p(x) E R[x]. If a(a *(p)) = a(p) and if a*(p(x)) is irreducible in F[x], then p(x) is not a product of two polynomials in Mx] each of degree < a(p).
Remark. Note that the degree condition is satisfied if p(x) is monic. Proof. Suppose that p(x) = f (x)g(x) in R[x] with a(f) < a(p) and a(g) < a(p). In F[x], we have a*(p) = a*(f)a  *(g)• Since a*(p) is irreducible, we may assume that a(a*(f))= 0. But
a(p) = a(0*(p)) = a(a*(i))+a(e(g))
= 807*(0 < a(g)
< a(p). This contradiction completes the proof. •
Example 15. Consider f (x) = 8x 3 — 6x — 1 in Z[x]. We will use the theorem by making a suitable choice of prime p and taking a : Z —> Zp to be the natural map; thus, a* reduces the coefficients of f (x) mod p. If we choose p = 2, then the degree condition is not satisfied because o *(f) has degree 0. If p = 3, then a*( f) = —x 3 —1 = —(x + 1)(x 2 —x + 1),
and a*(f) is not irreducible. If p = 5, then a*(f)= 3x3 —
x —1;
this is irreducible, by Exercise 49, for it has no roots in Z5. It follows from Theorem 34 that f (x) is not a product in Z[x] of polynomials of lower degree. Theorem 34 does not always apply. We shall see, in Exercise 67, that f (x) = x4 —10x2 +1 is irreducible in Q[x]; in Example 26, we will show that f (x) factors mod p for every prime p. We now have a way to see whether certain polynomials in Z[x] factor into polynomials of smaller degree, but we are really interested in whether polynomials are irreducible in Q[x]. A result of Gauss will solve this problem.
40
GALOIS THEORY
Definition. A polynomial f (x) = ao I aix ± • • • + anxn E primitive if the gcd5 of its coefficients is 1.
Z[X]
is called
If d is the gcd of the coefficients of f (x), then (11d) f (x) is a primitive polynomial. Observe that if 1(x) is not primitive, then there exists a prime p which divides each of its coefficients: if the gcd of the coefficients is d, let p be any prime divisor of d. Lemma 35 (Gauss's Lemma). The product of two primitive polynomials f (x) and g(x) is itself primitive.
Proof.6 Assume that the product f (x)g(x) is not primitive, so there is some prime p dividing each of its coefficients. Let a : Z > Zi, be the natural map, and consider the ring map a* : Z[x] > Z p [x I reducing coefficients modp. Now a*(fg) = o *(f)o *(g).
But a*(fg) = 0 in Zp [x] while o *(f) 0 0 and o *(g) 0 0, and this contradicts the fact that Z,, [x]is a domain. • Lemma 36. Every nonzero f (x) E Q[x] has a unique factorization f (x) = c(f)1*(x), where c(f) E Q is positive and f*(x) E Z[X] is primitive.
Remark. The positive rational c(f) is called the content of f (x). Proof. Let f (x) = (ao/bo) + (al I bi)x + • • • + (an Ibn )xn E Q[X]. Define B = bo  •  bn , so that g(x) = Bf (x) E Z[X]. Now define D = ±d, where d is the gcd of the coefficients of g(x); the sign is chosen to make D I B positive. Now (B I D)1(x) = (11 D)g(x) lies in Z [x], and it is a primitive polynomial. If we define c(f) = DI B and f *( x ) = (B I D) f (x), then f (x) = c(f) f* (x) is a desired factorization. Suppose that 1(x) = eh (x) is a second such factorization, so that e is a positive rational and h(x) E Z[x] is primitive. Now c(f) f*(x) = f (x) = eh(x), so that f*(x) = Re Ic(f)]h(x). Write e Ic(f) in lowest terms: elc(f)= ulv, where u and v are relatively prime positive integers. The equation vf* (x) = uh(x) holds in Z[x]; equating like coefficients, v is 5 Exercise 41 defines the gcd of finitely many integers and shows that it always exists. 6 This elegant proof of Gauss's lemma was shown me by Peter Cameron.
IRREDUCIBLE POLYNOMIALS
41
a common divisor of each coefficient of uh(x). Since (u, v) = 1, Euclid's lemma in Z shows that v is a (positive) common divisor of each coefficient of h(x). Since h(x) is primitive, it follows that v = 1. A similar argument shows that u = 1. Therefore, elc(f) = ulv = 1, so that d = c(f) and hence f* (x) = h(x). • Corollary 37. If f (x)
E Z[x], then c(f) E
Z.
Proof. If d is the gcd of the coefficients of f (x), then ( 1 /d) f (x) E Z[X] is primitive. Since f (x) = d[(11d) f (x)] is a product of a positive rational d (even an integer) and a primitive polynomial, the uniqueness in the lemma gives c(f) = d E Z. • Corollary 38. If f (x)
E Q[x] factors as f (x) = g(x)h(x) in Q[x], then
c(f) = c(g)c(h) and f* (x) = g* (x)h* (x).
Proof. We have f (x) = g(x)h(x) = [c(g)g* (x)][c(h)h* (x)] = c(g)c(h)g* (x)h* (x).
Since c(g)c(h) is a positive rational, and since the product of two primitive polynomials is primitive, the uniqueness of the factorization in the preceding lemma gives c( f) = c(g)c(h) and f * (x) = g* (x)h* (x). • Theorem 39 (Gauss). If p(x)
E Z[X] is not a product of two polynomials
in Z[x] each of degree < a(), then p(x) is irreducible in Q[x].
Proof. If f (x) = g(x)h(x) in Q[x], then f (x) = c(g)c(h)g* (x)h* (x) in Q[x], where g* and h* are primitive polynomials in Z[x]. But c(g)c(h) = c(f) E Z, by Corollary 37. Therefore, f (x) = [c(f)g*(x)]h* (x) is a factorization in Z[x]. • Remark. The proof of this last theorem can be adapted to a more general setting: replace Z and Q by a unique factorization domain and its field of fractions. This is the main ingredient of the proof that if R is a unique factorization domain, then so is R[xj; it follows that if F is a field, then xn ] is a unique factorization domain. Theorem 40 (Eisenstein Criterion). Let f (x) = ao I aix +• • • ± a n xn Z[x]. If there is a prime p dividing a i for all i < n, but with p not dividing an and p 2 not dividing 610, then f (x) is irreducible in Q[x].
E
42
GALOIS THEORY
Proof. Let f (x) = g(x)h(x) = (bo + bix + • • • + bmx m )(co + cix + • •  + ckx k );
by Theorem 39, we may assume that both g and h lie in Z[x]. By hypothesis, p I ao = boco so that p I bo or p I co, by Euclid's lemma in Z; since p2 does not divide ao, only one of them is divisible by p, say, p I c o but p does not divide bo. The leading coefficient a n = bm ck is not divisible by p, so that p does not divide ck (or bm ). Let cr be the first coefficient not , c,._ 1 ). If r < n, then p I a r , and divisible by p (so p does divide co , bocr = a,. — + • • • + brco) is divisible by p; hence p I bocr , contradicting Euclid's lemma (because p divides neither factor). It follows that r = n, hence k = 0, and h(x) is constant. Therefore, f (x) is irreducible. • Remark. The following more elegant proof of Gauss's lemma is due to Peter Cameron. Suppose that f (x) = g(x)h(x) in Z[x], where g(x) = bo bix + •   + b m xm has m < n and h(x) = + cix + • •  + ckx k has k < n. Consider the map a* : Z[x] > Z[x] reducing coefficients modp. In Z p [xJ, we have a*(f) = an xn; thus, a*(f ) is a constant times monic irreducibles all equal to x. But a* ( f) = o *(g)a*(h), so that unique factorization (see Exercise 51) shows that a*(g) and o *(h) have similar factorizations. Therefore, o *(g) = b m xn and p I b i for all i < m; similarly, cr*(h) = ck xk and p I c i for all j < k. In particular, p I 1)0 and p I co and so p2 I boco = ao , a contradiction. The Eisenstein criterion shows that x 5 — 4x + 2 is irreducible over Q; this polynomial does not surrender easily to our first criterion. Definition. If p is a prime, then the pth cyclotomic polynomial is cDp (x) = (xP — 1)/(x — 1) = xP 1 + X P2 ± • • • + X ± 1. Corollary 41. The pthcyclotomic polynomial for every prime p.
Op (V)
is irreducible in Q[x]
Proof. Recall Exercise 66: A polynomial f (x) is irreducible if and only if f (x c) is irreducible, where c is a constant. In particular,
is irreducible if and only if cto p (x + 1) = ((x + 1)P — 1)/x
IRREDUCIBLE POLYNOMIALS
43
is irreducible. The latter polynomial is xP I + pxP 2 1 ()xP 3 1 • • • + p, where () is the binomial coefficient. Since p is prime, Exercise 7 shows that Eisenstein's criterion applies; we conclude that Op CO is irreducible. • If n is not prime, then xn  I ± xn2 + • • • + x + 1 factors in Q[x]. For example, X 3 + X 2 + X + 1 = (X + 1)(X 2 + X + 1). Corollary 42. If an integer a is not a petfect square, then xn  a is irreducible in Q[x] for every n > 2. Proof. Since a 0 ±1, there is some prime p dividing a, and Eisenstein's
criterion applies with this prime. • This last corollary shows that there are irreducible polynomials over Q of arbitrary degree n. Exercises
63. Let f(x) = ao + aix +  • • + anxn E Z[Xl. If rls is a rational root of f(x), where rls is in lowest terms, i.e., (r, s) = 1, then r I ao and s I a.
Conclude that any rational root of a monic polynomial in Z[x] must be an integer. 64. Test whether the following polynomials factor in Q[x]: (i) 3x 2 — 7x
—
5;
(ii) 6x 3 — 3x — 18; (iii) x 3 — 7x + 1. (iv) x 3 — 9x — 9. 65. Let F be a field. Prove that if ao + aix + • • • +anxn E F[x] is irreducible, then so is a n + an—ix + •   + aoxn. 66. If c E R, where R is a ring, then the map f (x) 1 > f (x + c) is an isomorphism of the ring R[x] with itself. Conclude, when R is a field, that p(x) is irreducible if and only if p(x + c) is irreducible. —
67. Prove that f (x) = x 4 10x 2 + 1 is irreducible in Q[x]. (Hint. Use Exercise 63 to show that f (x) has no rational roots; then show that there are no rationals a, b and c with —
x4  10x 2 + 1 = (x 2 + ax + b)(x 2  ax + c).)
44
GALOIS THEORY
Classical Formulas We now derive the classical formulas for the roots of quadratics, cubics, and quartics. Definition. A polynomial f (x) of degree n is called reduced if it has no X n 1 term; that is, f(x) =
rnx n 4_ rn 2x n2 ± rn_3x n3 + . . .
Lemma 43. If f (X) = a n Xn +a„_ i Xn 1 1an_2X n2 1 • • • , then replacing X by x — a n _i 1 n gives a reduced polynomial
.7(x) = f (x 
a_/n);
moreover, if u is a root of :I(x), then u — a n_ i In is a root of 1(X).
Proof. The first statement is a straightforward calculation, and the second statement follows from the equation 0 = fiu) = f (u — an _ i 1 n). • The quadratic formula is usually proved by "completing the square," but we shall do it in a way that anticipates the derivations of its generalizations to cubics and quartics. Consider the quadratic X2 + bX + c. The substitution replacing X by x — lb gives the reduced quadratic x2 + c — 1 b2 having roots u = ±1N/b 2 — 4c. By Lemma 43, one obtains the familiar formula for the roots of the original quadratic: —lb ±1Vb2 — 4c. Before discussing formulas for the roots of a polynomial f (x) of higher degree, we must say that if f(x) E Z[X], one should first use Exercise 63 to see if it has any rational roots. If u is a root of a cubic polynomial f (x), for example, then its remaining roots are the roots of the quadratic polynomial f(x)/(x — u).
The reduced polynomial arising from a cubic X 3 + aX 2 + bX + c has the form g(x) =x 3
± qx + r;
CLASSICAL FORMULAS
45
by Lemma 43, a formula for the roots of g(x) will give a formula for the roots of the original one. The coming formula is essentially due to Scipio del Ferro (ca. 1515); a similar formula was discovered by Tartaglia about the same time, and both appeared in print for the first time in the book of Cardan (1545). Let u be a root of g(x), and choose numbers y and z with u = y + z. Then
u 3 = (y + z)3 = y3 + z 3 + 3(Y2z + Yz 2) = y3 + z 3 +3uyz. Therefore,
(1)
y3 +z 3 + (3yz + q)u + r = 0.
So far we have imposed only one constraint on y and z, namely, u = y +z. By Exercise 68, we may impose a second constraint:
(2)
yz =
—
q/3,
so that, in Eq. (1), the linear term in u vanishes. We now have
y3 ± Z3 =
—
r
and
y3 z 3 = q3 /27. —
These two equations can be solved for y 3 and z 3. In detail, y3 — q 3 /27y3 =
—
r,
and hence y 6 ± r y 3 — q 3 /27 = 0.
The quadratic formula gives
(3)
y 3 = (— r + A/ r 2 + 4q3 /27),
and Eq. (2) gives z = q/3y. Having found one root u = y + z of g(x), one can find the other two as the roots of the quadratic g (X) I (x — u). Here is an explicit formula for the other two roots, in contrast to the method just described for finding them. If co = e2mi/3 is a cube root of unity, then there are three values for y; one is given by Eq. (3); the other two are coy and co2 y. The corresponding "mates" are —
—q13coy = (11co)z = co2z
46
GALOIS THEORY
and
_q/360 2 y
= (1/(0 2 )z = wz.
We conclude that the roots of the cubic polynomial are given by the cubic
formula: Z; CO
y
(0 2z ;
(02 y ± an;
here y 3 = (— r
NrR)
and R = r 2 + 4q3 /27. Example 16. If f (x) = x 3 — 15x — 126, then f (x) is reduced [otherwise, one would reduce it via the substitution x 1—> x — b/31. Here, q = —15, r = —126, R = 15376, and VT? = 124. Hence, y3
= 1[—(126) + 124] = 125,
so that y = 5; moreover, z = —q/3y = 15/15 = 1, so that one root is
u = y + z = 6. The other two roots can be found either by using the quadratic formula on (x 3 — 15x — 126)/(x — 6) = x 2 + 6x + 21 (they are —3 ± or by using the cubic formula (they now appear as 5co + co 2 and 5w2 + co). Example 17. Consider f (x) = x 3 — 7 x + 6 = (x — 1)(x — 2)(x + 3),
whose roots are, plainly, 1, 2, and —3. The cubic formula gives ‘,3
1(
1/400 27 ) '
and so one root of f (x) is
11
1
2
/400) 27
'
13/1 (6 —14° ).
This expression is, thus, equal to 1, 2, or —3! It is not obvious that the value of the expression is real or rational, let alone an integer. It is plain that a similar phenomenon will occur whenever R = r 2 + 4q3 /27 is negative. Every cubic has a real root, and the cubic formula involves
CLASSICAL FORMULAS
47
Our 16th century ancestors were mystified by the phenomenon illustrated in Example 17. At that time, imaginary roots of quadratics (indeed, even negative roots of quadratics) were generally ignored. For example, to find the sides a and b of a rectangle having area A and perimeter p, one converts the equations A = ab
p = 2a + 2b , into the quadratic equation 2a 2 — pa + 2A = 0 having roots
a = 1 (13 ± V p 2 — 16 A) . If p2 — 16A is negative, then it is natural to say that there is no rectangle having the given area and perimeter. One would not invent complex numbers to find some ethereal rectangle living somewhere beyond the realm of the senses. But how can one explain square roots of negative numbers occurring in Example 17? The importance of the cubic formula in the history of mathematics is that such examples forced our ancestors to deal with complex numbers. We shall return to this point when we discuss the Casus Irreducibilis (Theorem 102).
Remark. There is a trigonometric solution to the cubic, due to Viete, that does give the roots of f (x) = x 3 + qx + r in recognizable form (there is a proof in [Rotman, A First Course in Abstract Algebra]). If all the roots of f (x) are real, then they are
t cos(a/3),
t cos(a/3 + 27r/3),
t cos(a/3 + 47r/3),
where t = 4/4q/3 [q must be negative in this case] and cos (a) = —4r I t 3 . If f (x) has complex roots, then there are two possibilities, depending on the sign of —4q/3. If —4q/3 > 0, then the real root of f (x) is
t cosh(P/3), where cosh(18) = —40 3 . If —4q/3 F[x]l(p*) with c + (p) 1> o  (c) + (p*) for all c E F and x + (p) 1—* x + (p*). Define as the composite
a
F (f3) 422* F[x]l (p)E* r[x]l(p*) —'L> 4 F'(,').
Using Theorem 47, it is easy to see that a is an isomorphism extending a that sends fi to r. The uniqueness of follows from Exercise 73. •
a
We now extend Lemma 50 so that it treats not necessarily irreducible polynomials. The second part of it introduces a new kind of polynomial. Definition. Let f (x) distinct) irreducibles:
E F[x]
have the factorization into (not necessarily
f (x) = api(x)   • Pt (x),
where a E F; then f (x) is separable if each p 1 (x) has no repeated roots. Let F be a field and let q(x) E F[x] be irreducible. If the derivative q' (x) is not the zero polynomial, then its degree is smaller than the degree of q(x); hence (q, q') = 1 and q(x) is separable, by Exercise 44. It follows that if F has characteristic 0, then every nonconstant polynomial is separable; if F has characteristic p, then it is possible that q' = 0 (see Example 21 below). Fields in which every nonconstant polynomial is separable are called perfect. Thus, every field of characteristic 0 is perfect; Exercise 77 asks the reader to prove that every finite field is perfect.
56
GALOIS THEORY
Definition. If E I F is an extension, then a E E is called separable if either it is transcendental or its irreducible polynomial is separable; an extension is called separable if every one of its elements is separable. Example 21. Here is an example of an inseparable extension. Let K = Zp (t), the field of all rational functions over Zp. The polynomial q(x) = xP — t E K[x] is irreducible over K (see Exercise 75). Its splitting field E I K is not separable: if a E E is a root of q(x), then xP — t = (x — a)P in E[x] because E has characteristic p. Note that q' (x) = pxP 1 = 0. Remark. When studying infinite fields of characteristic p > 0, it is important to know whether a field extension E I F is separable. One can prove that E, = [a E E : a is separable over F1, called the separable closure of F in E, is a subfield of E. An element a E E is called purely inseparable if its irreducible polynomial factors as (x — a )m over some splitting field (m must be a power of p); an extension K I F is called purely inseparable if every a E K is purely inseparable over F. Now EIE, is purely inseparable, and so every extension ELF is a separable extension E, I F followed by a purely inseparable extension EIE„. When EIF is finite, then the degrees of these extensions are useful (for proofs of these results, see van der Waerden, Modern Algebra). Theorem 51. Let a : F —> F' be an isomorphism of fields, let f (x) F[x], and let f* (x) = o  *(f (x)) be the corresponding polynomial in fy[x]; let E be a splitting field of f (x) over F and let E' be a splitting field of f* (x) over F'. (i) There is an isomorphism ii : E —> E' extending a. (ii) If f (x) is separable, then a has exactly [E: F] extensions
1 F
a
1,
a.
E
SPLITTING FIELDS
57
Proof. (i) The proof is by induction on [E : F]. If [E : F] = 1, then E = F and f (x) is a product of linear factors in F[x]; it follows that f*(x) is also a product of linear factors, and so E' = F'; therefore, we may define = a. If FE: F] > 1, choose an irreducible factor p(x) of f (x) having degree > 2, and choose a root fi of p(x), hence a root of f (x), which must be in E. Let p*(x) E P[x] correspond to p(x), and let /3' E E' be a root of p*(x). By Lemma 50, for each such /3' there is a unique isomorphism : F(3) F (8') extending a with &(,6) = fi'. Now E is a splitting field of f (x) over F(13) and E` is a splitting field of f*(x) over F(8'). Since [E : F] = [E : F(f4)][F(13) : F], and since [F(13) : F] > 2, it follows that [E : F(I3)] < [E : F]. By induction, there exists a : E —> E' extending a , hence extending a. (ii) This proof, a modification of that in part (i), also proceeds by induction on FE: F]. If [E : F] = 1, then E = F and there is only one extension a of a, namely, a itself. If FE: F] > 1, let f (x) = p(x)g(x), where p(x) is irreducible of degree d, say. If d = 1, then we may replace f (x) by g(x) without changing the problem. If d > 1, choose a root 13 of p(x). If a is any extension of a to E, then o (fi) is a root fi' of p* (x); since f *(x) is separable, p*(x) has exactly d roots 13' E El; by Lemma 50, there are exactly d isomorphisms a : F(fi) F'(fi') extending a, one for each /3'. Now E is a splitting field of f (x) over F(18) and E' is a splitting field of f * (x) over F'(13'). Since [E : F(,6)] = [E : F]/d, induction shows that each of the d isomorphisms a has exactly [E : F]Id extensions to E; therefore, a has exactly [E : F] extensions a , because every r extending a has r I F (fi) = some • Corollary 52. If f (x) E F[x], then any two splitting fields of f (x) over F are isomorphic by an isomorphism fixing F pointwise. Proof. In Theorem 51(i), choose F = F' and a the identity on F. • Corollary 53 (E.H. Moore). Any two finite fields of order q = pn are isomorphic. Proof. Any field F of order q is the splitting field of xq — x over Zp , as we saw in Theorem 33. • One calls the field of order pn the Galois field of this order and denotes it by GF(pn), although GF(p) is usually denoted by Z p . Another common notation for the field with q = pn elements is Fq . Both x3 +x +1 and x 3 + x 2 + 1 are irreducible in Z2[x], for neither has a root in Z2. Hence, 7Z2[x]/(x 3 + x + 1) and Z2lx[/(x 3 +x 2 + 1) are fields
58
GALOIS THEORY
of degree 3 over Z2; that is, both fields have 2 3 = 8 elements. By Moore's theorem, both of these fields are isomorphic. More generally, one sees that if f (x) and g(x) are irreducible polynomials over Z i, which have the same degree, then Zp [x ]/ ( f (x)) and Zp [x ] 1 (g (x)) are isomorphic.
