M345P11 Galois Theory

Transcription

1 M345P11 Galois Theory Lectured by Prof Alessio Corti, notes taken by Wanlong Zheng Comments or corrections should be sent to Last updated: April 20, 2018

2 Contents 1 Introduction What is Galois theory Galois theory when dealing with equations A first example The Eisenstein s criterion Field theory Algebraic extensions Fundamental theorem of Galois theory first half Normal and separable extensions Splitting fields Normal extensions Separable degree Separable extensions Fundamental theorem of Galois theory Galois theory with biquadratic extensions Examples of Galois correspondence Galois groups of cubics Galois groups of biquadratics Quartics equations Solubilities in radicals Soluble polynomials Soluble groups S n not soluble for n The theorem Generic equations of degree n

3 1 Introduction week 2 lecture 1 Galois theory is about fields which we denote by K. A field is a ring where 1 = 0, and where for all x = 0, there exists y with xy = 1. Example. 1. Q = {rational numbers}. 2. the fraction field in n variables: k(t 1,..., t n ) = Frac(k[t 1,..., t n ]). 3. R, C. 4. finite field F = Z/(p) for a prime number p. 5. F = k[x]/(f) for a prime (irreducible) polynomial f k[x]. The last two examples are quotient rings of the form R/m for some maximal ideal m. 6. quadratic surds like Q( 2) = {a + b 2 : a, b Q}. 7. cyclotomic field: Q(ζ n ), where ζ n = e 2πi/n. It is defined to be the smallest subfield of C that contains ζ n. For example, F = Q(ω) = Q( 3). Why? If ω F, then ω ω 2 = 3 F. If 3 F, then ω = ( 1 + 3)/2 F. 1.1 What is Galois theory week 2 lecture 2 Definition. Let K, L be fields. A field homomorphism f : K L is a set-theoretic function such that f(a + b) = f(a) + f(b), f(ab) = f(a)f(b) for all a, b K, and f(0) = 0, f(1) = 1. Lemma. Every field homomorphism is injective. Proof. If a = 0, then a has an inverse b. Now f(ab) = f(a)f(b) = 1, and therefore f(a) = 0. Then if f(x 1 ) = f(x 2 ), then f(x 1 x 2 ) = 0, and from above x 1 x 2 = 0. Therefore, given a field homomorphism f : K L, we can always identify K as the image of f (subset) in L. We call the field homomorphism a field extension, and write as K L. Notation. If k K, k L, then Emb k (K, L) = {field homomorphisms f : K L : f k = id}. Example. Q Q( 2), R C. Galois theory studies field extensions, and, in particular, those extensions K L where a polynomial f K[X] has no roots in K but has roots in L. For example, X 2 2 Q[X] has roots in Q( 2). We will learn the following: Given a field K, f K[x], there exists an optimal extension K L such that f(x) splits completely in L, called the splitting field. Remark. (Very important) If K L is a field extension, then L is naturally a K-vector space. Definition. An extension K L is finite if L is finite dimensional as a K-vector space. The degree of L over K is the dimension of L, denoted as [L : K] = dim K L. We will prove the tower law: suppose K L M are extensions, then K M is finite if and only if K L and L M are finite, and [M : K] = [M : L][L : K]. Lemma. Suppose K L is finite. Then every f Emb k (L, L) is an isomorphism. Proof. L is a finite dimensional vector space. An injective linear map from L to itself must also be surjective. Thus we have an inverse. Verify the inverse is a homomorphism. It follows that for a finite extension k L, Emb k (L, L) = Aut k (L) is the group {field isomorphism f : L L : f k = id}. This group is called the Galois group and it is denoted by G k (L)

4 Fundamental theorem of Galois theory If k L is finite, normal and separable, then there is a 1-to-1 correspondence between {intermediate fields K M L} {subgroups H G = Aut k L} M = {a L : h H, h(a) = a} Normal and separable: we will see that M H = {σ G : σ M = id} 1. K L normal there exists f K[X] such that the splitting field of f is L. 2. If ch(k) = 0 (i.e. K contains Q), then all extensions are separable. Example. We start with Q, and consider the splitting field of x 3 2 = (x 3 2)(x ω 3 2)(x ω 2 3 2). So the splitting field must contain ω (or 3 equivalently) and 3 2, i.e. consider Q K = Q( 3 2, 3). Clearly, G = Aut Q (K) permutes the three roots of x 3 2. Suppose α is a root, so α 3 2 = 0, and σ G = Emb Q (K, K). Then, 0 = σ(0) = σ(α) 3 σ(2). But σ G so it fixes all rationals, and therefore σ(α) 3 2 = 0. So we have a group homomorphism ρ : G S 3. This map, in this case, is actually an isomorphism 1. Here we represent S 3 by S 3 = σ = (123), τ = (23) σ 3 = τ 2 = e, τστ = σ 2. We draw a diagram with partial order (subgroup) along the vertical direction, and with the correspondence at the top of this page in mind: H K = Q( 3 2, 3) e F 1 F 2 F 3 F 4 (23) (13) (12) (123) Q S 3 Here, F 1 to F 4 are the intermediate fields, and we will use the correspondence above to figure out what they are. Recall that for each subgroup H, the intermediate field is defined to be F = {a Q( 3 2, 3) : h H, h(a) = a}. Recall that if (23) is in G, it corresponds to switching the roots a 2, a 3. a 2 = ω 3 2 a 1 = 3 2 a 3 = ω The reason for this will become clear later in the course - 3 -

5 Let s start with H = (23). This corresponds to the complex conjugation, i.e. it fixes 3 2. So F 1 = Q( 3 2). For σ = (123), notice that σ(ω) = σ( a 2 ) = a 3 = ω. So σ fixes ω, and F 4 = Q(ω) = Q( 3). So the a 1 a 2 complete picture is the following: K = Q( 3 2, 3) e Q( 3 2) Q(ω 3 2) Q(ω 2 3 2) Q( 3) (23) (13) (12) (123) Q S Galois theory when dealing with equations 1 Suppose we want to solve y 3 + 3py + 2q = 0 over some field K ω, and that it splits as (y a 1 )(y a 2 )(y a 3 ) over L (note we have a 1 + a 2 + a 3 = 0). Suppose the Galois group G of the extension K L contains σ = (123). Consider the two quantities: u = ω 2 a 1 + ωa 2 + a 3, v = ωa 1 + ω 2 a 2 + a 3. Then we have σ(u) = ω 2 a 2 + ωa 3 + a 1 = ωu, and σ(v) = ω 2 v. We can recover the a i s from u, v: a 1 = ωu + ω2 v 3, a 2 = ω2 u + ωv 3, a 3 = u + v. 3 So technically, if we can solve u, v, we can then solve the cubic equation. Notice even further that σ(u 3 ) = u 3, σ(v 3 ) = v 3, i.e. σ fixes u 3, v 3. Now suppose that G is actually the symmetric group on the three roots, and G contains τ = (12). Note that τ(u) = v, τ(u 3 ) = v 3. It follows that u 3 + v 3 and u 3 v 3 are G-invariant. Exercise. Find expressions for u 3 + v 3, u 3 v 3 in terms of p, q and hence write a quadratic equation with coefficients in K, of which u 3, v 3 are the two roots. 1 cf. Q1 on Ex1-4 -

