An Introduction to Galois Theory Andrew Baker [23/01/2013] School of Mathematics & Statistics, University of Glasgow. E-mail address: [email protected] URL: http://www.maths.gla.ac.uk/∼ajb √ Q( 3 2,ζ ) √ ggggggggggg√ggg2ggggngnngngngng2ngngnngnn √HH 2 F3FFFFFFFFFF Q( 3 2) Q( 3 2ζ ) Q( 3 2ζ2) FF3F NN OOOOOOOOOOOOOONNOO3OA3AOAOAOAOAAOAOAOAOAAO3AOAOAOAAOAOAOAOA II 3k3kkkkkkkkkk2kkkFkFkFFkFkFkFkFFQ(ζQQ3) Q SS (cid:11)(cid:11) ∼ Gal(E/Q) = S {id,((cid:16)(cid:16)2 3)}ooWooWoWoWoWoWo{WoiWodWo,Wo(Wo(cid:16)(cid:16)1W3o2Wo3}Wo})oW}}oPW}o}PW}oPW}oPW}oP3}W2o}PW}oPW}oP}WoP}WoP}W}oP}{WoP}WPi}d},23((cid:21)(cid:21)(cid:21)(cid:21) 1 2w)w}wwSwSw3SwSwSwwS3wSwS2wSwSw{wSiwSdwSw,Sw(w1ww2 3)(cid:13)(cid:13) ,(1 3 2)} {id} √ The Galois Correspondence for Q( 3 2,ζ )/Q 3 Introduction: What is Galois Theory? Muchofearlyalgebracentredaroundthesearchforexplicitformulaeforrootsofpolynomial equations in one or more unknowns. The solution of linear and quadratic equations in a single unknown was well understood in antiquity, while formulae for the roots of general real cubics and quartics was solved by the 16th century. These solutions involved complex numbers rather than just real numbers. By the early 19th century no general solution of a general polynomial equation ‘by radicals’ (i.e., by repeatedly taking n-th roots for various n) was found despite considerable effort by many outstanding mathematicians. Eventually, the work of Abel and Galois led to a satisfactory framework for fully understanding this problem and the realization that the general polynomial equation of degree at least 5 could not always be solved by radi- cals. At a more profound level, the algebraic structure of Galois extensions is mirrored in the subgroups of their Galois groups, which allows the application of group theoretic ideas to the studyoffields. ThisGalois Correspondence isapowerfulideawhichcanbegeneralizedtoapply to such diverse topics as ring theory, algebraic number theory, algebraic geometry, differential equations and algebraic topology. Because of this, Galois theory in its many manifestations is a central topic in modern mathematics. In this course we will focus on the following topics. • Thesolutionofpolynomialequationsoverafield,includingrelationshipsbetweenroots, methods of solutions and location of roots. • The structure of finite and algebraic extensions of fields and their automorphisms. We will study these in detail, building up a theory of algebraic extensions of fields and their automorphism groups and applying it to solve questions about roots of polynomial equations. The techniques we will meet can also be applied to study the following some of which may be met by people studying more advanced courses. • Classictopicssuchassquaring the circle, duplication of the cube, constructible numbers and constructible polygons. • Applications of Galois theoretic ideas in Number Theory, the study of differential equations and Algebraic Geometry. There are many good introductory books on Galois Theory, some of which are listed in the Bibliography. In particular, [2, 3, 8] are all excellent sources and have many similarities to the present approach to the material. ⃝ A. J. Baker (2012) BY: ii Contents Introduction: What is Galois Theory? ii Chapter 1. Integral domains, fields and polynomial rings 1 Basic notions, convention, etc 1 1.1. Recollections on integral domains and fields 1 1.2. Polynomial rings 6 1.3. Identifying irreducible polynomials 12 1.4. Finding roots of complex polynomials of small degree 16 1.5. Automorphisms of rings and fields 19 Exercises on Chapter 1 23 Chapter 2. Fields and their extensions 27 2.1. Fields and subfields 27 2.2. Simple and finitely generated extensions 29 Exercises on Chapter 2 33 Chapter 3. Algebraic extensions of fields 35 3.1. Algebraic extensions 35 3.2. Splitting fields and Kronecker’s Theorem 39 3.3. Monomorphisms