GALOIS THEORY (2012, M24) NOTES CONTENTS

GALOIS THEORY (01, M4) NOTES TERUYOSHI YOSHIDA CONTENTS Intro 1: Cubics, Quartics Intro : Circles 5 1. Classical Galois Theory (as Galois did it) 6 1.1. Basic notions 6 1.. Simple extensions 7 1.3. Finite extensions 8 1.4. K-homomorphisms 10 1.5. K-homomorphism into C 11 1.6. Galois extensions 15 1.7. Galois correspondence 17 1.8. Insolvability of quintics I: radical extensions 0 1.9. Insolvability of quintics II: general equations 1.10. Solving by radicals 4 1.11. Discriminants, and revisiting Intro 1 5. General Fields and Applications 8.1. General remarks 8.. Splitting fields and algebraic closures 9.3. Example I: Finite fields 30.4. Application I: Cyclotomic fields 33.5. Separability 36.6. Example II: Symmetric Function Theorem 39.7. Application II: Galois groups over Q 41 3. Modern Galois Theory (linear algebraic approach) 44 3.1. Dedekind s and Artin s lemmas 44 3.. Towers of extensions 45 3.3. Traces and norms 47 3.4. Infinite extensions, etc. 50 Date: January 1, 013. Note: 3.4 and the Appendices are not examinable. 1

TERUYOSHI YOSHIDA Appendix 1: Roots of Unity, Radical / Soluble Extensions ( 1.8, 1.11) 5 Appendix : Why General Fields, and How? (.1) 54 Appendix 3: Zorn s Lemma and Algebraic Closures (.) 56 Appendix 4: Gauss Lemma (from Groups, Rings & Modules;.4,.7) 58 Appendix 5: Algebraic Independence of Elementary Symmetric Polynomials (.6) 59 Appendix 6: Normal Basis Theorem ( 3.1) 60 Appendix 7: Transitivity of Traces / Norms ( 3.3) 61 Appendix 8: What Next? 6 Index 64 INTRO 1: CUBICS, QUARTICS Lecture 1 (4 Oct, Th.) Quadratics. (1) X + b = 0 = X = ± b. () X ax + b = (X α)(x β) = 0. Note a = α + β, b = αβ. We reduce to (1) i.e. the case a = 0. Set: α := α a, β := β a. Then: α + β = 0, α β = αβ a a (α + β) + 4 = b a + a 4 = b a 4, hence α, β are roots of X + (b a ) a = 0, i.e. ± 4 4 b by (1). Thus α, β = a a ± 4 b. Cubics. (3) X 3 c = 0 = X = 3 c, 3 cζ, 3 cζ, where 1, ζ, ζ are the roots of X 3 1 = 0. As X 3 1 = (X 1)(X + X + 1), we have ζ, ζ = 1 ± 3 4 by (). [If one is ζ then the other is ζ. Note ζ + ζ + 1 = 0.] (4) X 3 + bx c = (X α)(x β)(x γ) = 0. By expanding: 0 = α + β + γ, b = αβ + βγ + γα, c = αβγ. The Lagrange resolvents of this cubic are defined as: x := α + βζ + γζ, xζ = αζ + βζ + γ, xζ = αζ + β + γζ y := α + βζ + γζ, yζ = αζ + β + γζ, yζ = αζ + βζ + γ. By (3), the first row gives the roots of X 3 x 3 = 0, second X 3 y 3 = 0. Using ζ + ζ + 1 = 0 and adding these expressions (remember α + β + γ = 0), we get x + y = 3α, xζ + yζ = 3β, xζ + yζ = 3γ,

hence ( x + y (α, β, γ) = 3 Now by (note ζ ζ = 1 and ζ + ζ + 1 = 0) x, y determine each other. Also, GALOIS THEORY 3, xζ + yζ, 3 xζ + ) yζ. 3 xy = α + β + γ αβ βγ γα = (α + β + γ) 3b = 3b, x 3 + y 3 = (x + y)(x + yζ)(x + yζ ) = (x + y)(xζ + yζ )(xζ + yζ) = 3α 3β 3γ = 7c x 3 y 3 = 7b 3 shows that x 3, y 3 are the two roots of X 7cX 7b 3 = 0, solved in (). (5) X 3 ax + bx c = (X α)(x β)(x γ) = 0. Reduced to (4), i.e. the case a = 0 as before, by setting α := α a 3, β := β a 3, γ := γ a 3. A computation (exercise) shows that α, β, γ are the three roots of X 3 + (b a ) ( X 3 7 a3 ab ) 3 + c = 0, solved in (4). We get α, β, γ by adding a to them. 3 Quartics. (6) X 4 ax 3 + bx cx + d = (X α)(x β)(x γ)(x δ) = 0. Similarly we subtract a from α, β, γ, δ and assume a = α + β + γ + δ = 0. 4 Now we will reduce to (5). Let: x := α + β = (γ + δ), y := α + γ = (β + δ), z := α + δ = (β + γ). Then we have ( x + y + z (α, β, γ, δ) = hence it suffices to find x, y, z. First note:, x y z xyz = (α + β)(α + γ)(α + δ), x + y z = α 3 + (β + γ + δ)α + (βγ + γδ + δβ)α + βγδ, x y + z ). = (α + β + γ + δ)α + (βγα + γδα + δβα + βγδ) = c. As x = (α + β)(γ + δ), y = (α + γ)(β + δ), z = (α + δ)(β + γ),

4 TERUYOSHI YOSHIDA we have x + y + z = (αβ + αγ + αδ + βγ + γδ + γδ) = b, x y + y z + z x = b 4d x y z = c (above). Hence x, y, z are roots of the resolvent cubic: (exercise), X 3 + bx + (b 4d)X c = 0, which is solved in (5). Now we get x, y, z by (1). If we choose signs for x, y, then z is determined by xyz = c, hence there are 4 choices. Summary of the symmetries. (1): b b, S = C. (1) (): rational (+,,, ). 3 (3): c 3 cζ, A 3 = C 3. 3 cζ (4): Note S 3 A 3 (normal subgroup) and S 3 /A 3 = S : x xζ, y yζ A 3 xζ yζ x 3 y 3 S (4) (5): rational (+,,, ). (6): S 4 {α, β, γ, δ} (group acting by permutation); this induces S 4 S 3 {x, y, z }. The kernel of this surjective homomorphism S 4 S 3 is: S 4 {permutations of α, β, γ, δ fixing each of x, y, z } = {id, (α β)(γ δ), (α γ)(β δ), (α δ)(β γ)} = V 4 = C C, i.e. S 4 V 4 and S 4 /V 4 = S3. Note that V 4 = {x x, y y, z z change signs of x, y, z but not of xyz} {±x, ±y, ±z}. Recall: Field = a set closed under +,,,.

GALOIS THEORY 5 (i) Rational operations (+,,, ) occur within the same field ((1) (), (4) (5)). (ii) But whenever the field has changed (been extended), an additional symmetry (will be called Galois groups) was introduced ((1),(3),(4),(6)). Galois insight ignore (i), and keep track of (ii). We exploited the subgroups partially symmetric polynomials of the roots. S 5 (or S n for n 5) has no proper normal subgroups other than A 5 (or A n ) which is simple, so similar reduction seems impossible. INTRO : CIRCLES Lecture (6 Oct, Sa.) (a) Abstract = Intuitive. By a circle, we mean: (Euclid, BC 300) P : centre, r: radius, C := {Q plane P Q = r}. (Descartes, 17c) C := {(x, y) R x + y = r }. ( ) (Galois/Lie, 19c) A set on which the following group acts: T θ : rotation of angle θ, s.t. R/πZ = (i) T 0 = id, (ii) T θ T θ = T θ +θ, (iii) T θ = T θ θ θ πz ( ) captures the essence of circular shapes of cylinders, cones, wine glasses... (b) Dictionaries. Intuitions Manipulating symbols Geometry Algebra Groups Pythagoras P Q = r r = ( ) ( ) ( ) x + y trigs x cos θ sin θ x T θ = y sin θ cos θ y (x, y) C = T θ (x, y) C gives cos θ + sin θ = 1, T θ T θ = T θ +θ gives the addition formula. The symmetry group ( ) was hidden beneath the equation ( )... not so obvious! Similarly, beneath the equation X 3X 5 = 0 lies X 3 X, because (3 X) 3(3 X) 5 = X 3X 5. (check!) Beneath the equation X 4 + 5X 3 6X 1X + 1 = 0 lies the group C 4 : X 4X (1 X) (1 X)(1 + 3X) 4X 1 X (...check! [from Gauss diary 1797]) 1 + 3X