Exercises
72.
(i) Let El F be an extension, and let a, 13 E E be algebraic elements over F. If a 0 0, prove that a + 13, afi, and a 1 are all algebraic over F. (Hint. Use Lemma 49 to prove that F(a,13) is a finitedimensional vector space over F.) (ii) If E / F is an extension, prove that the subset K = {a
E
E : a is algebraic over F)
is a subfield of E containing F. (iii) Define the algebraic numbers A to be the set of all those complex numbers that are algebraic over Q. Prove that A/Q is an algebraic extension that is not finite.
73. Let F be a field. Prove that if a is an isomorphism of F(a 1 , .. . , an ) with itself such that o (cri) = ai for i = 1, . . . , n, and a (c) = c for all c c F, then a is the identity. Conclude that if E is a field extension of F and if a, r : F(ai, ... , a n )  E fix F pointwise and a(a1 ) = t(ai) for all i, then a = T. 74. If FcB cE are fields and E1F is finite, then both EIB and B/F are finite, and [E: F] = [E: 13][13 : F]. 75. If K = Z p (t), prove that f (x) = xi' — t is irreducible in K[x]. (Hint. If E I K is a splitting field of f(x), then xP — t = (x — a)P for some a E E.) 76. Show that a field F of characteristic p is perfect if and only if every element of F has a pth root in F. 77. Show that every finite field F is perfect. (Hint: The function a 1> aP is always an injection F 4 F.)
THE GALOIS GROUP
59
The Galois Group We now set up an analogy with symmetries of polygons in the plane even though some of the algebraic analogues have not yet been defined. polygon P plane Vert(P) = {v1,  • • , Vn} linear transformation orthogonal transformation E(P) regular polygon8
polynomial f (x) E F[x] splitting field E of f (x) roots a 1 , .. • , an automorphism of E automorphism of E fixing F Galois group Gal( f) = Gal(E/F) irreducible polynomial
In the geometric setting, we saw that if P has n vertices, then E( P) is isomorphic to a subgroup of Sn , but we did not, in general, compute I E(P)I more precisely. We did see, in Theorem 4, that different types of triangles have nonisomorphic symmetry groups. Definition. If E is a field, then an automorphism of E is an isomorphism of E with itself. If E I F is a field extension, then an automorphism a of E fixes F pointwise if a(c) = c for every c E F. The next lemma, though very easy to prove, is fundamental; it is the analogue of Lemma 2. Lemma 54. Let f (x)
E F[x] and let E I F be an extension field of F. If
a : E > E is an automorphism fixing F pointwise, and ifo t E E is a root of f (x), then a (a) is also a root of f (x).
Proof. Let f (x) = co ± cix ± •  • ± c nxn , so that co ± cia ± • • • ± cn an = O. Applying a gives a(co ) + a (cl)a (a) ± • • • ± a (cn )a (a)n = co ± cia (a) ± • • • ± cn a (a)n = 0,
because a fixes F. Therefore, o (a) is a root of f (x). • 7 Since splitting fields of various polynomials do not all have the same dimension, this
analogy can be improved by considering polyhedra in higher dimensional euclidean space as well as polygons in the plane. 8 See Exercise 2 and Exercise 79.
60
GALOIS THEORY
Definition.9 Let E I F be a field extension. Its Galois group is Gal(E/ F) = (automorphisms a of E fixing F pointwise} under the binary operation of composition. If f (x) E F[x] has splitting field E, then the Galois group of f (x) is Gal(E/F). It is easy to check that Gal(E/F) is a group; it is a subgroup of the group of all automorphisms of E. Theorem 55. If f (x) E F[x] has n distinct roots in its splitting field E, then Gal(E/F) is isomorphic to a subgroup of the symmetric group S n, and so its order is a divisor of n!. Proof. Let X = {al, ... , a n } be the set of all the roots of f (x) in E. By Lemma 54, if a E Gal(E/F), then a (X) = X. The map Gal(E/F) —* Sx defined by a 14 a IX is easily seen to be a homomorphism; it is an injection, by Exercise 73. Finally, Sx L' Sn . • For example, the Galois group of a quartic polynomial is a subgroup of 54, and the Galois group of a quintic polynomial is a subgroup of 55. Example 22. The splitting field of x 2 ± 1 over R is, of course, C, and I Gal(C/R)I _< 2, by Theorem 55. In fact, I Gal(C/R)I = 2 because the group contains the automorphism o : z = a ± ib i—> z = a — ib.
Notice that a : i 1> —i and —i 1÷ i, and so it interchanges the roots. One sees that the elements of the Galois group should be regarded as generalizations of complex conjugation. Theorem 56. If f (x) E F[x] is a separable polynomial and if E I F is its splitting field, then I Gal(E I F)I = [E: F]. Proof. By Theorem 51(ii) with F = F',E=E',anda:F> F the identity, there are exactly [E: F] automorphisms of E that fix F. • Since x2 ± 1 is separable and [C : WI = 2, we see once again that I Gal(C/R) I = 2. 9 This is not the definition of Galois; it is the modern version introduced by E. Artin around 1930, and it is isomorphic to the original version.
THE GALOIS GROUP
61
Example 23. Let f (x) = x 3 — 1 E Q[x]; f (x) is separable, because Q has characteristic 0. Now f (x) = (x — 1)(x2 ± x ± 1) is a factorization of f (x) into irreducibles. If E is the splitting field of f (x) over Q, then E = Q(1, co, w2) = Q(co), where w is a primitive cube root of unity, that is, co is a root of x 2 ± x ± 1. Since x 2 + x +1 is an irreducible polynomial of degree 2, we have 2 = [Q(co) : Q] = [E : (I)] = I Gal(E/Q)I, by Theorem 56. Therefore, the Galois group is cyclic of order 2. Its generator takes wi—>. co2 = Fo; that is, the generator is complex conjugation. Example 24. Let g(x) = x 3 — 2 E Q[x]. The roots of g(x) are a, coa, and co2a, where a = .ii is the real cube root of 2 and co = e23 is a primitive cube root of unity, and so the splitting field E of g(x) is E = Q(a, wa, co 2a). We claim that E = Q(a, co): E c Q(a, w) because a, wa, co 2a E Q(a, w); Q(a, w) c E because co = wa la E E. Since g(x) is irreducible over Q, we have [Q(a) : Q] = 3. But Q(a) consists wholly of real numbers, and so it cannot be the splitting field E of g(x). Hence, [E : Q(a)] > 1, and I Gal(E/Q)I = [E : Q] = [E : Q(a)][Q(a) : Q] = 3[E : Q(u)] > 3; it follows that Gal(E/Q) L' S3, by Theorem 55. Lemma 57. Let FCBc E be a tower of fields with B/F the splitting field of some polynomial f (x) E F[x]. If a E Gal(E/F), then a IB Gal(B/F).
E
Proof. It suffices to prove that a (B) = B. If al, ... , an are the distinct roots of f (x), then B = F (al, . . . , an ). Now a (F) = F, and a(a1) E B for all i, by Lemma 54; it follows that a(B) = cr(F (al, • • • , an)) = F(a (al), ... , a(a n )) = B, as desired. • Theorem 58. Let Fc Bc E be a tower of fields with B/F the splitting field of some polynomial f (x) E F[x] and E I F the splitting field of some g(x) E F[x]. Then Gal(E/B) is a normal subgroup of Gal(E/F), and Gal(E/F)/ Gal(E/B)P.=. _%' Gal(B/F).
62
GALOIS THEORY
Proof. Define * : Gal(E/F) > Gal(B/F) by a 1* a113; Lemma 57 says that * does take its values in Gal(B/F). It is easily seen that * is a homomorphism with ker * = Gal(E/B) [if a 1B = identity, then a is an automorphism of E fixing 13], so that the latter is a normal subgroup of Gal(E/F). If r E Gal(B/F), then Theorem 51 shows that there is an automorphism f of E with *(f ) = f IB = r. Hence * is surjective, and the first isomorphism theorem for groups gives the result. • Remark. The hypothesis that E/ F is a splitting field enters only in showing that * is surjective. Without this hypothesis, one can prove only that the quotient group is isomorphic to a subgroup of Gal(B/F). Example 25. The Galois group of f (x) = x 3 — 2 E Q[x] was computed in Example 24. If a = ,N3,h and w = e27ri /3 , then we have seen that E = Q(a, co) is a splitting field of f (x) over Q, that [E : Q] = 6, and that Gal(E/Q)C._. S3. Let us now view this example in light of Theorem 58. Consider the tower of fields
Q C Q(w) c E . Since Q(w)/Q is a splitting field (of x 2 + x + 1), Theorem 58 gives Gal(E/Q(w)) < Gal(E/Q) and Gal(E/Q)/ Gal(E/Q(w))L' Gal(Q(co))/Q). Now Gal(Q(co)/Q) has order 2, because x 2 +x +1 is an irreducible polynomial of degree 2 (it has no rational roots). We claim that Gal(E/Q(co)) has order 3. A moment's thought shows that none of the roots a, coot, and co2a of x3 — 2 lies in Q(co). Since a cubic is irreducible over a field F if it has no roots in F, we see that x 3 —2 is irreducible over Q(co). By Theorem 45, [E : Q(co)] = 3, and by Theorem 56, I Gal(E/Q(a)))1 = 3. Therefore, in this case, the isomorphism of Theorem 58 is just S3/A 3 25 Z2. Note that A3 = (a), where a(w) = co and a : a 1> coa . Hence, a (coot) = co2a and a (co2a) = a, so that a is a 3cycle.
ROOTS OF UNITY
63
Exercises
78. Let f (x) E F[x] be an irreducible polynomial of degree n, and let E / F be a splitting field of f (x). (i) Prove that n I [E: F]. (ii) Prove that if f (x) is separable, then n I I Gal(E/F) I.
79. Let f (x)
F[x], let ELF be a splitting field, and let G = Gal(E/F) be the Galois group. E
(i) If f (x) is irreducible, then G acts transitively on the set of all roots of f (x) (if a and /3 are any two roots of f (x) in E, there exists a E G with a(a) = 13). (Hint: Lemma 50.) (ii) If f (x) has no repeated roots and G acts transitively on the roots, then f (x) is irreducible. Conclude, after comparing with Exercise 2, that irreducible polynomials are analogous to regular polygons. (Hint: If f (x) = g(x)h(x), then the gcd (g(x), h(x)) = 1; if a is a root of g(x) such that o (a) is a root of h(x), then a (a) is a common root of g(x) and h(x).)
80. Let E be the splitting field of 1(x) = x 4 — 10x 2 +1 over Q. Find Gal(E/Q). (Hint. See Exercise 67 and Example 20. The roots of f (x) are
Vi+An, ..n../5, ,./I+,1J, .5A./J.)
Roots of Unity The simplest field extensions of a field F are those in which we adjoin an nth root of an element c E F. To investigate these, it will be valuable for us to consider roots of unity. After all, if an = c, then the other nth roots of c are of the form coa, where w is some nth root of unity, and so it will be relevant whether or not F contains such roots of unity. Note that R, for example, contains the square roots of unity, namely, 1 and —1, but it contains no higher roots of unity (other than 1) because they are all complex. We begin with a preliminary discussion from group theory. Lemma 59. JIG = (a) is a cyclic group of order n and generator a, then C has a unique subgroup of order d for each divisor d of n, and this subgroup is cyclic.
64
GALOIS THEORY
Proof. If n = cd, we show that ac has order d (and so (a') is a subgroup of order d). Clearly (c/c)d = acd = a n = 1; we claim that d is the smallest such power. If (ac)r = 1, then n I cr [Theorem G.2(i)]; hence cr = ns = dcs for some integer s, and r = ds > d. To prove uniqueness, assume that (x) is a subgroup of order d (recall that every subgroup of a cyclic group is cyclic, by Theorem G.1). Now x = am and 1 = X d = amd ; hence md = nk for some integer k. Therefore, x = am = ( an/c 1)k = ( ay , so that (x) c (ac ). Since both subgroups have the same order d, it follows that (x) = (ac). • Recall Theorem G.2(ii): If C is a cyclic group with generator x and order n, then X k is also a generator of C if and only if k and n are relatively prime. It follows that if g(C) denotes the set of all generators of C, then Ig(C)I =
where v is Euler's function. Theorem 60. If n is a positive integer, then n=
E v (d). din
Proof. If G is a group, then it is easy to see that it is the disjoint union G=U g(C),
where C ranges over all the cyclic subgroups of G, because each element of G generates a unique cyclic subgroup. If G has order n, then counting gives n = E Ig(C)I = E co(d), where the summation ranges over all cyclic subgroups C of G, while if G is cyclic, then the lemma gives
E Ig(C)1 = E go(d), C
din
for there is exactly one cyclic subgroup of G for every divisor d of I GI. • Theorem 61. A group G of order n is cyclic if and only if for each divisor d of n, there is at most one cyclic subgroup of order d.
Proof. If G is cyclic, then the result follows from Lemma 59. Conversely, write G as a disjoint union (as in the preceding proof): G =U g(C). Hence n = IGI = E Ig (C)I, where the summation is over all cyclic subgroups
ROOTS OF UNITY
65
C of G. Since G has at most one cyclic subgroup of order d, Theorem 60
gives n=
E Ig(C)1 < E v (d) = n.
Therefore, G has exactly one cyclic subgroup of order d for every divisor d of n; in particular, there is a cyclic subgroup of order n, and so G is cyclic. • Theorem 62. If F is afield with multiplicative group 0 = F — {0}, then every finite subgroup G of F * is cyclic.
Proof. Suppose that IGI = n and d I n. If C is a cyclic subgroup of G of order d, then Lagrange's theorem gives xd = 1 for each of the d elements x c C. Were there a second cyclic subgroup of order d, then it would have at least one element not in C, so that G would contain at least d+1 elements x with X d = 1. But the polynomial X d — I has at most d roots in a field, and so G has at most one cyclic subgroup of order d. Theorem 61 now shows that G is cyclic. • Corollary 63. If n is a fixed positive integer, then all the nth roots of unity in afield F form a cyclic multiplicative group.
Corollary 64. If F is a finite field, then F * is cyclic and F = Z,, (a)for some a.
Proof. If IFl = q, take a to be a primitive (q — 1)st root of unity. • Remark. Exercise 81 shows that if F is an infinite field, then 0 is never a cyclic group. Definition. If F is a finite field of characteristic p, then an element a c F is called a primitive element if F = Zp (a). It follows that any generator of the multiplicative group F # is a primitive element. Let us call an element a in a field F a square if there is u E F with a = u2 ; i.e., a has a square root in F. In R, an element is a nonsquare if and only if it is negative, and so the product of two nonsquares, being positive, is a square. On the other hand, neither 2 nor 3 is a square in Q, and their product 6 is also not a square. The next corollary shows that, in this respect, finite fields behave more like the reals than the rationals. Corollary 65. If F is a finite field and a, b product ab is a square.
E
F are not squares, then their
66
GALOIS THEORY
Proof. If a is a primitive element of F, then every nonzero element of F has the form ak for some integer k, and ak is a square if and only if k is even. Since a and b are not squares, we have a = ak and b = am where both k and m are odd. Therefore, ab = a k±m is a square, because k + m is even. • There is a sophisticated proof of the next example which uses an analysis of Galois groups in algebraic number theory; the following elementary proof is due to G. J. Janusz. Example 26. The polynomial f (x) = x4 — 10x2 + 1 E Q[x] is irreducible, but it factors in Zp [xl for every prime p. We know that f (x) is irreducible in Q[x], by Exercise 67. Completing the square gives x4 — 10x 2 + 1 = x4 — 10x2 + 25 — 24 = (x 2 — 5) 2 — 24. Now regard the coefficients of f (x) as lying in Zp for some prime p. If V2,71. = 2,./6 lies in Z p , i.e., there is /3 E Zp with /32 = 6, then f (x) factors in Z[x] and we are done: 1°
f (x) = (x 2 — 5 + 2fi) (x 2 — 5 — 2/1) . We may now assume that the quadratic x 2 — 6 has no roots in Z p , and so x 2 — 6 E Z[x] is irreducible. By Theorem 45, E = Z p [x]l (x 2 6) is a field containing an element 13 with /32 = 6, and {1, /3} is a basis of E viewed as a vector space over Z. We claim that 5 + 2/3 is a square in E. Every u E E has a unique expression of the form u = a + b13, where a, b E Zp . If u2 = 5 +2,6, then substituting and equating coefficients gives —
a 2 + 6b 2 = 5 and 2ab = 2. We may assume that p 0 2 because f (x) = x 4 + 1 does factor in Z2 [x], so that ab = 1 and b = a 1 . Hence, a2 + 6a 2 = 5, which we rewrite as 0 = a4 — 5a2 + 6 = (a 2 ___ 2)(a 2 _ 3).
If neither 2 nor 3 is a square in Z p , then Corollary 65 would give 6 a square in Zp , contrary to hypothesis. Therefore, at least one of 2 or 3 is a square. 10 1n Z5, the element [6] = [1] is a square, but [6] is not a square in 17, for [1], [4], and [2] = [9] are the only squares in Z7.
ROOTS OF UNITY
67
We have shown that there exists a E Zp with (a + a 1 /3)2 = 5 + 2/3. It now follows that (a  a 1 13)2 = 5  2/3. Therefore, f (x) = (x 2  5 + 2fi)(x 2  5  2/3) [x 2 ___ (a _ a 1 0) 2 ][x 2 _ (a + a 1 fi) 2] = (x + a  a 1 13)(x  a + a 1 13)(x + a + a 1 fi)(x  a (x  a  a  1 fl)(x  a + a  1 fl)(x + a + a  1 13)(x + a  a  1 fl) = [ (x _ 0 2 _ a2fi2] [(x + a)2 _ a 2fl2] = [(x _ 02 _ 6a 21[(x + 12)2 —6a2] We have factored f (x) in Z p [x]. • Lemma 66. If a is a primitive element of GF(pn), then a is a root of an irreducible polynomial in Z[x] of degree n. Proof. If the irreducible polynomial of a over Zp has degree d, then Zp (a) has order p'. But this subfield is all of GF(pn) because a is a primitive element; hence d = n. • It follows from the existence of G F (pn) that there exist irreducible polynomials in Zp [x] of degree n for every n > 1. Theorem 67. Gal(GF(pn)1 GF(P)) a = Z n with generator u 1> u 1 . Remark. This generator is called the Frobenius automorphism. Proof. Denote GF(pn) by K and denote the Galois group by G. If a is a primitive element, then Lemma 66 says its irreducible polynomial q(x) has degree n, and so K contains at most n of its roots. If a E G, then a is completely determined by a (a) [because every nonzero element of K has the form a i and a (ai ) = a (a) i ]. But a (a) is a root of q(x) [which has at most a(q) = n roots], by Lemma 54; it follows that IG I < n. On the other hand, a : u 1> uP does lie in G, by Lemma 32; moreover, if j < n, then a j 0 1 (otherwise uP3 = u for all u, and K would contain pn roots of xr''  x, a contradiction). Thus, a has order at least n, and so G = (a) is cyclic of order n. • Lemma 68. Let n be a positive integer and let F be a field. If the characteristic of F is either 0 or is a prime not dividing n, then xn  1 has n distinct roots in a splitting field.
68
GALOIS THEORY
Proof. If f (x) = xn 1, then its derivative r (x) = nxn 1 . By hypothesis, this is not zero, and so the gcd (f, f') = 1; therefore, f (x) has no repeated roots. • Lemma 68 fails in a field F of characteristic p when p I n; for example, xP  1 = (x  1)P has at most one root (of multiplicity p) in any extension field of F, so that xkP  1 = (x"  1)" has repeated roots for all k > 1. Definition. Let n be a fixed positive integer and let F be a field. A generator of the group of all nth roots of unity is called a primitive root of unity. A primitive nth root of unity in C is e27ti/ 71 . Recall that if R is a ring, then U (R) denotes its multiplicative group of units. In particular, U(Z) = f [i] E Zn : (i, n) = 1). When p is a prime, therefore, U(Z) = rp , the multiplicative group of all nonzero elements. Theorem 69. If F is afield and E = F (a), where a is a primitive nth root of unity, then Gal(E/F) is isomorphic to a subgroup of U (Z n ), and hence Gal(E/F) is an abelian group. Proof. Since E = F (a), each a E Gal(E/F) is completely determined by its value on a. Now a (a) = a i for some i which is unique mod n, and we denote a by ai , where 0 < i < n  1. Theorem 0.2(ii) says that i must be relatively prime to n, for a I(a) is an automorphism of (a). Therefore, the function * : a, > [i] is a function * : Gal(E/F) U(Z n ). Now * is a homomorphism: oj oi (a) = oi(a i ) =
Hence, * (aj ai ) = [ji] = (ai )* (oi ). This map is injective, by Exercise 73. Therefore Gal(E/F) is isomorphic to a subgroup of U(4). • The multiplicative group U (Zn ) need not be cyclic; for example, U (Z8 ) consists of the congruence classes [1], [3], [5],[7], and it is isomorphic to the 4group. There is a deep partial converse of Theorem 69. The KroneckerWeber Theorem states that every finite abelian extension of Q [that is, a finite extension E/Q with Gal(E/Q) abelian] can be imbedded in a cyclotomic extension Q(co), where co is some root of unity.