6 2 A first example week 3 lecture 1 Example. Let K = Q( 3 2) C be the smallest subfield of C that contains 3 2. (Note that we already know Q( 2) = {a + b 2 : a, b Q}) Clearly, 3 2 K, = 3 4 K, and = 2 K. So we ask the question: is it true that K is the same as X = Q[ 3 2] := {a + b c 3 4 : a, b, c Q}? This is the same as asking: is X a field? Remark. Let L be a field, and X L be a subset. Then X is a field iff 0, 1 X, and for all a, b X, a + b, a b, a b, a/b (if b = 0) X. Addition, subtraction and multiplication are easy to verify for X. So we need to find, for a given a + b c 3 4 = 0, some x, y, z Q with (a + b c 3 4)(x + y z 3 4) = 1. Question/Exercise. Are 1, 3 2, 3 4 linear independent over Q? Let us first assume they are, and we will go back to this at the end. We expand out the terms in the product, and we have (ax + 2cy + 2bz) + (bx + ay + 2cz) (cx + by + az) 3 4 = 1. Assuming linear independence, this is equivalent to solving the linear equation: a 2c 2b b a 2c x y = 1 0. c b a z 0 And let s denote the coefficient matrix by A. Then det(a) = a 3 6abc + 2b 3 + 4c 3. Question: if a, b, c Q, does det(a) = 0 implies a = b = c = 0? Let us denote a possible zero to det(a) by (a, b, c). First, since every term in det(a) is of degree 3 (homogeneity), we can assume a solution a, b, c Z. Now we can further assume they do not have common factor larger than 1 (we can always divide out). Immediately we see a must be even, as every other term is even. So suppose a = 2a. But then we have 8a 3 12a bc + 2b 3 + 4c 3 = 0 4a 3 6a bc + b 3 + 2c 3 = 0, and now b must also be even. We write b = 2b and 2a 3 6a b c + 4b 3 + c 3 = 0. Now this forces c to be even as well. This contradicts our assumption that a, b, c have no common factor larger than one. So in our case, a + b c 3 4 = 0 implies that not all of a, b, c are 0, which further implies that det(a) = 0. So there is a unique solution of x, y, z, and we can write down the formulae using Cramer s rule: x = a2 2bc det(a), y = 2c2 ab det(a), z = b2 ac det(a). The upshot is that X is indeed a field, and K = X. We can of course generalise this result with the following proposition. Proposition. Suppose k R L where k, L are fields, and R is a ring. If R is finite dimensional as a k-vector space, then R is a field. Example. Apply this to k = Q, R = X = Q[ 3 2], and L = C. Then X is a field. Proof. Fix any 0 = a R. Consider the map T : R R defined as T(x) = ax. This is obviously a k-linear transformation. Since R L, ax = 0 iff x = 0, and so T is injective. But an injective linear map from a finite dimensional vector space to itself must also be surjective. Thus there exists x R with ax =

7 Now we go back to the question: are 1, 3 2, 3 4 linear independent over Q? Suppose not, i.e. there exist a, b, c Q not all 0 such that a + b c( 3 2) 2 = 0. This is to say that 3 2 is a root to the quadratic equation a + bx + cx 2 = 0 where a, b, c Q. Could this possibly happen? We know 3 2 does satisfy x 3 2 = 0 though. Claim: x 3 2 Q[x] is irreducible. We will be dealing with the proof later. Let s first assume it is the case. On the other hand, if k is a field, f k[x] has a degree, deg(f) N. We then have the Euclid s algorithm: for all f, g k[x], there exist q, r k[x] such that f = q g + r and deg(r) < deg(g). The existence of Euclid s algorithm implies that: hcf always exists; irreducible and prime elements are the same; k[x] is a unique factorisation domain. Now let f Q[x] be any polynomial. The polynomial x 3 2 is irreducible implies that either x 3 2 f or hcf(f, x 3 2) = 1. In the latter case, we know there will exist A(x), B(x) Q[x] such that Af + B(x 3 2) = 1. This is an identity, so in particular it holds for x = 3 2, i.e. A( 3 2)f( 3 2) = 1, and therefore, 3 2 is not a root of f. Equivalently, for any f Q[x], if 3 2 is a root of f, then x 3 2 must divide f. Now we can conclude that 1, 3 2, 3 4 are linear independent over Q, since clearly x 3 2 does not divide a + bx + cx The Eisenstein s criterion week 3 lecture 2 The last example suggests that we need some criteria for irreducibility of polynomials in Q[x]. Remark. Let p(x) = c 0 + c 1 x + + c d x d Z[x]. Suppose x 0 = p/q Q with p, q Z and hcf(p, q) = 1, then x 0 is a root of p implies that p c 0 and q c d. Proof. clearing up the denominators and we get q d c 0 + q d 1 c 1 p + + c d p d = 0. Now clearly p q d c 0, and therefore p c 0. Similarly q c d. Example. The polynomial x 3 2 Q[x] is irreducible. off the record By the remark above, x 3 2 does not have any rational roots. But if x 3 2 were reducible in Q[x], it would split as x 3 2 = (x x 0 )(x 2 + ax + b), and therefore would have a rational root x 0. How about polynomials of degree 4 or more, for example, f = 1 + x + x 2 + x 3 + x 4? First notice f = (x 5 1)/(x 1), and therefore f splits in C as f = 4 k=1 (x e 2πki/5 ). So the only way to factorise f in R would be f = (x 2 2 cos 2π 5 x + 1)(x2 2 cos 4π x + 1). So f is irreducible 5 in Q[x] iff cos 2π 5 = ( 5 1)/4 and cos 4π 5 are rationals, which is clearly not the case. In the case I do not know the exact value of cos 2π 5, I can consider the field extension Q K = Q(e 2πi/5 ), and the Galois group acts as automorphisms of the group of 5-th roots of unity. So the Galois group is C 4. From this I know the value of cos (2π/5) should be in some quadratic surd. Theorem. (Eisenstein s criterion) Let f(x) = a 0 + a 1 x + + a d x d Z[x]. Suppose there exists a prime number p such that p a d, p a i for all i < d, and p 2 a 0, then f(x) Q[x] is irreducible. Example. For any prime number p, the polynomial f(x) = x p 1 + x p Q[x] is irreducible