6 TERUYOSHI YOSHIDA 1. CLASSICAL GALOIS THEORY (AS GALOIS DID IT) 1.1. Basic notions. The symbols K, L, F will always denote fields. Definition 1. Let L be a field. If a subring K of L is a field, it is called a subfield of L. We say L is an extension of K. We refer to the pair as an extension L/K. (Read L over K, not a quotient in any sense.) Note that if K, L are both subfields of F and K L, then K is a subfield of L, hence we have extensions F/K, F/L and L/K; sometimes L/K is called a subextension of F/K. In this section ( 1) we are mainly interested in subfields of C, although until 1.4 all will be valid for general fields. Example. Q R C. Recall the notation Q( ) := {a + b a, b Q}, similarly for Q( 1) etc. Then Q Q( ) R and Q( 1) R. Recall that C was an R-v.s. (vector space). In general, if L/K is an extension. The multiplication in L makes L into a K-v.s. (L is an additive group; 1 x = x, (ab)x = a(bx), (a + b)x = ax + bx, a(x + y) = ax + ay by the axiom for rings.) We always consider L as a K-v.s. in this way. If K L F, then L is a sub-k-v.s. of F. Definition. We say an extension L/K is finite if L is a finite dimensional K-v.s., otherwise infinite. Its dimension is called the degree of L/K, denoted by [L : K]. (We also say L is finite over K, L is a finite extension of K.) Example. Q( )/Q : finite, with basis {1, }. [Q( ) : Q] =, [C : R] =, R/Q : infinite. Recall K[X] := the ring of polynomials in X with coefficients in K. [Polynomials are formal K-linear combinations of the monomials 1, X, X,...; they form a ring and a K-v.s., but usually not considered as functions.] Definition 3. Let L/K be an extension, and α L. Let I α := {P (X) K[X] P (α) = 0} K[X], the set of all polynomial which has α as a root. We say α is algebraic over K if I α {0}, and transcendental over K if I α = {0}. We say L/K is algebraic if every α L is algebraic /K (reads over K ); otherwise transcendental. Example., 3 : algebraic /Q. π, e: transcendental /Q. Proposition 4. Every finite extension is algebraic.

GALOIS THEORY 7 Proof. If [L : K] = n and α L, then the elements 1, α, α,..., α n L are linearly depedent over K, hence P (α) = 0 for some nonzero P K[X], i.e. I α 0. Now note that I α in Def. 3 is the kernel of the ring homomorphism f α : K[X] P (X) P (α) L ( plug in α ; it is also a K-linear map), hence an ideal of K[X]. (Also directly checked from P (α) = Q(α) = 0 = (P + Q)(α) = 0, R(α)P (α) = 0 R K[X].) Lecture 3 (9 Oct, Tu.) Definition 5. Let L/K be an extension, and α L be algebraic /K. As K[X] is a PID, we have I α = (P α ) = {multiples of P α } for a unique monic P α K[X]. This P α is called the minimal polynomial of α over K. Note that deg P α is minimal among deg P for P 0 in I α. Example. (i) Min. poly. of α = over Q: P α = X Q[X]. (ii) Min. poly. of α = over R: P α = X R[X]. (iii) Min. poly. of α = 3 over Q: P α = X 3 Q[X]. Consider f α : Q[X] C, and: Q(α) := Im f α = {a + bα + cα a, b, c Q} C (in fact R). This is a field: it is a ring, and every nonzero element is invertible. How do you find (1 + α) 1? Use (1 + α)(1 α + α ) = 1 + α 3 = 3, hence (1 + α) 1 = 1 3 (1 α + α ) Q(α). 1.. Simple extensions. Note: the intersection of subfields is a subfield; the unions are not in general. Definition 6. Let L/K be an extension and α L. We denote by K(α) the intersection of all subfields of L containing K and α, i.e. the minimal such subfield. Then K(α)/K is called the extension generated by α. We say L/K is simple if L = K(α) for some α L. Proposition 7. Let L/K be an extension and α L be algebraic /K. (i) Its minimimal polynomial P α over K is irreducible in K[X]. (ii) Im f α = K(α) and [K(α) : K] = deg P α. In particular K(α)/K is finite. Proof. (i): If P α (X) = P (X)Q(X), then P (α)q(α) = P α (α) = 0, hence P (α) = 0 or Q(α) = 0. Say P (α) = 0; then P I α = (P α ), i.e. P α P (reads: P α divides P ), hence Q is a unit in K[X]. (ii): (1) Im f α is a subfield of L.... ) It is a ring (being the image of a ring hom. fα ). Every x Im f α is of the form P (α) for some P K[X]. If x 0, then P / (P α ), i.e. P is not divisible by P α. Hence Q K[X] with P Q 1 mod P α, therefore P (α) 1 = Q(α) Im f α.

8 TERUYOSHI YOSHIDA () Im f α = K(α).... ) As Im fα is a subfield of L containing K and α, and any such subfield must contain Im f α, we have Im f α = K(α). (3) If deg P α = n, then {1, α, α,..., α n 1 } gives a basis of K(α).... ) For every x = P (α) Im fα, there exists Q, R K[X], with P = P α Q+R and deg R < n, hence x = P (α) = R(α) is a K-linear combination of {1, α, α,..., α n 1 }. If R(α) = 0 for some R K[X] with deg R < n, then P α R, hence R = 0. (1) Remark. (i) Different elements can generate the same field, i.e. we can have K(α) = K(α ) with α α (even P α P α ), e.g. Q( ) = Q(1 + ). (ii) By Prop. 4 and 7(ii), for an extension L/K and α L: α : algebraic /K K(α)/K : finite. (iii) If K L F and α F, then K[X] L[X] implies: (1) α : algebraic /K = α : algebraic /L. () its min. poly. Q α over L divides its min. poly. P α over K. We will see in 1.3 that the converse of (i) holds if L/K is finite. (iv) Related question. 3 : alg./q (root of X ), : alg./q (root of X 3 ). Is + 3 alg./q? (Root of what?) Note + 3 Q( )( 3 ), a tower of simple extensions. 1.3. Finite extensions. Note: If [L : K] = 1, then L = K (1-dim. K-v.s.). Proposition 8. (Tower Law) Let K L F. If F/K is finite, then so are L/K and F/L. Conversely, if L/K and F/L are finite, then so is F/K, and [F : K] = [F : L][L : K]. Proof. We have dim K L dim K F since L is a sub-k-v.s. of F, and dim L F dim K F since a spanning set of F as a K-v.s. is also spanning as a L-v.s. If {a 1,..., a n } L is a basis of L/K (i.e. a basis of L as a K-v.s.), and {b 1,..., b m } F is a basis of F/L, then every x F is written as x = x 1 b 1 + + x m b m with x j L, and each x j L is written as x j = x 1j a 1 + + x nj a n with x ij K. Hence x = ( ) x ij a i bj = x ij a i b j. If x = x ij a i b j = 0, then j i i,j i,j x ij a i = 0 j by the L-linear independence of {b j }, therefore x ij = 0 i, j by the K-linear i independence of {a i }. Thus {a i b j 1 i n, 1 j m} is a basis of F/K. () (1) (optional) A slicker proof of Prop. 7. By hom. thm. K[X]/(Pα ) = Im f α as rings and K-v.s.; as RHS is a subring of a field L, hence an integral domain, (P α ) is prime; being non-zero it is maximal (K[X] : PID), hence LHS is a field; so is RHS. As a K-v.s. the LHS has basis {1, X, X,..., X n 1 }, where X = X mod P α, we get dim = n. This LHS is an ext n of K in the abstract sense (a field K, but not C). Its element X is an abstract root of P α. () More generally, any fin. dim. L-v.s. V can be considered as a K-v.s. by restricting the scalar multiplication, and the same proof shows dim K V = [L : K] dim L V.

