Spring 1993 Algebra Prelim Solutions

  1. We use the structure theorem for finitely generated modules over a PID. This tells us that M is a direct sum of modules of the form R/Rt where t is either zero or a power of an irreducible element of R. If this direct sum has more than one factor, then we may write M = A $ \oplus$ B for some nonzero submodules A, B of M, and then A $ \cap$ B = 0. Therefore there is exactly one factor in the direct sum, and the result follows.

    1. Since the action is transitive, all the stabilizers are conjugate and we see that the stabilizer of any element of X has no element of finite order other than 1. Thus if f $ \in$ G has finite order larger than 1, then f cannot be in the stabilizer of any element of X, and it follows that f cannot fix any element of X.

    2. Suppose now that f has prime order q. Then the orbits of < f > have order dividing q. Since q is prime, these orbits have order 1 or q. But if one of the orbits has order 1, then f fixes the element in that orbit which contradicts the above. Therefore all orbits of < f > have order q, and it follows that q divides | X| as required.

  2. Since S/(P1 $ \cap$ ... $ \cap$ Pt) is finite, we see that S/Pi is finite for all i. Now S/Pi is an integral domain because Pi is prime, and finite integral domains are fields. Therefore S/Pi is a field and we deduce that Pi is a maximal ideal for all i, as required.

  3. Suppose K is a nontrivial extension field of F with degree which is not a power of p. Note that if L is any finite extension field of K, then the degree of L over F is also not a power of p, because [L : F] = [L : K][K : F]. Since we are in characteristic zero, everything is separable so by taking a splitting field containing K, we may assume that K is a Galois extension of F. Let G = Gal(K/F) and let P be a Sylow p-subgroup of G. Then [KP : F] = [G : P] and [K : F] = | G|. Since [G : P] has order prime to p, it follows that [KP : F] has order prime to p and we conclude that KP is a nontrivial extension field of F with degree prime to p.

  4. This question depends on what we are allowed to assume; some people take the given property as the definition of projective module. Also the question does not require that R be commutative. Let us use the definition that an R module P is projective if and only if it is a direct summand of a free R-module. Suppose first that we have the given property. Choose an epimorphism f : F - > M where F is a free R-module. Since the map f* : Hom(M, F) - > Hom(M, M) is surjective, there exists an R-module map g : M - > F such that fg is the identity map on M. Then g is a monomorphism and so M $ \cong$ gM. Also F = kerf $ \oplus$ gM, which shows that gM and hence also M are projective.

    Conversely suppose M is a direct summand of a free module F. First we show that F satisfies the given condition. Let f : N - > N' be a surjection of R-modules and let h : F - > N' be any R-map. Let {ei | i $ \in$ I} be an R-basis for F, where I is some indexing set. Since h is surjective, we may choose ni $ \in$ N such that f (ni) = h(ei) for all i. Now we can define g $ \in$HomR(F, N) by g(ei) = ni for all i, and then fg(ei) = h(ei) for all i. Thus fg = h, and we have proved the result in the case M = F.

    For the general case, write F = M $ \oplus$ P as R-modules, and let y: M - > F be the natural monomorphism Then we have a commutative diagram
     HomR(F, N) f* - > HomR(F, N')
     y * $ \downarrow$   $ \downarrow$y*
     HomR(M, N) f* - > HomR(M, N')
    where (y *g)(m) = g(ym) for all m $ \in$ M and g $ \in$HomR(F, N) or HomR(F, N'). Note that the right hand (and also the left hand) y* is surjective: if h $ \in$HomR(M, N), we may extend h to an R-map F - > N by defining it to be 0 on P and then y*h = g. By the previous paragraph the top f* is surjective and we deduce that the bottom f* is surjective as required.

  5. The dihedral group has a cyclic subgroup C of index 2. Let c denote the character of the regular representation of C. Since C is abelian, we may write c=a 1 + ... +an for some integer n, where the ai are degree one characters of C. Let y denote the character of V. Since cD is the character of the regular representation of D, we see that <y,c D > $ \ne$ 0. Therefore <y,a iD > $ \ne$ 0 for some i. Since y is irreducible, we deduce that aiD =y+f for some character f of D. Therefore y has degree at most that of aiD. But ai has degree 1, consequently aiD has degree 2 and the result follows.

  6. Let G denote the Galois group of X10 - 1 over $ \mathbb {Q}$, and let w be a primitive 10th root of 1 (so w = epi/5). Then the roots of X10 - 1 are wr, where r = 0, 1,..., 9, which shows that the splitting field for X10 - 1 is $ \mathbb {Q}$[w]. Since X10 - 1 = (X5 - 1)(X + 1)(X5 + 1)/(X + 1) and w does not satisfy (X5 - 1)(X + 1), we see that w is a root of (X5 + 1)/(X + 1). By making the substitution Y = X + 1 and using Eisenstein's criterion for the prime 5, we see that (X5 + 1)/(X + 1) is irreducible over $ \mathbb {Q}$. This shows that [$ \mathbb {Q}$[w ] : $ \mathbb {Q}$] = 4 and we deduce that | G| = 4. Finally we can define q$ \in$ G by q(w) =w3, because w3 is also a primitive 5th root of 1, and since q2(w) =w9$ \ne$w, we deduce that q2$ \ne$1. We conclude that G has an element of order 4 and hence G is cyclic of order 4.

Peter Linnell