ALGEBRA HW 9 CLAY SHONKWILER

ALGEBRA HW 9 CLAY SHONKWILER 1 Let F = Z/pZ, let L = F (x, y) and let K = F (x p, y p ). Show that L is a finite field extension of K, but that there are infinitely many fields between K and L. Is L = K[α] for some α L? Is L separable over K? Proof. Note that x satisfies the polynomial Z p x p K[Z], so the minimal polynomial of x over K must divide this polynomial. Since Z p x p = (Z x) p, this polynomial is inseparable and, hence, the minimal polynomial of x must be separable as well, meaning it must be a polynomial in Z p. However, this in turn implies that the minimal polynomial of x must be Z p x p. Therefore, [K(x) : K] insep = p. On the other hand, since the minimal polynomial of x has a unique root and since any K-embedding of K(x) into K must take x to one of the roots of its minimal polynomial, we see that there is only one such embedding; hence, [K(x) : K] sep = 1, and so [K(x) : K] = [K(x) : K] insep = p. Now, consider L = K(x, y) = K(x)(y) as an extension of K(x). Clearly, y satisfies the polynomial Z p y p K(x)[Z], so the minimal polynomial of y over K(x) must divide Z p y p. By the same argument as above, this implies that [L : K(x)] insep = p and, furthermore, since Z p y p = (Z y) p, we see that [L : K(x)] sep = 1, so [L : K(x)] = [L : K(x)] insep = p. Finally, this means that so L is finite over K. Define Now, for any K i, [L : K] = [L : K(x)][K(x) : K] = p p, K 1 = K(x p+1 + y) K 2 = K(x +1 + y). K n = K(x np+1 + y). (x ip+1 + y) p (x p(ip+1) + y p ) = (x p(ip+1) + y p ) (x p(ip+1) + y p ) = 0 1

2 CLAY SHONKWILER since we re in characteristic p, so x ip+1 + y satisfies the polynomial Z p (x p(ip+1) + y p ) K[Z]. Now, in K (or even L), f i (Z) = Z p (x p(ip+1) + y p ) = Z p (x ip+1 + y) p = (Z (x ip+1 + y)) p, so this polynomial is inseperable. Since the minimal polynomial of x ip+1 + y over K must divide this polynomial, it must also be inseparable and, therefore, a polynomial in Z p. The only such polynomial is Z p (x p(ip+1) + y p ), we see that f i (Z) is the minimal polynomial of x ip+1 + y over K and, hence, [K i : K] insep = p. Since the minimal polynomial of x ip+1 + y, f i (Z), has a unique root and any K-embedding of K i must map x ip+1 to another root of its minimal polynomial, we see that there is only on K-embedding of K i into K, so [K i : K] sep = 1. Hence, K i is purely inseparable over K, and so [K i : K] = p. In particular, this means that K i L for all i N. Now, suppose K i = K j for some i j. Then x ip+1 + y, x jp+1 + y K i = K j, so (x ip+1 + y) (x jp+1 + y) = x ip+1 x jp+1 K i = K j. Now, since K i K, we can divide by a multiple of x p ; in particular, x ip+1 x jp+1 x ip = x x (j i)p+1 K i = K j. Now, since 1, x (j i)p K K i, we can divide by 1 x (j i)p : x x (j i)p+1 1 x (j i)p = x K i = K j. Hence, x ip+1 K i = K j, so (x ip+1 + y) x ip+1 = y K i = K j, and so K i = L. However, since this is impossible, we conclude that K i K j for all i j, and so there are infinitely many distinct, non-trivial intermediate extensions K K i L. By the Primitive Element Theorem, since L is finite over K, L = K[α] if and only if there are only finitely many intermediate fields M such that K M L. Since we just saw that there are infinitely many such intermediate fields, we conclude that there is no such α L such that L = K[α]. Furthermore, if L is separable over K, then the Primitive Element Theorem tells us that there is such a primitive element; since there isn t one, we conclude that L is not separable over K. 2 Let ζ n = e 2πi/n C and let Φ n (x) be the minimal polynomial of ζ n over Q. (a): Find the roots of Φ n (x). Show that deg Φ n (x) = φ(n), where φ(n) = #{m Z 1 m n, (m, n) = 1}.