GALOIS THEORY 9 Lecture 4 (11 Oct, Th.) Definition 9. Let L/K be an extension and α 1,..., α n L. We denote by K(α 1,..., α n ) the intersection of all subfields of L containing K and α 1,..., α n, i.e. the minimal such subfield. Then K(α 1,..., α n )/K is called the extension generated by α 1,..., α n. The order of α 1,..., α n is irrelevant, and K(α 1,..., α n ) = K(α 1,..., α n 1 )(α n ) (both inclusions are immediate from the definitions). Proposition 10. (i) If L/K is an extension and α 1,..., α n L are algebraic over K, then K(α 1,..., α n )/K is finite. (ii) Conversely, every finite extension L/K is generated by finitely many elements, i.e. there exist α 1,..., α n L with L = K(α 1,..., α n ). Proof. (i): As α n is alg. /K, it is a fortiori alg. /K(α 1,..., α n 1 ), hence K(α 1,..., α n ) = K(α 1,..., α n 1 )(α n ) is finite over K(α 1,..., α n 1 ) by Prop. 7(ii). Repeat for each K(α 1,..., α i ) with 1 i n and use Tower Law (Prop. 8). (ii): Take a basis {e 1,..., e n } of L/K. Then K(e 1,..., e n ) = L because every x L is a K-linear combination of e 1,..., e n. Example. The min. poly. of 3 over Q( ) is X 3, as it is irreducible in Q( )[X] (else its root Q( ) generates a field of deg. 3 /Q), hence [Q(, 3 ) : Q( )] = 3 by Prop. 7(ii). Q(, 3 ) + 3... alg. /Q 3 Q( ) 6 Q Thus [Q(, 3 ) : Q] = 6 (Prop. 8). What is the min. poly. of + 3 over Q? (It has to have degree 6 because [Q( + 3 ) : Q] 6. In fact it is 1,,3 or 6 by Tower Law.) Example. The fields of the form K( a) and K( a, b) for non-square elements a, b K are called quadratic and biquadratic fields over K. Q(, 1) Q( 1) Q( ) Q( ) Q fin. ext. of degree 4 /Q, with a basis {1,, 1, }

10 TERUYOSHI YOSHIDA Remark. Finite extensions are algebraic (Prop. 4), but infinite algebraic extensions. (3) 1.4. K-homomorphisms. Lemma 11. Let L be a field and L be a ring with L {0}. Then any ring homomorphism τ : L L is injective. Proof. As Ker τ is an ideal of L not containing 1 L (... τ(1) = 1), hence {0}. Definition 1. Let L/K, L /K be two extensions of K. A K-homomorphism from L to L is a ring hom. τ : L L such that τ K = id. The set of all K-hom s from L to L is denoted by Hom K (L, L ). Note: All K-hom s are injective (also called embeddings), and they are K-linear. Here we re mainly interested in the set Hom K (L, C) (when K L C) and its cardinality. Example. (i) C/R. R-hom. τ : C C... of them (note C = R( 1)): id : 1 1 Hom R (C, C) = {id, cpx conj}, cpx conj : 1 1 (ii) Q( )/Q. Q-hom. τ : Q( ) C... of them: τ 1 = id :, τ :. Note: being a ring hom., it must send a root of P to a root of P if P K[X]. (iii) Q( 3 )/Q. Let ζ 3 = 1, ζ 1. 3 τ 3 Q-hom s τ : Q( 3 1 = id : 3 ) C: 3 τ : 3 ζ... 3 roots of X 3 in C. 3 3 τ 3 : ζ Note: in contrast to (i),(ii), here their images Im τ i are all different subfields of C. We see that K-hom s from K(α) are closely related to the roots of the minimal polynomial P α of α over K. Definition 13. (i) For a nonzero P K[X] and an extension L/K, we denote by Root P (L) the set of all roots of P in L. (ii) Let α L be algebraic /K. A root of its minimal poly. P α in L, i.e. an element of Root Pα (L) is called a conjugate of α in L over K. (3) Take all α C which are algebraic /Q (countable!), and order them as α1, α, α 3,.... We get a sequence Q(α 1) Q(α 1, α ) Q(α 1, α, α 3), so form their union Q := n Q(α 1,..., α n). It is a subfield of C, and contains all α C alg. /Q. Conversely every α Q belongs to some finite ext n of Q, hence is alg. /Q, i.e. Q/Q is algebraic. It is not finite, as irreducible poly. /Q of degree n for all n N.

We denote the cardinality of the set X by X. GALOIS THEORY 11 Proposition 14. (Roots and Hom s I) Let F/K, E/K be two extensions of K and α F be algebraic /K. Then we have a bijection: Hom K ( K(α), E ) τ = τ(α) RootPα (E). In particular we have Hom K ( K(α), E ) [K(α) : K]. Remark. The case we have in mind in this 1 is when K C and F = E = C. Proof. (1) We have a map.... ) As Pα (α) = 0 and τ is ring hom. with τ K = id, we have P α (τ(α)) = τ(p α (α)) = 0, i.e. τ(α) Root Pα (E). (i.e. every K-hom. sends α to its conjugate over K.) () It is injective.... ) Recall K(α) = Im fα, where f α : K[X] P P (α) E (Prop. 7(ii)), i.e. all el ts in K(α) are poly. in α with coeff. in K, the K-hom. τ is determined by τ(α) E. (3) It is surjective.... ) Let β RootPα (E). We will define τ : K(α) E satisfying τ(α) = β. Every x K(α) is written as x = P (α) with P K[X], and P is unique up to adding multiples of P α (any other choice of P is of the form P + P α Q) and P α (β) = 0, hence P (β) E is well-defined. So let τ(x) := P (β), i.e. τ : K(α) P (α) P (β) E. This is clearly a ring hom. with τ K = id. (4) Hom K ( K(α), E ) = RootPα (E) deg P α Prop.7(ii) = [K(α) : K] (Lem. 53(ii)). 1.5. K-homomorphism into C. Lecture 5 (13 Oct, Sa.) Note: subfields of C are always extensions of Q. For K C and α C, by conjugates of α over K we will always mean its conjugates in C, e.g. conj. of 3 over Q are 3, 3 ζ, and 3 ζ. Proposition 15. Let K be a subfield of C. (i) Let P K[X] an irreducible poly. in K[X]. Then Root P (C) = deg P. (This property will be called the separability of P.) (ii) Let α C be algebraic /K. Then Hom K ( K(α), C ) = [K(α) : K]. Proof. (i): A multiple root α C of P would also be a root of P (X) := d P (X) K[X], but dx deg P < deg P and P is irreducible, hence P, P are coprime in K[X], i.e. Q, R K[X] with P Q + P R = 1 in K[X], hence also in C[X]. Thus P, P are coprime in C[X], hence all roots are distinct (deg P of them, by the fundamental th m of alg. ). ( ) Prop.14 (ii): Hom K K(α), C = Root Pα (C) = (i) Prop.7(ii) deg P α = [K(α) : K].

1 TERUYOSHI YOSHIDA Next goal: generalise Prop. 15(ii) to: Hom K (F, C) = [F : K] finite F/K (called the separability of F/K). (4) Method: break down to simple ext ns & stack up. If K L F and ρ Hom K (F, C), then ρ L Hom K (L, C) (ring hom., id on K), i.e. we have a restriction map: Hom K (F, C) ρ ρ L Hom K (L, C). To climb up, count the # of ρ s with a fixed ρ L, i.e. the fibres of this map. (In other words, the # of ways to extend a given τ Hom K (L, C) to F. The essential case will be where F = L(α), i.e. F is simple over L.) Example. (i) i := 1, L = Q( ). Hom K (L, C) = {τ 1 = id, τ : }. F = Q(, i) L = Q( ) K = Q Let ρ Hom K (F, C). ρ L = τ 1, i.e. ρ( ) = ρ(i) = i = ρ 1 = id ρ(i) = i = ρ : (, i) (, i) ρ L = τ, i.e. ρ( ) = ρ(i) = i = ρ 3 : (, i) (, i) ρ(i) = i = ρ 4 : (, i) (, i) (ii) The min. poly. of α := over Q is X 4. F = Q( 4 ) L = Q( ) K = Q = we know Hom K (F, C): ρ 1 = id : 4 4 ρ : 4 4 ρ 3 ρ 4 : 4 4 i : 4 4 i conjugates of α /K: { 4, 4, 4 i, 4 i}. First two: roots of X =: P α (the min. poly. of α over L), Restrict them to L. Note ρ( ) = ρ(α ) = ρ(α). Thus: ρ 1, ρ :, i.e. ρ 1 L = ρ L = τ 1 = id, ρ 3, ρ 4 :, i.e. ρ 3 L = ρ 4 L = τ Start with ρ L Hom K (L, C) and try to extend: ρ L = τ 1 = ρ must map α to a root of P α, Last two: roots of X + = τ P α. ρ L = τ = ρ must map α to a root of τ P α, as ρ maps P α to τ P α = X +. i.e. to extend τ : L C to F = L(α), map α to a root of τp α C[X] (defined below). (4) In fact we will see that every F/K inside C is simple (Th.0), but for that we need this first.