ROOTS OF UNITY
69
Example 27. If p is a prime, then 4 = e2"j /P is a primitive pth root of unity over Q. As in the proof of Theorem 69, Q(4') is the splitting field of the cyclotomic polynomial O p (x) over Q. Since O p (x) is irreducible over Q, by Corollary 41, we have I Gal(Q()/Q)1 = p — 1. The remark after Theorem 69 shows that Gal(Q(4')/Q) is isomorphic to a subgroup of U(Z) = Z. By Corollary 64, the latter group is cyclic, being the multiplicative group of a field. Indeed, Ga1(Q()/Q) rp , by Theorem 56. The homomorphism fr occurring in Theorem 69 takes values in the multiplicative group U (Zn ). Here is a variant which takes values in the additive group Z. Theorem 70. Let F contain a primitive nth root of unity, and let f (x) = xn — c E F[x]. If E I F is a splitting field of f (x), then there is an injection 99 : G = Gal(E/F) —> Zn . Moreover, 1(x) is irreducible if and only if is surjective.
Proof. If co is a primitive nth root of unity and if a is a root of f (x), then an = c, and the list of all the roots of 1(x) is a, aco, . . . ,acon 1 . If a c G, then a. (a) = acoi , and a is completely determined by i; define (p(cr) = [i] if a (a) = a co i . We now show that q) : G —> Zn is a homomorphism, where Zn is the additive group. If r e G, then “‘I)) = co (because CO E F) and r (a) = acoi for some j. Hence, 0" : a 1—> a W i 1—>
r (aw l ) r(a)r(d) [acoi ]col a ct) j+i ,
so that q)(rcr) = [j + i] = (p(r) + cp(a). Therefore is a homomorphism; it is an injection, by Exercise 73. Now (p is surjective if and only if G acts transitively on the roots of f (x). By Exercise 79, this is equivalent to the irreducibility of f (x). • Corollary 71. Let p be a prime, let F be afield containing a primitive pth root of unity, and let f (x) = xP — c E F[x] have splitting field E. Then either f (x) splits and Ga1(E/F) = 1 or it is irreducible and Gal(E/F)
Z.
70
GALOIS THEORY
Proof. Consider the map Gal(E/F) —> Z p of the theorem. If f (x) splits, then Gal(E/F) = 1 and its image is trivial; if f (x) does not split, then its image is a nontrivial subgroup of Z. But Z p has no proper nontrivial subgroups, so that the map must be surjective, Gal(E/F) :=1' Z p , and f (x) is irreducible. • There is an elementary proof of this corollary that does not assume F contains roots of unity. Corollary 72. If p is a prime, F is afield of characteristic 0, and f (x) = xP — c E F[x], then either f (x) is irreducible in F[x] or c has a pth root in F. Proof." A splitting field E I F of f (x) contains an element a with aP = c as well as a pth root of unity co, for if the roots a, coa, . . . , coPl a of f (x) lie in E, then co = (coot) I a lies in E as well. If f (x) is not irreducible in F[x], then there is a factorization f (x) = g(x)h(x) in F[x] with each factor g(x) and h(x) having degree less than p; let 0 (g) = k < p. The constant term b of g(x) is, to sign, the product of some of the roots of f (x) (perhaps with multiplicity), so that ±b = a k co' E F for some integer m. Since corn is a pth root of unity and aP = c, we have
(±b)" = akil = ck . Since p is prime and k < p, we have (k, p) = 1, and so there are integers s and t with sk ± tp = 1. Therefore, C = C sk+IP = C ks C Pt = (+b) Ps C Pt = R±b) s Ct 1 P
,
and c has a pth root in F. •
Exercises
81. Prove that if F is an infinite field, then its multiplicative group Flt is never cyclic. (Hint. To eliminate the possibility PI = (u), consider the cases of characteristic 0 and characteristic p > 0 separately; the latter case should be further subdivided into cases: u transcendental over the prime field Z p u algebraic over Zr.) and 11 The proof works if F has characteristic q 0 p.
SOLVABILITY BY RADICALS
71
Solvability by Radicals We now show how the Galois group takes account of there being a formula for the roots of a polynomial that involves only the field operations and taking square roots, cube roots, etc.
Definition. A field extension B/F is a pure extension of type m if B = F (a), where am E F for some positive integer m. A tower of fields F= Bo C Bi C
Br
is a radical tower if each Bi+1 1 Bi is a pure extension. In this case, we call Bt I F a radical extension of F.
Definition. If f (x) E F[x], then f (x) is solvable by radicals over F if there is a radical extension B/F which contains a splitting field E of f (x) over F. We illustrate the definition of solvability by radicals by showing that quadratics, cubics, and quartics over fields of characteristic 0 are solvable by radicals (these formulas are not true for arbitrary fields; for example, the quadratic formula cannot hold when the field of coefficients has characteristic 2). If f (x) = x2 + bx + c E C[X], define
F = Q(b, c) and B = F(1b2
—
4c).
Then B/F is a pure extension of type 2, and B is the splitting field of f (x) over F; therefore, f (x) is solvable by radicals over F. If f (x) = x 3 +qx +r c C[x], define F = Q(q, r), define
B1 = F (■,/r 2 + 4q3 /27), which is pure of type 2, and define B2 = B1 (y), where y3
= 1(—r + ‘,/r2 + 4q3 /27),
which is pure of type 3. The cubic formula says that the roots of f (x) are y + z, coy + w 2 z, and co2 y + coz, where yz = —q/3 (so that Z E B2), and co is a primitive cube roots of unity. Therefore, if we define B3 =
72
GALOIS THEORY
B2(w), which is pure of type 3, then the splitting field E of f (x) is contained in B3, and f (x) is solvable by radicals. Note that it is possible that E is a proper subfield of B3, for E need not contain w; for example, f (x) may have three real roots in which case E is a subfield of R. If f (x) = x 4 + qx 2 + rx + s E C[X], define F = Q(q, r, s). In
the discussion of the quartic formula, we saw that it suffices to find three numbers k, f, and m. Now k 2 is a root of a certain cubic polynomial in F[x], so that there is a radical tower F c B1 C B2 C B3 with k2 E B3. Define B4 = B3(k). Since 2m = k2 + qlrlk and 21 = k2 + q — rlk, B4 contains./ and m. The quartic formula gives the roots of f (x) as the roots of (x 2 + kx + f)(x 2 — kx + m),
so that the radical tower can be lengthened two steps, each of type 2, by adjoining N/k 2 — 41 and ,,,/k 2 — 4m, with the last extension, B6, containing the splitting field of f (x). Therefore, f (x) is solvable by radicals. It should be plain that, conversely, if a polynomial f (x) is solvable by radicals, then there is an expression for its roots in terms of its coefficients, the field operations, and extraction of roots. Recall that a finite group is solvable if it has a normal series with abelian factor groups; moreover Theorem 0.20 shows that every quotient and every subgroup of a solvable group is itself solvable. Lemma 73. Let F be afield of characteristic 0,.let f (x) E fix] be solvable by radicals, and let E be a splitting field of f (x) over F. (i) There is a radical tower F = Ro C Ri C • • • C Rt with E C R t , with R t a splitting field of some polynomial over F, and with each R i I Ri_i is a pure extension of prime type p i . (ii) If R i I F is a radical extension as in part (i), and if F contains the pi th roots of unity for all i, then Gal(E/F) is a solvable group.
Proof. (i) Since f is solvable by radicals, there is a radical tower F = Bo C Bi C c B e
with E c B. By Exercise 83, there is an extension K/B t which is a splitting field of some polynomial in F[x), and by Exercise 85, K I F is also a radical extension. Of course, E C Bt C K. Finally, Exercise 82 says that a radical tower from F to K can be refined so that each step is a pure extension of prime type.
SOLVABILITY BY RADICALS
73
(ii) Let
F = Ro C
C • . • C Rt
be a radical tower as in part (i), and define
G i = Gal(Rt /Ri ). By hypothesis, F contains all the pi th roots of unity, so that each Ri is a splitting field of a polynomial over Ri_1. Thus, the hypothesis of Theorem 58 holds and Gal(Rt /F) = Go D Gi D
D G t = {1}
is a normal series. The factor groups Gal( Rt 1 R 1 _i)1 Gal(Rt I Ri ) of this normal series are isomorphic to Gal(Rt /Ri _ ), by Theorem 58, and these last groups are cyclic of prime order, by Corollary 71. Hence, Gal(R t /F) is a solvable group. Finally, applying Theorem 58 to the tower of fields
FcEcRi , we see that Gal(E/F) is a quotient group of the solvable group Gal(R t /F), and so it, too, is solvable, by Theorem G.20. • We now remove the hypothesis that the base field F contain certain roots of unity.
Let f (x) c F[x] be solvable by radicals over a field F of characteristic 0, and let E I F be its splitting field. Then Gal(E/F) is a solvable group. Theorem 74.
Proof. By hypothesis, there is a radical tower
F = Ro C Ri C • • C Rt, with E C R t . By Lemma 73(i), we may assume that each R/R_ 1 is of prime type pi and that Rt I F is a splitting field of some polynomial h(x) E F[x]. Let m be the lcm of the pi 's, and let co be a primitive mth root of unity. The tower can be lengthened by Rt C = R 1 (w), and then refined so that each pure extension in it has prime type. Observe that R' is a splitting field of (V" — 1)h(x) E F[x]. Construct a new tower by adjoining co first:
F = Ro F(co) c RI(w) c
c R t (co) = .
74
GALOIS THEORY
Notice that each extension in this tower is pure and that E, the splitting field of f (x), is contained in R'. Since F (w) F is a splitting field, Theorem 58 gives Gal(R' I F (co)) 1 Gal(R'/F) and Gal(k/F)/ Gal(R7F(w))
Gal(F(co)1 F).
Now Gal(F ((D)/ F) is abelian, hence solvable, by Theorem 69. Each step of the truncated tower F(w) C R 1 (w) C . . . C R(w) = is a pure extension of prime type, so that Theorem 73(ii), which applies because F(w) contains the necessary roots of unity, shows that the normal subgroup Gal(RVF(co)) of Gal(R'/F) is solvable. Therefore, Gal(R'/F) is solvable, by Theorem G.21. Finally, Theorem 58 applies to show that Gal(E/F) is a quotient group of the solvable group Gal(R'/F), hence is solvable, by Theorem G.20. • Of course, this last theorem gives the etymology of the word solvable in group theory. Theorem 75 (Abel—Ruffini). There exists a quintic polynomial f (x) Q[x] that is not solvable by radicals.
E
Proof. If f (x) = x 5 — 4x ± 2, then f (x) is irreducible over Q, by Eisenstein's criterion. Let E/Q be the splitting field of f (x) contained in C, and let G = Gal(E/Q). If a is a root of f (x), then [Q(a) : Q] = 5, and so [E Q] = [E Q(a)][Q(a)
Q] = 5[E : Q(0)1
By Theorem 56, I G I = [E : Q] is divisible by 5. We now use some calculus; f (x) has exactly two critical points, namely, +.946, and f (4/Tir5) 0; it follows easily that f (x) has exactly three real roots (they are, approximately, —1.5185, 0.5085, and 1.2435; the complex roots are —.1168 ± 1.4385i.) Regarding G as a group of permutations on the 5 roots, we note that G contains a 5cycle (it contains an element of order 5, by Cauchy's theorem, and the only elements of order 5 in S5 are 5cycles). The restriction of complex conjugation, call it a, is a transposition, for a interchanges the two complex roots while it fixes the three real roots. By Theorem G.39, S5 is generated by any transposition and any 5cycle, so that G = Gal(E/Q)
55
SOLVABILITY BY RADICALS
75
is not a solvable group, by Theorem G.34, and Theorem 74 shows that f (x) is not solvable by radicals. • Remark. Abel and Ruffini proved that the general quintic is not solvable by radicals; that is, there is no formula that works for an arbitrary quintic when one specializes its coefficients (the classical quadratic, cubic, and quartic formulas are of this sort). Theorem 75 is thus a stronger result than what they had shown.
Exercises
82. If EIF is a radical extension over F, then there is a radical tower F = 130 C Bi C ... C Br
with each [Bi± i : Bii a pure extension of prime type. (Hint. If a" E F and n = pm, then there is a tower of fields F c F(aP) C F(a).) 83. Let B/F be a finite extension. Prove that there is an extension KIB so that K IF is a splitting field of some polynomial f(x) c F[x]. 12 (Hint. Since B/F is finite, it is algebraic, and there are elements al , . . . , an with B = F(ai, ... , an ). If pi (X) E F[x] is the irrreducible polynomial of ai, take K to be a splitting field of f(x) = pi(x) • • • pn(x).)
84.
(i) If B and C are subfields of a field E, then their compositum B v C is the intersection of all the subfields of E containing B and C. Prove that if al , . . . , an E E, then F(al ) v •   v F(an) = F(ai , • • • , an). (ii) Prove that any splitting field K I F containing B (as in Exercise 83) has the form K = Biv ...v B r , where each Bi is isomorphic to B via an isomorphism which fixes F. (Hint: If Gal(K/F) z {ai ,    , ad, then define Bi = cri(B).)
85. Using Exercise 84, prove that any splitting field K I F containing a radical extension R1 1 F (as in Exercise 83) is itself a radical extension. Conclude that, in the definition of solvable by radicals, one can assume that the last field Bt is a splitting field of some polynomial over F. KIF is called a split closure of B/F; if f(x) is a separable polynomial, then a split closure KIF is called a normal closure of B/F. 12 A smallest such extension
76
GALOIS THEORY
Independence of Characters This section introduces the important notion of a fixed field, and characters are used to compute its degree over a base field. Definition. A character of a group G in a field E is a homomorphism 0: G —> E st , where E# = E — {0} is the multiplicative group of E. Definition. A set lab , an} of characters of a group G in a field E is independent 13 if there do not exist a1,... , a„ E E, not all 0, with
E ai ai (x) = 0 for all x e G. Lemma 76 (Dedekind). Every set (al,
, an } of distinct characters of
a group G in afield E is independent.
Proof. The proof is by induction on n. If n = 1, then alai(x) = 0 implies that al = 0 because o i(x) 0 0. Let n > 1 and assume there is an equation (1)
alai (x)
• • + a„an (x) = 0 for all x
E G,
where not all ai = 0. We may assume that every a i 0 0 lest induction apply; multiplying by an1 if necessary, we may further assume that an = 1. Since an 0 al, there exists y E G with on(Y) 0 al (Y). In Eq. (1), replace x by yx to obtain a l ai (y)o 1 (x)  • • ± an_iani(Y)an1(x) an(Y)an(x) = 0
(in this equation, y E G is fixed while x is an arbitrary element of G). Multiply by 0n (y) 1 to obtain an equation alan(Y) l a' (Y)al(x) + • • • +a(x) = 0; subtract this from Eq. (1) to obtain a sum of n — 1 terms al [1 — an(Y) i ai (Y)lat(x)
• • =0.
By induction, each of the coefficients is 0. Since al 0 0, we have 1 = (y)  1 al (y); hence on (y) = r1 (y), a contradiction. • 13 A11 the characters in a field
E form a vector space V (G, E) over E in which a+r:
x
a (x) r(x)
and ca : x
ca(x)
for a scalar c. Independence of characters is linear independence in V (G, E).
INDEPENDENCE OF CHARACTERS
77
Corollary 77. Every set {a1, . . . , an} of distinct automorphisms of a field E is independent.
Proof. An automorphism a of E restricts to a (group) homomorphism a: E#  E#, hence is a character. • Definition. Let Aut(E) be the group of all the automorphisms of a field E. If G is a subset of Aut(E), then E G = {a
E E : o (a) = a
for all o E G}
is called the fixed field. It is easy to see that E G is a subfield of E. The most important instance of this definition is when G is a subgroup of Aut(E), but there is an application when G is only a subset. Note that HcG
implies
EG c EH :
if a € E and a(a) = a for all a E G, then o (a) = a for all a EHc G. Example 28. If E/ F is a field extension with Galois group G = Gal(E/ F), then F c E G C E; we shall presently consider whether EG/F is a proper extension. Example 29. Let E = F(xi, ... , xn ) be the field of rational functions in several variables over a field F. Then G = Sn can be regarded as a subgroup of Aut(E); it acts by permuting the variables. The elements of the fixed field EG are called symmetric filtictions 14 over F. Lemma 78. If G = {al, . . . an } is a set of automorphisms of E, then [E : E G ] > n. 14 Symmetric functions arise naturally: if
f (x) = 11(x — a i) =xn +sn _ixn1 + •  •
+six
+SØ,
then each of the coefficients sj is a symmetric function of the roots al, ... , a n . This observation is the starting point of Lagrange and Galois (see Appendix D).
78
GALOIS THEORY
Proof. Otherwise [E: EG ] = r n; let {co l , . . . , 04, 44 } be linearly independent vectors in E over E G . Consider the system of n equations in n + 1 unknowns: a t (w 1 )x i +... + at (con+t)xn+i = 0 an (COI)X1 ± ' • ' ± an(COn+1)Xn+1 =
0.
There is a nontrivial solution (xi, . . . , xn+t) over E; we proceed to normalize it. Choose a solution having the least number r of nonzero components, say (a 1 , . . . , ar , 0, ... , 0); by reindexing the co i , we may assume that all nonzero components come first. Note that r 0 1 lest al (w1)a1 = 0 imply a l = 0. Multiplying by its inverse if necessary, we may assume that a,. = 1. Not all ai E E G lest the row corresponding to the identity of G violate the linear independence of {col , ... , con). Our last assumption is
GALOIS EXTENSIONS
79
that al does not lie in EG (this, too, can be accomplished by reindexing the co,). There thus exists ak with crk(a1) 0 al. The original system has jth row (1)
oi (co i )a l + • • • + oj (cor _ i )ar _ i + aJ (0)r ) = O.
Apply ak to this system to obtain alai (coi)ak(ai) + • • • + akai(wr 1)ak(ar 1) + aka; (wr) =
0.
Since G is a group, akal , . . . , ak on is just a permutation of al , . . . , a,. Setting ak a; = ai , the system has ith row ai(wi)ak(ai) + • • • + ai (wri)ak(ari) + ai(wr) = O. Subtract this from the ith row of Eq. (1) to obtain a new system with ith row: ai(coi)[al — crk (ai)] + • • • + ai(Wrl)[ar1 — ak(ar1)] = 0.
Since al — ok (a i ) 0 0, we have found a nontrivial solution of the original system having fewer than r nonzero components, a contradiction. • Corollary 80. If G, H are finite subgroups of Aut(E) with E G = E H , then G = H. Proof. If a E G, then clearly a fixes E G . To prove the converse, suppose a fixes E G and a il G. Then EG is fixed by the n +1 elements in G U to  }, so Lemma 78 and Theorem 79 give the contradiction: n = IGI = [E : E G ] > [E : E Gu la i] > n + 1.
Therefore, if a fixes EG, then a E G. If a E G, then a fixes E G = E H , and hence a sion is proved the same way. •
E
H; the reverse inclu
Galois Extensions Our discussion of Galois groups began with a pair of fields, namely, an extension E/ F that is a splitting field of some polynomial f(x) E F[x]. We
80
GALOIS THEORY
are now going to characterize those extension fields of F that are splitting fields of some polynomial in F[x]. Suppose that G = Gal(E/F); it is easy to see that F c EG c E. A natural question is whether F = EG; in general, the answer is no. For example, if F = Q and E = Q(a), where a is the real cube root of 2, then G = Gal(E/F) = Gal(Q(a)/Q) = {1} (if a E G, then o (a) is a root of x 3 —2; but E does not contain the other two (complex) roots of this polynomial). Hence E G = E 0 F. Theorem 81. The following conditions are equivalent for a finite extension E I F with Galois group G = Gal(E/F). (i) F EG ; (ii) every irreducible p(x) E F[x] with one root in E is separable and has all its roots in E; that is, p(x) splits over E; (iii) E is a splitting field of some separable polynomial f (x)
E
F[x].
Proof. (i) = (ii) Let p(x) E F[x] be an irreducible polynomial having a root a in E, and let the distinct elements of the set {a (a) : a E GI be al, , an . Define g(x) E E[x] by g(x) =i1(x — as ). Now each a E G permutes the ai ; so that each a fixes each of the coefficients of g(x) (for the coefficients are symmetric functions of the roots); that is, the coefficients of g(x) lie in EG = F. Hence g(x) is a polynomial in F[x] having no repeated roots. Now p(x) and g(x) have a common root in E, and so their gcd in E[x] is not 1; it follows from Corollary 18 that their gcd is not 1 in F[x]. Since p(x) is irreducible, it must divide g(x). Therefore p(x) has no repeated roots, hence is separable, and it splits over E. (ii) = (iii) Choose al E E with al fl F. Since EIF is a finite extension, a l must be algebraic over F; let pi (X) E F[x] be its irreducible polynomial. By hypothesis, p i (x) is a separable polynomial which splits over E; let K1 C E be its splitting field. If Ki = E, we are done. Otherwise, choose a2 E E with a2 0 ICI. By hypothesis, there is a separable irreducible p2(x) E F[x] having a2 as a root. Let K2 C E be the splitting
GALOIS EXTENSIONS
81
field of pi (x)P2(x), a separable polynomial. If K2 = E, we are done; otherwise, iterate this construction. This process must end with K m = E for some m because E I F is finite. (iii) = (i) By Theorem 56, I G I = [E: F]. But Theorem 79 gives I G I = [E : EG ], so that [E : F] = [E : EG ]. Since F c E', it follows that F = EG . • Definition. A finite field extension E I F is Galois (or normal) if it satisfies any of the equivalent conditions in Theorem 81. Remark. Terminology is not yet standard; some authors call Galois extensions normal, while others call an extension normal if it is the splitting field of any, not necessarily separable, polynomial.