8 Proof. We prove an equivalent statement: f(x + 1) is irreducible. Now notice that f(x) = xp 1 x 1, so f(x + 1) = (x + 1)p 1 x ( ) ( ) p p = x p 1 + x p p 1 Obviously now a d = 1, and p a d ; p a i for all i < d; and a 0 = p, p 2 a 0. off the record Consider the roots to x 12 1, and find an irreducible polynomial f(x) Q[x] that f(e 2πi/12 ) = 0. We could try to factorise the polynomial: x 12 1 = (x 6 + 1)(x 6 1) = (x 4 x 2 + 1)(x 2 x + 1)(x 2 + 1)(x 2 + x + 1)(x + 1)(x 1). Now for every factor c 12, we look at the primitive c-th roots (i.e. not an n-th root for any n < c) of unity, and define Φ c = (x ξ) Q[x]. Then the factorisation we get corresponds to ξ primitive c-th root of unity x 12 1 = Φ 12 Φ 6 Φ 4 Φ 3 Φ 2 Φ 1. For example, the 4-th primitive root i, i are characterised by the term x So our final answer is Φ 12 = x 4 x Exercise. Show that x 4 x is irreducible over Q. Exercise. More generally, for all n, define Φ c for all c n such that x n 1 = c n Φ c; also prove that Φ n is irreducible for all n, and is of degree φ(n), the Euler s totient function. Eisenstein s criterion turns out to be a consequence of the following Gauss lemma. Lemma. Let f(x) = a a d x d Z[x] with hcf(a 0,..., a d ) = 1. Suppose f(x) splits non-trivially as f(x) = g(x)h(x) in Q[x], then f splits in Z[x]. To prove this, we need some ring theory: Lemma. Let p Z be a prime number. Then p Z[x] is a prime element

9 This says if p a(x)b(x), then either p a(x) or p b(x). Naively let s consider a, b to be linear functions and p (a 0 + a 1 x)(b 0 + b 1 x) = (a 0 b 0 + (a 0 b 1 + a 1 b 0 )x + a 1 b 1 x 2 ). Then we must have p a 0 b 0, p a 0 b 1 + a 1 b 0, and p a 1 b 1. Without loss of generality, suppose p a 0. Then the second condition becomes p a 1 b 0. Either p a 1 and we are done, or p b 0, in which case from p a 1 b 1 we are always done. Of course, alternatively, Z[x]/p = F p [x] is an integral domain, and therefore the ideal (p) is prime. Proof. (Gauss lemma) We know if p is prime in Z, then p is prime in Z[x]. We first kill the denominators: there exists a smallest c N such that cf = g(x) h(x) with g, h Z[x]. We claim that c = 1, and if so, we have a split in Z[x]. If c = 1, then there exists a prime number p c, and therefore p cf. By the ring theory above, wlog p g. Then c p f = g p h is another split in Z[x] but with c/p < c, contradicting our assumption that c is minimal. Same idea shows that if R is a UFD, then R[x] is a UFD. So for a field k, k[x 1,..., x n ] is a UFD. Proof. (Eisenstein s criterion) By Gauss, if f is not irreducible in Q[x], then there exist h(x), g(x) Z[x] such that f(x) = h(x)g(x). Modulo p, we get 0 = a d x d f(x) h(x)g(x) mod p Since F p [x] is a UFD, there exist d 1, d 2 with d 1 + d 2 = d, and b d1 c d2 = a d such that Then we get h b d1 x d 1, g c d2 x d 2 mod p h(x) = b d1 x d 1 + b d1 1x d b 1 x + b 0 p b i i < d 1 g(x) = c d2 x d 2 + c d2 1x d c 1 x + c 0 p c i i < d 2 But then, the constant term of f = hg is b 0 c 0, which is divisible by p 2, and this is a contradiction

10 3 Field theory 3.1 Algebraic extensions Now that we know how to make many irreducible polynomials in Q[x], we go back to field theory. Suppose K L, then for all a L, we write K(a) to be the smallest subfield of L that contains K and a. Remark. If I is some indexing set, and K i L for all i I, then i I K i is a subfield of L, and K(a) = F. KFL a F Definition. A field extension K L is primitive if there exists an a L such that K(a) = L. Definition. The evalutation homomorphism at a is the map φ a : K[x] K(a) L, f(x) f(a). Proposition. Consider a field extension K L and a L, and the evaluation homomorphism φ a : K[x] L. Then there are two exclusive alternatives: 1. φ a is injective, and φ a extends to φ a : K(x) K(a) L. 2. φ a is not injective, and the kernel is the principal ideal generated by a unique monic irreducible polynomial f(x) K[x]. In this case, φ a passes through the quotient to define a field homomorphism: φ a : K[x]/(f) K(a) L. If (1) holds, we say a is transcendental; if (2), we say a is algebraic over K, and f is the minimal polynomial of a. week 4 lecture 1 The last example we have talked about illustrated all of these with K = Q, L = C, a = 3 2 as an algebraic extension. We know that a = e, π are transcendental, and Q(e) = Q(π) are isomorphic to Q(x), so this proposition doesn t say much about transcendental extensions. In fact, we don t really care about transcendental extensions in this course. Proof. We know K[x] is a Euclidean domain (ED), and thus a principal ideal domain (PID). Hence every ideal, ker(φ a ) in particular, is of the form (f) for some f K[x], and clearly there is a unique one that is monic. We need to prove f is irreducible. However, this comes automatically. Consider the inclusion: R = K[x]/(f) φ a L. Since L is a field, R contains no zero-divisors. In terms of elements in K[x], this translates to for all g, h K[x], if f gh, then f g or f h, which is precisely to say f is prime (or irreducible, same in a ED). It remains to show φ a is an isomorphism from R to K(a). Clearly it is injective. To show it is surjective, first note that R is actually a field (the ideal f is maximal if you took algebra III, or for all g K[x], either f g, in which case g is 0 in R, or hcf(f, g) = 1 as f is irreducible, and therefore the image of g in R is invertible). So R L is a subfield containing a and K. On the other hand, K(a) is defined to be the smallest field in L that contains a and K. Together φ a is surjective. Corollary. Suppose K is a field and f(x) K[x]. Then there exists a field L and a L such that f(a) = 0. Proof. Let g f be a prime factor of f, and consider L = K[x]/g. Corollary. If K L 1, K L 2 are two field extensions, and a 1 L 1, a 2 L 2 have the same minimal polynomial f(x) K[x], then there exists a unique isomorphism φ : K(a 1 ) K(a 2 ) such that φ(a 1 ) = a 2. Proof. both are isomorphic to K[x]/f. Corollary. Let K L = K(a) and a is algebraic with minimal polynomial f(x) K[x]. Let K Ω be another field extension. Then Emb K (K(a), Ω) = {roots of f(x) in Ω}. Proof. Suppose ξ Ω is a root to f. Then consider the evaluation homomorphism at ξ: φ ξ : K[x] Ω