ALGEBRA HW 9 3 Proof. We want to show that Φ n (x) = Ψ n (x), where Ψ n (x) := (x ζn m ). 1 m n (m,n)=1 Now, if ψ : Q(ζ n ) Q, then ψ must fix Q and is completely determined by ψ(ζ n ). Now, since ζ n n = 1, it must be the case that 1 = ψ(1) = ψ(ζ n n) = ψ(ζ n ) n, so ψ(ζ n ) is an nth root of unity; that is, ψ(ζ n ) = ζn k for some 1 k n. On the other hand, suppose (k, n) 1. Then n = kd for some 1 < d < n, and so (ζn) k d = ζn n = 1; that is, k is of order d < n. Since ζn d 1, it cannot be the case that ψ(ζ n ) = ζn. k Hence, the only possible images of ζ n under embeddings of Q(ζ n ) Q are ζn m for (m, n) = 1. Since all conjugates of ζ n must be roots Φ n (x), we see that Φ n (x) divides Ψ n (x). On the other hand, suppose (m, n) = 1 and we define the map ψ m : Q(ζ n ) Q(ζ n ) by 1 1 ζ n ζ m n and extend as usual. Then ψ m is certainly a homomorphism. On the other hand, since m (Z/nZ), m has a multiplicative inverse m (Z/nZ), which is also relatively prime to n. Then ψ m : Q(ζ n ) Q(ζ n ) given by 1 1 ζ n ζ m n is also a homomorphism. Now, for α = a 1 ζ n + + a n 1 ζn n 1 + a n Q(ζ n ), ψ m ψ m (α) = ψ m (ψ m (a 1 ζ n ) + + ψ m (a n )) = ψ m (a 1 ζ m n + + a n 1 ζ m(n 1) n + a n ) = ψ m (a 1 ζ m n ) + + ψ m (a n 1 ζ m(n 1) n ) + ψ m (a n ) = a 1 ζ mm n + + a n 1 ζ mm (n 1) n = a 1 ζ n + + a n 1 ζn n 1 = α, + a n + a n so ψ m ψ m id. A parallel computation shows that ψ m ψ m id, so we see that ψ m is an inverse of ψ m. Therefore, ψ m is an isomorphism and, thus, is an embedding Q(ζ n ) Q. Since ψ m (ζ n ) = ζn m, this implies that ζn m must be a root of Φ n (x). Since our choice of m < n such that (m, n) = 1 was arbitrary, we see that all such m must be roots of Φ n (x) and, thus, Ψ n (x) must divide Φ n (x).

4 CLAY SHONKWILER Since each divides the other and each is monic, we conclude that Φ n (x) = Ψ n (x). Clearly, the degree of Φ n (x) is φ(n). (b): Show that d n Φ d (x) = x n 1, and deduce that d n φ(d) = n. Proof. Since the roots of x n 1 are precisely the nth roots of unity, we see that, in Q, we can factor x n 1 as x n 1 = n (x ζn). j j=1 Thus, simply grouping appropriately and using the fact that Φ d = (x ζd m ), which we showed in (a) above, we see that 1 m d (m,d)=1 n x n 1 = ζn) j=1(x j = d n 1 m d (m,d)=1 (x ζ n d ) = d n Φ d (x). Obviously, deg(x n 1) = n. On the other hand, since x n 1 = Φ d (x), d n n = deg(x n 1) = deg Φ d (x) d n = d n deg Φ d (x) = d n φ(d) by the result proved above. (c): Let K n = Q(ζ n ). What is [K n : Q]? Answer: Since Φ n (x) is the minimal polynomial of ζ n, K n Q[x]/(Φ n (x)), which is a degree φ(n) extension of Q, since Φ n is irreducible of degree φ(n). Hence, [K n : Q] = φ(n). (d): Show that K n is Galois over Q. Proof. Since Q is characteristic 0, it suffices to show that K n is normal over Q. However, this is clear, because the roots of Φ n (x) are simply powers of ζ n, so Φ n (x) splits in K n ; that is, K n is the splitting field of Φ n (x) (and, in fact, of x n 1), so K n is normal over Q.