There is a relative version of Galois extension. Definition. Given a field extension EIF,an intermediate field is a field B withFcBc E. Definition. Let E/ F be a Galois extension and let B and C be intermediate fields. If there exists an isomorphism B —> C fixing F, then C is called a conjugate of B. Theorem 82. Let E I F be a Galois extension, and let B be an intermediate field. The following conditions are equivalent.
(i) B has no conjugates (other than B itself); (ii) If a E Gal(E/F), then ajB
E
Gal(B/F);
(iii) B/F is a Galois extension. Proof. (i) = (ii) Obvious. (ii) = (iii) Let p(x) E F[x] be an irreducible polynomial having a root fi in B. Since B c E and E I F is Galois, Theorem 81 says that p(x) is a separable polynomial having all its roots in E. Let fi' E E be a root of p(x)• By Lemma 50, there exists an isomorphism r : F(9) —* F (13') fixing F and with r(/3) = /3', and r extends to a E Gal(E/F) because EIF is Galois (Theorem 51). By (ii), a(B) = B, so that fi' = a(fl) E o (B) = B. Therefore, B contains all the roots of p(x), and so p(x) splits in B. Theorem 81 shows that B I F is a Galois extension.
82
GALOIS THEORY
(iii) = (i) B/F is a splitting field of some polynomial f (x) over F, so that B = F (ai , . . . , an ), where al, ... , an are all the roots of f (x). By the proof of Lemma 54, every injective map 6 : B —* E fixing F must permute the roots of f (x). It follows that 6(B) = 6(F(ot1, • • • , an)) = F(Oai, ... , IMO = B. •
Example 30. Consider f (x) = x 3 — 2 E Q[x]. As we have seen in Example 24, a splitting field for f (x) is E = Q(a, co), where a = :n. and co = e27ri/3 . Since E/Q is a splitting field of a separable polynomial (Q is a perfect field), E/Q is a Galois extension. If g(x) = x 3 — 3x 2 ± 3x — 3, then g(x) is irreducible in Q[x], by Eisenstein's criterion, but it has a root in E, namely, fi = 1 + a. It follows that g(x) splits in E[x], as the reader may check. The intermediate field B = Q(co) is a Galois extension over Q, for it is a splitting field of x 3 — 1. We have seen in Example 24 that Gal(E/Q) L"' S3. It follows that a (B) = B for every a E Gal(E/Q). On the other hand, if C = Q(a), then Q(a 2) is a conjugate of C, and Q(a2) C. Exercises 86. If EIF is a Galois extension and Bisan intermediate field, then EIB is a Galois extension. 87. If F has characteristic 0 2 and EIF is a field extension with [E: Fl= 2, then E I F is Galois. 88. Show that being Galois need not be transitive; that is, ifFcBcEand EIB and B/F are Galois, then EIF need not be Galois. (Hint: Consider Q c Q(a) c 0(3), where a is a square root of 2 and /3 is a fourth root of 2.) 89. Let E = F (xi, . . . , xn ) and let S be the subfield of all symmetric functions. Prove that [E : S] = n! and Gal(E/S) a= Sn . (Hint: Show that E I S is a splitting field of the separable polynomial f (t) = 11(t xi).) —
90. Let E I F be a Galois extension and let p(x) c F[x] be irreducible. Show that all the irreducible factors of p(x) in E[x] have the same degree. (Hint: Use Exercise 84.) 91. Given a field F and a finite group G of order n, show that there is a subfield K c E = F(xi, . .. ,x n ) with Gal(E/K)"2",' G. (Hint: Use Exercise 89 and Cayley's theorem (Theorem G.24).)
THE FUNDAMENTAL THEOREM OF GALOIS THEORY
83
The Fundamental Theorem of Galois Theory Given a Galois extension EIF, the fundamental theorem will show a strong connection between the subgroups of Gal(E/F) and the intermediate fields between F and E. Definition. A lattice is a partially ordered set (L, ) in which each pair of elements a,b c L has a least upper bound a v b and a greatest lower bound a A b. Recall that a nonempty set L is a partially ordered set if ‹ is a reflexive, transitive, and antisymmetric binary relation on L. An element c is an upper bound of a and b if a < c and b < c; an element d is a least upper bound of a and b if it is an upper bound with d < c for every upper bound c. Greatest lower bound is defined analogously, reversing the inequalities. Example 31. If X is a set, let L be the family of all the subsets of X, and define A < B to mean A c B. Then L is a lattice with AvB=AUB
and AAB=Ail B.
Example 32. If G is a group, let Sub(G) be the family of all the subgroups of G, and define H ‹ K to mean H C K. Then Sub(G) is a lattice with H v K the subgroup generated by H and K, and HAK=H il K. Example 33. Let EIF be a field extension, let Lat(E/F) be the family of all intermediate fields, and define B < C to mean B C C. Then Lat(E/F) is a lattice with B v C their compositum and BAC=Bn C. Example 34. Let L be the set of all integers n > 1, and define n < m to mean n I m. Then L is a lattice with n v m = lcm{n, m} and n A m = gcd{n, m}. The next result generalizes the De Morgan laws, where L = L' = P (X), the power set of a set X, and y is complementing. Lemma 83. If L and L' are lattices and y : L > L' is an order reversing bijection [a < b implies y(b) < y(a)], then y (a v b) = y(a) A y(b) and y (a A b)= y(a) v y(b).
84
GALOIS THEORY
Proof. Now a,b < a v b implies y (a), y(b) > y(av b); that is, y (a v b) is a lower bound of y (a), y (b). It follows that y (a) A y(b) › y (a v b); since y is surjective, there is c E L with y (a) A y(b) = y (c) . Apply y 1 a, b < c < av b. (whicsealyntobrdevsigalo)tbn Hence c =avb and y (a v b) = y (c) = y (a) A y (b). A similar argument proves the other half of the statement. • Theorem 84 (Fundamental Theorem of Galois Theory). Let E/ F be a Galois extension with Galois group G = Gal(E/F). (i) The function y : Sub(G) > Lat(E/F), defined by H i> E", is an order reversing bijection with inverse 3 : B 1.> Gal(E/B). Go EGai(E/B) = B and Gal(E I E H ) = H. EHv K ElinK
= =
n EK; EH v EK;
EH
Gal(E/B v C)
= Gal(E/B) 11Gal(E/C);
Gal(E/B 11 C)
= Gal(E/B) v Gal(E/C).
(iv) [B: Fl = [G : Gal(E/B)] and [G: H] = [E H : F]. (v) B/F is a Galois extension if and only if Gal(E/B) is a normal subgroup of G. Proof. (i) It is easy to see that y is order reversing: K < H implies E H < E". That y is injective is precisely the statement of Corollary 80. To see that y is surjective, consider the composite Lat(E/F)
5 > Sub(G)> ) Lat(E/F),
where 3 is the map B 1 Gal(E/B). Then yS : B i> Gal(E/B) 1> EGal(E/13) . By Exercise 86, E/ F Galois implies that E/ B is Galois for every intermediate field B; hence Theorem 81 gives B = E(al(E/B) hence y3 is the identity and y is a surjection. It follows that y is a bijection with inverse 3. (ii) This is just the statement that 3y and y3 are identity functions. (iii) The first pair of equations follows from Lemma 85 because y is an order reversing bijection; the second pair follows because 3 = y 1 is also an order reversing bijection. ,
(iv) [B: Fl = FE: FlI[E : B] = IGI/I Gal(E/B)I = [G : Gal(E/B)],
APPLICATIONS
85
that the degree of B/F is the index of Gal(E/B) in G. The second equation follows from setting B= E H , because Gal(E/E") = H. (v) If B/F is Galois, then we have seen, in Theorem 58, that Gal(E/B) is a normal subgroup of G. Conversely, suppose that H is a normal subgroup of G; is E H IF a Galois extension? If a E G, r E H, and a E EH, (a) for some t' E H, by normality of H in G, and then ra (a) = r' (a) = o (a) because r' fixes a. Therefore a E E H implies a (a) c E H ; that is, a(E H ) c E H , indeed, o (E") = E H because both have the same dimension over F. By Theorem 82, E H IF is a Galois extension. • SO
Applications Corollary 85. A Galois extension E I F has only finitely many intermediate fields. Proof. The Galois group Gal(E/F), being finite, has only finitely many subgroups. • Theorem 86 (Steinitz). A finite extension E I F is simple if and only if it has only finitely many intermediate fields. Proof. Assume that E = F (a) and let p(x) be the irreducible polynomial of a over F. Given an intermediate field B, let g(x) be the irreducible polynomial of a over B. If B' is the subfield of B generated by F and the coefficients of g(x), then g(x) is also irreducible over B'. Since E = B(a) = B'(a), it follows that [E : B] = [B(a) : B] and [E : B'] = [B' (a) : B']; hence [E : B] = [E : B'], for both equal the degree of g(x). Therefore, B = B' and B is completely determined by g(x). But g(x) is a divisor of p(x); as there are only finitely many monic divisors of p(x) over E, there are only finitely many intermediate fields B. Assume that El F has only finitely many intermediate fields. If F is finite, then Corollary 64 shows that E I F is simple: just adjoin a primitive element. We may, therefore, assume that F is infinite. Now E = ,an ); by induction on n, it suffices to prove that E = F(a, fi) is a simple extension. Consider all elements y of the form y = a + tfi, where t c F; there are infinitely many such y because F is infinite. Since there are only finitely many intermediate fields, there are only finitely many fields of the form F(y). There thus exist distinct elements t, t' c F with
86
GALOIS THEORY
F(y) = F(y'), where y' = a + t'13. Clearly, F(y) C F(a, fi). For the reverse inclusion, F(y) = F(y') contains y —y' = (t tpfi. Since t we have 46 E F(y). But now a = y —t13 E F(y), so that F(a, 13) c F(y), as desired. • Corollary 87. If ElF is a finite simple extension and B is an intermediate field, then B I F is simple. Corollary 88 (Theorem of the Primitive Element). Every Galois extension E I F is simple. Proof. Immediate from Corollary 85 and Theorem 86. • Using the proof of Theorem 86, it is easy to show that one may choose a primitive element of F (al, , an ) of the form tiaid• • •itnan for ti E F. Corollary 89. The Galois field G F (pn) has exactly one subfield of order p d for every divisor d of n. Proof. We have seen in Theorem 67 that Gal(G F(pn)1 G F(P)) moreover, Lemma 59 shows that a cyclic group of order n has exactly one subgroup of order d for every divisor d of n. Now a subgroup of order d has index nld, and so the Fundamental Theorem says that the corresponding intermediate field has degree nId. But the numbers nld,asd varies over all the divisors of n, themselves vary over all the divisors of n. • Even more is true. The lattice of all intermediate fields is the same as the lattice of all subgroups of Zn , and this, in turn, is the same as the lattice of all the divisors of n under lcm and gcd (a sublattice of the lattice of Example 34). Corollary 90. If E F is an abelian extension, i.e., a Galois extension whose Galois group Gal(E/F) is abelian, then every intermediate field B is a Galois extension. Proof. Every subgroup of an abelian group is a normal subgroup. • Corollary 91. Let f (x) E F[x] be a separable polynomial, and let E I F be a splitting field. Let f (x) = g(x)h(x) in F[x], and let B/F and C I F be splitting fields of g(x), h(x), respectively, contained in E. If B fl C = F (such fields are called linearly disjoint over F), then Gal(E/F) Gal(B/F) x Gal(C/F).
APPLICATIONS
87
Proof. Recall that if H and K are subgroups of a group G, then G is their direct product, denoted by G = H x K, if both H and K are normal subgroups, H CI K = { l }, and H v K = G. Since B/F and CIF are Galois extensions, both Gal(E/B) and Gal(E/C) are normal subgroups of Gal(E/F). The hypothesis gives B v C = E, so that Gal(E/B) il Gal(E/C) = Gal(E/B v C) = Gal(E/E) = {l}. Also, linear disjointness gives Gal(E/B) v Gal(E/C) = Gal(E/B 11C) = Gal(E/F). Hence Gal(E/F) is a direct product: Gal(E/F) = Gal(E/B) x Gal(E/C). Finally, a general fact about arbitrary groups H and K, namely, (H x K)III K,
gives Gal(E/F)/ Gal(E/B) 1 Gal(E/C), while Theorem 58 gives Gal(E/F)/ Gal(E/B)2__' Gal(B/F). Therefore, Gal(E/C)''=' Gal(B/F). Similarly, Gal(E/B) Ls Gal(C/F), as desired. . The fundamental theorem can also suggest counterexamples, for it translates problems about fields (which are usually infinite structures) into problems about finite groups. For example, let El F be a Galois extension, and let B and C be intermediate fields of degree 2" and 2c, respectively; is the degree of their compositum B v C also a power of 2? If G = Gal(E/F) and H and K are the subgroups corresponding to B and C, respectively, then the fundamental theorem gives [B v C : F] = [G : H n K].
The translated question is: If both [G : H] and [G : K] are powers of 2, must [G : H 11 K] be a power of 2? In Exercise 89, we saw that there is a Galois extension El F with Gal(E/F) '' S4. Let H be the subgroup of all permutations of (1, 2, 3} (that is, all a E S4 with a (4) = 4) and let K be the
88
GALOIS THEORY
subgroup of all permutations of 12, 3, 4). Now [54 : H] = 4 = [S4 : K], but [54: H n K] = 12 (because H n K = 1(1), (23)) has order 2). We are now going to prove the fundamental theorem of algebra, first proved by Gauss (1799). Assume that R satisfies a weak form of the intermediate value theorem: if f (x) c r [x] and there exist a, b E R such that f (a) > 0 and f (b) < 0, then f (x) has a real root. Here are some preliminary consequences. (1) Every positive real r has a real square root.
If f (x) = x 2 — r , then f (1 I r) > 0 and f (0) < 0. (2) Every quadratic g(x)
E C[x] has a complex root.
First, every complex number z has a complex square root. Write z in polar form: z = re i° , where r > 0, and = Nfieje l2 . It follows that the quadratic formula can give the (complex) roots of g(x). (3) The field C has no extensions of degree 2.
Such an extension would contain an element whose irreducible polynomial is a quadratic in C[x], and (2) shows that no such polynomial exists. (4) Every f (x) E R[x] having odd degree has a real root.
Let f (x) = . Fan_IX" ±Xn E Now Iai I < t — 1 for all i, and la0
alt
[x]. Define t = 1+E lai I.
• • • + an1 01 1 0 (for any not necessarily odd n) because the sum of the early terms is dominated by t n . When n is odd, f (—t) < 0, for (—O n = (1) n t n < 0,
and so the same estimate as above now shows that f (—t) < 0.
APPLICATIONS (5)
89
There is no field extension E IR of odd degree > 1.
If a E E, then its irreducible polynomial must have even degree, by (4), so that [R(a) : R] is even. Hence [E : RI = [E : R(a)][R(a) : R] is even. Theorem 92 (Fundamental Theorem of Algebra). Every nonconstant f (x) E C[x] has a complex root. Proof. If f (x) = E ai xi E C[x], define 7(x) = E iii x i , where ai is the complex conjugate of ai . If f(x)f(x) = E ck xk, then ck = Ei + j=k ai d j ; it follows that Ck = Ck, so that f (x) .1)(x) E R[x]. Since f (x) has a complex root if and only if f(x)f(x) has a complex root, it suffices to prove that every real polynomial has a complex root. Let p(x) be an irreducible polynomial in R[x}, and let E/R be a splitting field of (x 2 I 1)p(x) which contains C. Since R has characteristic 0, E IR is a Galois extension; let G be its Galois group. If I GI = 2mk, where k is odd, then G has a subgroup H of order 2'n , by the Sylow theorem (Theorem G.13); let B = E H be the corresponding intermediate field. Now the degree [B : RI equals the index [G : 11] = k. But we have seen above that R has no extension of odd degree > 1; hence k = 1 and G is a 2group. By Theorem G.23, the subgroup Gal(E/C) of G (corresponding to C) has a subgroup of index 2 provided I Gal (E IC)I > 1; its corresponding intermediate field is an extension of C of degree 2, and this contradicts (3) above. We conclude that Gal(E/C) = {1} and E = C. • Corollary 93. Every f (x) f (x) has a factorization
E C[X]
of degree n > 1 splits over C; that is,
f (x) = c(x — ai) • • • (x — an), where c, al, . . . , a n E C.
Proof. An easy induction on n > 1. • Remark. A field K is called algebraically closed if every f (x) E K[x] has a root in K (thus, C is algebraically closed). It can be proved that every field F has an algebraically closed extension; indeed, it has a smallest such, which is called its algebraic closure.
90
GALOIS THEORY
Exercises
92. Let EIF be a Galois extension with [E: F] > 1. (i) Must there be an intermediate field of prime degree over F? (Hint: The alternating group A6 has no subgroups of prime index [see Theorem G.37].) (ii) Same question as in (i) with the added hypothesis that Gal(E/F) is a solvable group. 93. Show that Zp (X , y) is a finite extension of its subfield not a simple extension.
Zp (X P,
yP), but it is
94. Let K = Zp (t) be the field of rational functions, let f (x) = xP x K[x], and let EIK be a splitting field of f (x). Prove that Gal(E/K) but that f (x) is not solvable by radicals. —
—
tc
Galois's Great Theorem We prove the converse of Theorem 74: Solvability of the Galois group of f (x) E F[x], where F is a field of characteristic 0, implies f (x) is solvable by radicals. We begin with some lemmas; the first one has a quaint name signifying its use as a device to get around the possible absence of roots of unity in the ground field. Lemma 94 (Accessory Irrationalities). Let E/ F be a splitting field of f (x) E F[x] with Galois group G = Gal(E/F). If F* I F is an extension and E* I F* is a splitting field of 1(x) containing E, then restriction a 1> a IE is an infective homomorphism
Gal(E*/F*) —> Gal(E/F). Proof. The hypothesis gives
E = F(ai, ... ,an ) and E* = F*(a1, ... ,an), where al, ... , an are the roots of f (x). If a E Gal(E*/F*), then a permutes the at 's and fixes F*, hence F; therefore, o IE E Gal(E/F). Using Exercise 73, one sees that a 1—> alE is an injection. •
GALOIS'S GREAT THEOREM
91
Definition. If E I F is a Galois extension and a E Ell = E — {0}, define its norm N (a) by N(a) = a (a). a EGal(E/F)
11
Here are some preliminary properties of the norm whose simple proofs are left as exercises. In (i) and (iv), G = Gal(E/F). (i) If a E E* , then N (a) E F* (because N (a) E EG = F).
(ii) N(4) = N (a)N (13), so that N: E* ÷ F* is a homomorphism. (iii) If a E
0, then N (a) = a', where n = [E: F].
(iv) If a E G and a E E*, then N (a (a)) = N(a). Given a homomorphism, one asks about its kernel and image. The image of the norm is not easy to compute; the next result (which was the ninetieth theorem in an exposition of Hilbert (1897) on algebraic number theory) computes the kernel of the norm in a special case. Lemma 95 (Hilbert's Theorem 90). Let E I F be a Galois extension whose Galois group G = Gal(E/F) is cyclic of order n; let a be a generator of G. Then N (a) = 1 if and only if there exists fi E E* with
a= Proof. If a = Pa (f3) 1 , then
N(a) = M46a(13 ) 1 )= N(P)N(a( 8 ) 1 ) = N(46)N(a( 3)) 1 = N(I3)N( 5) 1 = 1. For the converse, define "partial norms":
So = a, 81 = ao (a), 82 = ao (a)o2 (a),
8n 1
=
ao (a) • • • an 1 (a) = N (a) = 1.
It is easy to see that (1)
aa (Si ) = 8i+ i
for all 0 < i < n — 2.
92
GALOIS THEORY
By independence of the characters ii , a, a 2 , with
... , a n111, there
exists y
E E
SoY + ia(Y) + • • • + 3a' (y) ± • • . + 3n2a n2 (Y) + a n1 (y) 0 0; call this sum /3. Using Eq. (1), one easily checks that a(3) = ce i [Sia(Y) + • • • + Si a i (y) ± . . . ± 8, i _ 1 an 1 (y)14 a n( y ). But an = 1, so that the last term is just y = a 1 30y . Hence a (p) = a 1 13, as desired. • Corollary 96. Let E I F be a Galois extension of prime degree p. If F has a primitive pth root of unity, then E = F(P), where is a pure extension.
p"
E F, and so E I F
Proof. If co is a primitive pth root of unity, then N(co) = coP = 1, because CO E F. Now G = Gal(E/F);.`: Zp , by Corollary 71, hence is cyclic; let a be a generator. By Hilbert's Theorem 90, we have co = 18c:1(18r' for some p E E. Hence a(f3) = Pco1 . It follows that a(PP) = (pay ')P = pP, and so pP E E G = F because a generates G and E I F is Galois. Note that /3 ci F, lest co = 1, so that F(/3) 0 F is an intermediate field. Therefore E = F(3), because [E: F] = p, and hence E has no proper intermediate fields. • Here is another proof of this last corollary that uses neither Hilbert's Theorem 90 nor the norm. The existence of an element /3 E E with co = 13a (/3) 1 is shown by an elegant application of linear algebra. 15 (We have given the first proof because the norm is a very important tool in algebraic number theory, and Hilbert's Theorem 90 itself is a useful result that is one of the early theorems involving homological algebra.) Corollary 97. Let E I F be a Galois extension with Gal(E/F) = (a) .1' Z p , where p is a prime. If F contains a primitive pth root of unity co, then there is an element p E E with co = po(pri.
Proof. View E as a vector space over F and a : E > E as a linear transformation. Since a P = 1, we see that a satisfies the polynomial xP — 1. Now a satisfies no polynomial of smaller degree, lest we contradict independence of the characters 1, a, ... ,aP 1 . Therefore, xP — 1 is the minimal polynomial of a; indeed, xP — 1 is the characteristic polynomial of a. 15 E. Houston, A Linear Algebra Approach to Cyclic Extensions in Galois Theory,
Math. Monthly 100 (1993), 6466.