GALOIS THEORY 13 Definition 16. Let L be a field and τ : L L be a ring hom. For P L[X], we denote by τp L [X] the poly. obtained by applying τ to the coefficients of P. In the following proposition, generalising Prop. 14 (where L = K and τ = id), we allow arbitrary fields. But again we have the case K C and F = E = C in mind. Proposition 17. (Roots and Hom s II) Let F/K, E/K be two extensions of K. Let K L F and α F be algebraic /L with min. poly. P α over L. Then for every τ Hom K (L, E) we have a bijection: Proof. (1) map. { ρ HomK ( L(α), E ) ρ L = τ } ρ = ρ(α) RootτPα (E). ρ(p α (α)) = 0, i.e. ρ(α) Root τpα (E).... ) Pα (α) = 0, ρ is ring hom. with ρ L = τ, we have τp α (ρ(α)) = () Inj.... ) As all elements in L(α) are poly. in α with coeff. in L, the map ρ is determined by ρ L = τ and ρ(α) E. (3) Surj.... ) Let β RootτPα (E), and we ll define ρ with ρ(α) = β. Every x L(α) is written as x = P (α) with P L[X], and P is unique up to adding multiples of P α. As τp α (β) = 0, the element τp (β) E is well-defined. So let ρ(x) := τp (β), i.e. ρ : L(α) P (α) τp (β) E. This is clearly a ring hom. with ρ K = τ. (5) Now we can prove our first big results. Lecture 6 (16 Oct, Tu.) Theorem 18. (Separability) Let F/K be a finite ext n inside C. Then: Hom K (F, C) = [F : K]. Proof. Let F = K(α 1,..., α n ) (Prop. 10(ii)). If n = 1 this is Prop. 15(ii). Use induction on n. Let L := K(α 1,..., α n 1 ), F = L(α) with α := α n. Consider the restriction map: Hom K (F, C) ρ ρ L Hom K (L, C). (5) (optional) Proofs of Prop. 14, Prop. 17 in fancier diagrams: τ X K[X]/(P α ) L[X] L [X] L := τ(l) = f α fβ = τ : L = L τ α K(α) E β X L[X]/(P α ) L [X]/(τP τ := f β fα 1 α ) X = f α f β = ρ := f β τ fα 1 α K(α) E β

14 TERUYOSHI YOSHIDA By Prop. 17, the inverse image of each τ Hom K (L, C) has cardinality Root τpα (C). Now τp α is irred. in τ(l)[x], being the image of P α L[X] (irred. by Prop. 7(ii)) under the ring isom. L[X] = τ(l)[x] extending τ : L = τ(l). Hence Root τpα (C) Prop.15(i) = deg τp α = deg P α Prop.7(ii) = [L(α) : L], thus Hom K (F, C) = [L(α) : L] Hom K (L, C) = [F : L] [L : K] (ind. hyp.) = [F : K]. (Tower Law Prop. 8) We have also proved: Lemma 19. Let F/K be a finite ext n inside C and K L F. Then the map Hom K (F, C) ρ ρ L Hom K (L, C). is surjective, i.e. one can extend every K-hom. τ : L C to F. Theorem 0. (Primitive Element Theorem) Let F/K be a finite ext n. Then we have Hom K (F, E) [F : K] for any ext n E/K. Moreover if it is an equality for some E/K, then F/K is simple. In particular, every finite ext n inside C is simple by Th. 18. Proof. When K < (finite fields), the simplicity will be proved directly (Th. 68) and the first claim follows by Prop. 14. So let K = (e.g. any K C, in which case Q K). Let F = K(α 1,..., α n ) (Prop. 10), and τ 1,..., τ d Hom K (F, E) be distinct. If we find α F such that τ 1 (α),..., τ d (α) are all distinct, then τ j K(α) Hom K ( K(α), E ) (1 j d) are all distinct, hence ( ) Prop.14 d Hom K K(α), E [K(α) : K] [F : K], and d = [F : K] forces K(α) = F, so we win. n Let P = α i X n F [X], and we try α F of the form α = P (x) with x K. Since i=1 α 1,..., α n generate F/K, if j j then we cannot have τ j (α i ) = τ j (α i ) for all 1 i n. Thus τ j P E[X] are all distinct poly s, so ( τj P (X) τ j P (X) ) E[X] j j is a non-zero poly., hence x K which is not its root, since K =. Then τ j P (x) τ j P (x) for any j j, in other words for α := P (x) the el ts τ j (α) are all distinct. Example. (i) Q(, 3) = Q( + 3). (ii) Q(, 3 ) = Q( + 3 ). (6) (6) All but finitely many unlucky poly s in the generators would work, so PET is a loose th m.

GALOIS THEORY 15 1.6. Galois extensions. Definition 1. Let L/K, L /K be extensions. If a K-hom. τ : L L is a bijection, then τ 1 : L L is also a K-hom. (ring hom., id on K), & we say τ is a K-isomorphism. An K-isom. L L is called a K-automorphism of L, and the set of all K-aut. of L is denoted by Aut K (L), a subset of Hom K (L, L). It is a group under composition (... id L Aut K (L); σ, τ Aut K (L) = στ, σ 1 Aut K (L)). Lemma. (i) If there is a K-hom. τ : L L, then [L : K] [L : K]. (ii) If [L : K] = [L : K] <, then every τ Hom K (L, L ) is a K-isom. (iii) If L/K is a finite ext n, then Hom K (L, L) = Aut K (L) and Aut K (L) [L : K]. Proof. (i),(ii): Linear Algebra. Recall K-hom s are injective (Lemma 11). Let V, V be K-v.s. If K-lin. inj. V V, then dim K V dim K V. An injective K-lin. map V V is bij. if dim K V = dim K V < by Rank-Nullity. (iii) follows from (ii) and Th. 0. Remark. If L C, then the last part of (iii) follows from Hom K (L, L) Hom K (L, C) and Hom K (L, C) = [L : K] (Th.18). Definition 3. A finite ext n L/K is called a Galois extension if Aut K (L) = [L : K]. In this case Aut K (L) is called the Galois group of L/K, and denoted by Gal(L/K). Remark. Every Galois ext n is simple by Th. 0. If L = K(α), then L/K is Galois iff Root Pα (L) = deg P α, where P α is the min. poly. of α over K, by Prop. 14 and 7(ii). Proposition 4. Let L/K be a finite ext n inside C. TFAE: (i) L/K : Galois. (ii) Every K-hom. τ : L C maps L into itself. (iii) α L, every conjugate of α over K is in L. (iv) L = K(α 1,..., α n ) and every conjugate of α i over K is in L (1 i n). Proof. (i) (ii): By the remark after Lem., L/K is Galois iff Hom K (L, L) = Hom K (L, C). (ii) (iii): Let β be a conjugate of α, i.e. β Root Pα (C). Then by Prop. 14 we have τ ( ) Hom K K(α), C with τ(α) = β, and it extends to ρ HomK (L, C) by Lem. 19. Then (ii) says β = ρ(α) L. (iii) (iv) is clear. (iv) (ii): Let τ Hom K (L, C). As every α L is a poly. in α 1,..., α n with coeff. in K, τ(α) is a poly. in τ(α 1 ),..., τ(α n ) with coeff. in K. But τ(α i ) is a conj. of α i over K (Prop. 14), hence in L by (iv), therefore τ(α) L. Example. (i) C/R : Galois. (ii) Q( )/Q : Galois. (iii) Q( 3 )/Q : NOT Galois. (7) (7) For K L F, even if F/L, L/K are Galois F/K may not be Galois, e.g. Q Q( ) Q( 4 ).