ALGEBRA HW 9 5 3 Let K be a field, and let x 1,..., x n be transcendentals over K. For i = 1,..., n, let s i be the ith symmetric polynomial in x 1,..., x n. (a): Show that K(x 1,..., x n ) is the splitting field over K(s 1,..., s n ) of the polynomial Z n s 1 Z n 1 + s 2 Z n 2 + ( 1) n s n. Proof. Consider the product n (Z x i ). i=1 Then, when we multiply out this product, the coefficient on x k will be ( 1) k s k, so n (Z x i ) = Z n s 1 Z n 1 + s 2 Z n 2 + ( 1) n s n. i=1 Hence, we see that the roots of this polynomial are precisely the x i, so K(x 1,..., x n ) is the splitting field over K(s 1,..., s n ) of this polynomial. (b): Show that the extension K(s 1,..., s n ) K(x 1,..., x n ) is Galois. Proof. Since the minimal polynomial of x i must divide Z n s 1 Z n 1 + s 2 Z n 2 + ( 1) n s n, which has all distinct roots, so the minimal polynomial of x i is seperable. Since this is true for all i = 1,..., n, we see that the extension K(s 1,..., s n ) K(x 1,..., x n ) is separable. On the other hand, since we saw in (a) that K(x 1,..., x n ) is the splitting field of the polynomial Z n s 1 Z n 1 +s 2 Z n 2 +( 1) n s n, this extension is also normal and, therefore, Galois. (c): Show that the Galois group is the symmetric group S n. Proof. First, note that any permutation σ S n corresponds to an automorphism φ σ of K(x 1,..., x n ) where φ σ (x i ) = x σ(i) and φ σ K = id K, since the x i are algebraically independent. Hence, if we can show that for each σ S n, φ σ K(s1,...,s n) is the identity map, then this will imply that there is a copy of S n contained in G. However, this is clear, because, since s i is symmetric for all i, φ σ (s i ) = s i for all i = 1,..., n and all σ S n. Thus, S n G. Note that this implies that #G #S n = n!. On the other hand, consider the notation L = K(s 1,..., s n ). Then [L(x 1 ) : L] n, since the minimal polynomial of x 1 must divide f(z) = Z n s 1 Z n 1 +s 2 Z n 2 +( 1) n s n = (Z x 1 ) n i=2 (Z x i). In turn, [L(x 1, x 2 ) : L(x 1 )] n 1, since the minimal polynomial of x 2 over L(x 1 ) divides f(z) Z x i, which has degree n 1. Iterating in