Amer.
GALOIS'S GREAT THEOREM
93
Since co 1 is a root of xP — 1, it is an eigenvalue of a (remember that co 1 E F). If 6 is an eigenvector of a corresponding to co 1 , then a (13) = co 1 6. Therefore, co = /3a (13) 1 . • Theorem 98 (Galois). Let F be afield of characteristic 0, and let E I F be a Galois extension. Then G = Gal(E/F) is a solvable group if and only if E can be imbedded in a radical extension of F. Therefore, the Galois group of f (x) E F[x], where F is afield of characteristic 0, is a solvable group if and only if f (x) is solvable by radicals. Proof. Sufficiency is Theorem 74, and we now prove the converse. Since G is solvable, Corollary 0.17 provides a normal subgroup H of prime index, say, p. Let co be a primitive pth root of unity, which exists because F has characteristic 0. We first prove the theorem, by induction on [E : F], assuming that co E F. The base step is obviously true. For the inductive step, consider the intermediate field E 11 . Now E/EH is a Galois extension (by Exercise 86), Gal(E/E H ) is a solvable group (being a subgroup of the solvable group Gal(E/F) = G), and [E : E H ] < [E : F]. By induction, there is a radical tower E H C Ri C • • • C R m , where E C R m . Now E H I F is a Galois extension, because H < G, having degree [E H : F] = [G : H] = p. Since we are assuming that F contains co, Corollary 96 gives E H = F(13), where OP E F; that is, E H /F is a pure extension. Hence, the radical tower can be lengthened by adding the prefix F c EH , thereby displaying Rm IF as a radical extension. For the general case, define F* = F(w) and E* = E (co). Observe that E*I F is a Galois extension, for if EIF is a splitting field of f(x) E F[x], then E* I F is a splitting field of the necessarily separable polynomial f (x)(xP — 1). It follows that E* I F* is also a Galois extension; let G* = Gal(E*/F*). By Lemma 94, there is an injection G* + G = Gal(E/F), so that G* is solvable (being isomorphic to a subgroup of a solvable group). Since co E F*, we know that E*, and hence its subfield E, can be imbedded in a radical extension R* I F*; there is a radical tower F* c RI' c • • • c I?: = R* . But F* = F(co) is a pure extension, so that the radical tower can be lengthened by adding the prefix F C F*, thereby displaying R*IF as a radical extension. •
94
GALOIS THEORY
This theorem implies the classical theorems.
Corollary 99. If F is afield of characteristic 0, then every polynomial in F[x] of degree n < 4 is solvable by radicals. Proof. The Galois group is a subgroup of S4. But S4 is solvable, by Theorem G.34, and every subgroup of a solvable group is itself solvable. • An earlier theorem of Abel states, when translated into group theoretic language, that a polynomial with a commutative Galois group is solvable by radicals; such groups are called abelian because of this theorem. Abel's theorem is a special case of Galois's, for every abelian group is solvable. A deep theorem of Feit and Thompson (1963) says that every group of odd order is solvable. It follows that if F is a field of characteristic 0 and f (x) E F (x) is a polynomial whose Galois group has odd order, equivalently, whose splitting field has odd degree over F, then f (x) is solvable by radicals. Suppose one knows the Galois group G of a polynomial f (x) E Q[x] and that G is solvable. Can one, in practice, use this information to find the roots of f (x)? The answer is affirmative; we suggest the reader look at the books of [Edwards] and [Gaal] to see how this is done.
Exercises
95. Let EIF be a finite separable extension with Galois group G. Define the trace T : E —> E by T (a) =
EuEG
(i) Prove that im T c F and that T (a + fi) = T (a) + T() for all a, /3 E E. (ii) Show that T is not identically zero. (Hint: Independence of characters.) 96. Assume that EIF is a separable extension of degree n and cyclic Galois group G = Gal(E/F) = (a). (i) If a E G, define r = a — identity, and prove that ker T = imt. (Hint: Use EIF being a Galois extension to show that ker r = F and hence dim(im r) = n — 1; show that dim(ker r) = n — 1 as well.)
DISCRIMINANTS
95
(ii) Prove the Trace Theorem: If E I F is a Galois extension with cyclic Galois group Gal(E/F) = (a), then ker T = la EE:a=a (I3)
—
13 for some /3 E El.
97. Let F be a field of characteristic p > 0. (i) Let f (x) = XP—X—CE F[x] and let u be a root of f (x) in some splitting field EIF. Show that every root of f(x) has the form u+i, where 0 < i < p. (ii) Show that xP —X—CE F[x] either splits or is irreducible.
98. Let F be a field of characteristic p > 0, and let E I F be a Galois extension with cyclic Galois group (a) of order p. (i) Prove there is a E E with a (a) — a = 1. (Hint. Use the trace theorem.) (ii) Prove that E = F (a), where a is a root of an irreducible polynomial in F[x] of the form xP —x — c. 99. Here is a proof of Exercise 98, similar to that in Corollary 97, which does not use the trace theorem. Let E I F be a Galois extension, where F is a field of characteristic p > 0, and let a E Gal(E/F) have order p. View a as a linear transformation, and define T = a — identity. (i) Prove that V' = 0. (ii) Prove that if a E ker r I im r, then TP 1 (a) = 0. Using the fact that p and p — 1 are relatively prime, prove that t(a) = 0. (iii) Prove that ker r = F and that im r n ker r 0 {0}. (Hint. Show that E = im .1 + ker r if im r n ker r = {0)) (iv) Prove that 1 E 1M T. (Hint. Prove that im r n ker r = F, and so
Fc
imt.)
Discriminants Let F be a field of characteristic 0, let f (x) E F[x] be a polynomial of degree n having splitting field EIF, and let G= Gal(E/F). If f (x) = c(x
—
a 1)  • 
(x
define A = 11(oti i<j
—
a' ).
—
an ),
96
GALOIS THEORY
Although the number A does depend on the indexing of the roots, a new indexing of the roots can only change some factors ai  ai to a./  ai =  (ai  aj ). Therefore, the sign of A depends on the listing of the roots, whereas D = A2 depends only on the set of roots. Remark. There is a connection between A and the alternating group A. If n E Sn , let ir act on A = fli ,./ (ai  a3 ) by permuting the subscripts: cf(A) = il i ,i (aci  aa j). Now 7r(A) = +A; define 0 : Sn * Z2 by 0(n. ) = [0] if 7r(A) = A, and 0(2r) = [1] if 7r(A) = A. It is easy to see that 0 is a surjective homomorphism with kernel A n , for the alternating group is the unique subgroup of Sn having index 2 (Theorem G.29). Therefore, 7r (A) = A for 7r even, and 7r(A) = A for 7r odd. Definition. The discriminant of a polynomial f (x) E F[x] is D = A2 .
It is clear that f (x) has repeated roots if and only if D = 0. Each a e G permutes al , — , an , so that a (A) = +A; hence A 2 = D E E G = F. If f (x) = x 2 + bx + c, then the quadratic formula gives the roots of
a = 1 ( b + ../ b 2 

4c) and
13 = 1(b  , V b 2  4c).
It follows that D
= 6,2 = (cr _ fi) 2 = b2 _..... 4c.
If f(x) is a cubic with roots a, 13,y, then
it is not obvious how to compute D from the coefficients of f (x). Definition. A polynomial f (x) = xn + cn _ 1 xn 1 +... + co is reduced if c„_ 1 = 0. If f (X) is a monic polynomial of degree n and if n 0 0 in F, then its associated reduced polynomial is f (x) = f (x  c n _ i I n).
Recall Lemma 43: If f (x) = xn + cn _ix" + • • • ± co E F[x] and ,f3 E F, then /3 is a root of f (x) if and only if /3  cn _ i 1 n is a root of 1(x).
DISCRIMINANTS
97
Theorem 100. (i) A polynomial f (x) and its associated reduced polynomial '.I(x) have the same discriminant. (ii) The discriminant of a reduced cubic f (x) = x 3 + qx ± r is D = —4q 3 — 27r2 .
Proof. (i) If the roots of f (x) are a l , . . . , an , then the roots of f (x) are #1 ,  •  ,fin, where
pi = ai
—
an _ i I n. Therefore
11(a, —cej) =11(fit — i<j i 0. In this case, either NU) E Q and GLi Z3 or Niii ig Q and G L' S3.
Proof. Note first that D 0 0: since Q has characteristic 0, it is perfect, and hence irreducible polynomials over Q have no repeated roots. If f (x) has three real roots, then A is real and D = A2 > 0. Conversely assume that f (x) has one real root a and two complex roots: /3 = u+i v and fi = u —i v. Since 13 — /3 = 2i v and a = E, we have A=
=
(a — 13)(a — T3)(13 — 13) (a —8)(a — MO —8)
= la  fil2(2iv),
and so D = A 2 = —4v2 la — /31 4 < O.
102
GALOIS THEORY
Let E/Q be the splitting field of f (x). If f (x) has exactly one real root a, then E # Q(a). Hence I G I > 3 and G ''.._' S3. If f (x) has three real roots, then D > 0 and ,s/T) is real. By Lemma 103, Gg_". A31=." Z3 if and only if/T) is rational; hence G 2= S3 if Vi) is irrational. •
Example 35. The polynomial x 3 —2
E Q[x] is irreducible, and its discriminant is D = —108. Therefore, its Galois group is 53. The polynomial x 3 — 4x +2 E Q[x] is irreducible, by Eisenstein's crite
rion, and its discriminant is D = 148. Since 112i is irrational, the Galois group is S3. The polynomial x 3 — x + A E Q[x] is irreducible, by Exercise 63, and its discriminant is D = 1. Since VI is rational, the Galois group is Z3. Consider a (reduced) quartic f (x) = x4 + qx2 + rx ± s E (Q[x]; let E/Q be its splitting field and let G = Gal(E/Q) be its Galois group. (By Exercise 101, it is no loss in generality to assume f (x) is reduced.) If f (x) has a rational root u, then f (x) = (x — u)c(x), and its Galois group is the same as that of its cubic factor c(x); but Galois groups of cubics have already been discussed. Suppose that f (x) = h(x)k(x) is the product of two irreducible quadratics; let a be a root of h(x) and let /3 be a root of k(x). If Q(a) n Q(3) = Q, that is, if these fields are linearly disjoint, then Corollary 91 shows that G ..', V, the four group; otherwise, a E Q(/3), so that Q(/3) = Q(a, /3) = E, and G has order 2. We are left with the case f (x) irreducible. The basic idea now is to compare G with the four group, namely, the normal subgroup of S4: V = ((1), (12)(34), (13)(24), (14)(23)), so that we can identify the fixed field of V n G. If the four (distinct) roots of 1(x) are al, a2, a3, a4, then consider the numbers: u
= (a1 + a2)(a3 + a4),
v =
(at + a3)(a2 + a4),
w=
(a1 + a4)(a2 + a3).
It is clear that if a E Vn G, then a fixes u, v, and w. Conversely, checking each of the 24 permutations shows that if a E S4 fixes (ai ± a j)(ak ± ae), then a E VU {(ij), (Id), (ikj f), (i Ej k)}. It follows that a E G fixes each of u, v, w if and only if a E V n G, and so Q(u, v, w) is the fixed field of V n G.
GALOIS GROUPS OF QUADRATICS, CUBICS, AND QUARTICS
103
Definition. The resolvent cubic 16 of f (x) = x 4 + qx 2 + rx + s is g(x) = (x — u)(x — v)(x — w).
Theorem 105. If g(x) is the resolvent cubic of f (x) = x 4 + qx 2 + rx + s, then ex) = x 3 — 2qx 2 + (q 2 — 4s)x + r 2 . Proof. In our discussion of the classical quartic formula, we saw that f (x) = (x 2 + kx + f)(x 2 — kx + m) and k2 is a root of h(x) = x3 + 2qx2 + (q 2 — 4s)x — r 2 , a polynomial differing from the claimed expression for g(x) only in the sign of its quadratic and constant terms. Thus, a number # is a root of h(x) if and only if —0 is a root of g(x). Let the four roots al , a2, «3, «4 of 1(x) be indexed so that al, a2 are roots of x 2 +kxF1 and a3, a4 are roots of x 2 —kx+m. Then k = —(a 1 Fa2) and —k = — (a3 + a4); therefore u = (a1 + a2)(a3 + a4) = —k 2 and —u is a root of h(x) since h(k 2) = 0. Now factor 1(x) into two quadratics, say, f (x) = (x 2 + icx + i)(x 2 — lcx + riz), where a1, a3 are roots of the first factor and a 2 , a4 are roots of the second. The same argument as above now shows that V = (a l + a3)(a2 + a3) = —/c 2 , 16 There is another resolvent cubic in the literature which arises from another combination
of the roots invariant under V. Define r
U = a ia2 + Ce3CX4,
r v = a la3 + a2a4, w
r
= ala4 + a2a3 ,
and define h(x) = (x — — v')(x — w'). This cubic (which is distinct from g(x) above) behaves much as g(x) does. The reason for our preference for g(x) is Exercise 103; one can use g(x) to compute the discriminant of a quartic.
104
GALOIS THEORY
hence —v is a root of h(x). Similarly, —w = —(a 1 of h(x). Therefore
+ a4)(a2 +a3) is a root
h(x) = (x + u)(x + v)(x + w),
and so g(x) = (x — u)(x — v)(x — w)
is obtained from h(x) by changing the sign of the quadratic and constant terms. • Theorem 106. Let f (x) E Q[x] be an irreducible quartic with Galois group G, and let m be the order of the Galois group of its resolvent cubic. (i) If m = 6, then G
S4.
(ii) If m = 3, then G
A4
(iii) If m = 1, then G
V.
(iv) If m = 2, then G
D8 or G
Z4
Remark. Note that, in the ambiguous case (iv), the two possible groups have different orders. See Exercise 113. Proof. We have seen that Q(u, v, w) is the fixed field of V n G. By the Fundamental Theorem,
IG/v n GI = [G : V n G] = [Q(u, v, w) : Q]
= I Gal(Q(u, v, w)/12) I. Since f (x) is irreducible, G acts transitively on its roots, by Exercise 79, hence I G I is divisible by 4 (Theorem G.10), and the theorem follows from Exercise 106 and Exercise 107. • Example 36. Let f (x) = x 4 — 4x + 2 E Q[x]; f (x) is irreducible, by Eisenstein's criterion. The resolvent cubic is g(x) = x 3 — 8x + 16.
Now g(x) is irreducible, for if one reduces mod 5, one obtains x 3 + 2x + 1, and this polynomial is irreducible over Z5 because it has no roots. The discriminant of g(x) is —4864, so that Theorem 104 shows that the Galois group of g(x) is S3, hence has order 6. Theorem 106 now shows that G S4.
GALOIS GROUPS OF QUADRATICS, CUBICS, AND QUARTICS
Example 37. Let f (x) = x 4 — 10x2 Exercise 67. The resolvent cubic is
+
1E
105
Q[x]; f (x) is irreducible, by
x 3 + 20x 2 + 96x = x(x + 8)(x + 12). In this case, Q(u, v, w) = Q and m = 1. Therefore, G V. (This should not be a surprise if one recalls Example 20 where we saw that f (x) arises as the irreducible polynomial of a = + Na where Q(a) = (1)(N/2, N/j).) Remark. If d is a divisor of I S41 = 24, then it is known that 54 has a subgroup of order d. If d = 4, then V and Z4 are nonisomorphic subgroups of order d; for any other divisor d, any two subgroups of order d are isomorphic. We conclude that the Galois group G of a quartic is determined to isomorphism by its order unless I GI = 4.
Exercises
103. If f (x) is a quartic, then its discriminant is the discriminant of its resolvent cubic. (Hint: = —(ai — a4)(a2 a3) = =
— a3)(a2 a4) a2)(a3 a4).)
104. If f (x) = x4 + ax 2 bx + c, prove that the discriminant of f (x) is D = —16a 4c + 4a 3b2 + 128a 2c2 — 144ab2c + 27b4 — 256c3 . 105. Show that x 3 + ax + 2 E R[x] has three real roots if and only if a  F with D(xy) = x D(y) + D(x)y; an ordered pair (F, D) is called a differential field. Given a differential field (F, D) with F a (possibly infinite) extension of C, its differential Galois group is the subgroup of Gal(F/C) consisting of all a commuting with D. If this group is suitably topologized and if the extension F/C satisfies conditions analogous to being a Galois extension (it is called a Picard—Vessiot extension), then there is a bijection between the intermediate differential fields and the closed subgroups of the differential Galois group. The latest developments are in Magid (1994). There is Galois theory in algebraic topology. A covering space of a topo
108
GALOIS THEORY
logical space X is an ordered pair (X, p), where p : X > X is a certain type of continuous map. The elements of the group Cov(X/ X) defined as ihomeomorphisms h : X —> X : ph = p} are dual to the elements of a Galois group in the following sense. If i : F > E is the inclusion, where E I F is a Galois extension, then an automorphism a of E lies in the Galois group if and only if o i = i. When X is simply connected, then Cov(X/ X) ._ '__ 71(X), the fundamental group of X; moreover, there is a bijection between the family of all covering spaces of X and the family of all subgroups of the fundamental group. I am awed by the genius of Galois (18111832). He solved one of the outstanding mathematical problems of his time, and his solution is beautiful; in so doing, he created two powerful theories, group theory and Galois theory, and his work is still influential today. And he did all of this at the age of 19; he was killed a year later.
Appendices
Appendix A Group Theory Dictionary
Abelian group. A group in which multiplication is commutative. Alternating group A n . The subgroup of Sn consisting of all the even per
mutations. It has order in!. For all x, y, z, one has (xy)z = x(yz). It follows that one does not need parentheses for any product of three or more factors.
Associativity.
Automorphism.
An isomorphism of a group with itself.
Commutativity.
For all x, y, one has xy = yx.
Two elements x and y in a group G are called conjugate if there exists g E G with y = gxg 1 .
Conjugate elements.
Two subgroups H and K of a group G are called conjugate if there exists g E G with
Conjugate subgroups.
K = gHg 1 = {ghg 1 : h E H}. Coset of H in G. A subset of G of the form gH = {gh : h E H}, where H is a subgroup of G and g E G. All the cosets of H partition G; moreover, gH =g'H if and only if g' g' E H.
110
APPENDICES
A group G which contains an element g (called a generator) such that every element of G is some power of g.
Cyclic group.
Dihedral group D2n . A group of order 2n containing an element a of order n and an element b of order 2 such that bab = a 1 .
A permutation that is a product of an even number of transpositions. Every r cycle, for r odd, is an even permutation.
Even permutation.
Factor groups. Given a normal series G = Go D GI D ... D Gn = {I},
its factor groups are the groups Gi/Gi + i for i > 0. Four group V. The subgroup of S4 consisting of
1, (12)(34), (13)(24), and (14)(23); it is a normal subgroup. Generator of a cyclic group G. An element g E G whose powers give all the elements of G; a cyclic group may have several different gener
ators. Group. A set G equipped with an associative multiplication such that there is a unique e E G (called the identity of G) with ex = x = xe for all x E G, and, for each x E G, there is a unique y E G (called the inverse of x) with yx = e = xy. One usually denotes e by 1 and y
by x 1 . (Some of these axioms are redundant.) Homomorphism. A function f: G —> H, where G and H are groups, such that f (xy) = f (x) f (y) for all x, y E G. One always has f(l) = 1 and f (x 1 ) = f (x) 1 . Image. Given a homomorphism f : G —> H, its image im f is the subgroup of H consisting of all f (x) for x E G. Index [G: 11]. The number of (left) cosets of a subgroup H in G; it is equal to I GI/IH I when G is finite. Isomorphism.
A homomorphism that is a bijection.
Kernel. Given a homomorphism f : G —> H, its kernel ker f is the (necessarily) normal subgroup of G consisting of all x E G with f (x) = 1. One denotes this by H < G.
GROUP THEORY DICTIONARY
111
Natural map. If H is a normal subgroup of G, then the natural map is the homomorphism r :G—> GIH defined by r(x)=xH.
A sequence of subgroups
Normal series of G.
G = Go
Gi
Gn = {1}
with each Gi ± i a normal subgroup of G,. (A subgroup G i may not be a normal subgroup of G.) Normal subgroup.
A subgroup H of a group G such that, for all g E G, gHg 1 = {ghg 1 : h E H} = H.
Order of an element x E G. The least positive integer m, if any, such that
= 1; otherwise infinity. Order IGI of a group G.
The number of elements in G.
pgroup. A group in which every element has order some power of the prime p. If G is finite, the I GI is a power of p. Permutation. A bijection of a set to itself; all the permutations of a set X form a group under composition, denoted by Sx.
If H is a normal subgroup of G, then GI H is the family of all cosets gH of H with multiplication defined by
Quotient group GI H.
gHg'H = gg'H;
the order of GI H is [G : H]; the identity element is 1H = H; the inverse of gH is g 1 H. Simple group G. A group G and G.
{1} whose only normal subgroups are {1}
Solvable group. A group having a normal series with abelian factor groups. Subgroup H of G. A subset of G containing 1 which is closed under mul
tiplication and inverse. Subgroup generated by a subset X. The smallest subgroup of G containing X, denoted by (X), consists of all the products x1ax2 b • • • xn z , where x, E X and the exponentsa, b, , z = ±1.
112
APPENDICES
Sylow psubgroup of a finite group G. A subgroup of G of order pi' , where pn is the highest power of p dividing I G I. Such subgroups always exist, and any two such are conjugate, hence isomorphic. Symmetric group S n . The group of all permutations of (1, 2, . . . , n} under composition; it has order n!.