16 TERUYOSHI YOSHIDA Lecture 7 (18 Oct, Th.) Definition 5. Let P K[X] with K C, and Root P (C) = {α 1,..., α n }. Then K(α 1,..., α n ) C is called the splitting field of P over K. Corollary 6. Let P K[X] with K C. Its splitting field over K is Galois over K. Proof. It is a finite ext n of K by Prop. 10(i), and as all conjugates of α i over K belong to Root P (C) (... min. poly. of α i divides P ), it is a Galois ext n of K by Prop. 4(iv). Example. (i) Q( )/Q is the splitting field of X over Q. (ii) The splitting field of X 3 over Q is Q( 3, 3 ζ, 3 ζ ) = Q( 3, ζ), where ζ 3 = 1. (iii) If K(α)/K is Galois, then it is the splitting field of P α. (... If Root Pα (C) = {α 1,..., α n }, they are all in K(α) by Prop. 4(iii), hence K(α) = K(α 1,..., α n ).) Big example 1: Cyclotomic ext ns. ( πi ) Definition 7. Let N 1, and ζ = ζ N := exp C. Then: N µ N := Root X N 1(C) = {1, ζ, ζ,..., ζ N 1 } (the set of N-th roots of unity) is a multiplicative group, cyclic of order N. Its element ζ i is a generator of this group iff (i, N) = 1; they are called the primitive roots. For K C, the splitting field of X N 1 over K is denoted by K(µ N ) and called a cyclotomic extension of K. Note K(µ N ) = K(1, ζ, ζ,..., ζ N 1 ) = K(ζ) = K(ζ i ) ( i N with (i, N) = 1). Proposition 8. Let K C and N 1. We have an injective group hom.: Gal ( K(µ N )/K ) ( Z/(N) ) (τ : ζ ζ i ) i mod N, where ( Z/(N) ) := {i mod N (i, N) = 1} is the multiplicative group of units in the ring Z/(N) = {i mod N i Z}. Proof. Let τ Gal ( K(µ N )/K ). As ζ µ N and ζ has order N in µ N, τ(ζ) µ N and has order N, i.e. τ(ζ) = ζ i with (i, N) = 1, and i is well-def d modn so we have a map. As τ(ζ) determines τ (... K(µ N ) = K(ζ)), this map is injective. If τ(ζ) = ζ i and σ(ζ) = ζ j, then στ(ζ) = σ(ζ i ) = (ζ j ) i = ζ ij, hence it is a group hom. Corollary 9. K(µ N )/K is abelian, i.e. a Galois ext n with an abelian Galois group. Remark. We will see that this injection is actually bijective when K = Q (Irreducibility of Cyclotomic Polynomials, Th. 71).

GALOIS THEORY 17 Big example : Kummer ext ns. Definition 30. Let N 1 and µ N K C. Let a K, and if N a C is a root of X N a, then Root X N a(c) = { N a, N aζ, N aζ,..., N aζ N 1 }, where ζ = ζ N. The splitting field of X N a over K is called a Kummer extension of K, and is equal to K( N a) for any choice of N a. Proposition 31. If K( N a)/k is as above, we have an injective group hom.: Gal ( K( N a)/k ) Z/NZ (τ : N a N aζ i ) i mod N, where Z/NZ is the additive group of Z/(N). In particular K( N a)/k is abelian. Proof. Let τ Gal ( K( N a)/k ). Then τ( N a) = N aζ i for some i, well-def d mod N and independent of the choice of N a (... τ( N aζ j ) = ( N aζ i )ζ j = ( N aζ j )ζ i ). As τ( N a) determines τ, this map is injective. If τ( N a) = N aζ i and σ( N a) = N aζ j, then στ( N a) = σ( N aζ i ) = ( N aζ j )ζ i = N aζ i+j, i.e. it is a group hom. Example. Let ζ 3 = 1 with ζ 1. Q( 3, ζ)... spl. field of X 3 /Q ( Gal./Q) Z/3Z : Kummer 6 Q(ζ) C = ( Z/(3) ) : cyclotomic Q What group is Gal ( Q( 3, ζ)/q )? 1.7. Galois correspondence. Example. ζ = ζ 8 = (1 + i)/ : root of X 4 ρ + 1. 1 = id, Gal ( Q(µ 8 )/Q ) ρ : (, i) (, i), = ρ 3 : (, i) (, i), Q(µ 8 ) = Q(, i) ρ 4 : (, i) (, i). Prop.8 = ( Z/(8) ) = {1, 3, 5, 7 mod 8} = C C. Q( ) Q(i) Q( ) Q ρ 1 fixes all elements in Q(µ 8 ). ρ fixes Q( ), ρ : ζ ζ 7 ρ 3 fixes Q(i), ρ 4 fixes Q( order ρ 3 : ζ ζ 5 = ζ ). ρ 4 : ζ ζ 3 Note: = ζ + ζ 7, i = ζ = ζ ζ 5, = ζ + ζ 3.

18 TERUYOSHI YOSHIDA Lecture 8 (0 Oct, Sa.) Lemma 3. Let F/K be an ext n and G := Aut K (F ). (i) Let K L F. Then Aut L (F ) is a subgroup of G, i.e. Aut L (F ) = {σ G σ L = id} = {σ G σ(α) = α α L} G (ii) Let H be a subgroup of G. The subset F H := {α F σ(α) = α σ H} F of all elements in F fixed by all autom s in H is a subfield of F, called the fixed field of H. We have K F G F H F and H Aut F H (F ). Proof. (i): An L-autom. F F is nothing other than a K-autom. satisfying σ L = id. (ii): As every σ G is a ring hom., if σ(α) = α and σ(β) = β, then σ(α + β) = α + β, σ(αβ) = αβ and σ(α 1 ) = α 1 when α 0. Hence F H is a field. As σ K = id σ G we have K F G. As σ F H = id σ H we have H Aut F H (F ). Proposition 33. Let F/K be a simple finite ext n (e.g. any finite ext n inside C, cf. Th. 0). (i) If F/K is Galois and G := Gal(F/K), then F G = K. (ii) If F G = K for a subgroup G of Aut K (F ), then F/K is Galois and G = Aut K (F ). Proof. (i): Note K F G F. As G Aut F G(F ), we have: G Aut F G(F ) Lem.(iii) [F : F G ] [F : K]. But since F/K is Galois G = [F : K], hence all are equalities & F G = K. (ii): Let F = K(α), and P α be the min. poly. of α. Set: Q α := σ G( X σ(α) ) F [X]. It has α as a root, and its coeff. are the elementary symmetric poly. (Def. 81) of {σ(α) σ G}, hence in F G = K (since el ts of G just permute {σ(α) σ G}). Hence P α Q α. Thus [F : K] Prop.7(ii) = deg P α deg Q α = G Aut K (F ) Lem.(iii) [F : K], hence all are equalities, F/K is Galois & G = Aut K (F ). (8) Remark. By Prop. 33, a finite F/K inside C is Galois iff F Aut K(F ) = K. Theorem 34. (Fundamental Theorem of Galois Theory) Let F/K be a Galois ext n. Then the following maps (Galois correspondence), defined by Lem. 3: { } { } subfields L w/ L AutL (F ) subgroups H of K L F F H H G := Gal(F/K) are bijections, inverse to each other. If L H, then [F : L] = H and [L : K] = G / H. (8) Prop. 33(ii) holds without assuming F/K simple/finite, as long as G is finite (Artin s Lemma, Prop. 88).

