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1 Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups Problem Problem Problem Rings Problem Problem Problem Fields Problem Problem Problem

2 1 Groups 1.1 Problem 1 Suppose G is a finite group whose automorphism group Aut(G) has prime order. Then Aut(G) is cyclic, and so every subgroup of Aut(G) is cyclic. In particular, the inner automorphism group Inn(G) := {Φ g Aut(G) Φ g (x) = gxg 1, g G} Aut(G) is cyclic. Since G/Z(G) = Inn(G), and G/Z(G) is cyclic if and only if G is abelian, it follows that G must be abelian. By the fundamental theorem of finitely generated abelian groups, G = Z/p k 1 1 Z Z/p kn n Z for some n N, primes p 1,..., p n and positive integers k 1,..., k n. There is an injective morphism Aut(Z/p k 1 1 Z) Aut(Z/p kn n Z) Aut(Z/p k 1 1 Z Z/p kn n Z) = Aut(G) sending (f 1,..., f n ) to the product morphism f 1 f n, so H := Aut(Z/p k 1 1 Z) Aut(Z/p kn n Z) is a subgroup of Aut(G); in particular, H divides Aut(G). Letting ϕ denote the Euler totient function, Aut(Z/p k i i Z) = ϕ(pk i i ) = pk i 1 i (p i 1), so H = n i=1 p k i 1 i (p i 1) If any p i > 5, then p i 1 is necessarily composite, so we must have p i {2, 3}. If p i = 3 and k i > 1, then 6 p k i 1 i (p i 1), so we must have k i = 1 if p i = 3. Similarly, if p i = 2, and k i > 2, then 4 H, so we must have k i {1, 2} if p i = 2. If G contains any of Z/3Z Z/3Z, Z/3Z Z/4Z, Z/4Z Z/4Z, then 4 H, so G must be of one of the following forms: (Z/2Z) n, (Z/2Z) n Z/3Z, (Z/2Z) n Z/4Z where n 0. If n > 1, then Aut(G) contains Aut(Z/2Z Z/2Z) = GL 2 (F 2 ) = S 3, whence Aut(G) is not prime. Thus, the only possibilities are Z/2Z, Z/3Z, Z/2Z Z/3Z, Z/4Z, Z/2Z Z/4Z It is straightforward to see that if G is any of Z/3Z, Z/4Z and Z/6Z, then Aut(G) = 2. If G = Z/2Z, then Aut(G) is trivial, and hence is not of prime order. The only case that remains is G = Z/2Z Z/4Z. Since Aut(Z/2Z) Aut(Z/4Z) is a subgroup of Aut(G), it suffices to show that Aut(G) > 2, in which case Aut(G) is not prime, since 2 Aut(G). Indeed, it is easy to check that the assignments (1, 0) (1, 2), (0, 1) (0, 1) on generators define an automorphism of Z/2Z Z/4Z which is not in the image of Aut(Z/2Z) Aut(Z/4Z). Hence, if Aut(G) is prime, then G is isomorphic to one of 1.2 Problem 2 Z/3Z, Z/4Z, Z/6Z Let G be a finite group, and let H be a non-normal subgroup of G of index n > 1. 2