Appendix B Group Theory Used in the Text
All groups in this appendix are assumed to be finite even though several of the theorems hold (perhaps with different proofs) in the infinite case as well. Definitions of terms not defined in this appendix can be found in the dictionary, Appendix A. Theorem G.1. Every subgroup S of a cyclic group G = (a) is itself cyclic. Proof. If S = f 1 1, then S is cyclic with generator 1. Otherwise, let m be the least positive integer for which am E S; we claim S = (am). Clearly (am) c S. For the reverse inclusion, take s = ak E S. By the division algorithm, there are integers q and r with 0 < r < m and k =qm+r.
But ak = aqm±r = (am)q ar gives ar C S. If r > 0, the minimality of m is )q E (a m). • contradicted; therefore r = 0 and ak = (m )q Theorem G.2. (i) If a E G is an element of order n, then am = 1 if and only if n I m. (ii) If G = (a) is a cyclic group of order n, then ak is a generator of G if and only if (k, n) = 1. (iii) If x E G has order n, then the order of x is 1(x) I. Proof. (i) Assume that am = 1. The division algorithm provides integers q and r with m = nq ± r, when 0 < r < n. It follows that ar = am nq = am a nq = 1. If r > 0, then we contradict n being the smallest positive integer with a n = 1. Hence r = 0 and n I m. Conversely, if m = nk, then a m = a nk = ( an)k = ik = 1.
GROUP THEORY USED IN THE TEXT
113
(ii) Recall that two integers are relatively prime if and only if some integral linear combination of them is 1. If ak generates G, then a E (a'), so that a = al" for some t E Z. Therefore a k" = 1; by (i), n I kt — 1, so there is V E Z with nv = kt — 1. Therefore, 1 is a linear combination of k and m, and so (k, n) = 1. Conversely, if (k, n) = 1, then nt ± ku = 1 for t, u E Z; hence a = a nt+ku = a nt aku
= a lcu E (a") .
Therefore every power of a also lies in (a" ) and G = (a k ). (iii) The list 1, a, a 2 , . . . , an 1 has no repetitions: if there are i < j with al = ai, then ai i = 1, contradicting n being the smallest exponent for which an = 1. Now (1, a, a 2 , .. . , an 1 ) c (a), and we let the reader prove the reverse inclusion. It follows that I (a)I = Ill , a, a 2 , . . . , a"'} = n. •
Theorem G.3 (Lagrange). If H is a subgroup of a group G, then IGI = [G : H]IHI.
Proof. The relation on G, defined by x — y if y = xh for some h E H, is an equivalence relation whose equivalence classes are the cosets of H. Therefore, the cosets of H in G partition G. Moreover I HI = Ix H I for every x E G (because h 1—> xh is a bijection), so that I GI is the number of cosets times their common size. • It follows that [G : II] = IGIIIH I. In particular, if H is a normal subgroup of a group G (so that the quotient group G/ H is defined), then
IG/H I = [G : Hl = IGI/IHI when G is finite. Another consequence of Lagrange's theorem is that the order of x E G is a divisor of IG I, for Theorem G.2 shows that the order of x is the order of the subgroup (a). Hence, aI G I = 1 for all a E G. If f : G —) H is a homomorphism, denote the image of f by im f and the kernel of f by ker f. Lemma G.4. Let f : G —> I I be a homomorphism. Then f is an injection if and only if ker f = I 1 }.
114
APPENDICES
Proof. If f is an injection, then x 0 1 implies f (x) 0 f(1) = 1, and so x V ker f. Conversely, assume ker f = {1} and that f (x) = f (y) for x, y E G. Then 1
= f (x) f (y) 1 = f (x) f (y 1 ) = f (x.Y 1 )
and xy 1 E ker f = {1}. Hence x = y and f is an injection. • Theorem G.5 (First Isomorphism Theorem). Iff:G > Hisahomomorphism, then ker f is a normal subgroup of G and G / ker f1." im f. Proof. Let K = ker f. Let us show K is a subgroup. It does contain 1 (because f(1) = 1); if x,y E K (so that f (x) = 1 = f (y)), then f (xy) = f(x)f(y) = 1 and xy E K; if X E K, then f (x 1 ) = f(x) 1 = 1 and x 1 E K. Furthermore, the subgroup K is normal: if x E K and g E G, then f (gxg 1 ) = f (g) f (x) f (g) 1 = f (g) f (g) 1 = 1 and so gxg 1 E K. Define v : G/ K —> im f by v(xK) = f (x). Now v is well defined: if x' K = xK, then x' = xk for some k E K, and f(x') = f (xk) = f (x) f (k) = f (x). It is routine to check that v is a homomorphism (because f is) with imp = im f. Finally, v is an injection, by Lemma 0.4, because 99(x K) = 1 implies f (x) = 1, hence x E K and xK = K. • If K, H are subgroups of G, then K v II is the smallest subgroup of G containing K and H; that is, K v H is the subgroup of G generated by K U H. Lemma G.6. If K and H are subgroups of G with K normal in G, then KvH=KH=Rh:kEKandhell)=HK. Proof. Clearly KH c K v H. For the reverse inclusion, it suffices to prove that K H is a subgroup, for it does contain K U H. Now khk 1 h 1 = k(hk i h l )hh i = (kk2)(hh1) E K H for some k2 E K (because K is normal). Also (kh)  1 = hlk1 = (hlkloh1 = Vh 1 E K H for some k' E K (again, because K is normal). Therefore, K H is a subgroup. If hk E HK, then hk = (hkh 1 )h = k'h c K H for some k' c K, and so HKcK H; the reverse inclusion is proved similarly. • If K and H are subgroups of G with K normal, then the family of those cosets hK of K with h E H is easily seen to be a subgroup of G1K. Indeed, one may check, using Lemma G.6, that this subgroup is KHIK.
GROUP THEORY USED IN THE TEXT
115
Theorem G.7 (Second Isomorphism Theorem). If K and H are subgroups of G with K normal in G, then K 11 H is a normal subgroup of H and HI(Kr1H)=1= KHIK.
Proof. Let 7 : G—> G/K be the natural map, defined by g(x) = xK, and let f : H —> GI K be the restriction r 1 H. Now ker f = KnH and im f is the family of all cosets xKinGIK with x E H (hence im f = KH/K). The first isomorphism theorem now gives the result. • Theorem G.8 (Third Isomorphism Theorem). If S C K are normal subgroups of G, then K / S is a normal subgroup of GIS and (G/S)/(K/S) 2=' G/K.
Proof. The function f :G/S—> GIK given by x S i xK is well defined because S C K. One checks easily that f is a surjective homomorphism with kernel KIS, and so the theorem follows from the first isomorphism theorem. • Theorem G.9 (Correspondence Theorem).
Let K be a normal subgroup of G, and let S* be a subgroup of G* = GI K. (i) There is a unique intermediate subgroup S. i.e., KcSc G, with SIK =S*; (ii) If S* is a normal subgroup of G*, then S is normal in G; (iii) [G* : S*] = [G : S]; (iv) If T* is normal in S*, then T is normal in S and S*IT*La SIT.
Proof. (i) Define S = Ix EG:xK E S * 1. (ii) If a E G, and x E S, then axa 1 K = aKxKa 1 K E S*, because S* is normal in G*; therefore axa 1 C S. (iii) [G* : Si = IG * 111S * 1= IGIKIIISIKI = OG01010111 10 = IGIIISI = [G : S]. (iv) T is normal in S, by (ii), and S*IT* =(SIK)/(TIK):._ SIT,
by the third isomorphism theorem. •
116
APPENDICES
Definition. A group G acts on a set X if there is a function G x X X,
denoted by (g, x)
F> g • x,
(i) 1 . x = x for all x
such that:
E X,
where 1 is the identity in G;
(ii) (gh)  x = g• (h  x) for all x E X and for all g, h E G.
Definition. If G acts on X and x C X, then the orbit of x is 0(x) = {gx:gEG}c X,
and the stabilizer of x is the subgroup G x =fgEG:g•x=x}c G.
A group G acts transitively on X if, for each x, y with g • x = y. In this case, 0(x) = X.
E X,
there exists g
E G
Every group G acts on itself (here X = G) by conjugation: define
g • x = gxg 1 . The orbit 0(x) of x
E G is
its conjugacy class:
fy E G : y = gxg 1 for some g E Gl;
the stabilizer of x is IgEG:x=gx=gxg 1 1=fgEG:gx=xg}
(this last subgroup, called the centralizer of x in G, is denoted by CG(x)). The reader may check that the family of all orbits partitions X, for the relation x — y on X, defined by y = g•x for some g E G,isan equivalence relation on X whose equivalence classes are the orbits.
Theorem G.10. If G acts on a set X and if x
E X, then
10(x)I = [G : G.x] = IGIIIGA• In particular, if G acts transitively on X, where IXI = n, then IGI = nIG xi.
GROUP THEORY USED IN THE TEXT
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Proof. Define q : 0(x) —> {the family of all cosets of G, in G} by co(g • x) = gG x .
Now (p is well defined, for if g•x =h•x (where g,h c G), then Fr i g • x = x,h 1 g E G„, and gG, = hG x . Reversing this argument shows that is an injection: if Og  x) = Oh • x), then gG, = hG„, h 1 g E G„, and g  x = h  x. Finally, q is surjective, for a coset gG„ is co(g  x). Hence, q) is a bijection. If G acts transitively, then 0(x) = X and 10(x)I = n = In hence n = [G : G x ] = IGIIIG x i, and IGI = niG x l. • Corollary G.11. If x
E
G, then
the number of conjugates of x = [G : C G (x)].
Proof. This is the special case of G acting on itself by conjugation. • Lemma G.12. If p is a prime not dividing m (p f m) and if k > 1, then i p ic nA P{
P
k)•
Proof. Write the binomial coefficient as follows: p
km) _ p k m (pkm _ 1) . . . (p k m _ i) . . . ( p k m _ p k ± 1)
pk
pk ( p k _ 1) . . . ( p k _ i) . . . ( p k _ p k + 1)

(
By Euclid's lemma, any factor p of the numerator (or of the denominator) arises from a factor of pkm—i (or of p k  i). If (m, p) = 1 and 1 < i < p/C , then 13' I mp g' — i if and only if pt I i. Therefore, the highest power of p dividing pm — i is the same as the highest power of p dividing p k  i (because p { m). Every factor of p upstairs is thus canceled by a factor of p downstairs, and hence the binomial coefficient has no factor p. • Theorem G.13 (Sylow). If G is a group of order p k m, where p is a prime not dividing m, then G contains a subgroup of order p'.
Proof. (Wielandt) If X is the family of all subsets of G of cardinality p " , then Lemma G.12 shows that p { IXI. Let G act on X by left translation: if B c G and I BI = p k , then g  B = {gb : b E B}.
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There is some orbit 0(B) with p { 10(B)I (otherwise p divides the cardinality of every orbit, hence p divides I X I). Choose such a subset B E X. Now IGI/IG B I = [G : G B] = 10(B)I is prime to p; it follows that I GBI = pk m > p k for some m' I m. On the other hand, if bo E B and g E GB, then gbo E g•B=B (definition of stabilizer); moreover, if g and h are distinct elements of GB, then gbo and hbo are distinct elements of B. Therefore I GB I 0. Corollary G.14 (Cauchy). If p is a prime dividing IGI, then G contains an element of order p.
Proof. Let H be a Sylow psubgroup of G and choose x E H # = H — {1}. By Lagrange's theorem, the order of x is pt for some t. If t = 1, we are done; if t > 1, then it is easy to see that xP l 1 has order p. • Lemma G.15. Every finite abelian group G 0 {1 } contains a subgroup of prime index.
Proof. We first prove that if G has composite order rs, then G has a proper subgroup. Choose x E G with x 0 1. If x has order < rs, then (x) is a proper subgroup; otherwise, x has order rs and (xr) is a proper subgroup. The proof of the lemma is by induction on the number k of (not necessarily distinct) prime factors of I GI. If k = 1, then G has prime order and {1} has prime index. If k> 1, the first paragraph gives a proper subgroup H, necessarily normal (because G is abelian), and so the quotient group G I H is defined. By induction, G I H has a subgroup S* of prime index, and the correspondence theorem gives a subgroup S of G of prime index. • Theorem G.16.
A group G 0 {1} is solvable (it has a normal series with abelian factor groups) if and only if G has a normal series with factor groups of prime order
Proof. Sufficiency is obvious; we prove necessity by induction on IGI. Assume that G = Go D GI D • • • D G n = { 1)
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is a normal series with G/ G,+1 abelian for all i ; we may further assume that G 0 G1. By Lemma G.15, the abelian group GIG ' has a (necessarily normal) subgroup S* of prime index; the correspondence theorem gives an intermediate subgroup S (G D S D G1) with S normal in G and with [G : S] = [G / GI : 5*] prime. Now S is a solvable group (consider the normal series SD Gi D G2 D •  • D G n = Ill; SIG ] is abelian because it is a subgroup of the abelian group G/ G1), and
induction provides a normal series of it with factor groups of prime order. •
Corollary G.17. Every solvable group has a normal subgroup of prime index.
Recall that the commutator of elements x, y c G is  . [x, y] = xyx  1 y1
The commutator subgroup G' of G is the subgroup generated by all the commutators (the product of two commutators may not be a commutator). Note that G' is a normal subgroup of G, for if a c G, then ak, yla 1 = [axa 1 , aya 1 ];
moreover, GIG' is abelian.
Lemma G.18. If H is a normal subgroup of G, then G I H is abelian if and only if G' c H.
Proof. If G/H is abelian, then for all x, y
E
G,
xyH = xHyH = yHxH = yxH,
and so xyx 1 y i E H; it follows that G' C H because every generator of G' lies in H. Conversely, if G' C H, then the third isomorphism theorem shows that G/H is a quotient group of the abelian group G/G', and hence it is abelian. •
Definition. The higher commutator subgroups are defined inductively: G (°) = G;
G+ 1 ) =
that is, G (i + I) is the commutator subgroup of GO).
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APPENDICES
Lemma G.19. A group G is solvable if and only if G (n) = { 1} for some n. Proof. If G is solvable, then there is a normal series G = Go D GI D D G n = (1)
with each factor group G i l G i+ i abelian. We prove, by induction on i, that G i G" ) ; this will give the result. If i = 0, then Gi = Go= G. Assume, by induction, that Gi D G" ) ; then G: D G (i)' = G (i+1) . But G i /Gi+ I abelian implies G 1+1 D q, by Lemma G.18, and so Gi + 1 Conversely, if G (n) = {1} (of course, G (1) = G'), then G = G (13) D G (I) D G (2) D • • • D O n) = 11)
is a normal series with abelian factor groups; hence G is solvable. • Theorem G.20. If G is a solvable group, then every subgroup and every quotient group of G is also solvable.
Proof. If H is a subgroup of G, then it is easy to prove by induction that Ho) c G (1) for all i. Hence, On ) = {1} implies H (n) = {1} and H is solvable. If ço : G K is a surjective homomorphism, then w(G') = K': if uvu 1 v 1 is a commutator in K, choose x,y c G with 99(x) = u and v(y) = v; then (xyx 1 y 1 ) = uvu 1 v  1 . One proves easily, by induction, that q(G) = K (i) for all i. Hence, if G is solvable, then G (n) = {1 } for some n and K (n) = {1}; therefore K is solvable. Now take K =GIN, where N is any normal subgroup of G, and take ço to be the natural map G —> GIN. • Theorem G.21. Let G be a group with normal subgroup H. If H and G I H are solvable groups, then G is solvable.
Proof. Let G H = G* = Gt) D GT D • • • D
= (1)
be a normal series with abelian factor groups. By the correspondence theorem, there is a series G=G0DG1D•DG,„=H
with each G i normal in Gi _ 1 and with abelian factor groups. Since H is solvable, there is a normal series H = Ho D Hi D • D lin = 111
GROUP THEORY USED IN THE TEXT
121
with abelian factor groups. Splicing these two series together gives a normal series for G with abelian factor groups. • One can also prove this result using the criterion in Lemma G.19. Definition. The center of a group G is Z(G) = {g E G : gx = xg
for all x E GI.
It is easy to see that Z(G) is an abelian normal subgroup of G. It is also easy to prove that g E Z(G) if and only if the conjugacy class of g is {g}, so that I Z (G)I is the number of conjugacy classes of cardinality 1. There are groups G with Z(G) = {1}; for example, Z(S3 = { 1}. )
Lemma G.22. I f p is a prime and G 0 {1} is a pgroup, then Z(G) #
1 1 1.
Proof. Partition G into its conjugacy classes: using our remark above about conjugacy classes of cardinality 1, there is a disjoint union G = Z(G)U C i U    U C1 ,
where the Ci are the conjugacy classes of cardinality larger than 1. If we choose xi E Ci, then Corollary G.11 gives IGI = IZ(G)I + EiG : CG(xi)]. By Lagrange's theorem, [G : CG(xi).[ is divisible by p for all i (if xi ft Z(G), then CG (Xi) 0 G and [G : CG(Xj)1 0 1), and so p divides I Z(G)I • • Theorem G.23. Every pgroup G is solvable, and hence it has a normal subgroup of index p if G 0 {1}. Proof. We prove that G is solvable by induction on I GI. If I G I 0 1, then Z(G) 0 {1}, by Lemma G.22. If Z(G) = G, then G is abelian, hence solvable. If Z(G) 0 G, then GIZ(G) is a pgroup of order < IG I, hence it is solvable, by induction. Since Z(G) is solvable, being abelian, Theorem G.21 shows that G is solvable. As G is solvable, the second statement follows from Corollary G.17. • Let us pass from abstract groups to permutation groups; Cayley's theorem shows that this is no loss in generality. Recall that Sx , the symmetric group on a set X, is the set of all permutations (bijections) of X under composition. If X = {xi, . .. , x n }, then there is an isomorphism Sx >. Sn (namely, a i >. 0a0  ', where 0(x1 ) = i) and one usually identifies these two groups. 
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APPENDICES
Theorem G.24 (Cayley). Every group G of order n is (isomorphic to) a subgroup of S. Proof. If a E G, then the function A a : G > G, defined by x ax, is a bijection, for its inverse is A a 1 : x ax; hence Aa E SG (of course, Aa . It remains to prove that A is an SG Sn ). Define A: G > SG by a injective homomorphism. If a, b E G are distinct, then Aa Ai) (because these two functions have different values on 1 E G). Finally, A is a homomorphism: Aci Ab : x
bx
a(bx)
and
A.ab : x
(ab)x,
so the associative law implies Aa b = Aa Ab, as desired. • Lemma G.25. The alternating group A n is generated by the 3cycles. Proof. If a E An , then a = r1 • • • rm , where each ri is a transposition and m is even; hence a = (r1r2)(T3r4) • • • (rmi rm). If r2k_1 and r2k are not disjoint, then their product is a 3cycle: r2k_1t2k = (ab)(ac) = (acb); 17 if they are disjoint, then r2k r2k = (ab)(cd) = (ab)(bc)(bc)(cd) = (bca)(cdb). Therefore a is a product of 3cycles. • Lemma G.26. The commutator subgroup of S n is A. Proof. Since Sn / A n is abelian (it has order 2), Lemma G.18 gives Sin C A. Since An is generated by the 3cycles, it suffices to prove every a = (ijk) is a commutator. Since a has order 3, a = a4 = (a 2)2 . But a 2 = (ikj) = (ij)(ik), 17 We multiply permutations from right to left:
(a r)a = a (r (a)) because we are composing functions: that is, ar : a (ab)(ac) = (acb) because (ab)(ac) : a
c
c; b
b
ra
a (ra). In particular,
a; c > a 1> b.
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123
so that a = a4 = (ii)(ik)(ii)(ik); this is a commutator because (if) = (ij) 1 and (ik) = (ik) 1 . • Lemma G.27. If y = (io, il, ... , i) is a kcycle in S n and a E Sn, then aya 1 is also a kcycle; indeed, aYa
1
= (aio, ail, ... ,
Conversely, if y' = (4 ) , i i ,... , ik' _ I ) is another kcycle, then there exists a E Sn with y' = aya I .
Proof. If f 0 aid , 0 < j < k 1, then a 1 f 0 ij and so y(a l t) = therefore a ya": f 1> a 1 f 1> a 1 f i> f; that is, a ya 1 fixes f. If f = aid , then aya 1 : f = aii i) ii }> i i+1 }> aii+i (read subscripts mod k). Hence aya 1 and (aio, ail, ... , aik_i) are equal. Conversely, given y and y', choose a permutation a with aii = i; for all j. Then the first part of the proof shows that y' = aya 1 . • Remark. The same technique proves the lemma with y a cycle replaced by y a product of disjoint cycles. Lemma G.28. If H is a subgroup of a group G of index 2, then H is a normal subgroup of G.
Proof. If a e G and a 0 H, then aH n H = 0 and, by hypothesis, aH U H = G; hence aH is the complement of H. Since Han H = 0, it follows that Ha c aH; that is, after multiplying on the right by a 1 , H c aHa 1 .
This inclusion holds for every a E G, so we may replace a by a to obtain H c a 1 Ha; that is, a H a 1 c H. Therefore, H is a normal subgroup of G. • Theorem G.29. The alternating group A n is the only subgroup of S n having index 2. Proof. If [Sn : H] = 2, then H is normal in Sn , by Lemma 0.28, and Lemma G.18 gives A n = S'n C H (for Sn / H has order 2, hence is abelian). But Iii n I = n!12 = IHI, and so H = A n . • We are going to prove that A5 is a simple group.