between extensions 42 3.4. Algebraic closures 45 3.5. Multiplicity of roots and separability 48 3.6. The Primitive Element Theorem 52 3.7. Normal extensions and splitting fields 54 Exercises on Chapter 3 55 Chapter 4. Galois extensions and the Galois Correspondence 57 4.1. Galois extensions 57 4.2. Working with Galois groups 58 4.3. Subgroups of Galois groups and their fixed fields 60 4.4. Subfields of Galois extensions and relative Galois groups 61 4.5. The Galois Correspondence and the Main Theorem of Galois Theory 62 4.6. Galois extensions inside the complex numbers and complex conjugation 64 4.7. Galois groups of even and odd permutations 65 4.8. Kaplansky’s Theorem 68 Exercises on Chapter 4 71 Chapter 5. Galois extensions for fields of positive characteristic 73 5.1. Finite fields 73 iii 5.2. Galois groups of finite fields and Frobenius mappings 77 5.3. The trace and norm mappings 79 Exercises on Chapter 5 80 Chapter 6. A Galois Miscellany 83 6.1. A proof of the Fundamental Theorem of Algebra 83 6.2. Cyclotomic extensions 84 6.3. Artin’s Theorem on linear independence of characters 88 6.4. Simple radical extensions 90 6.5. Solvability and radical extensions 92 6.6. Symmetric functions 96 Exercises on Chapter 6 97 Bibliography 101 iv CHAPTER 1 Integral domains, (cid:12)elds and polynomial rings Basic notions, convention, etc In these notes, a ring will always be a unital ring, i.e., a ring with unity 1 ̸= 0. Most of the rings encountered will also be commutative. An ideal I ▹R will always mean a two-sided ideal. An ideal I ▹R in a ring R is proper if I ̸= R, or equivalently if I R. Under a ring homomorphism φ: R −→ S, 1 ∈ R is sent to 1 ∈ S, i.e., φ(1) = 1. 1.1. Definition. Let φ: R −→ S be a ring homomorphism. • φ is a monomorphism if it is injective, i.e., if for r ,r ∈ R, 1 2 φ(r ) = φ(r ) =⇒ r = r , 1 2 1 2 or equivalently if kerφ = {0}. • φ is an epimorphism if it is surjective, i.e., if for every s ∈ S there is an r ∈ R with φ(r) = s. • φ is an isomorphism if it is both a monomorphism and an epimorphism, i.e., if it is invertible (in which case its inverse is also an isomorphism). 1.1. Recollections on integral domains and (cid:12)elds The material in this section is standard and most of it should be familiar. Details may be found in [3, 5] or other books containing introductory ring theory. First we recall some important properties of elements in a ring. 1.2. Definition. Let R be a ring. An element u ∈ R is a unit if it is invertible, i.e., there is an element v ∈ R for which uv = 1 = vu. We usually write u−1 for this element v, which is necessarily unique and is called the (multi- × plicative) inverse of u in R. We will denote the set of all invertible elements of R by R and note that it always forms a group under multiplication. 1.3. Definition. Let R be a commutative ring. Then a non-zero element z ∈ R is a zero-divisor if there is a non-zero element w ∈ R for which zw = wz = 0. A commutative ring R in which there are no zero-divisors is called an integral domain or an entire ring. This means that for u,v ∈ R, uv = 0 =⇒ u = 0 or v = 0. 1.4. Example. The following rings are integral domains. (i) The ring of integers, Z. 1 (ii) If p is a prime, the ring of integers modulo p, F = Z/p = Z/(p). p (iii) The rings of rational numbers, Q, real numbers, R, and complex numbers, C. (iv) ThepolynomialringR[X],whereRisanintegraldomain;inparticular,thepolynomial rings Z[X], Q[X], R[X] and C[X] are all integral domains. 