GALOIS THEORY 19 Proof. Let F = K(α) by PET (Th. 0). The remark after Def. 3 says Root Pα (F ) = deg P α, where P α is the min. poly. of α /K (it splits into distinct linear factors in F [X]). Take L. The min. poly. Q α of α over L divides P α, hence Root Qα (F ) = deg Q α. As F = L(α), the same remark says F/L is Galois. Now Prop. 33(i) shows F Aut L(F ) = L. Take H. By above F/F H is Galois, hence simple (Th. 0). Since H Aut F H (F ) by Lem. 3(ii), the Prop. 33(ii) shows H = Aut F H (F ). If H = Aut L (F ), then [F : L] = H as F/L is Galois, and hence [L : K] = G / H by Tower Law (Prop. 8). Corollary 35. Let F/K be a Galois ext n inside C with G := Gal(F/K), and K L F. (i) F/L is Galois and H := Gal(F/L) is a subgroup of G. (ii) L/K : Galois H G (normal). If this holds, we have an isom. of groups: Proof. (i): Shown in the proof of Th. 34. G/H σh = σ L Gal(L/K). (ii): If σ G, then K σ(l) F, and the subgp. corresponding to σ(l) is σhσ 1, because for ρ G we have ρ L = id σρσ 1 σ(l) = id. Hence: H G σhσ 1 = H σ G By the surjection (Lem. 19): σ(l) = L σ G ( ) (by Galois corresp. Th.34). G = Hom K (F, C) σ σ L Hom K (L, C), ( ) is equivalent to say that every τ Hom K (L, C) maps L into L, i.e. L/K is Galois (Prop. 4(ii)). In this case, the surjection above is a gp. hom. G Gal(L/K) with kernel = {σ G σ L = id} = H. ( πi ) Example. (5th roots of unity) ζ = ζ 5 := exp : a root of 5 X 5 1 = (X 1)(X 4 + X 3 + X + X + 1). Lecture 9 (3 Oct, Tu.) We know (will prove later) X 4 + X 3 + X + X + 1 is irred. in Q[X], hence is the min. poly. of ζ over Q. Recall Q(µ 5 ) = Q(ζ) and Gal ( Q(ζ)/Q ) = {id, σ, σ, σ 3 }, where: and G = ( Z/(5) ) = C4 (Prop. 8): id : ζ ζ, σ : ζ ζ, σ : ζ ζ 4, σ 3 : ζ ζ 3, ζ σ ζ 3 σ σ ζ σ ζ 4 (ζ 5 = 1). It has a subgp H := {id, σ } = C, and σ interchanges ζ ζ 4.

0 TERUYOSHI YOSHIDA {id} H G F = Q(ζ) L =? K = Q = H fixes ζ + ζ 4, and σ(ζ + ζ 4 ) = ζ + ζ 3. As L := F H must be quadratic, ζ + ζ4, ζ + ζ 3 must be roots of a quad. eq n /Q. (ζ + ζ 4 ) + (ζ + ζ 3 ) = ζ 4 + ζ 3 + ζ + ζ = 1, (ζ + ζ 4 )(ζ + ζ 3 ) = ζ 3 + ζ 4 + ζ + ζ = 1. = they are roots of X + X 1, i.e. 1 ± 5... ( 1 ± 5 ). L = Q = Q( 5). ( 1 + 5 ) Now ζ, ζ 4 are roots of X X + 1 = 0 [... Re (ζ + ζ 4 ) = ζ + ζ 4 > 0.] = ζ, ζ 4 = ( 1 + 5)/ ± ( 1 + 5) /4 4 = ( 1 + 5) ± 10 5 4... ζ = ( 1 + 5) + 10 5 4 (... Im ζ > 0). 1.8. Insolvability of quintics I: radical extensions. Example. ζ 3 = 1, ζ 1. F/Q : spl. field of P := X 3. By Prop. 8 and 31: F = Q( 3, ζ) Kummer, Z/3Z cyclo., C L = Q(ζ) K = Q = ρ = id, ρσρ 1 : { Gal(F/L) = {id, σ, σ }, σ( 3 ) = 3 ζ, σ L = id, Gal(L/K) = {id, τ}, τ(ζ) = ζ. extend τ to ρ Gal(F/K), say: ρ(ζ) = ζ, ρ( 3 ) = 3. ( 3 choices for roots of τp = X 3.) 3 ρ 1 3 σ ζ ρ 1 ζ σ 3 ζ ζ ρ ρ 3 ζ = ρσρ 1 = σ.... Gal(F/K) = D 6 = S3 (non-abelian). (9) Idea: Solving radicals (cyclo/kummer ext ns) can only produce a tower of abelian ext ns a limited class of Galois ext ns. Definition 36. Let F/K be a finite ext n inside C. It has a finite set of generators (Prop. 10(ii)), and the ext n E/K generated by all conjugates (over K) of all of these generators is Galois (Prop. 10(i), 4(iv)). As every Galois ext n of K inside C containing F must contain E by Prop. 4(iii), (9) Exercise: write out the 3 order subgps & the corresponding cubic subfields (see Example in.). ζ

GALOIS THEORY 1 it is the minimal Galois ext n of K containing F. In particular it is independent of the choice of generators. We call E/K the Galois closure of F/K. Example. (i) Q( 3, ζ) = Q( 3, 3 ζ, 3 ζ ) is the Galois closure of Q( 3 )/Q. (ii) If F = K(α) (PET, Th. 0), then its Galois closure is E = K(α 1,..., α n ), where α = α 1,..., α n are conj. of α over K (i.e. the spl. field of P α ). Remark. By Galois theory (Th. 34), if G := Gal(E/K) then G := Gal(E/F ) is its subgroup; F/K corresponds to the pair (G, G ). Definition 37. We say a pair (G, G ) of a finite group G and its subgroup G is soluble if there is a sequence (G i ) of subgroups G = G 0 G 1 G n 1 G n = G with G i 1 G i (normal) and G i 1 /G i cyclic for 1 i n. We say G is soluble if (G, {id}) is. A Galois ext n is called soluble if its Galois group is soluble. Example. The symmetric groups S 3, S 4 are soluble: E F K {id} G G S 3 A3 3 {id}, S4 A4 3 V4 C {id}. Lemma 38. (i) Let G be a finite group. If G H G, then: (G, G ) : soluble G/H, (H, G ) : both soluble. (ii) Finite abelian groups are soluble. Proof. (i): Let p : G G/H be the canonical surjection σ σh. ( ): If (G i ) is a sequence for (G, G ), then ( p(g i ) ), (H G i ) give sequences for G/H, (H, G ): (H G i 1 )/(H G i ) cyclic G i 1 /G i p(g i 1 )/p(g i ) = cyclic = cyclic ( ): If (G i ), (H i ) are sequences for G/H, (H, G ), then combine ( p 1 (G i ) ) and (H i ): G/H = G 0 G 1 G m = {id} p G = p 1 (G 0 ) p p 1 (G 1 ) p 1 (G m ) = H = H 0 p cyclic p 1 (G i 1 )/p 1 (G i ) = H 1 H n = G p cyclic G i 1 /G i (ii): Induction on G. If G σ id and H = σ, then H is cyclic and G/H < G, so use (i)( ). [or: Str. Th m.] Example. S n is not soluble for n 5. (... S n A n and A n (n 5) is simple non-abelian hence not soluble; use Lem. 38(i)( ).)

TERUYOSHI YOSHIDA Lecture 10 (5 Oct, Th., 01st birthday of Évariste Galois) Definition 39. We say a Galois ext n L/K inside C is radical if there is a finite ext n F of L such that F/K is a succession of cyclotomic and Kummer ext ns, i.e. K = K 0 K 1 K n = F, with K i /K i 1 : cyclo. or Kummer. Theorem 40. Radical extensions are soluble. (10) Proof. Let L/K be radical and take F/L as in Def. Let E/K be the Galois closure of F/K, and G := Gal(E/K), H := Gal(E/L) and G := Gal(E/F ). As cyclo./kummer ext ns are abelian, (G, G ) is soluble by the Galois correp. (Th. 34, Cor. 35) and Lem. 38(ii),(i)( ). Hence G/H = Gal(L/K) (Cor. 35) is soluble by Lem. 38(i)( ). E cyclo. (abelian) or Kummer (cyclic) K n {id} = F G = G n. L K 0 = Remark. (oral) (i) Recall the Example in the beginning (S 3 ). H. K G = G 0 soluble (ii) There are many equations whose spl. fields have Galois group S n (examples later). (iii) Next we prove that general eqn s (the roots α, β, γ,... are transcendental numbers) of deg. n have Galois group S n. Hence there cannot be a formula to solve quintics (or higher) by radicals (... S n not soluble). 1.9. Insolvability of quintics II: general equations. Definition 41. Let K C and P K[X]. The Galois group Gal(P ) of P is defined as the Galois group Gal(F/K) for the splitting field F of P over K. Next: For general equation P of degree n, we have Gal(P ) = S n. Proposition 4. Let K C and P K[X]. (i) Then Gal(P ) is a subgroup of the autom. group Aut ( Root P (C) ) of the finite set Root P (C). In particular, a choice an ordering of the roots Root P (C) = {α 1,..., α n } gives an injection Gal(P ) S n = Aut ( {1,..., n} ). (ii) If P is irred. in K[X] with deg P = n, then Gal(P ) is isomorphic to a transitive subgroup G S n i.e. for every i, j {1,..., n}, there exists σ G with σ(i) = j. (10) We can extend the definition of radical ext ns to any (non-galois) finite ext n, and prove that its Galois closure is soluble (see Appendix 1).