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APPENDICES
Lemma G.30. (i) There are 20 3cycles in S5, and they are all conjugate in S5. (ii) All 3cycles are conjugate in A5. Proof. (i) The number of 3cycles (abc) is 5 x 4 x 3/3 = 20 (one divides by 3 because (abc) = (bca) = (cab)). The conjugacy of any two 3cycles follows at once from Lemma G.27. (ii) Given 3cycles y, y', one must find an even permutation a with y' = aya 1 . This can be done directly, but it involves consideration of various cases; here is another proof. If a = (123) and Cs (a) is the centralizer of a in S5, then Corollary G.11 gives 20 = [S5 : Cs(a)]; hence ICs(a)I = 6. But we can exhibit the six elements that commute with a: 1, • a,
a2 ,
( 45),
(45)a,
(45)a 2 .
Only the first three of these are even permutations, and so ICA (a)1 = 3, where CA (a) is the centralizer of a in A5. By Corollary G.11, the number of conjugates of a in A5 is [A5 : CA (a)] = I A51/ICA (a)I = 60/3 = 20. Therefore, all 3cycles are conjugate to a = (123) in A5. • Theorem G.31. A5 is a simple group. Proof. If H 0 (1) is a normal subgroup of A5 and if a E H, then every conjugate of a in A5 also lies in H. In particular, if H contains a 3cycle, then it contains all 3cycles, by Lemma G.30(ii); but then H = A5, by Lemma G.25. Let a E H, a 0 1. After a harmless relabeling, we may assume either a = (123), a = (12)(34), or a = (12345) (these are the only possible cycle structures of (even) permutations in A5). If a = (123), then H = A5, as we have noted above. If a = (12)(34), define r = (12)(35); then rat 1 = (r1 r2)(r3 r4) = (12)(45) and ro r 1 a 1 = (354) E H. Finally, if a = (12345), define t = (132); then
ar l o1 = (al a2a3) = (234) and ro r 1 a 1 = ( 134). In each case, H must contain a 3cycle. Therefore, A5 contains no proper normal subgroups 0 {1} and hence it is simple. •
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One can prove, by induction, that A„ is simple for all n > 5. The next counting lemma is useful. Lemma G.32. If A and B are subgroups of a finite group G, then
IA n RABI = IAMBI, where AB is the subset lab :aEAandbEBI. Proof. We are going to use the following fact. If X and Y are finite sets and v : X ÷ Y is a surjection for which Iv  ' (y)I = IV' (y')I for all y, y' E Y, then IYI = IXI/IV 1 (Y)IDefine co : A x B  AB by (a, b) 1* ab; of course, q) is a surjection. We claim that 1 (ab)
= {(ac, c l b) :ce An B}.
It is clear that (ac, c ' 13) c 99 1 (ab). Conversely, if (a, 6) E V 1 (ab), then ab = a/3, where a E A and 0 E B. Hence, cc l a = f3b 1 E A n B, and so (a, /3) = (a (a  'a), (b 1 )',6) = (a, b). Therefore, IV' (ab)I =
IA n Bland
IABI = IA X BI/IA
n B. •
Corollary G.33. The only normal subgroups of S5 are {1}, A5, and Ss. Proof. Let H 0 {1} be a normal subgroup of S5. The second isomorphism theorem gives H ll A5 a normal subgroup of A5; as A5 is a simple group, either H n A5 = A5 or H n A5 = {1}. In the first case, A5 C H and H = A5 or H = S5. In the second case, there is h E H with h V A n , so that HA5 = S5. Since H n A5 = {1}, Lemma G.32 gives I HI = IS5111A5l= 2. If h E H, h 0 1, then h = (ab) (the only other elements of order 2 have the form (ab)(cd), and they are even permutations). It is easy to find a conjugate distinct from h, and this contradicts the normality of H. • Theorem G.34. Sn is solvable for n < 4, but it is not solvable for n > 5. Proof. If m < n, then S„, is (isomorphic to) a subgroup of S. Since every subgroup of a solvable group is itself solvable (Theorem G.20), it suffices to show that S4 is solvable and S5 is not solvable. Here is a normal series of S4 that has abelian factor groups: S4 D A4 i V 3 (1),
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APPENDICES
where V is the four group (the factor groups have orders 2, 3, 4, respectively, hence are abelian). Were S5 solvable, then its subgroup A5 would also be solvable. Since A5 is simple, its only normal series is A5 D {1}, and the (only) factor group is the nonabelian group A 5 /{1} ' A5, • We now discuss Exercise 106, the group theoretic basis of the computation of the Galois groups of irreducible quartic polynomials over Q. First of all, we list the subgroups G of S4 whose order is a multiple of 4. If I GI = 4, then the only abstract groups G are Z4 and Z2 X Z2, and both occur as subgroups of S4 (in particular, V'_',' Z2 X Z2). There is a subgroup of order 8 isomorphic to the dihedral group Dg, namely, the symmetries of a square regarded as permutations of the 4 corners; since a subgroup of order 8 is a Sylow 2subgroup of S4, all subgroups of order 8 are isomorphic to Dg. Theorem G.29 shows that A4 is the only subgroup of order 12 and, of course, S4 itself is the only subgroup of order 24. If G C S4 and V is the four group (which is a normal subgroup of S4), then the second isomorphism theorem gives Gn V.:1G and G/GrIV ;5.GVIV CS41V. Define
m
=
IG/G
n vi;
it follows that m is a divisor of {,S4 : V] = 24/4 = 6 (S4/ V ' 53, but we do not need this fact.) Theorem G.35 (Exercise 106). Let G C 54 have order a multiple of 4 and let m = IG I G n VI. (i) tf m = 6, then G = S4; (ii) if m = 3, then G = A4; (iii) if m = 1, then G = V; (iv) ifm = 2, then G:: D8 or Z4 or V. Proof. If m = 6 or 3, then 'GI > 12 (IGI is divisible by 3 and, by hypothesis, 4). By Theorem G.29, A4 is the only subgroup of S4 of order 12, and so A4 C G in either case. But V C A4. It follows easily that m = 6 forces G = 54 and m = 3 forces G = A4. If m = 1, then G=GnV and G c V; since 'GI is a multiple of 4, it follows that G = V. If m = 2, then IG1 = 2IG n vi; since IV 1 = 4, we have IG n VI = 1, 2, or 4. We cannot have IGn VI = 1 lest 'GI = 2, which is not a multiple
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127
of 4. If 1 G 11 V 1 = 4, then I G 1 = 8 and G2" D8 (as we remarked above, D8 is a Sylow 2subgroup). If IG n vI = 2, then IG1 = 4 and G .L' Z4 or V (these are the only abstract groups of order 4). • The possibility m = 2 and G7'= ' V can occur. Let G be the following isomorphic copy of V in S4: G = {1, (12)(34), (12), (34)).
Note that G n v = {1, (12)(34)1 and m = IG/ G n vI = 4/2 = 2. This group G does not act transitively on (1, 2, 3, 4) because, for example, there is no g E G with g(1) = 3. Exercise 107 invokes the extra hypothesis of G acting transitively to eliminate the case G L V from the list of candidates for G when m =2.
Lemma G.36. If G is a group and H is a subgroup of index n, then there is a homomorphism ç9: G > Sn with ker v c H.
Proof. Let X be the family of all cosets of H in G; since 'XI = n, it is easy to see that Sx S, (where Sx is the group of all permutations of X). For g E G, define go(g) : X > X by v(g) : a H t> gaH (where a E G); note that Og) is a bijection, for its inverse is v(g 1 ). To see that v is a homomorphism, compute: Ogg') : aH 1—> (gg')aH; co(g)v(g t ) : aH 1—> g taH i g(g taH).
If 99(g) is the identity on X, then v(g) : aH 1> aH for all a E G; in particular, c(g) : HI> H, so that gH = H and g c H. •
Theorem G.37. A6 has no subgroups of prime index. Proof. Now A6 is a simple group of order 360 = 23 . 32 . 5 (in fact, An is a simple group of order In! for all n > 5). If H is a subgroup of prime index, then [A6 : H] = 2, 3, or 5. By Lemma G.36, there is a homomorphism q : A6 —> Sn where n = 2, 3, or 5, with ker v c H; in particular, ker v is a normal subgroup of A6 with ker v 0 A6. Since A6 is simple, ker v = {1) and v is an injection. But this is impossible because IS51 = 120 < 360. • ,
Lemma G.38. S5 has no subgroups of order 30 or of order 40. Proof. If H is a subgroup of order 30, then H has index [S5 : H] = 120/30 = 4. Lemma G.36 gives a homomorphism v :
S5 —> S4
with
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APPENDICES
ker q) c H. But ker q) is a normal subgroup of S5, and so its order must be 1, 60, or 120 (Corollary G.33). Since I HI = 30, it follows that ker q) = (1}, and S5 is isomorphic to a subgroup of S4, a contradiction. A similar argument shows that S5 has no subgroup of index 3. • Theorem G.39. If a is a 5cycle in S5 and r is a transposition in S5, then (a, r) = 55.
Proof. Let H = (a, r) be the subgroup generated by a and r. We may assume that a = (1 2 3 4 5) and r = (1 i). Now some power of a, say, a k i into 1, so that Lemma G.27 gives a' (l i)a k = (j 1) for some j caries (actually, j = ak 1). Note that i 0 j because a" does not commute with (1i). But (1i)(1j) = (1 j i), an element of order 3. The order of H is thus divisible by 2, 3, and 5, hence I H I > 30. By Lemma G.38, I H I = 60 or 120. If I I/ I = 60, then H = A5, by Theorem G.29; but H # A5 because r E H is an odd permutation. Therefore H = S5. • A more computational proof shows first that every transposition can be obtained from a and r, and then that S5 is generated by the transpositions.
Theorem G.40. A subgroup H of S5 is solvable if and only if IH I < 24. Proof. We leave to the reader the fact that every group of order < 24 is solvable (whether or not it is a subgroup of S5; indeed, every group of order < 60 is solvable). Since 1 S5 1 = 120, the only divisors of I S51 larger than 24 are 30, 40, 60, and 120. Now S5 itself is not solvable, by Theorem G.34; also, A5 is the only subgroup of order 60 (Theorem G.29), and it is not solvable because it is simple and not abelian (Theorem G.31). Lemma G.38 completes the proof. • Theorem G.40 is used in Exercise 111. It is implicit in the second part of this exercise that S5 does have a subgroup of order 20; the normalizer of a Sylow 5subgroup is such a subgroup, where the normalizer NG (P) of a subgroup P of G is defined as: NG(P) = fg E G : gPg 1 = Pl.
Of course, S5 does have a solvable subgroup of order 24, namely, S4.
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129
Appendix C RulerCompass Constructions We are going to show that the classical Greek problems: squaring the circle, duplicating the cube, and trisecting an angle, are impossible to solve. As we shall see, the discussion uses only elementary field theory; no Galois theory is required. It is clear one that can trisect a 60 0 angle with a protractor (or any other device than can measure an angle); after all, one can divide any number by 3. Therefore, it is essential to state the problems carefully and to agree on certain ground rules. The Greek problems specify that only two tools are allowed, and each must be used in only one way. Let P and Q be points in the plane; we denote the line segment with endpoints P and Q by PQ, and we denote the length of this segment by I PQ I . A ruler (or straightedge) is a tool that can draw the line L(P, Q) determined by P and Q; a compass is a tool that draws the circle with radius IPQI and center either P or Q; denote these circles by C(P; Q) or C( Q; P), respectively. Since every construction has only a finite number of steps, we shall be able to define "constructible" points inductively. Given the plane, we establish a coordinate system by first choosing two distinct points, A and A; call the line they determine the xaxis. Use a compass to draw the two circles C (A; A) and C(A; A) of radius IAA I with centers A and A, respectively. These two circles intersect in two points; the line they determine is called the y axis; it is the perpendicular bisector of AA, and it intersects the xaxis in a point 0, called the origin. We define the distance I OA I to be 1. We have introduced coordinates in the plane; in particular, A = (1, 0) and A = (1, 0).
•• •
A
0
Figure 5
A
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APPENDICES
Informally, one constructs a new point T from (not necessarily distinct) old points P,Q,R, and S by using the first pair P,Q to draw a line or circle, the second pair R, S to draw a line or circle, and then obtaining T as one of the points of intersection of the two drawn lines, the drawn line and the drawn circle, or the two drawn circles. More generally, a point is called constructible if it is obtained from A and A by a finite number of such steps. Given a pair of constructible points, we do not assert that every point on the drawn line or the drawn circles they determine is constructible. Here is the formal discussion.
Definition. Let E,F,G, and H be (not necessarily distinct) points in the plane. A point Z is constructible from E, F, G, and H if either (i) Z E L(E, F) fl L(G, H), where L(E, F) L(G, H);
(ii) Z E L(E, F) n C(G; H); (iii) Z E C(E; F) fl C(G; H), where C(E; F) C(G; H). A point Z is constructible if Z = A or Z = A or if there are points P1 ,.. ,P with Z = P. so that, for all j > 1, the point Pi+i is constructible from points in {A , A, P1 , , Pi ).
Example 38. Let us show that Z = (0, 1) is constructible. We have seen above that the origin P1 = 0 is constructible. The points P2 = (0, .A and P3 = ( 0, .A are constructible, for both lie in C(A;A) n C(A; A), and so the yaxis L(P2, P3) can be drawn. Finally, Z = (0, 1) E L(P2, P3) 11 C(0; A).
In our discussion, we shall freely use any standard result of euclidean geometry. For example, every angle can be bisected with ruler and compass; i.e., if (cos 0, sin 0) is constructible, then so is (cos 0/2, sin 0/2).
Definition. A complex number Z = x + iy is constructible if the point (x, y) is a constructible point. Example 38 shows that the numbers 1, —1, 0, structible numbers.
Lemma R.1. A complex number Z =
X  I
and i are con
iy is constructible if and only if its real part x and its imaginary part y are constructible.
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131
Proof. If z is constructible, then a standard euclidean construction draws the vertical line L through (x, y) which is parallel to the yaxis. It follows that x is constructible, for the point (x, 0) is constructible, being the intersection of L and the xaxis. Similarly, the point (0, y) is the intersection of the yaxis and a line through (x, y) which is parallel to the xaxis. It follows that P = (y, 0) is constructible, for it is an intersection point of the xaxis and C(0; P). Hence, y is a constructible number. Conversely, assume that x and y are constructible numbers; that is, Q = (x, 0) and P = (y, 0) are constructible points. The point (0, y) is constructible, being the intersection of the yaxis and C(0; P). One can draw the vertical line through (x, 0) as well as the horizontal line through (0, y), and (x, y) is the intersection of these lines. Therefore, (x, y) is a constructible point, and so z = x + iy is a constructible number. • Definition. We denote by K the subset of C consisting of all the constructible numbers.
Lemma R.2. (i) If K nR is a subfield of R, then K is a subfield of C. (ii) lf K n R is a subfield ofR and if ,iii E K whenever a positive, then K is closed under square roots.
e K nift is
Proof. (i) If z = a +ib and w = c+id are constructible, then a, b, c, d E K n R, by Lemma R.1. Hence, a + c,b+d E K n R, because K n R is a subfield, and so (a + c) + i (b + d) E K, by Lemma R.1. Similarly, z w = (ac — bd) + i (ad + bc) E K. If z 0 0, then z' = (a I z z) — i (b I Cz) . Now a,bE Kr) r, by Lemma R.1, so that z z = a2 + b2 E K nR, because K n R is a subfield of C. Therefore, z" E K. (ii) If z = a + ib E K, then a,b E K il R, by Lemma R.1, and so r 2 = a2+b2 E KnR, as in part (i). Since r 2 is nonnegative, the hypothesis gives reKnr and ,fi: e K n R. Now z = re i° , so that ei° = r 1 z E K, because K is a subfield of C. That every angle can be bisected gives E K, and so Afi = .Nfie j°/2 E K, as desired. • Theorem R.3. The set of all constructible numbers K is a subfield of that is closed under square roots and complex conjugation.
C
Proof. For the first two statements, it suffices to prove that the properties of K n R in Lemma R.2 do hold. Let a and b be constructible reals.
132
APPENDICES
(i) —a is constructible. If P = (a, 0) is a constructible point, then (—a, 0) is the other intersection of the xaxis and C(0; P). (ii) a ± b is constructible.
0 Figure 6 Let I = (0, 1), P = (a, 0) and Q = (b, 1). Now Q is constructible:
it is the intersection of the horizontal line through I and the vertical line through (b, 0) [the latter point is constructible, by hypothesis]. The line through Q parallel to I P intersects the xaxis in S = (a I b, 0), as desired. Although Figure 6 is drawn with a, b positive, it is clear that this construction works for any choice of signs of a, b. (iii) ab is constructible.
Figure 7 By (i), we may assume that both a and b are positive. In Figure 7, A = (1, 0), B = (1 + a, 0), and C = (0, b). Define D to be the intersection of the yaxis and the line through B parallel to AC. Since the triangles OAC and 0 B D are similar,
10BI/10AI = 10D1/1 0 c1;
RULERCOMPASS CONSTRUCTIONS
133
hence (a + 1)/1 = (b + IC DI)lb, and ICDI = ab. Therefore, b + ab is constructible. Since —b is constructible, by (i), we have ab = (b + ab) — b constructible, by (ii).
(iv) If a 0 0, then a 1 is constructible.
Let A = (1, 0), S = (0, a), and T = (0, 1+a). Define B as the intersection of the xaxis and the line through T parallel to AS; thus, B = (1 + u, 0) for some u. Similarity of the triangles 0 SA and OT B gives 10T1/1 0 S1 = 10BI/10AI. Hence, (1 ± a)1 a = (1 + u)11, and so u = a 1 . Therefore, 1 ± a 1 is constructible, and so (1 ± a 1 ) 1 = a 1 is constructible. —
(v) If a > 0, then NAT2 is constructible.
Figure 9 Let A = (1,0) and P = (1 + a, 0); construct Q, the midpoint of 0 P. Define R as the intersection of the circle C(Q; 0) with the vertical line through A. The (right) triangles A 0 R and ARP are similar, so that
10 AI/I ARI = IAR I/ 1 A PI, and so I AR I =
134
APPENDICES
(vi) If z = a + ib
E K, then = a — ib is constructible.
By Lemma R.2, K is a subfield of C. Now a, b c K, by Lemma R.1, and i E K, by Example 38. Therefore, —bi E K, and so a — ib E K. • Corollary R.4. If a, b, c are constructible, then the roots of the quadratic ax2 + bx + c are also constructible.
Proof. This follows from the theorem and the quadratic formula. • We now consider subfields of C to enable us to prove an inductive step in the upcoming theorem. Lemma R.5. Let F be a subfield of C that contains i and that is closed under complex conjugation. Let z = a + ib, w = c + id E F, and let P = (a, b) and Q = (c, d). (i) If a + ib c F, then a E F and b E F. (ii) If the equation of L(P, Q) is y = mx + q, where m, q c R, then m, q E F. (iii) If the equation of C(P; Q) is (x _ 0 2 + (y —b)2 = r 2, where a, b, r E R, then r 2 E F.
Proof. (i) If z = a+ib E F, then a =1(Z1 i) E F and ib = 1(z i) E F; since we are assuming i E F, we have b E F. (ii) If L(P, Q) is not vertical, its equation is y — b = m(x — a). Now m = (d—b)1(a—c) E F, since a,b,c,d E F, and so q = —ma+ b E F. (iii) The circle C(P; Q) has equation (x — a) 2 ± (y — b)2 = r2 , and r2 = (c — a) 2 + (d — b) 2 E F. • Lemma R.6. Let F be a subfield of C that contains i and that is closed under complex conjugation. Let P, Q, R, S be points whose coordinates lie in F, and let a = u + iv E C. If either a
E
L(P, Q) n L(R, S), where L(P, Q) 0 L(R, S),
aE
L(P, Q) n C(R; S),
aE
C(P; Q) n C(R, S), where C(P; Q) 0 C(R; S),
or
then [F(a) : F] < 2.
Proof. If L(P, Q) is not vertical, then Lemma R.5(ii) says that L(P, Q) has equation y = mx + b, where m,b E F. If L(P, Q) is vertical, then its equation is x = b because P = (a, b) E L(P, Q), and so b E F, by
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135
Lemma R.5(i). Similarly, L(R, S) has equation y = nx+c or x = c, where m, b, n, c E F. Since these lines are not parallel, one can solve the pair of linear equations for (u, v), the coordinates of a E L(P, Q)11L(R, S), and they also lie in F. In this case, therefore, [F (a) : F] = 1. Let L(P, Q) have equation y = mx + b or x = b, and let C(R, S) have equation (x — c) 2 + (y — d) 2 = r 2 ; by Lemma R.5, we have m, q, r 2 E F. Since a = u + iv E L(P, Q) n C(R; S), r2
=
(u — c) 2 + (v — d) 2
= (u — c)2 + (mu + q — d) 2 , so that u is a root of a quadratic polynomial with coefficients in F 11 R. Hence, [F(u) : F] < 2. Since v = mu + q, we have V E F(u), and, since i E F, we have a E F(u). Therefore, a = u + iv E F(u), and so [F (a) : F] < 2. Let C(P; Q) have equation (x — a) + (y — b) 2 = r2 , and let C(R; S) have equation (x _ 02 ± (y d)2 = s2 . By Lemma R.5, we have r 2 , s2 E F fl R. Since a E C(P; Q) fl C(R; S), there are equations (u — a) 2 + (v — b) 2
= r 2 and (u — c) 2 + (v — d) 2 = s2 .
After expanding, both equations have the form u 2 + v 2 + something = 0. Setting the something's equal gives an equation of the form tu + t'v + t" = 0, where t,t' , t" E F. Coupling this with the equation of one of the circles returns us to the situation of the second paragraph. •
Theorem R.7. A complex number z is constructible if and only if there is a tower of fields Q = Ko C K1 C  •  C K n , where
ZE
K„ and [K i+1 : K J ] < 2 for all j.