1.5. Definition. Let I ▹R be a proper ideal in a commutative ring R. • I is a prime ideal if for u,v ∈ R, uv ∈ I =⇒ u ∈ I or v ∈ I. • I is a maximal ideal R if whenever J ▹R is a proper ideal and I ⊆ J then J = I. • I ▹R is principal if I = (p) = {rp : r ∈ R} for some p ∈ R. Notice that if p,q ∈ R, then (q) = (p) if and only if q = up for some unit u ∈ R. We also write p | x if x ∈ (p). • p ∈ R is prime (or is a prime) if (p) ▹ R is a prime ideal; this is equivalent to the requirement that whenever p | xy with x,y ∈ R then p | x or p | y. • Risaprincipal ideal domain ifitisanintegraldomainandeveryidealI▹Risprincipal. • A non-zero element p ∈ R is irreducible (or is an irreducible) if for u,v ∈ R, p = uv =⇒ u or v is a unit. 1.6.Example. EveryidealI▹Zisprincipal, soI = (n)forsomen ∈ Zwhichwecanalways take to be non-negative, i.e., n > 0. Hence Z is a principal ideal domain. 1.7. Proposition. Let R be a commutative ring and I ▹R an ideal. (i) The quotient ring R/I is an integral domain if and only if I is a prime ideal. (ii) The quotient ring R/I is a (cid:12)eld if and only if I is a maximal ideal. 1.8. Example. If n > 0, the quotient ring Z/n = Z/(n) is an integral domain if and only if n is a prime. For any (not necessarily commutative) ring with unity there is an important ring homomor- phism η: Z −→ R called the unit or characteristic homomorphism which is defined by    1|+·{·z·+1} if n > 0,    n η(n) = n1 = −(1+···+1) if n < 0,  | {z }    −n   0 if n = 0. Since 1 ∈ R is non-zero, kerη ▹Z is a proper ideal and using the Isomorphism Theorems we see that there is a quotient monomorphism η: Z/kerη −→ R which allows us to identify the quotient ring Z/kerη with the image ηZ ⊆ R as a subring of R. By Example 1.6, there is a unique non-negative integer p > 0 such that kerη = (p); this p is called the characteristic of R and denoted charR. 1.9. Lemma. If R is an integral domain, its characteristic charR is a prime. 2 Proof. Consider p = charR. If p = 0 we are done. So suppose that p > 0. The quotient monomorphism η: Z/kerη −→ R identifies Z/kerη with the subring imη = imη of the integral domainR. Buteverysubringofanintegraldomainisitselfanintegraldomain, henceZ/kerη is anintegraldomain. NowbyProposition1.7(i),kerη = (p)isprimeidealandsobyExample1.8, p is a prime. (cid:3) 1.10. Remark. When discussing a ring with unit R, we can consider it as containing as a subring of the form Z/(charR) since the quotient homomorphism η: Z/(charR) −→ R gives an isomorphism Z/(charR) −→ imη, allowing us to identify these rings. In particular, every integraldomaincontainsasasubringeitherZ = Z/(0)(ifcharR = 0)orZ/(p)ifp = charR > 0 is a non-zero prime. This subring is sometimes called the characteristic subring of R. The rings Z and Z/n = Z/(n) for n > 0 are often called core rings. When considering integral domains, the rings Z and F = Z/p = Z/(p) for p > 0 a prime are called prime rings. p Here is a useful and important fact about rings which contain a (cid:12)nite prime ring F . p 1.11. Theorem (Idiot’s Binomial Theorem). Let R be a commutative ring containing F p for some prime p > 0. If u,v ∈ R, then (u+v)p = up+vp. Proof. We have p1 = 0 in R, hence pt = 0 for any t ∈ R. The Binomial Expansion yields ( ) ( ) ( ) p p p (1.1) (u+v)p = up+ up−1v+ up−2v2+···+ uvp−1+vp. 1 2 p−1 Now suppose that 1 6 j 6 p−1. Then we have ( ) p p(p−1)! (p−1)! = = p× . j j!(p−j)! j!(p−j)! There are no factors of p appearing in (p−1)!, j! or (p−j)!, so since this number is an integer it must be divisible by p, i.e., ( ) p (1.2a) p | , j or equivalently ( ) p (1.2b) ≡ 0 (mod p). j Hence in R we have ( ) p 1 = 0. j Combining the divisibility conditions of (1.2) with the expansion of (1.1), we obtain the required equation in R, (u+v)p = up+vp. (cid:3) 1.12. Definition. A commutative ring | is a (cid:12)eld if every non-zero element u ∈ | is a unit. This is equivalent to requiring that |× = |−{0}. The familiar rings Q, R and C are all fields. 1.13. Example. If n > 1, the quotient ring Z/n is a field if and only if n is a prime. 1.14. Proposition. Every (cid:12)eld is an integral domain. 