11 Tautologically, ker φ ξ = (f) (f must be in ker φ ξ, and conversely we know ker must be generated by a prime g which divides f, but f is already prime). So the proposition above states that there is a map φ ξ : K(a) = K[x]/f Ω a x ξ. Conversely, if we write f(x) = b 0 + b 1 x + + x d, and φ : K(a) Ω Emb K (K(a), Ω), then at x = a, we apply φ to get: 0 = φ(0) = b 0 + b 1 φ(a) + + φ(a) d. week 4 lecture 2 So ξ = φ(a) is a root of f. The first corollary suggests we always have an extension field. However, f usually will only have one root in there. Example. Consider x 7 2 Q[x], L = Q( 7 2). In L[x], f(x) = (x 7 2)(x x ). Denote by g L[x] the second factor. Is g irreducible? (Yes, but you can try) If you are not convinced, x 3 2 will also do. In C, f factors as f = 7 k=1 (x ɛ k 7 2). We also have a 1-to-1 correspondence: So there are 7 embeddings of Q( 7 2) in C: {Emb Q (Q( 7 2), C)} {a C : a 7 = 2}. σ k : Q( 7 2) C 7 2 ɛ k 7 2 Definition. K L is algebraic if for all a L, a is algebraic over K. Naturally, we ask the two questions: Question A: Suppose L = K(a) where a is algebraic over K. Does this imply K L is algebraic? Question B: Given K L and a, b L algebraic. Are a + b, a b, a b, a/b (if b = 0) algebraic? To convince you these are not trivial questions, you can try playing around with these questions using only elementary tools. For example, if a satisfies some monic irreducible polynomial f(x), and b satisfies some g(x), what polynomial does a + b satisfy? It turns out these questions can be solved with one simple idea, which we will illustrate below. Definition. Let K L be a field extension. The degree of L over K, deg K L = [L : K], is the dimension of L as a K-vector space (also written as dim K L). K L is finite if dim K L <. Remark. Notice that finite implies algebraic (not the converse). If we let n = dim K L <, and pick a L. Consider the set {1, a, a 2,..., a n }. There are n + 1 elements, and therefore they must be linear dependent. In particular, there exist λ 0,..., λ n K such that a satisfies λ 0 + λ 1 x + + λ n x n = 0. Remark. If K L, and a L with minimal polynomial f(x) K[x] of degree n, then [K(a) : K] = n. Indeed, the set {1, a,..., a n 1 } must be a basis of K(a) as a vector space over K. Linear independence is obvious, as the minimal polynomial is of degree n. To see they generate K(a), first notice that the map φ a : K[x] K(a) is surjective (from the proposition). Thus for any b K(a), there exists some g(x) K[x] such that g(a) = b. We could apply Euclidean s algorithm to get g = qf + r where deg r < deg f = n. Then b = g(a) = q(a)f(a) + r(a) = r(a), and r(a) is in the span of the set. Theorem. (Tower law) If K L E, then [E : K] = [E : L][L : K]. In particular, K E is finite iff both K L and L E are finite

12 Surprisingly, this simply idea can directly leads to the solutions to both questions we asked. K K(a) K(b) Corollary. Both questions have answer yes. K(a, b) = K(a)(b) = K(b)(a) L Proof to A. We know K L = K(a) has degree n, thus this extension is finite. But finite implies algebraic. Proof to B. Consider K K(a) K(a, b). We know K K(a) is finite. The fact that b is algebraic over K surely implies b is algebraic over K(a), so K(a) K(a, b) is finite. From tower law, [K(a, b) : K] is finite, thus algebraic. Corollary. K L. Then the set {a L : a is algebraic over K} is a field. Proof. (Tower law) Suppose {a 1,..., a n } L is a basis of L as a K-vector space, and {b 1,..., b m } E a basis of E as a L-vector space. We claim that {a i b j } is a basis of E as a K-vector space. First we prove this is a generating set. For any x E, x can be written as x = λ 1 b λ m b m λ j L, b i E = (µ 11 a µ 1n a n )b µ ij K, a i L E = µ ij a j b i. To show the linear independence, suppose 0 = µ ij a j b i = j ( i µija i)b j. Then i µija i = 0 for all j, and then µ ij = 0 for all i, j. Theorem. Suppose K L is finite, and K E is finite. Then there exists a finite extension K Ω and embeddings: N.B. this theorem does not say Ω is unique. In fact, we have K L E Ω 7 2 e 2πi/7 7 2 Q Q( 7 2) Q( 7 2) Q( 7 2) or Q(e 2πi/7, 7 2) inclusion Proof. We prove by induction on [L : K]. If [L : K] = 1, then L = K, and we can simply take Ω = E. Otherwise, let a L \ K algebraic with minimal polynomial f(x) = x n + λ 1 x n λ n K[x]. We have two cases: f has a root in E or not. Case 1: if f has a root in E, by the third corollary to the proposition before, we know there is an embedding K(a) E. Of course we also have K(a) L. In a diagram: K K(a) L E Ω By the tower law, [L : K(a)] < [L : K], and by induction there exists such an Ω. Case 2: we know there exists E E such that f has a root in E. Then there also exists an embedding K(a) σ E. In a diagram:

13 K K(a) E L E Ω week 5 lecture 1 Now [L : K(a)] < [L : K], and we can use induction again to make Ω. We mentioned in one of our remarks that an extension is finite implies it is algebraic. Now we give an equivalent condition. Definition. K L is finitely generated if there exist a 1,..., a r L such that L = L(a 1,..., a r ). Corollary. K L is finite iff K L is algebraic and finitely generated. Proof. : We know finite implies algebraic. Now pick a L \ K. From the tower law, [L : K] = [L : K(a)][K(a) : K]. But a K implies [K(a) : K] > 1, and therefore [L : K(a)] < [L : K]. By induction K(a) is finitely generated, and L = K(a)(a 1,..., a r ) = K(a, a 1,..., a r ). The converse should be clear from the tower law, as [L : K] = [L : K(a 1,..., a r 1 )] [K(a 1,..., a r 1 ) : K(a 1,..., a r 2 )] [K(a 1 ) : K], which is a finite product of finite numbers, and is therefore finite. Theorem. Let K L be finite, and K Ω be another extension. Then Example. Emb Q (Q( 7 2), C) = 7. Emb K (L, Ω) [L : K]. Proof. If L = K(a), and f(x) is the minimal polynomial of a with degree n. Then we know there is a 1-to-1 correspondence between {Emb K (K, Ω)} and {roots of f in Ω}, and there are at most n roots in Ω. We do the general case by induction on the degree. Suppose K L. Choose a L \ K. Consider K K(a) L, and restriction defines a function of sets: Emb K (L, Ω) ρ Emb K (K(a), Ω). Now we want to estimate ρ 1 (φ) for a given φ Emb K (K(a), Ω). From the tower law, we know [L : K(a)] < [L : K], and we can therefore assume by induction that ρ 1 (φ) [L : K(a)]. Then and we are done. Emb K (L, Ω) [L : K(a)][K(a) : K] = [L : K], week 5 lecture Fundamental theorem of Galois theory first half We will finish off this section proving the easy half of the fundamental theorem of Galois theory. Definition. Suppose L is a field, and G Aut(L). The fixed subfield of G in L is the set K = {a L : g G, g(a) = a}. It is also written as L G. Theorem. L is a field, G Aut(L) a finite group, and K = L G the fixed field of G. Then Emb K (L, L) = G. Explanation: we previously mentioned that under certain conditions, a finite extension K 0 L will have the correspondence {intermediate fields K 0 E L} {subgroups G of Emb K0 (L, L) = Aut K0 L} K = L G E Emb E (L, L) G