Proof. If Z is constructible, there is a sequence of points 1, —1, zi, . • • ,zn = z with each z 3 obtainable from {1, —1, zi, — , z11); since i is constructible, we may assume that z 1 = i. Define Ki = Q(zI, • • . , zi). Given u = z i+ i, there are points E, F,G,H e K i with one of the following: u
E
L(E, F)n L(G, H);
u
E
L(E, F)n C(G; H);
u
E
C(E; F)n C(G; H).
136
APPENDICES
We may assume, by induction on j > 1, that Ki is closed under complex conjugation, so that Lemma R.6 applies to show that [K i+ 1 : KJ ] < 2. Finally, note that Ki+1 is also closed under complex conjugation, for if zi +i f (x) E Ki [X], then is the other root of f (x). isarotfqudc To prove the converse, it suffices to prove that if [B : F] = 2, where F c K, then B/F is a pure extension of type 2, say, B = F(I3), where fi E L(P, Q) n C(R; S) for P, Q, R, S E F; it will then follow that B C K. Since [B: F] = 2, there is a with B = F(a), where a is a root of some irreducible quadratic x 2 +bx +c E F[x]. If we define /3 = N/b 2 — 4c, then B = F(3) displays B/F as a pure extension of type 2. To see that /3 can be realized as a point in the intersection of a line and a circle, we use the construction in Theorem R.3(v). Let the line L be the vertical line through A = (1, 0) and let the circle have center Q = (1(1 + 13 2 ), 0) and radius 1 0 '/32). • 2\ Corollary R.8. If a complex number z is constructible, then [Q(z) : a power of 2.
Q] is
Proof. This follows from the theorem and Lemma 49. • Remark. The converse of this corollary is false. In Example 36, we saw that p(x) = x 4 —4x + 2 is an irreducible polynomial over Q whose Galois group Gal(E/Q) is S4, where E/Q is a splitting field of p(x). Were every root of p(x) constructible, then every element of E would be constructible, for all constructible numbers form a subfield of C, by Theorem R.3. If H is a Sylow 2subgroup of G S4, however, then [G : H] = 3; the intermediate field E H thus has degree [E H : = [G : H] = 3, and so none of its elements are constructible, by Corollary R.8. This contradiction shows that some root of p(x) is not constructible, even though every root has degree 4 over Q. It is now a simple matter to dispose of some famous problems. (1) It is impossible to "square the circle." The problem is to construct, with ruler and compass, a square whose area is equal to the area of a circle of radius 1; in other words, one asks whether N/Tr is constructible. But it is a classical result, proved by F. Lindemann in 1882, that 7, hence is transcendental over Q (see [Hadlock, p. 471), and so it does not lie in any finite extension of Q, let alone one of degree a power of 2.
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(2) It is impossible to "duplicate the cube." The problem is to construct a cube whose volume is 2; in other words, is the real cube root of 2, call it a, constructible? Now x 3 — 2 is irreducible over Q, by Eisenstein, and so [Q(a) : Q] = 3, which is not a power of 2. Corollary R.8 gives the result. This result was first proved by P. L. Wantzel in 1837. (3) It is impossible to trisect an arbitrary angle. An angle 0 is given by two intersecting lines; it is no loss in generality to assume the lines intersect at the origin and that one line is the xaxis. If we could draw the angle trisector, then the point (cos 0/3, sin 0/3), which is the intersection of the trisector and the unit circle, would be constructible; hence cos 0/3 would also be constructible, by Lemma R.1. We will now show that 60° cannot be trisected. Computing the real part of e310 = (cos 0 ± i sin 0) 3 gives the trigonometric identity: cos 30 = 4 cos 3 0 — 3 cos 0. Defining u = 2 cos 9 and 9 = 20°, we arrive at the equation u3 — 3u — 1 = 0. It is easy to see that this cubic is irreducible (it has no rational root, by Exercise 63), and so [Q(u) : Q] = 3. Corollary R.8 shows that u is not constructible. This result was also proved by P. L. Wantzel in 1837. (4) Regular pgons. Galois theory will be used in discussing this problem. Theorem R.9 (Gauss). If p is an odd prime, then a regular pgon is constructible if and only if p = 22: 1 for some t > 0. Proof. This is again a question of constructibility of a point on the unit circle, namely, z = e2 r 11 P. Now the irreducible polynomial of z over Q is the cyclotomic polynomial O p (x) of degree p — 1 (Corollary 41). Assume z is constructible. By Corollary R.8, p — 1 = 2s for some s. We claim that s itself is a power of 2. Otherwise, there is an odd number k > 1 with s = km. But X k + 1 factors over Z (because —1 is a root); setting x = 2in thus gives a forbidden factorization of p.
138
APPENDICES
Conversely, assume p = 22: + 1 is prime. Since z is a primitive pth root of unity, Q(z) is the splitting field of O p (x) over Q. Hence Gal(Q(z)/Q) has order 22', and so the Galois group is a 2group. But a 2group has a normal series in which each factor group has order 2 (this follows easily from Theorem G.23); by the fundamental theorem of Galois theory, there is a tower of fields Q = K0 C Ki C  • • C K. = Q(z) with [Ki+ i : K] = 2 for all i, that is, z is constructible, by Theorem R.7. • Remark. Primes of the form 2 2' + 1 are called Fermat primes. The values 0 < t < 4 do give primes (they are 3, 5, 17, 257, 65,537), the next few values of t do not give primes, and it is unknown whether any other Fermat primes exist. Gauss actually gave a geometric construction of the regular 17gon. Corollary R.10. It is impossible to construct a regular 7 gon, a regular 11gon, or a regular 13gon.
Proof. 7, 11, and 13 are not Fermat primes. • The following result is known (see ftladlock, p. 1061): Theorem R.11. A regular ngon is constructible if and only i fn is a product of a power of 2 and distinct Fermat primes. It follows that regular 9gons and regular 14gons are not constructible; on the other hand, a regular 15gon is constructible. It is possible that there are only finitely many constructible regular ngons with n is odd, for there may be only finitely many Fermat primes.
Appendix D Oldfashioned Galois Theory
Gimme that oldtime Galois theory; If it's good enough for Galois, then it's good enough for me!
OLDFASHIONED GALOIS THEORY
139
I am a creature of the twentieth century; algebraic systems and their automorphism groups are part of my mother's milk. When writing the definition of Galois group for this text, I asked myself an obvious question: how did such thoughts occur to Galois in the late 1820's? The answer, of course, is that he did not think in such terms; for its first century, 18301930, the Galois group was a group of permutations. In the late 1920's, E. Artin, developing ideas of E. Noether going back at least to Dedelcind, recognized that it is both more elegant and more fruitful to describe Galois groups in terms of field automorphisms (Artin's version is isomorphic to the original version). In 1930, van der Waerden incorporated much of Artin's viewpoint into his influential text "Moderne Algebra," and a decade later Artin published his own lectures. So successful have Artin's ideas proved to be that they have virtually eclipsed earlier expositions. But we have lost the inevitability of the definition; group theory is imposed on the study of polynomials rather than arising naturally from it. This appendix is an attempt to remedy this pedagogical problem by telling the story of what happened in the beginning. The reader interested in a more thorough account may read [Edwards] or [Tignol]. We use modern notation and terms even though they were unknown in the late eighteenth century. In particular, F shall denote a subfield of the complex numbers. Permutations arise simultaneously with the question of finding the roots of a polynomial. If n f (x) = Dx — ai) = x n ± bn _lx" ± •  • ± bix ± bo,
i=1
then one sees easily that bn_i is, to sign, the sum of all products of j roots ai: ail ai2 • ' . aii •
, 5, then Sn has no subgroups of index 3 or 4. Cauchy (1815)
148
APPENDICES
established the calculus of permutations, e.g., decomposition into disjoint cycles; he proved that, for n prime, Sn has no subgroups of index r with 2 < r < n. Galois knew that some polynomials are solvable by radicals and some are not; it was reasonable that it depends on the roots. The Lagrange resolvent A(x) is not sensitive to this. Indeed, it seems that Lagrange was seeking a formula for the roots of the general polynomial xn + b n_ix" + • • • + bo: the roots of any particular polynomial f (x) of degree n would be obtained from the "master formula" by substituting the specific coefficients of f (x). (The classical formulas for polynomials of degree < 4 are of this form.) If f (x) E F[x] has roots al, ... , an , Lagrange first regarded al, ... , an as indeterminates, then he formed *(xi , •  • , xn) = (x1 I cox2 ± co2x3 ± ... ± coni xon , symmetrized to obtain
**(x,x,,... ,xn ) = ilk —ai,frIxi,...
, xn)l,
a ES,,
defined A(x) to be the factor of ** of degree r (in x) which is the product over all distinct polynomials a Ili , and finally specialized (x 1 , . .. , xn ) back to (a1, — , an). But even if a*(xi, • • • , xn ) and r *(x1, . .. , xn ) are distinct polynomials, the numbers a*(ai, ... , an ) = c*(ai , ... , an) may be equal. As a polynomial over F (w)(x i , .. . , xn ), the Lagrange resolvent A(x) = gx; xl, . . . , xn ) has distinct roots; A.(x) = gx; al , . . . , an), as a polynomial over F, may have repeated roots. One can discard these extra roots but, unfortunately, fa
E Sn : (0 *) (a 1 , • • • , an) = lif (a i , • • • 
may not be a subgroup of Sn and this prevents the generalization of Lagrange's Rational Function Theorem from being true. Galois jettisoned *(x 1 , ... , xn ) which, after all, works best when the degree n is prime; he replaced it by an n !valued function V(x i , ... , xn) with an added property: all (a V) (ai , • . • , an) are distinct (of course, this forces all the ai to be distinct; this minor point is easily handled by Exercise 44). Let us call (after Edwards) such a function V a Galois resolvent 21 of f (x). Galois knew that such resolvents exist (Lagrange had proven it); indeed, there are such of the form V(xi , .. • , xn) = cixi +    + cnxn, for suitable cl,  . , Cn E F. Denote V(ai, . . • , an ) by v1. Since V is 21 Actually, I would prefer that the polynomial y (x) below be called the Galois resolvent,
for it is analogous to A(x) whereas V is analogous to *.
OLDFASHIONED GALOIS THEORY
149
n!valued, there are rational functions 01(x), . . . , O(x) in F(x) with ai = O(v1) for all i. The next step ought to be the symmetrization of V: define
V*(x; ,
[x  a V (xi,
, x,) =
,x,,)],
a€S7,
and then choose a factor of V* by discarding repeated roots. Galois did this indirectly. Let y(x) be the irreducible polynomial of vi over F, and let v1, , vni be the roots of y (x). 1 if and Recall Exercise 55: Let f (x), g(x) E F[x]. Then (f, g) only if there is a field E containing both F and a common root of f (x) and g(x)• It follows that if p(x) is irreducible, then p(x) divides h(x). Therefore, y(x) divides V (x) = V* (x; al, . . . , an), and so each root v i y(x) has the form a V (a , , an ) for some permutation a E Sn . Butof Galois wanted a more explicit description of a. Here is an easy generalization of Exercise 50: Let p(x) E F[x] be an irreducible polynomial and let (1)(x) E F(x) be a rational function; if (v) = 0 for some root v of p(x), then 0:1)(v) = 0 for every root v' of p(x).
Theorem H.9. Let f (x) E F[x] have distinct roots al, . . . , a n , and let V 1 ,... , v n, be as above; let a i = Oi (vi), where Oi(x) Then for each j = 1, . . . , m, the function = 01(111)
ei(Vi),
i = 1,
E F (x) for all i.
,
n,
is a permutation of the roots al, . . . a n .
Proof. Define c1(x) E F(x) by (1)(x) = f (t9i (x)). Now
cl) 04) = f (ei(v 1)) = f(a) = 0; since y(x) is irreducible, the generalized Exercise 50 shows that 0 = b (v i ) = f (19i (v i )); that is, Oi (vi ) is a root of f (x), hence is one of the a's. To see that ai is a permutation, it suffices to prove it is an injection. Suppose that 0, (v i ) = Ok (v i ). Now (1)(x) = 0, (x) 0k(x) is a rational function with (KO = 0; it follows that 0 = (111(v1) = 0i(v1) — ek(vi) = a — ak. Since all the roots of f (x) are distinct, i = k, as desired. • Galois defined the Galois group of f (x) as Gal( f) = {all a : ai = 0i(v1)
150
APPENDICES
This is the beginning of Galois's 1831 paper in which he characterizes polynomials solvable by radicals as those having a solvable Galois group. (For a proof that this definition is equivalent to the modern one in terms of automorphisms, see [Tignol, p. 329].) Subtle group theoretic clues were in the air, but only Galois recognized their significance; developing them, he invented group theory and solved the mystery of the roots of polynomials. This is even more impressive when we realize that this is no less than the birth of modern algebra.
References
[1] E. Artin, Galois Theory (second edition), Notre Dame, 1955. [2] G. Birkhoff and S. Mac Lane, A Survey of Modern Algebra, (fourth edition), Macmillan, 1977. [3] W. S. Burnside and A. W. Panton, The Theory of Equations, vol. II, Longmans, Green, 1899. [4] S. Chase, D. Harrison, and A. Rosenberg, Galois Theory and Cohomology of Commutative Rings,Mem. Amer. Math. Soc., 1965. [5] E. Dehn, Algebraic Equations, Columbia University Press, 1930. [6] H. M. Edwards, Galois Theory, Springer, 1984. [7] L. Gaal, Classical Galois Theory with Examples, (fourth edition), Chelsea, 1988. [8] C. R. Hadlock, Field Theory and Its Classical Problems, Math. Assn. Amer., 1978. [9] N. Jacobson, Structure of Rings, Amer. Math. Soc., 1956. [10] N. Jacobson, Basic Algebra I, Freeman, 1974. [11] I. Kaplansky, An Introduction to Differential Algebra, 1957.
Hermann,
[12] I. Kaplansky, Fields and Rings (second edition), University Chicago Press, 1974.
152
REFERENCES
[13] A. R. Magid, Lectures on Differential Galois Theory, Mathematical Society, 1994.
American
[14] G. A. Miller, H. F. Blichfeldt, and L. E. Dickson, Theory and Applications of Finite Groups, Dover, 1961, Originally published by Wiley, 1916. [15] E. Netto, Theory of Substitutions, Chelsea, 1961, Reprint of 1882 edition. [16] J. Rotman, A First Course in Abstract Algebra,
PrenticeHall, 1996.
[17] J.P. Tignol, Galois's Theory of Algebraic Equations, Wiley, 1988. [18] B. L. van der Waerden, Modern Algebra (fourth edition), 1966. [19] B. L. van der Waerden, A History of Algebra, Springer, 1985. [20] H. Weyl, Symmetry, Princeton, 1952.
Ungar,
Index
Abbati, P., 147 AbelRuffini theorem, 74 abelian group, 109 accessory irrationalities, 90 action of group, 116 transitive, 116 additive group, 8 adjoining, 51 algebraic element, 51 extension, 51 algebraic numbers, 58 algebraically closed field, 89 alternating group, 109 Artin, E., 60, 139 associated reduced polynomial, 96 associativity, 109 automorphism field, 59 Frobenius, 67 group, 109 Bezout, E., 143 bijection, 3 binomial theorem, 12 biquadratic polynomial, 10 Cameron, P., 40 cancellation law, 13 Cardan, G., 45 Cauchy's theorem, 118 Cayley's theorem, 122 center of group, 121 centralizer, 116
character, 76 characteristic p, 35 characteristic 0, 35 coefficients, 9 commutative ring, 8 commutativity, 109 commutator, 119 subgroup, 119 higher, 119 compass, 129 compositum, 75 congruence class, 8 conjugacy class, 116 conjugate elements, 109 intermediate fields, 81 subgroups, 109 constant polynomial, 10 constant term, 10 constructible number, 130 point, 130 subfield of all, 131 content of polynomial, 40 correspondence theorem groups, 115 rings, 23 coset, 109 cubic formula, 46 cubic polynomial, 10 cyclic group, 110 cyclotomic polynomial, 42 Dedekind, R., 76
154
INDEX
degree extension, 50 polynomial, 10 degree formula, 53 derivative, 12 Descartes, R., 48 dihedral group, 110 discriminant, 96 divides, 24 division algorithm, 17 quotient, 17 remainder, 17 domain, 13 principal ideal, 24 unique factorization, 37 Eisenstein criterion, 41 elementary symmetric function, 140 Euclid's lemma, 26 Euclidean algorithm, 26 Euler's function, 16 evaluation at a, 29 even permutation, 110 extension field, 50 factor groups, 110 FeitThompson theorem, 94 Fermat prime, 138 Ferrari, L., 48 field, 14 algebraic extension, 51 algebraically closed, 89 automorphism, 59 extension, 50 finite extension, 50 fixed, 77 Galois extension, 81 Galois field, 57 intermediate, 81 normal closure, 75 normal extension, 81 perfect, 55 prime field, 35 pure extension, 71 radical extension, 71
separable extension, 56 simple extension, 51 split closure, 75 splitting, 53 finite extension, 50 first isomorphism theorem groups, 114 rings, 23 fixed field, 77 fixes, 59 four group, 110 fraction field, 15 Frobenius automorphism, 67 fundamental theorem algebra, 89 Galois theory, 84 symmetric functions, 140 Galois extension, 81 Galois field, 57 Galois group field extension, 60 of polynomial, 60 Galois resolvent, 148 Galois theorem finite fields, 36 great theorem, 93 Galois, E., 148 Gauss theorem constructibility regular ngons, 137 irreducibility, 41 Gauss's lemma, 40 Gauss, C., 88 gcd, 24 generate ideal, 20 generator of cyclic group, 110 greatest common divisor, 24 group, 110 abelian, 109 alternating group, 109 cyclic, 110 cyclic group generator, 110 dihedral, 110
INDEX 155 four group, 110 pgroup, 111 simple, 111 solvable, 111 symmetric group, 112 group of units, 16 herring, 106 higher commutator subgroups, 119 Hilbert's Theorem 90 , 91 Hilbert, D., 107 homomorphism group, 110 ring, 17 Houston, E., 92 ideal, 18 generated by subset, 20 maximal, 33 prime, 32 principal, 20 proper, 18 ideal generated by subset, 20 image group homomorphism, 110 ring homomorphism, 18 independence of characters, 76 index, 110 injection, 3 integers mod n, 8 integral domain, 13 intermediate field, 81 inverse function, 3 irreducible polynomial, 31 of element, 53 isomorphic rings, 17 isomorphism group, 110 ring, 17 Janusz, G. J., 66 kernel group homomorphism, 110 ring homomorphism, 18 Kronecker theorem, 34
KroneckerWeber theorem, 68 Lagrange resolvent, 145 Lagrange's theorem groups, 113 rational function theorem, 146 lattice, 83 lcm, 28 leading coefficient, 10 least common multiple, 28 Lindemann, F., 136 linear polynomial, 10 linearly disjoint subfields, 86 maximal ideal, 33 monic polynomial, 10 Moore's theorem, 57 multiple, 24 natural map, 21, 111 norm, 91 normal closure, 75 normal extension, 81 normal series, 111 factor groups, 110 normal subgroup, 111 orbit, 116 order element, 111 group, 111 order reversing, 83 origin, 129 orthogonal group, 3 orthogonal transformation, 3 over R, 9 perfect field, 55 permutation, Ill even, 110 pgroup, 111 polynomial associated reduced polynomial, 96 biquadratic, 10 constant polynomial, 10
156
INDEX
constant term, 10 content, 40 cubic, 10 cyclotomic, 42 degree, 10 equality, 9 leading coefficient, 10 linear, 10 monic, 10 polynomial over R, 9 primitive, 40 quadratic, 10 quartic, 10 quintic, 10 reduced, 44,96 root, 10 rvalued, 145 separable, 55 solvable by radicals, 71 symmetric, 140 zero polynomial, 9 polynomial function, 29 polynomial ring over R, 9 prime field, 35 prime ideal, 32 primitive element, 65 primitive polynomial, 40 primitive root of unity, 68 principal ideal domain, 24 principal ideal generated by a, 20 proper ideal, 18 pure extension, 71 type m,71 quadratic polynomial, 10 quartic formula, 48 quartic polynomial, 10 quintic polynomial, 10 quotient, 17 quotient group, 111 quotient ring, 22 radical extension, 71 radical tower, 71 rational functions, 15
reduced polynomial, 44, 96 regular polygon, 7 relatively prime, 25 remainder, 17 repeated roots, 31 second version, 34 resolvent cubic, 48, 103 ring homomorphism, 17 ring map, 17 R modulo 1, 22 root, 10 ruler, 129 rvalued polynomial, 145 Scipio del Ferro, 45 second isomorphism theorem groups, 115 separable element, 56 extension, 56 polynomial, 55 several variables, 16 Shafarevich, I. R., 107 simple extension, 51 simple group, 111 solvable by radicals, 71 solvable group, 111 split closure, 75 splits, 34 splitting field, 53 square, 65 stabilizer, 116 Steinitz theorem, 85 subfield, 17 subgroup, 111 subgroup generated by subset, 111 subring, 12 surjection, 3 Sylow psubgroup, 112, 118 Sylow theorem, 117 symmetric functions, 77 elementary, 140 fundamental theorem, 140 symmetric group, 112 symmetric polynomial, 140
symmetries, 4 symmetry group, 4 Tartaglia, 45 theorem of primitive element, 86, 146 third isomorphism theorem groups, 115 trace, 94 trace theorem, 95 transcendental element, 51 transitive, 63 transitive action, 116 translation, 3 type of pure extension, 71 unique factorization domain, 37 unit, 14 Vandermonde, A.T., 141 Viete, F., 47 Wantzel, P. L., 137 zero polynomial, 9