6 CLAY SHONKWILER this fashion, we see that [L(x 1,..., x k ) : L(x 1,..., x k 1 )] n k + 1 and, in turn, that [L(x 1,..., x n ) : L] = [L(x 1,..., x n ) : L(x 1,..., x n 1 ] [L(x 1 ) : L] 2 3 (n 1) n = n!. Since L(x 1,..., x n ) = K(x 1,..., x n ) and since this is a Galois extension of L, we know that #G = [K(x 1,..., x n )] n! Since we ve shown that n! #G n!, we conclude that, in fact, #G = n!. Since #S n = n! and S n G, this in turn implies that G S n. 4 Let L be a normal field extension of K, and let K 0 be the maximal purely inseparable extension of K contained in L. View L as contained in a fixed algebraic closure K of K. (a): Let β L, and let β 1,..., β n K be the distinct images of β under the K-embeddings L K. Let f(x) = (x β i ). Show that f(x) K 0 [x]. Proof. Since L is normal over K, every K-embedding L K is an automorphism of L, so it must be that case that all of the β i are in L. Hence, f(x) L[x]. Now, as in Problem 3(a) above, the coefficients of f(x) are just the elementary symmetric polynomials s 1,..., s n in β 1,..., β n. Furthermore, again as in 3(a), any K-automorphism of L simply permutes the β i, so the s i are fixed by any such automorphism. Thus, since s i L for each i = 1,..., n and every s i is fixed by every K-automorphism of L, we see that each s i is purely inseparable over K and, therefore, must be contained in the maximal purely inseparable extension of K contained in L. Therefore, we conclude that f(x) K 0 [x]. (b): In part (a), show that f(x) is the minimal polynomial of β over K 0. Proof. Since the identity map is certainly a K-automorphism of L, β = β i for some i. Hence, f(β) = 0, so the minimal polynomial of β over K 0 must divide f(x) K 0 [x]. Now, note that if ψ is a K-automorphism of L, then ψ K0 defines a K-embedding of K 0 into L K; since K 0 is purely inseparable, there is only one K- embedding K 0 K, so we see that ψ K0 id. Therefore, ψ is also a K 0 -automorphism of L. Therefore, every K-conjugate of β is also a K 0 -conjugate of β, i.e. all the β i are images of β under K 0 -embeddings L K 0 = K. Hence, since every conjugate of β must be a factor of its minimal polynomial over K 0, we see that the minimal polynomial of β must be divisible by f(x). Therefore, since

ALGEBRA HW 9 7 the minimal polynomial and f mutually divide eachother and both are monic, we conclude that f(x) is the minimal polynomial of β over K 0. (c): Conclude that L is separable over K 0. Proof. Since our choice of β in part (a) was arbitrary, we see that the minimal polynomial over K 0 of every element of L is separable, which implies that L is separable over K 0. 5 Let K = F p (t) and let L = K [ t ], where p is an odd prime number. (a): Find the maximal separable extension K of K in L, and the maximal purely inseparable extension K 0 of K in L. Answer: As we ve seen, the minimal polynomial of t over K is x t. Let α = t. Now, in L, we can factor this as ( x t = x p ) ( t x p + ) t = (x p α p ) (x p + α p ) = (x α) p (x + α) p ; since p 2, α α, so we see that there are two distinct roots of the minimal polynomial of α over K. Hence, there are two distinct embeddings L K, and so [L : K] sep = 2. Thus, if we can find an intermediate separable extension of degree 2, this must be the maximal separable extension K of K in L. Our choice, is clear, however: since t = α p L, consider K [ t ]. t has separable minimal polynomial ( x ) ( t x + ) t = x 2 t K[x], so K = K [ t ] is the maximal separable extension of K in L. On the other hand, since [L : K] = [L : K] sep [L : K] insep and [L : K] =, [L : K] sep = 2, we know that [L : K] insep = p. Now, p t has minimal polynomial x p t = ( x p t ) p over K, so we see that K 0 = K [ p t ] is purely inseparable. Since it is an extension of degree p, we conclude that it is the maximal purely inseparable extension of K in L. (b): Show explicitly in this example that L is the compositum of K and K 0, by expressing t as a combination of elements from K and K 0. Proof. Certainly, it suffices to show that we can express t as a combination of elements from K and K 0. Now, t K and ( p ) p 1 t 2 K 0, since p is odd and thus p 1 is divisible by 2. In

8 CLAY SHONKWILER 1 turn, this implies that K ( p t) p 1 0. Now, 2 t t ( p ) p 1 = p t ( ) p 1 = p t t 2 t 2 2 t = p p 1 t p 1 = t. Hence, we see that, indeed, t is the combination of elements of K and K 0, so L is the compositum of K and K 0. (c): What if instead p = 2? Answer: If p = 2, then α t = t, and so the minimal polynomial for α, x t = (x α) p (x + α p ) = (x α) has only a single root. Since every K-embedding of L into K must map α to a root of this polynomial, this implies that there is only a single such K-embedding, so L is purely inseperable over K. Hence, K = K, K 0 = L and it s trivially true that L is the compositum of K and K 0. DRL 3E3A, University of Pennsylvania E-mail address: shonkwil@math.upenn.edu