14 So we are essentially saying that composing the second map with the first map gives identity with no extra condition at all. N.B. G Emb K (L, L) 1, so G Emb K (L, L) [L : K] from the last theorem. It suffices to prove [L : K] G. Proof. Let n = G, and pick any a 1,..., a n+1 L. We will show they are K-linear dependent, i.e. non-trivial λ i K such that λ 1 a λ n+1 a n+1 = 0. Let G = {σ 1,..., σ n }. Consider the matrix: σ 1 (a 1 ) σ 1 (a 2 ) σ 1 (a n+1 ) σ 2 (a 1 ) σ 2 (a 2 ) σ 2 (a n+1 ) A n,n+1 = M n,n+1(l). σ n (a 1 ) σ n (a 2 ) σ n (a n+1 ) In the L-vector space L n+1, the linear system A λ 1. λ n+1 = 0 will have non-trivial solutions since rk(a) < n + 1. But solutions occur in L. We need to show λ i s are in K, and we are done. By rearranging the columns of A if necessary, we may assume that 1. λ 1,..., λ m = 0, and λ i = 0 for all i > m; 2. λ m = 1; and such that m is the smallest number satisfying these conditions. Under these conditions the solution is unique. Suppose we have another solution λ 1,..., λ n+1, then λ 1 λ 1,..., λ n+1 λ n+1 will be another solution, and that the m-th term will be 0, thus contradicting the minimality of m. We show λ j K by definition, i.e. for all σ G, σ(λ j ) = λ j. Now for any σ G, we apply σ to our original linear equation to get (σ is a L-automorphism): σ(λ 1 ) σσ 1 (a 1 ) σσ 1 (a 2 ) σσ 1 (a n+1 ) σσ 2 (a 1 ) σσ 2 (a 2 ) σσ 2 (a n+1 ) σ(λ m ). 0 = 0. σσ n (a 1 ) σσ n (a 2 ) σσ n (a n+1 ). 0 Notice that the map G G, g σ g is injective. So the rows are the same with the ones in A with a different order. RHS is zero, so we can swap the rows to get a different solution: By uniqueness, σ(λ j ) = λ j for all j, and λ j K. σ(λ 1 ) σ 1 (a 1 ) σ 1 (a 2 ) σ 1 (a n+1 ) σ 2 (a 1 ) σ 2 (a 2 ) σ 2 (a n+1 ) σ(λ m ). 0 = 0. σ n (a 1 ) σ n (a 2 ) σ n (a n+1 ). 0 1 Fix x G. For any k K, by definition, any g G will fix k, and therefore x must also fix k

15 4 Normal and separable extensions 4.1 Splitting fields Definition. K L is a splitting field for f(x) K[x] (not necessarily irreducible) if 1. f(x) splits completely in L[x], i.e. f(x) = (x x i ) for some x i L; 2. If λ 1,..., λ m are the roots of f(x) in L, then L = K(λ 1,..., λ m ). Proposition. Given a field K, and f K[x], we have the properties: 1. There exists a splitting field. 2. If K L is a splitting field for f, and K Ω is any field in which f splits completely, then Emb K (L, Ω) =. Assuming these properties, we have the following corollary. Corollary. Any two splitting fields of f K[x] are isomorphic. Proof. We know there exist embeddings σ 1 : L 1 L 2 and σ 2 : L 2 L 1. Claim. σ 1 is an isomorphism. σ 1 is injective and K-linear. But dim L 1 = dim L 2, as the existence of σ 1 implies dim L 1 dim L 2, and the existence of σ 2 implies dim L 2 dim L 1. So σ 1 is an isomorphism. Example. Q( 2, 3) is the splitting field of (x 2 2)(x 2 3). But it is also the splitting field for f(x) = x 4 10x Question 1. How to show f splits completely without factorising it? Proof. (proposition) 1. Suppose f is irreducible. Then consider K L = K[x]/(f). Tautologically, a = x L is a root, as f(a) = f(x) = 0 L. So f has a linear factor X a in L[X]. Putting in 1 root at a time, and we get a splitting field for f. If f is not irreducible, then we do the same to each of its prime factor. Remark. the proof shows [L : K] n! where n is the degree of f. 2. Suppose a L is a root with minimal polynomial g f. In particular, g splits completely in Ω. So there exists some b Ω that is a root of g, and we have an embedding K(a) Ω, a b. Inductively, we could construct an embedding L = k(a 1,..., a n ) Ω. week 6 lecture 2 Example. A field with 8 elements. Consider F = F 2 [x]/(x 3 + x + 1). Note: x 3 + x + 1 is irreducible over F 2, as neither 0 or 1 is a root. Denote by F 2 (a) the field F where a 3 + a + 1 = 0. Question. Does x 3 + x + 1 splits completely in F 2 (a) (i.e. F = F 2 (a))? We can list all 8 elements in F, and make a multiplication table: 0 1 a a + 1 a 2 a a 2 + a a 2 + a + 1 a 0 a a 2 a 2 + a a a 2 + a + 1 a a 2 0 a 2 a + 1 a 2 + a + 1 a 2 + a a a Now make long division in F 2 (a) to get: x 3 + x + 1 = (x a)(x 2 + ax + a 2 + 1) and by noticing further that a 2 is another root, we split x 3 + x + 1 as x 3 + x + 1 = (x + a)(x + a 2 )(x + a 2 + a). 1 cf. Q8 on Ex1-14 -

16 Example. Fix a prime number p. For all integer m 0, there exists a field F with q = p m elements, and it is unique up to isomorpihsm (denoted by F q ). Proof. (1) Uniqueness: Suppose F = q, then F must be the splitting field for f(x) = x q x F p [x] over F p [x]. Indeed, for all a F, we must have a q = a, since either a = 0, and a q = 0 = a, or if a = 0, a is in the group F q \ {0} which is of order q 1, and thus a q 1 = 1. But since F only has q elements, we must have x q x = a F(x a). Thus uniqueness. (2) Existence: Let F be the splitting field for x q x. Claim: F = q. Notice that X := {a F : a q a = 0} F actually satisfies X = F, as X is a field, and F is defined to be the smallest field containing all the roots. The fact that X is closed under multiplication is obvious. To see it is closed under addition, notice that in a characteristic p field, (α + β) p = α p + β p, and then the rest follows. Therefore if x q x has q distinct roots, then the claim is true Normal extensions Definition. A finite extension K L is normal if for all extensions K Ω, and for all σ 1 : L Ω, σ 2 : L Ω in Emb K (L, Ω), we have σ 1 (L) σ 2 (L). Remark. K L E. If K E is normal, then L E is normal, as any σ Emb L (E, Ω) also fixes K. Theorem. For a finite extension K L, the following are equivalent: (I) K L is normal (II) for all irreducible f K[x], either f has no root in L, or f splits completely in L. (III) there exists f K[x] such that K L is a splitting field for f. Proof. (I) (II): suppose f K[x] irreducible and has one root in L, say a L. Claim. There exists a field extension K Ω such that 1. f splits completely in Ω, i.e. there exist λ 1,..., λ m Ω with f(x) = m i=1 (x λ i ). 2. for each i = 1,..., m, the embedding σ i : K(a) Ω extends to σ i : L σ i K(a) K σ i Ω where σ(a) = λ i which we know to exist and to be unique. If this claim is true, then by normality, σ i (L) = σ 1 (L) for all i. In particular, λ i σ 1 (L) for all i = 1,..., m. So we can write f(x) = m i=1 (x σ 1 1 (λ i)), and f splits completely in L. Now we prove the claim. Let K E be a splitting field for f. So we have the following diagram: L σ i Ω i K(a) σ i E 1 cf. Jacobian criterion