3 (a) Suppose that H is divisible by a prime p n. Note G acts by left multiplication on the set G/H of cosets of H in G; since [G : H] = n, note G/H = n. One can view this action as a homomorphism ϕ: G S(G/H) = S n, where ϕ(g )(gh) = g gh for any g G, gh G/H. Note that ϕ(g)(h) = gh = H if and only if g H, so K := ker(ϕ) is a subgroup of H. Since K is normal in G, it must be normal in H as well. As H is not a normal subgroup of G, K must be a proper subgroup of H. If K is nontrivial, then H is not simple, and we re done. If K is trivial, then ϕ is injective, so G is isomorphic to a subgroup of S n and thus G divides S(G/H) = S n = n! by Lagrange. But G = [G : H] H = n H which is divisble by np, so n! is divisible by np, a contradiction. (b) Apparently, this is false. 1.3 Problem 3 (a) Let G = x be cyclic group and H a subgroup of G. Since every element of H is a positive power of x, there exists minimal positive k N such that x k H. We claim H = x k. Indeed, for any n k N such that x n H, apply Euclidean division to write n = kq + r for some k N and 0 r < k. Then x n (x k ) q = x n kq = x r H, and since r < k, the minimality of k implies r = 0, i.e. k n. (b) The same counting methods in S06.G2(a) show that GL 2 (Z/pZ) = (p 2 1)(p 2 p) elements and hence SL 2 (Z/pZ) = GL 2(Z/pZ) (Z/pZ) = (p 1)2 p(p + 1) p 1 = (p 1)(p)(p + 1) (c) Let l be an odd prime. By the second Sylow theorem, any two Sylow l-subgroups of SL 2 (Z/pZ) are conjugate, hence isomorphic, so it suffices to exhibit a Sylow l-subgroup which is cyclic for each odd prime l. Note that p, p 1, p + 1 have no odd prime factors in common, so if l q divides SL 2 (Z/pZ) = (p 1)p(p + 1), then l q divides one of p 1, p or p+1. Since the subgroups of a cyclic group are bijective with the divisors of the order of that cyclic group, it is enough to find cyclic subgroups of SL 2 (Z/pZ) of order p 1, p and p + 1, as this will exhibit the required cyclic Sylow l-subgroup for all possible l. Note that (Z/pZ) is a cyclic group of order p 1 which embeds diagonally into SL 2 (Z/pZ), so this takes care of the case p 1. Any Sylow p-subgroup has order p and thus must be cyclic. The case of p + 1 is the trickiest. Let l q be the largest power of l dividing p + 1, and note that GL 2 (Z/pZ) = (p 1) 2 p(p + 1), so l q must also be the largest power of l dividing GL 2 (Z/pZ), i.e. any subgroup of order l q of GL 2 (Z/pZ) is a Sylow l- subgroup of GL 2 (Z/pZ). Let F = F p 2 be the finite field with p elements; recall F is a 2-dimensional vector space over Z/pZ. Then F is cyclic (see F06.F1 below) of order p 2 1 = (p 1)(p + 1). For any fixed α F, the left multiplication map L α : F F sending x F to αx is a vector space isomorphism, so after fixing a basis B for F over Z/pZ, F can be embedded in GL 2 (Z/pZ) by sending α F to the invertible matrix representing L α with respect to B. Since SL 2 (Z/pZ) is the kernel of 3

4 det: GL 2 (Z/pZ) (Z/pZ), it is a normal subgroup of GL 2 (Z/pZ). The subgroup of GL 2 (Z/pZ) isomorphic to F has a cyclic subgroup P of order l q, which as remarked above must be a Sylow l-subgroup of GL 2 (Z/pZ). By F05.G2, P SL 2 (Z/pZ) is a Sylow l-subgroup of SL 2 (Z/pZ) which must also be cyclic, so we re done. 2 Rings 2.1 Problem 1 Let m be a prime ideal of Z[X], and let ϕ: Z Z[X] be the inclusion map. The preimage of any prime ideal under a ring homomorphism is a prime ideal, so ϕ 1 (m) = Z m is a prime ideal of Z, whence Z m = {0} or Z m = pz for some prime p Z. Suppose Z m = pz. Then p m, so pz[x] m, whence p, m = m. In particular, Z[X]/m = Z[X]/ p, m = (Z[X]/pZ[X])/m = (Z/pZ)[X]/m where m is the image of the ideal m under the projection π : Z[X] Z[X]/pZ[X] (this is indeed an ideal since π is surjective). Since m is a prime ideal, Z[X]/m is a domain, so m must also be a prime ideal of (Z/pZ)[X]. As Z/pZ is a field, Z/pZ[X] is a PID, and any nonzero prime ideal of a PID is generated by an irreducible element. Hence, either m = {0}, in which case m = pz[x], or m is principal, generated by a (monic) irreducible polynomial g(x) (Z/pZ)[X]. In the latter case, let f(x) m such that π(f(x)) = g(x). Note f must exist, since m is the image of m under π. We may also assume f is monic, as pz[x] m and the leading coefficient of f is congruent to 1 mod p since g(x) is monic. Further, f(x) is irreducible over Z[X], or else g(x) would admit a nontrivial factorization over (Z/pZ)[X] by reducing the given factorization of f(x) mod p. We claim m = p, f(x). Indeed, the inclusion p, f(x) m is trivial by the choice of f. Let q(x) m; then π(q(x)) = j(x)g(x) for some j(x) (Z/pZ)[X]. Taking l(x) Z[X] such that π(l(x)) = j(x), it follows then that π(q(x) l(x)f(x)) = π(q(x)) j(x)g(x) = 0 (Z/pZ)[X], i.e. there exists s(x) Z[X] such that q(x) l(x)f(x) = p s(x). Hence, q(x) = f(x) l(x) + p s(x) p, f(x), so m = p, f(x). Now suppose Z m = {0}. Since Z[X] is a domain, {0} is a prime ideal of Z[X], so suppose m {0} and let g(x) m nonzero. Since Z[X] is a UFD, g(x) can be uniquely factored into a product of irreducibles. Using the primality of m inductively on this factorization implies that m contains some irreducible factor f(x) of g(x). As m Z = {0}, we may further assume that f is non-constant, and since f is irreducible, the content of f must be 1. We claim that m = f(x). Let 0 h(x) m. Note that f(x) divides h(x) over Z[X] if and only if f(x) divides h(x) over Q[X]. Indeed, content is multiplicative by Gauss lemma, and so f(x)q(x) = h(x) for some q(x) Q[X] implies cont(f)cont(g) = cont(h). Since cont(h) Z and cont(f) = 1, this implies cont(g) = cont(h) Z. Since cont(k(x)) Z implies k(x) Z[X] for any k Q[X], this proves the claim. Suppose, then, that f(x) does not divide h(x) over Q[X]. Since Q[X] is a PID and f(x) is irreducible and h(x) is nonzero, this means that f, h generate the unit ideal in Q[X], i.e. there exist p(x), r(x) Q[X] such that 4