GALOIS THEORY 3 Remark. As a reordering in (i) amounts to a conjugation in S n, we can consider Gal(P ) as a subgroup of S n, well-def d up to conjugation. Proof. (i): Let F be the spl. field of P over K. An el t σ Gal(P ) = Gal(F/K), being a K-hom., maps Root P (C) into itself. As σ is injective (Lem. 11) and Root P (C) is a finite set, σ : Root P (C) Root P (C) is a bijection (autom.). As F/K is generated by Root P (C), the action of σ on Root P (C) determines σ. (ii): As P is irred. of deg. n, Root P (C) = n by Prop. 15(i). If Root P (C) = {α 1,..., α n }, then P is the min. poly. of α i for all i, hence for every i, j there exists τ Hom K ( K(αi ), C ) with τ(α i ) = α j by Prop. 14. Extending this τ to σ Hom K (F, C) = Gal(F/K) by Lem. 19, we get σ(α i ) = α j. Example. (i) If a cyclic subgroup of S n is transitive, then it has order n. (ii) Transitive subgroups of S 3 are: A 3 = C3, S 3. (iii) Transitive subgroups of S 4 are, up to conj.: C 4, V 4, D 8, A 4, and S 4. (Here V 4 := {id, (1)(34), (13)(4), (14)(3)} = C C, the Klein 4-group.) (iv) Transitive subgroups of S 5 are, up to conj.: C 5, D 10, F 0, A 5, and S 5. (Here F 0 := (1345), (1)(453), the Frobenius group of order 0.) Remark. (oral) All transitive subgroups of S n appear as Gal(P ) for some P K[X] for some K, but Gal(P ) is the whole S n for most irred. eq ns P of deg. n. Definition 43. Let K be a field, X 1,..., X n be indeterminates, and K[X 1,..., X n ] be the ring of poly. in X 1,..., X n with coefficients in K (an integral domain). Its field of fractions is denoted by K(X 1,..., X n ), the field of rational functions in n variables /K. Proposition 44. Let n 1. (i) There exist α 1,..., α n C, transcendental over Q, such that the field F := Q(α 1,..., α n ) C is isom. to Q(X 1,..., X n ) by X i α i for 1 i n. (ii) The symmetric group G := S n acts on F by permuting α 1,..., α n, and F/F G is Galois with the Galois group G. Proof. (i): Use induction on n. As L := Q(α 1,..., α n 1 ) is isom. to Q(X 1,..., X n 1 ) (ind. hyp.), it is countable, hence only countably many el ts in C are alg. /L. So choose α n which is transcendental over L. Then there is no non-zero P (X 1,..., X n ) Q[X 1,..., X n ] with P (α 1,..., α n ) = 0, hence the ring hom. f : Q[X 1,..., X n ] P P (α 1,..., α n ) C is injective, and extends to a ring hom. f : Q(X 1,..., X n ) P Q P (α 1,..., α n ) Q(α 1,..., α n ) C because C is a field. Then F := Im f is isom. to Q(X 1,..., X n ), and it is the minimal field containing α 1,..., α n, i.e. Q(α 1,..., α n ). C

4 TERUYOSHI YOSHIDA (ii): G acts on Q(X 1,..., X n ) by permuting X i, hence also on F, i.e. σ G acts as fσf 1, by permuting α i. Let K := F G. Then F = Q(α 1,..., α n ) implies F = K(α 1,..., α n ), and n G Aut K (F ) by Lem. 3(ii). The coefficients of P = (X α i ) F [X] are the elementary symmetric poly. s in α i, hence in K. Hence α i are the roots of P K[X], i.e. alg. /K. Thus F/K is finite (Prop. 10(i)), and Prop. 33(ii) shows that F/K is Galois and G = Gal(F/K). Theorem 45. (Insolvability of Quintics) For n 5, there is no formula, involving only radicals and rational functions, which expresses α 1,..., α n in terms of their elementary symmetric polynomials. i=1 Proof. The Galois ext n F/F G in Prop. 44 is not soluble, hence not radical by Th. 40. Lecture 11 (7 Oct, Sa.) 1.10. Solving by radicals. Next goal: Converse of Th. 40, i.e. all soluble ext ns are radical. Recall: soluble groups are built out of cyclic groups. Definition 46. A Galois ext n is called cyclic if its Galois group is cyclic. Recall: for N 1 and µ N K C, Kummer ext ns of K are K( N a)/k for a K. Their Galois groups inject to Z/NZ (Prop. 31), hence they are cyclic. Example. For N =, as µ = {±1} K for any K C, every quadratic ext n is Kummer (i.e. every quad. eq n is solved by ). Theorem 47. (Kummer Theory) Let N 1 and µ N K C. Then every cyclic ext n of K with degree N is a Kummer ext n. Proof. Let L/K be cyclic of deg. N. Choose a generator σ of the Galois group: Gal(L/K) = {id, σ, σ,..., σ N 1 }. Let ζ = ζ N µ N. Suppose we found α L (= L \ {0}) with σ(α) = αζ. Then the conjugates of α over K are σ i (α) = αζ i for 1 i N (Prop. 14), and these are all distinct. Hence [K(α) : K] = N (Prop. 7(ii)), therefore K(α) = L. Let a := α N. Then σ(a) = σ(α N ) = σ(α) N = (αζ) N = a, so a is fixed by all σ i, hence a K (Prop. 33(i)).... L = K( N a). So STP: considering σ as a K-linear transformation of L as a K-v.s., ζ is an eigenvalue of σ. Let Q K[X] be the min. poly. of σ (as in Linear Algebra). Then Λ := Root Q (K) is the set of all eigenvalues of σ. We want ζ Λ. As σ N = id, we have Q X N 1, hence Λ µ N. Now Λ is a multiplicative subgroup of µ N, because if λ, µ Λ and σ(α) = λα, σ(β) = µβ for α, β L, then as σ is a ring hom. σ(αβ) = (λα)(µβ) = (λµ) αβ and σ(α 1 ) = (λα) 1 = λ 1 α 1, i.e. λµ, λ 1 Λ.

GALOIS THEORY 5 Hence Λ = µ d for some d N, i.e. Q = X d 1. But σ has order N, so σ d id unless d = N.... Q = X N 1, Λ = µ N and ζ Λ. Corollary: soluble ext ns are radical as long as K contains µ N for large enough N. Lemma 48. Let K L F and α F. If K(α)/K is Galois, then L(α)/L is Galois, and Gal ( L(α)/L ) σ σ K(α) Gal ( K(α)/K ) is an injective group hom. Proof. As the min. poly. of α over L divides that of α over K, by Prop. 4(iii) all conj. of α over L are in K(α) L(α), hence L(α)/L is Galois (Prop. 4(iv)). The map is clearly a group hom., and injective as σ is determined by σ(α), hence by σ K(α). Remark. The image corresponds to K(α) L by Galois theory, i.e. Gal ( L(α)/L ) = ( ) Gal K(α)/K(α) L. Theorem 49. Soluble ext ns inside C are radical. L(α) K(α) L K(α) L K Proof. Let L/K be Galois with Gal(L/K) = G 0 G 1 G n = {id} and G i 1 /G i cyclic. Let K i := L G i be the corresponding subfields under Galois theory (Th. 34), so K = K 0 K 1 K n = L with K i /K i 1 cyclic, Gal(K i /K i 1 ) = G i /G i 1 (Cor. 35). Let N i := G i 1 /G i for 1 i n, and N := N 1 N n. Then we have a tower of fields K K(µ N ) K 1 (µ N ) K n (µ N ) = L(µ N ), with L L(µ N ). Applying Lem. 48 to K i := K i 1 (α) (PET, Th. 0), we see that K i (µ N ) = K i 1 (µ N )(α) is cyclic over K i 1 (µ N ) of degree dividing N i (... its Galois group is isom. to a subgroup of G i 1 /G i ), hence N. Thus K i (µ N )/K i 1 (µ N ) is Kummer by Kummer Theory (Th. 47). 1.11. Discriminants, and revisiting Intro 1. Any formula for solving eq ns by radicals? We need normal subgps of Gal(P ) to climb up the splitting field of P, but the only non-trivial normal subgps of S n are A n and V 4. Definition 50. Let n 1, K C and P K[X] with deg P = Root P (C) = n. Let i : Gal(P ) S n be the injection in Prop. 4, defined up to conjugation in S n. If H S n (normal), then we have a well-defined normal subgroup Gal(P ) H := i 1 (H) of Gal(P ). First consider H = A n. Let P as in Def. 50, F/K its spl. field and Root P (C) = {α 1,..., α n } Gal(P ). Let: P := i<j (α i α j ) = ( 1) n(n 1) (α i α j ) F. (called the discriminant of P ) i j