17 where σ i (a) = λ i, and Ω i is some other field that L, E embed in. Now by previous theorem, E Ω i will give E Ω. This proves the claim. (II) (III): Since K L is finite, thus it is finitely generated, say L = K(a 1,..., a l ) for some a i L \ K. Let f i (x) K[x] be the minimal polynomial for a i. Manifestly L is the splitting field for l i=1 f i (x). (III) (I): We know L = K(λ 1,..., λ m ), where f(x) = (x λ i ). Suppose L Ω, and σ Emb K (L, Ω). K L L Ω σ For all i, σ(λ i ) Ω is a root for f, i.e. σ sends {λ i : i = 1,..., m} to {λ i : i = 1,..., m}. However, σ(l) is generated by σ(λ i ), and therefore (I) follows. Remark. For the last part of the proof, think of L = Q( 3 2) embeds in C in two different ways. But x 3 2 splits in Ω = C, so σ : Q( 3 2) C is actually Q( 3 2) Q( 3 2). Corollary. For every finite extension K L, there exists K L Ω where K Ω is normal. Idea of proof. There exist a i such that L = K(a 1,..., a l ). Let f i be the minimal polynomial for a i, and take Ω to be the splitting field of l i=1 f i (x). 4.3 Separable degree Suppose K L finite. We define the separable degree as [L : K] s = Emb K (L, Ω), where L Ω is some extension, and K Ω is normal. Remark. Back in a theorem we proved a while ago, [L : K] s [L : K]. Theorem. 1. [L : K] s is well defined, i.e. independent of the choice of Ω. 2. If K L E, then [E : K] s = [E : L] s [L : K] s. Lemma. Suppose K L E with K E normal. E Ω any extension and σ : L Ω embedding. E K L Ω σ E σ Ω Then σ(l) E. Proof. There exists a bigger field Ω such that σ : L Ω extends to σ : E Ω. But K E is normal. So σ(e) E, and σ(l) E. Proof. (1) Suppose we pick two normal extension K Ω 1 and K Ω 2. We can always make a bigger field Ω: L Ω 1 K Ω Ω

18 week 7 lecture 1 We will argue that the degree measure in Ω 1 is the same as in Ω, and same for Ω 2. We may as well assume Ω 1 Ω 2 = Ω. Then by the previous lemma, any embedding from L to Ω 2 has image inside Ω 1, thus is an embedding from L to Ω 1. So [L : K] s is well defined. (2) Choose E Ω such that K Ω is normal. Then every embedding E Ω can be restricted to an embedding L Ω. Claim: the restriction map res : Emb K (E, Ω) Emb K (L, Ω) is surjective, i.e. every embedding σ : L Ω extends to σ : E Ω. First, we can do this in a bigger field: E σ Ω L σ Ω But recall we chose E Ω, so we can apply the previous lemma to K E Ω σ Ω to get σ(e) Ω. QED claim. Now for all σ, res 1 (σ) = Emb K (E, Ω), and therefore: So we have the tower law for the separable degree. [E : K] s = [E : L] s [L : K] s. Question. Could it be that the separable degree is the same as the old degree we defined? N.B. We definitely know [L : K] s [L : K]. If L = K(a), where a has minimal polynomial f K[x], then Emb K (L, Ω) = {α Ω : f(α) = 0}, i.e. [L : K] s = deg f. So our question becomes: is there a field K and f K[x] irreducible such that in some extension Ω, f splits completely but the roots are not distinct? In a characteristic 0 field, for f K[x], λ L is a repeated root iff f (λ) = 0. So f has no repeated roots iff hcf(f, f ) = 1. If f is irreducible, it implies that f = 0, then f has no repeated roots. However, in a characteristic p field, f = 0 with f = 0 is possible. So we have some problems to deal with here. Example. K = F p (t), the fraction field in t with coefficients in F p. Consider f = x p t irreducible. If L = K(a) where a p = t, then f(x) = (x a) p. In this case, [L : K] s = 1, [L : K] = p. 4.4 Separable extensions week 7 lecture 2 Definition. Let K be a field. A degree n polynomial f(x) K[x] is separable if it has n distinct roots in some extension of K. If K L, an element a L is called separable over K if the minimal polynomial of a is separable. Example. K = F p (t), a = p t is not separable. Indeed, the minimal polynomial of a is x p t, since 1. a K. If [P(x)/Q(x)] p = t, then P p = tq p. This is an identity in the ring F p [t], which is a UFD. So we use the usual argument to show t P and t Q. 2. x p t = (x a) p in F p (a) F p (t). If (x a) k F p (t)[x] for some k < p, then expand this, we have the second term having coefficient ka F p (t), which is impossible

19 Let K be a field. We can define a derivative operator D : K[x] K[x], D( a n x n ) = na n x n 1. This map has the properties that: 1. D(λf + µg) = λd(f) + µd(g), 2. D(fg) = D(f)g + fd(g) for all f, g K[x], λ, µ K. Proposition. 1. f K[x] is separable iff hcf(f, Df) = If f is irreducible, then f is not separable iff Df = 0. Proof. (i) Let f K[x], and K L where L splits: f(x) = m i=1 (x λ i ) L[x]. Then Df = m m ( (x λ j )). i=1 j =i If f is separable, we claim hcf(f, Df) = 1, as otherwise there exists an i with x λ i Df. This is impossible because x λ i divide all but the i-th summand in the sum above. This shows hcf(f, Df) = 1 in L[x], so surely also in K[x]. On the other hand, if f is inseparable, then there exists λ L such that (x λ) 2 f, and f = (x λ)g(x). Then x λ Df, and f, Df have a common factor x λ in L[x]. We claim they must have a common factor in L[x] as well. Otherwise, we can write φf + ψdf = 1, but then this is an identity also holds in L[x], which is a contradiction. (ii) If f is irreducible, then hcf(f, Df) = 1 iff hcf = f, and then f Df, which is true iff Df = 0. Remark. In a char 0 field, all irreducible polynomials are therefore separable. However, in a char p field, for f K[x], if Df = 0, then there must exist some g K[x] such that f = g(x p ). Indeed, if we write f = a 0 + a 1 x +..., then Therefore a i = 0 for all p i. Df = a 1 + 2a 2 x + + pa p x p pa 2p x 2p Definition. A field K of characteristic p is perfect if K p = K, or equivalently, for all a K, there exist b K such that b p = a. Example. For all q = p m, the finite field F q is perfect. This is because the group homomorphism F q F q defined by b b q in injective, thus also surjective. Proposition. If K is perfect, then every irreducibly polynomial is separable. week 8 lecture 1 Proof. If the irreducible polynomial f is not separable, then Df = 0, and by the remark above, f = h(x p ). We write h(x) = x n + a 1 x n a n. Since K is perfect, there exist b i s such that b p i = a i. Then consider: g(x) = x n + b 1 x n b n. Therefore g p = x np + b p 1 (xp ) n b p n = h(x p ) = f(x). This contradicts the fact that f is irreducible. This proposition explains why we picked F p (t) in our first example, as it is the simplest infinite field with characteristic p, so that there would exist irreducible non-separable polynomial. We would like to describe the separable extension without referring to elements of the fields. Definition. A finite extension K L is separable if: for all K F 1 F 2 L, if [F 2 : F 1 ] s = 1, then F 1 = F

