Algebra Prelim Solutions, Fall 2003

  1. Let G be a simple group of order 2256 = 16*3*47. The number of Sylow 47-subgroups is congruent to 1 mod 47 and divides 48. Since G is simple, the number of Sylow 47-subgroups is not 1 (otherwise G would have a normal Sylow 47-subgroup), consequently G has 48 Sylow 47-subgroups. Next the number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 16*47. This number cannot be 1 because G is simple. Nor can it be 4, because then G would be isomorphic to a subgroup of A4 which only has order 12. We conclude that G has at least 16 Sylow 3-subgroups. Finally G has at least two Sylow 2-subgroups. Note that two distinct Sylow 47-subgroups P, Q intersect trivially, because | P /\ Q| divides 47 and is not 47. Similarly two distinct Sylow 3-subgroups intersect trivially. On the other hand we cannot assert that two distinct Sylow 2-subgroups intersect trivially.

    We now count elements. The number of elements of order 47 is 46*48; the number of elements of order 3 is at least 2*16, and the number of elements of order a power of two is at least 17. We now have at least 17 + 2*16 + 46*48 = 2257 elements in G, which is too many. This proves that there is no simple group of order 2256.

  2. (i)
    Every element of G can be written in the form xa1yb1...xanybn, where ai, bi $ \in$ Z. Using x7 = y3 = 1, we may assume that ai, bi> 0 for all i. Now whenever there is a subword yx, we can replace it with xy2, because yxy-1 = x2, so we may push all the x's to the left and all the y's to the right. This proves the result.

    Using (i) and x7 = y3 = 1, we may write every element of G in the form xiyj where < i < 6 and < j < 2. It follows that | G< 7*3 = 21.

    This is because q respects the relations. Specifically, (1 3 4 5 6 7)7 = 1, ((2 3 5)(4 7 6))3 = 1,

    ((2 3 5)(4 7 6))(1 2 3 4 5 6 7)((2 3 5)(4 7 6))-1 = (1 3 5 7 2 4 6).

    We see from (iii) that imq has elements of order 7 and 3. Therefore 21 divides | imq| and hence 21 divides | G|. Since | G< 21 from (ii), we deduce that | G| = 21 as required.

  3. We may assume that a, b =/=  0. Let d be the greatest common divisor of a, b. We claim that a/d, b/d are coprime. If r divides a/d, b/d, then rd divides a, b and hence rd divides d, say srd = d for some s $ \in$ R. Therefore sr = 1, which means that r is a unit and the claim is established. By hypothesis (a/d )R + (b/d )R = xR for some x $ \in$ R. Then aR + bR = xdR and the result follows.

  4. Note that M @ Z/2Z as R-modules, so we need to prove Z/2Z $ \otimes_{R}^{}$ Z/2Z @ Z/2Z. Define q : Z/2Z $ \otimes_{R}^{}$ Z/2Z - > Z/2Z by q(x, y) = xy. It is easily verified that q is an R-balanced map and therefore q induces an R-module map q' : Z/2Z $ \otimes_{R}^{}$ Z/2Z - > Z/2Z satisfying q(x $ \otimes$ y) = xy. Now define an R-map f : Z/2Z - > Z/2Z $ \otimes_{R}^{}$ Z/2Z by fx = x $ \otimes$ 1. Then q'fx = q'(x $ \otimes$ 1) = x, so q'f is the identity. Also fq'(x $ \otimes$ y) = f(xy) = xy $ \otimes$ 1 = x $ \otimes$ 1. Therefore fq' is the identity on a set of generators of Z/2Z $ \otimes_{R}^{}$ Z/2Z (in this case these generators x $ \otimes$ y are the whole of Z/2Z $ \otimes_{R}^{}$ Z/2Z, but that won't always be the case) and we deduce that fq' is also the identity. This proves that Z/2Z $ \otimes_{R}^{}$ Z/2Z @ Z/2Z as required.

  5. Note that N is finitely generated. This is because M is finitely generated implies M $ \oplus$ M is finitely generated, which implies (M $ \oplus$ M)/(N $ \oplus$ 0) @ N is finitely generated.

    We now apply the structure theorem for finitely generated modules over a PID. We may write

    M  @ Ra $\displaystyle \oplus$ (R/q1R)a1 $\displaystyle \oplus$ ... $\displaystyle \oplus$ (R/qnR)an    
    N  @ Rb $\displaystyle \oplus$ (R/q1R)b1 $\displaystyle \oplus$ ... $\displaystyle \oplus$ (R/qnR)bn    

    where n, a, b, ai, bi are non-negative integers and the qi are nonassociate prime powers. Since M $ \oplus$ M @ N $ \oplus$ N, we see that

    R2a $\displaystyle \oplus$ (R/q1R)2a1 $\displaystyle \oplus$ ... $\displaystyle \oplus$ (R/qnR)2an @ R2b $\displaystyle \oplus$ (R/q1R)2b1 $\displaystyle \oplus$ ... $\displaystyle \oplus$ (R/qnR)2bn.

    By the uniqueness part of the structure theorem, we deduce that 2a = 2b, 2a1 = 2b1, ..., 2an = 2bn. Therefore a = b, a1 = b1, ..., an = bn and the result follows.

  6. (i)
    Note that f has degree 4. Since ai satisfies f, we have [K(ai) : K< 4. Therefore [K(ai,aj) : K< 4*4 = 16, in particular K(ai,aj) =/= L as required.

    Assume that f is monic. Then if a $ \in$ K is the coefficient of x3 of f, we have a1 + a2 + a3 + a4 = - a. Since L is the splitting field over K of the polynomial f and we are in characteristic zero, L is a Galois extension of K. Let G denote the Galois group of L over K. Then the subfields of L containing K are in a one-to-one correspondence with the subgroups of G, and G permutes the ai faithfully. If K(a1 + 2a2 + 3a3) =/= L, then K(a1 + 2a2 + 3a3) corresponds to a nontrivial subgroup of G, in particular a1 + 2a2 + 3a3 is fixed by some nontrivial permutation of the ai. In other words there is 1 =/= s $ \in$ S4 such that

    a1 + 2a2 + 3a3 = as1 + 2as2 + 3as3.

    By inspection, this will yield a different linear relation from a1 + a2 + a3 + a4 = - a from above. (Clearly this relation is not 0 = 0. If we subtract the right hand side from the left hand side, then the sum of the coefficients of the ai is 0.) Therefore we may express the ai in terms of just two of the ai; in other words, L = K(ai,aj) for some i, j, which contradicts (i). Finally K[a1 + 2a2 + 3a3] = K(a1 + 2a2 + 3a3) because a1 + 2a2 + 3a3 is algebraic over K and the result follows.

  7. If the assertion is false, then I(V1) + I(V2) is contained in a maximal ideal M of k[x1,..., xn]. By Hilbert's Nullstellensatz, M is of the form (x1 - a1,..., xn - an), where ai $ \in$ k for all i. Since f (a1,..., an) = 0 for all f $ \in$ M, we see that f (a1,..., an) = 0 for all f $ \in$ I(V1) and f $ \in$ I(V2). Therefore (a1,..., an) $ \in$ V1 /\ V2 which is a contradiction, and the result follows.

Peter Linnell