5 f(x)p(x) + h(x)r(x) = 1. But clearing the denominators of p(x), r(x) by multiplying by a sufficiently large integer n > 0 implies nf(x)p(x) + nh(x)r(x) = n m, a contradiction since m Z = {0} by hypothesis. Thus, f(x) divides h(x) over Q[X], so f(x) divides h(x) over Z[X] as argued above, whence m = f(x). Hence, the prime ideals of Z[X] are of the following form: (1) {0} (2) pz[x] for some prime p Z (3) f(x) for some irreducible polynomial f(x) Z[X] (4) p, f(x) for some prime p Z and irreducible polynomial f(x) Z[X] such that π(f) (Z/pZ)[X] is also irreducible 2.2 Problem 2 Let R be a Noetherian domain. (a) Let x R be a nonzero non-unit. Recall that y R divides x if and only if y x. Suppose x is not irreducible; then we may write x = yz for y, z R nonzero non-units, whence x y. Since R is a domain, this containment is proper. Indeed, if y x, then y = vx for some v R, so x = zvx. Hence, (1 zv)x = 0, and as x 0, this implies 1 zv = 0, i.e z is a unit, a contradiction. If y and z are both irreducible elements of R, then we are done; if not, suppose that (WLOG) y is reducible. Then we can similarly write y as a product of nonzero non-units wv, extending the chain of ideals x y w. Continuing inductively, this division process must terminate in finitely many steps for any divisor of x, as the chain of ideals thus furnished must stabilize since R is Noetherian. This proves the claim. (b) Let Ω be the set of nonzero ideals I of R such that I does not contain a product of nonzero prime ideals. If Ω is nonempty, then it contains a maximal element since R is Noetherian; denote this element A. A cannot be prime by hypothesis, so there exist x, y R such that xy A, but x, y / A. Then (A + x )(A + y ) = A + xa + ya + xy A. Since x, y / A, A is properly contained in both A x := A + x and A y := A + y. By the maximality of A, then A x, A y / Ω, so both A x and A y contain a finite product of nonzero prime ideals. The total (finite) product of these nonzero prime ideals is then contained in A x A y A, a contradiction, so Ω must be empty. (c) Let x R be an irreducible element, and let I = x. By part (b), I contains a finite product of nonzero prime ideals P 1,..., P k, and by hypothesis each P i contains a principal prime ideal p i for some p i R. Then in particular, p 1 p 2 p k P 1 P 2 P k I, so p 1 p 2 p k I. Then p 1 p 2 p k = rx for some r R, whence rx p 1, so x p 1 or r p 1 by the primality of p 1. In the latter case, this implies x = sp 1 for some s R. By the irreducibility of R and primality of p 1, this must mean s R, so I = p 1, which proves the claim. Suppose that r p 1, i.e. r = sp 1 for some s R. Since R is a domain, cancellation implies p 2 p 3 p k = sx. If we continue this process, then either 5