6 TERUYOSHI YOSHIDA As the RHS is clearly fixed by Gal(P ), it is in K by Prop. 33(i). Note that P 0 because we assumed Root P (C) = n. For example: P = X ax + b = P = (α β) = a 4b, P = X 3 + bx c = P = (α β) (β γ) (γ α) = 4b 3 7c, where we used α, β, γ for α 1, α, α 3. Lecture 1 (30 Oct, Tu.) Proposition 51. For P as in Def. 50, we have Gal(P ) A n P : square in K ( ), and the subgroup Gal(P ) A n of Gal(P ) corresponds to K ( ) P by Galois theory. Proof. Take one of the square roots of P (depending on the ordering of roots!) P := (α i α j ), i<j and consider the action of σ Gal(P ) S n on it. As a transposition (ij) changes signs of α i α j, α i α m, α m α j for i < m < j, it sends P to P. Hence we have σ Gal(P ) A n σ ( ) P = P. ( ) Now we prove ( ). ( ): If P K, then every σ Gal(P ) is in Gal(P ) A n by ( ). ( ): As ( ) says P is fixed by all σ Gal(P ), it is in K by Prop. 33(i). The latter claim is clear if either side of ( ) is true. If not, then Gal(P ) A n has index in Gal(P ), and its fixed field L satisfies [L : K] = by Th. 34. Now ( ) says [K ( ) P : K] = and ( ) says K ( ) ( ) P L, hence K P = L. Example. Let P = X 3 + bx c K[X] be irred. with Root P (C) = {α, β, γ} (distinct). Then Gal(P ) is A 3 or S 3 (Prop. 4(ii)), and Prop. 51 tells you which, e.g. when K = Q: P = X 3 3X + 1 : P = 81 = Gal(P ) = A 3, P = X 3 + X + : P = 140 = Gal(P ) = S 3. Now we revisit Lecture 1. To make the Kummer theory work for cyclic cubic extensions, we assume ζ = ζ 3 K. For a cubic with distinct roots P = X 3 + bx c = (X α)(x β)(x γ) K[X] and its Langrange resolvents x = α + βζ + γζ, y = α + βζ + γζ, the el ts x 3, y 3 were the roots of X 7cX 7b 3. Let L := K(x 3 ) = K(y 3 ).

GALOIS THEORY 7 K(α, β, γ) = F = K(x) {id} Solving the quadratic shows L = K ( ) 7 P, but 7 = (3 3) = (3(ζ + 1)) is a square in K, hence L = K ( ) P. For a quartic with distinct roots 3 L = K(x 3 ) Gal(P ) A 3 K Gal(P ) P = X 3 + bx cx + d = (X α)(x β)(x γ)(x δ) K[X], recall we had x := α + β = (γ + δ), y := α + γ = (β + δ), z := α + δ = (β + γ), xyz = c, x, y, z : roots of X 3 + bx + (b 4d)X c. For G := Gal(P ), K(α, β, γ, δ) = F = K(x, y, z) {id} G G A 4 G V 4 {id}, coming from the solubility of S 4 : S 4 A 4 V 4 C {id}, with S 4 /V 4 = S3, V 4 = C C, and S 3 permutes x, y, z. biquad. 4 L = K(x, y, z ) 3 K ( P ) Gal(P ) V 4 Gal(P ) A 4 K Gal(P ) Exercise. (Appendix 1) Every cyclo. ext n is contained in a tower of Kummer ext ns (do induction on N for K(µ 1, µ,..., µ N )). So cyclo. ext ns are not needed in the def. of radical ext ns. Solving eq ns is related to ruler-and-compass constructions: in terms of Cartesian coord. (x, y) R or x + yi C, it can only do +,,, and solve quadratic eq ns (intersection with circles), so all coord. of the obtained points lie in a successive quadratic ext ns of the field generated over Q by the coord. of given points. Example. (i) Cannot trisect a general given angle cos α generates a cubic ext n of Q(cos 3α); a successive quadratic ext ns have degree n, hence cannot contain a cubic ext n by Tower Law (Prop. 8). Similarly, cannot double a given cube solving X 3 is also cubic (the altar of Apollo at Delphi; oracle to the Delians). (ii) Constructing regular N-gons essential cases are when N = p prime. Need to solve (X p 1)/(X 1) = X p 1 + + X + X + 1 Q[X], which is irreducible (set X = Y + 1 and use Eisenstein s criterion; we prove a more general theorem later). This is impossible unless p 1 is a power of ; only known such primes are, 3, 5, 17, 57 and 65537 (it has to be of the form m +1, called Fermat primes, as b +1 ab +1 if a is odd). We ll see that these cases are constructible (Gauss).

8 TERUYOSHI YOSHIDA Lecture 13 (1 Nov, Th.). GENERAL FIELDS AND APPLICATIONS (oral) In this section, we generalise 1 to arbitrary fields (pre-war 0c; see Appendix ), and discuss applications to Galois groups over Q..1. General remarks. Definition 5. Let K be a field. The kernel of the unique ring hom. f : Z n 1 + (n times) + 1 K from Z is a prime ideal in Z. Hence Ker f = (p) for p = 0 or a prime number p, the characteristic of K, denoted by char K. If char K = 0, then f is injective and extends to Q K, hence its image is a subfield isom. to Q (unique isom.). If char K = p > 0, then Im f = Z/(p) =: F p is a subfield isom. to F p (again unique isom.). In both cases Im f is the smallest subfield (prime field) of K. Remark. When we have a fixed ring hom. τ : K L of fields (hence τ : inj. by Lem. 11), we may want to identify the isomorphic fields K and τ(k), and consider L as an extension of K. We do this only in the following two cases: (11) (i) Every field can be considered as an ext n of Q or F p in a unique way, as its prime field is uniquely isom. to Q or F p. (ii) If P K[X] is an irred. poly., then (P ) is a maximal ideal of K[X] and K P := K[X]/(P ) is a field. The canonical injection K K[X] gives a ring hom. K K[X] K[X]/(P ), hence K K P (Lem. 11, or: non-zero constants are not in (P )). We consider K K P, and call K P /K the ext n obtained by adjoining a root of P (the root X := X mod P K P is an abstract root of P, which generates K P /K). Example. X + 1 F 3 [X] : irred. (... 0 = 0, 1 = = 1 1 in F 3 ). = F 3 [X]/(X + 1) = {0, 1,, X, X + 1, X +, X, X + 1, X + mod X + 1} quadratic ext n of F 3 (a field with 9 el ts, isom. to Z[i]/(3)). (iii) (non-example) K = Q( 3 ) has three Q-hom s τ 1, τ, τ 3 : K C. Apart from τ 1 = id, we d rather not identify K with τ (K) or τ 3 (K) because K is given as a subfield of C, and τ (K), τ 3 (K) are different subsets of C, though isomorphic as fields. Remark. (oral) Once we start considering ext ns of K which are not subsets of a fixed set (C in 1) but sticking out into nowhere, the notion of K-isom s gets more important. K-isomorphic ext ns share many properties (they look the same from the K-point of view): algebraicity; finiteness; degree; structure of subfields; Hom sets with other ext ns; Root sets of poly. s. But we don t identify K-isomorphic ext ns (e.g. (iii) above), and we also care about how many K-isom s there are (e.g. Galois groups). (11) In (i) τ is unique, and in (ii) τ is canonical, i.e. uniquely fixed by the context.