20 Remark. For K F L, if K L is separable, then K F and F L are separable. In fact the converse is also true. Theorem. 1. K L is separable iff [L : K] s = [L : K]. 2. For K F L, K L is separable iff K F and F L are both separable. Proof. Suppose [L : K] s = [L : K], then by two tower laws, we have [L : K] = [L : F 2 ][F 2 : F 1 ][F 1 : K] [L : K] s = [L : F 2 ] s [F 2 : F 1 ] s [F 1 : K] s Now the left sides are equal, and every term in the top row is at least as large as the corresponding term in the bottom row, so they must equal to each other. In particular, [F 2 : F 1 ] s = [F 2 : F 1 ]. But [F 2 : F 1 ] = [F 2 : F 1 ] s = 1, so F 1 = F 2. For the converse, we first prove the claim: Claim: Suppose K K(a) is separable, then a is separable over K (i.e. its minimal polynomial is separable over K), and therefore [K(a) : K] s = [K(a) : K]. Let f(x) be the minimal polynomial of a. Assume for a contradiction that f(x) is not separable. Then by previous remark, f = h(x p ), where h(x) K[x] must also be irreducible (if h = h 1 h 2 then f = h 1 (x p )h 2 (x p )). Now b = a p satisfies h, and thus h is the minimal polynomial of b. So we can consider the tower K K(b) K(a) where K K(b) has degree the degree of h, and K K(a) has degree the degree of f. Thus [K(a) : K(b)] = p. We now claim the minimal polynomial of a over K(b) must be x p b. Indeed, a satisfies this polynomial, and it is of degree p. However, this polynomial has p identical roots, and [K(a) : K(b)] s = 1. This contradicts the fact that K K(a) is separable. qed claim. Now suppose K L is separable, then pick a L \ K, and consider K K(a) L. By the remark, both extensions are separable. By claim, [K(a) : K] s = [K(a) : K]. On the other hand, we know [L : K(a)] < [L : K] and K(a) L is separable. So we assume by induction that [L : K(a)] s = [L : K(a)]. By two tower laws, we have [L : K] = [L : K] s. The second part is a simple corollary of the first part. We leave it as an exercise. The following corollaries are the usual ways of presenting this topic on a textbook. They are direct consequences of the theorems above. Corollary. An element a is separable over K iff K K(a) is separable. Proof. We proved in the previous claim. For, if a is separable, then [K(a) : K] s = [K(a) : K], and by theorem above, K K(a) is separable. Corollary. K L is separable iff for all a L, a is separable over K. Proof. follows from previous corollary. : first notice that given K F L, and if a L is separable over K, then it is separable over F. Therefore, if we pick a and consider the tower K K(a) L, we know both extensions are separable, thus K L is separable. As a fact, [L : K] s [L : K] s, so we can write [L : K] = [L : K] s [L : K] i, where [L : K] i is called the inseparable degree. Moreover, the inseparable degree is a power of p, the characteristic of field K

21 5 Fundamental theorem of Galois theory week 8 lecture 2 Definition. K L is a finite Galois extension if it is finite, normal, and separable. Normal says there are enough roots in L, and separable says all roots are unique. Theorem. (Fundamental theorem of Galois theory, aka Galois correspondence) Suppose K L is finite Galois, then there is a canonical G-equivariant inclusion-reversing bijection between the following sets: {intermediate fields K F L} {subgroups H of G = Emb K (L, L)} {a L : φ(a) = a φ H} = H = L H F F = Emb F (L, L) Inclusion reversing: the bigger the field, the smaller the group (vice versa): H. F 1 F 2, then F 2 F 1. G-equivariant: notice G acts on both sets (on fields by applying the map, and on groups by conjugation), and the correspondence is compatible with these actions and, e.g.: (gf) = (ghg 1 ). We already showed that if H is a finite subgroup of field automorphisms of L, and H its fixed field, then Emb H (L, L) = H, i.e. (H ) = H, and [L : H ] = H. In fact, it is clear that H (H ) Emb H (L, L). Note: K L is normal, so H L is normal (K H L) 1. Proof. Since F (F ), it suffices to prove inclusiong in the other direction. We know K L is finite, normal and separable, therefore F L is also finite, normal and separable. Recall that F = {φ : L L : φ F = id} = Emb F (L, L). Now F L is normal, by the definition of separable degree, Emb F (L, L) = [L : F] s, i.e. F = [L : F] S. Note also that 2 F Emb (F ) (L, L), and therefore F Emb (F ) (L, L) = [L : (F ) ] s. Now apply tower law of separable degree to F (F ) L: [L : F] s = [L : (F ) ] s [(F ) : F] s, where LHS= F, and [L : (F ) ] s F. So [(F ) : F] s must be 1. But the extension is separable, so [(F ) : F] = 1, and F = (F ). 5.1 Galois theory with biquadratic extensions Definition. Let K be a field with char K = 2. L/K is biquadratic if it is a splitting field of a polynomial f(x) = (x 2 a) 2 b for some a, b K. Of course, the roots are ± a ± b. There are many possible intermediate fields between K and L. Potentially something like the diagram below: 1 This might worth mentioning: K H is a normal extension H is a normal subgroup of G, hence the name normal. cf PS2 Q3,8,9. 2 In fact we have equality here, i.e. F = ((F ) ) (by taking H = F ), but we only need inclusion in this direction

22 K( b, a + b, a b) K( b, a b) K( b, a 2 b) K( b, a + b) K( b) K( a 2 b) K Of course, in K with char = 2, if α is not a square, then [K( α) : K] = 2. But for β K not a square, could β be a square in K( α)? Say γ = a + b α with γ 2 = β. Then a 2 + b 2 α + 2ab α = β. So ab = 0, and a = 0 or b = 0. But β not a square in K, so we must have b 2 α = β. E.g. 18 not a square in Q, but 18 = (3 2) 2. Now let s apply Galois theory. Let K be a field with char = 2, K α, β = 0 not squares, and β/α also not a square in K. We just saw x 2 β = 0 has no roots in K( α). Consider the splitting field of (x 2 α)(x 2 β), K( α, β). K( α, β) 2 K( α) 4 K This extension is Galois, and G = Emb K (L, L) has size 4. But obviously we have 4 different subgroups of G. Since C 4 only has 3 subgroup, G must be C 2 C 2, which has 5 subgroups. So we have one more subfield. The complete picture is: 2 K( α, β) {1} K( α) K( αβ) K( β) C 2 1 diagonal subgroup 1 C 2 K C 2 C