3 Proof. Let | be a field. Suppose that u,v ∈ | and uv = 0. If u ̸= 0, we can multiply by u−1 to obtain v = u−1uv = 0, hence v = 0. So at least one of u,v must be 0. (cid:3) 1.15. Lemma. Let R be an integral domain. If p ∈ R is a non-zero prime then it is irre- ducible. Proof. Suppose that p = uv for some u,v ∈ R. Then p | u or p | v, and we might as well assume that u = tp for some t ∈ R. Then (1−tv)p = 0 and so tv = 1, showing that v is a unit with inverse t. (cid:3) Now let D be an integral domain. A natural question to ask is whether D is isomorphic to a subring of a field. This is certainly true for the integers Z which are contained in the field of rational numbers Q, and for a prime p > 0, the prime ring F is itself a field. p 1.16. Definition. The fields Q and F where p > 0 is a prime are the prime (cid:12)elds. p Of course, we can view Z as a subring of any subfield of the complex numbers so an answer to this question may not be unique! However, there is always a ‘smallest’ such field which is unique up to an isomorphism. 1.17. Theorem. Let D be an integral domain. (i) There is a field of fractions of D, Fr(D), which contains D as a subring. (ii) If φ: D −→ F is a ring monomorphism into a (cid:12)eld F, there is a unique homomorphism φe: Fr(D) −→ F such that φe(t) = φ(t) for all t ∈ D ⊆ Fr(D). φ // D << F inc (cid:15)(cid:15) ∃! φe Fr(D) Proof. (i) Consider the set P(D) = {(a,b) : a,b ∈ D, b ̸= 0}. Now introduce an equivalence relation ∼ on P(D), namely (a′,b′) ∼ (a,b) ⇐⇒ ab′ = a′b. Of course, it is necessary to check that this relation is an equivalence relation; this is left as an exercise. We denote the equivalence class of (a,b) by [a,b] and the set of equivalence classes by Fr(D). We define addition and multiplication on Fr(D) by [a,b]+[c,d] = [ad+bc,bd], [a,b][c,d] = [ac,bd]. ′ ′ We need to verify that these operations are well defined. For example, if [a,b] = [a,b] and ′ ′ [c,d] = [c,d], then ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ (ad +bc)bd = adbd+bcbd = abdd+bbcd = (ad+bc)bd, 4 and so (a′d′ +b′c′,b′d′) ∼ (ad+bc,bd); hence addition is well defined. A similar calculation shows that (a′c′,b′d′) ∼ (ac,bd), so multiplication is also well defined. It is now straightforward to show that Fr(D) is a commutative ring with zero 0 = [0,1] and unit 1 = [1,1]. In fact, as we will soon see, Fr(D) is a field. Let [a,b] ∈ Fr(D). Then [a,b] = [0,1] if and only if (0,1) ∼ (a,b) which is equivalent to requiring that a = 0; notice that for any b ̸= 0, [0,b] = [0,1]. We also have [a,b] = [1,1] if and only if a = b. Now let [a,b] ∈ Fr(D) be non-zero, i.e., a ̸= 0. Then b ̸= 0 and [a,b],[b,a] ∈ Fr(D) satisfy [a,b][b,a] = [ab,ba] = [1,1] = 1, so [a,b] has [b,a] as an inverse. This shows that Fr(D) is a field. We can view D as a subring of Fr(D) using the map j: D −→ Fr(D); j(t) = [t,1] which is a ring homomorphism; it is easy to check that it is a monomorphism. Therefore we may identify t ∈ D with j(t) = [t,1] ∈ Fr(D) and D with the subring imj ⊆ Fr(D). (ii) Consider the function Φ: P(D) −→ F; Φ(a,b) = φ(a)φ(b)−1. If (a′,b′) ∼ (a,b), then Φ(a′,b′) = φ(a′)φ(b′)−1 = φ(a′)φ(b)φ(b)−1φ(b′)−1 = φ(a′b)φ(b)−1φ(b′)−1 = φ(ab′)φ(b′)−1φ(b)−1 = φ(a)φ(b′)φ(b′)−1φ(b)−1 = φ(a)φ(b)−1 = Φ(a,b), so Φ is constant on each equivalence class of ∼. Hence we may define the function φe: Fr(D) −→ F; φe([a,b]) = Φ(a,b). It is now easy to verify that φeis a ring homomorphism which agrees with φ on D ⊆ Fr(D). (cid:3) The next three corollaries are left as an exercise. 1.18. Corollary. If F is a (cid:12)eld then F = Fr(F). 1.19. Corollary. If D is a subring of a (cid:12)eld F, then Fr(D) ⊆ Fr(F) = F and Fr(D) is the smallest sub(cid:12)eld of F containing D. 1.20. Corollary. Let D and D be integral domains and let φ: D −→ D be a ring 1 2 1 2 monomorphism. Then there is a unique induced ring homomorphism φ∗: Fr(D1) −→ Fr(D2) which satis(cid:12)es φ∗(t) = φ(t) whenever t ∈ D1 ⊆ Fr(D1). φ // D D 1 2 inc inc (cid:15)(cid:15) (cid:15)(cid:15) // Fr(D ) Fr(D ) 1 φ∗ 2 Moreover, this construction has the following properties. 5