6 I = p i for some 1 < i < n, or we eventually obtain p k = lx for some l R. If x / p k, then l p k, so l = qp k for some q R. Then cancellation once more implies xq = 1, i.e. x is a unit, a contradiction, so x p k. Applying the same reasoning as above, x = p k, so x is prime. 2.3 Problem 3 Let R be a commutative ring, and let M be a finitely generated R-module. Suppose there exists a positive integer n and a surjective R-module homomorphism ϕ: M R n. Then there is an exact sequence 0 ker(ϕ) M ϕ R n 0 which is split, because R n is a free R-module, hence projective. Thus, M = R n ker(ϕ). Since M is finitely generated, there exists a positive integer k and a surjective R-module homomorphism R k M. Hence, there is a sequence of surjective R-module homomorphisms R k M π ker(ϕ) where π is the canonical projection M = R ker(ϕ) ker(ϕ). Hence, ker(ϕ) is finitely generated by k elements, namely the images of the k generators of R k under this sequence of morphisms. 3 Fields 3.1 Problem 1 Let F be a finite field of positive characteristic p; recall, then, that F = p k for some k N. To show that F is cyclic, two lemmas are useful here: Lemma Let ϕ denote the Euler totient function. For any n N, d n ϕ(d) = n. Proof. Since Z/nZ has exactly one (cyclic) subgroup of order d for each d n, and there are ϕ(d) generators of this cyclic subgroup, it follows that Z/nZ has ϕ(d) elements of order d for each d dividing n. Then d n ϕ(d) = Z/nZ = n. Lemma Suppose G is a finite group such that there are at most d elements in G of order dividing d for each d N +. Then G is cyclic. Proof. Let G := n, and let ρ(d) denote the number of elements of order d in G. By Lagrange s theorem, the order of any element of G must divide n, so ρ(d) = 0 if d does not divide n. We therefore have n = d n ρ(d) = d n ρ(d). If ρ(d) is nonzero, then G has at least one element of order d, which generates a cyclic subgroup C d of G of order d. Since all d elements of C d have order dividing d, these must be all elements of order dividing d in G. In particular, the ϕ(d) generators of C d constitute all elements of order d in G, so ρ(d) = ϕ(d) or ρ(d) = 0. By the lemma above, n = d n ρ(d) d n ϕ(d) = n 6

7 so ρ(d) = ϕ(d) for every d n. In particular, ρ(n) = ϕ(n) > 0, so there is at least one element of G with order n, whence G is cyclic. Corollary Any finite multiplicative subgroup of a field F is cyclic. In particular, if F is finite, then F is cyclic. Proof. The polynomial X d 1 F [X] has at most d roots, so any finite multiplicative subgroup of F meets the hypotheses of the lemma above. The argument above is due to J.P. Serre: see A Course in Arithmetic for more details. Now, by Lagrange s theorem, every element of F has multiplicative order dividing F = p k 1, so every nonzero element of F is a root of X pk 1 1 F [X], whence every element of F is a root of f(x) = X pk X F [X]. Since f has at most p k = F roots over F, every element of F is a root of f of multiplicity of one, and these are all roots of f. Hence, F is a splitting field for the separable polynomial f over Z/pZ, so F is a Galois extension of Z/pZ. 3.2 Problem 2 See S04.F1, part (c). 3.3 Problem 3 Let f(x) be an irreducible polynomial over F, and K/F a normal extension. Let L/K/F be the Galois closure of K over F, and let G := Gal(L/F ). Suppose f(x) = f 1 (x) f n (x) K[x], where f i is irreducible over K for each i = 1,..., n. Let α be a root of f 1, and β be a root of f i for some fixed i 1. Since G acts transitively on the roots of f, there exists σ G such that σ(α) = β. Since K/F is normal, σ K is an automorphism of K, so σ(f 1 (x)) K[x] is an irreducible polynomial having σ(α) = β as a root. Hence, f i (x) divides σ(f 1 (x)) over K[x], but both are irreducible, and so must be equal. Then deg(f i (x)) = deg(σ(f 1 (x))) = deg(f 1 (x)), so every irreducible factor of f(x) has the same degree, namely deg(f 1 (x)) = [K(α) : K]. As an easy corollary, this shows that if f has a root in K, then f splits over K, since f has an irreducible factor of degree 1 over K. 7

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