23 6 Examples of Galois correspondence week 9 lecture 1 For the remaining part of this course, we will be mainly dealing with some specific examples of Galois correspondence, including the famous insolubility of quintic equations. 6.1 Galois groups of cubics We study f(x) = x 3 σ 1 x 2 + σ 2 x σ 3 K[x]. Question. Assume f(x) is irreducible, separable, and L the splitting field of f, then what is the group Gal(L/K)? Proposition. Suppose f(x) = x n σ 1 x n 1 + σ 2 x n 2 ± ± σ n K[x] irreducible, separable, and L the splitting field. Then G S n = {group of permutations on the set of roots of f}, and it acts transitively on the roots. Proof. There exists a group homomorphism ρ : G S n, i.e. if g G, λ L a root, then g(λ) is also a root. Indeed, since g Emb K (L, L), it fixes all elements of K, and therefore 0 = g(f(λ)) = f(g(λ)). ρ is injective: suppose λ 1,..., λ n L are roots of f, and suppose g(λ i ) = λ i for all i, then g is identity on K(λ 1,..., λ n ), which is precisely L. Thus ρ has trivial kernel. G acts transitively, i.e. given λ 1, λ i L roots, there exists a g G such that g(λ 1 ) = λ i : here we use f is irreducible. K[x]/f φ 1 L φ 2 K L φ There exist two embeddings φ 1, φ 2 : K[x]/f L where φ 1 ([x]) = λ 1, and φ 2 ([x]) = λ i. Since L/K is normal, there exists φ : L L such that φ K = Id, φ φ 1 = φ 2, and φ(λ 1 ) = λ i (We can always make bigger field Ω, and K L Ω is normal, thus the image of φ 2 contains the image of φ 1 ). Now back to cubics, G can only be S 3 or a cyclic group of order 3. Theorem. Assume char K = 2, and consider the discriminant = σ 2 1 σ2 2 4σ3 1 σ 3 4σ σ 1σ 2 σ 3 27σ 2 3 K. Then is not a square in K iff G = S 3. Corollary. If char K = 3, then we can complete the cube and assume f(x) = x 3 + ax + b, and = 4a 3 27b 2. If we make a table of the discriminant of the polynomials x 3 nx 1 (for n = 2), we would find that is very rarely a square, i.e. almost always the case that G = S 3. Definition. A polynomial f(x 1,..., x n ) K[x 1,..., x n ] is symmetric if for all σ S n, f(x σ(1),..., x σ(n) ) = f(x 1,..., x n ). The elementary symmetric polynomial σ i k[x 1,..., x n ] are defined by the formula: (x x 1 )(x x 2 ) (x x n ) = x n σ 1 x n 1 + σ 2 x n 2... In particular, σ i are homogeneous of degree i. Let s write out a few: σ 1 = x 1 + x 2 + x x n σ 2 = i<j x i x j σ k = i 1 <i 2 < <i k x i1 x i2... x ik Theorem. Every symmetric polynomial is a function of the elementary symmetric polynomials, i.e. an element of K[σ 1, σ 2,..., σ n ]

24 Now let s go back to discriminant. Let the roots of f be λ 1, λ 2, λ 3 L, and consider d = (λ 1 λ 2 )(λ 2 λ 3 )(λ 1 λ 3 ) = = 0 If char K = 2, then d is not symmetric, as (12) S 3 will give d. However, D = d 2 will be symmetric. Exercise: D = (λ 1, λ 2, λ 3 ). Now we can prove the theorem regarding discriminant. Proof. Observe that G A 3 = C 3 iff d K. Hopefully it should all make sense too you if you go back to section 1.2 at this stage. 6.2 Galois groups of biquadratics week 10 lecture 2 Consider f(x) = x 4 2ax 2 + (a 2 b) K[x] where K is a field with char = 2. Let L be the splitting field. We know that if none of b, a 2 b b(a 2 b) is a square in K, then [L : K] = 8, and the roots are ± a ± b. Fix β L with β 2 = b, and α, α L with α 2 = a + β, α 2 = a β. L K(α) K(α ) K(β) Task: determine the Galois group G and all subgroups/intermediate fields. 2 2 K We know G = 8, and G {group of permutations of the 4 roots}. Now take any σ G. Notice that σ 2 (β) = σ(b) = b, and therefore σ(β) = ±β. If σ(β) = β, then σ 2 (α) = σ(α 2 ) = a + β, and thus σ(α) = ±α. Similarly σ(α ) = ±α. If σ(β) = β, then σ 2 (α) = σ(α 2 ) = a β, and thus σ(α) = ±α, and σ(α ) = ±α. There are 8 possibilities. So in fact all 8 cases occur. 2 α 2 2 α α α By direct verification, G is the symmetry group of this square, i.e. G = D 8 = σ, τ σ 4 = τ 2 = e, τστ = σ 1, where σ is the rotation counterclockwise by 90 degrees, and τ is the reflection along the horizontal axis. The lattice of subgroups of D 8 Consider γ = α α, δ = α + α, δ = α α. Then γ 2 = a 2 b, and therefore K(γ) is degree 2 over K. Now σ 2 (α) = α, σ 2 (α ) = α, and so σ 2 fixes γ. Similarly στ fixes γ as well. So the fixed fixed field corresponding to σ 2, στ is K(γ). Also (βγ) 2 = b(a 2 b), so degree 2 over K. Same reasoning applies

25 {e} τ σ 2 τ σ 2 στ σ 3 τ σ 2, τ σ σ 2, στ D 8 L K(α) K(α ) K(β, γ) K(δ) K(δ ) K(β) K(βγ) K(γ) K 6.3 Quartics equations Let f(x) = x 4 + ax 3 + bx 2 + cx + d K[x], and L the splitting field. Let s just say the Galois group G = S 4. There is a surjective homomorphism ρ : S 4 S 3 with kernel V = C 2 C 2 (here ρ has a section and splits, and S 4 = V S 3 ). ρ is defined by the actions on the set {A, B, C}, where A = {{1, 4}, {2, 3}}, B = {{1, 3}, {2, 4}}, C = {{1, 2}, {3, 4}}, and S 4 is the usual actions on the numbers For example, (12)A = B, (12)B = A, and thus we can identify (12) by (AB) S 3. The kernel is {e, (14)(23), (13)(24), (12)(34)} = V. Given groups K, H, and actions of H on K, i.e. a group homomorphism ρ : H Aut(K), we can make a new group, called the semidirect product G = K H, where G as a set is just the Cartesian products, and the group structure is by: (k 1, h 1 )(k 2, h 2 ) = (k 1 ρ(h 1 1 )(k 2), h 1 h 2 ). Nevertheless, we found a normal subgroup V of S 4, and S 4 /V = S 3. By Galois correspondence, we have a tower: ( K = S 4 ) ( F = V ) ( L = {e} ) where F L has Galois group V = C 2 C 2 and has degree 4, and K F has Galois group G/V and has degree 6. Notice that since V is a normal subgroup, K F is a normal extension, and therefore F is a splitting field. We expect: 1. There is a cubic equation such that its roots generate F. 2. Go from F to L, we take two more square roots. We want to understand F. Now inside L, V S 4 acts on the ring R = K[x 1, x 2, x 3, x 4 ], and F = V. It is logical to look for V invariants in the ring R. The most common two sets of invariants (equivalent) are Y 1 = (x 1 + x 4 )(x 2 + x 3 ) Z 1 = x 1 x 4 + x 2 x 3 Y 2 = (x 1 + x 3 )(x 2 + x 4 ) or Z 2 = x 1 x 3 + x 2 x 4 Y 3 = (x 1 + x 2 )(x 3 + x 4 ) Z 3 = x 1 x 2 + x 3 x