January 1999 Algebra Prelim Solutions

1. Let 0$\ne$a $\in$ R. We must prove that a is invertible, so suppose to the contrary that a is not invertible. Then a²R is an ideal of R and since a is not invertible, we see that aR$\ne$R and consequently a²R$\ne$R. By hypothesis a²R is a prime ideal of R and since a² $\in$ a²R, we deduce that a $\in$ a²R. Therefore a = a²r for some r $\in$ R. Since 0 is a prime ideal of R, we see that R is an integral domain and we deduce that 1 = ar. Thus a is invertible and we have a contradiction. This completes the proof.

2. By hypothesis G has a normal subgroup of K of order p³. Then G/K is a group of order q³. A nontrivial q-group (where q is a prime) has a normal subgroup of order q, so G/K has a normal subgroup H/K of order q. Then H is a normal subgroup of order p³q, as required.

3. Suppose R has an element a which is neither a zero divisor nor a unit. Then aR is a proper submodule of R, because a is not a unit. Also the map r | - > ar is an R-map from R onto aR which has kernel 0, because a is a nonzero divisor. This shows that R is isomorphic to the proper R-submodule aR.

   Conversely suppose R is isomorphic to the proper R-sumodule M. Then there is an R-isomorphism q : R - > M. Set a = q1. Then aR = (q1)R = q(1R) = qR = M, so aR$\ne$R and we see that a is not a unit. Finally if ar = 0, then qr = q(1r) = (q1)r = ar = 0 and we deduce that r = 0, because q is an isomorphism. Therefore r is a nonzero divisor and the result follows.

4. Since a satisfies X² - a² $\in$ K(a²)[X], we see that [K(a) : K(a²)] = 1 or 2. Also [K(a) : K] = [K(a) : K(a²)][K(a²) : K]. Since [K(a) : K] is odd, we deduce that [K(a) : K(a²)] = 1 and the result follows.

5. By the fundamental structure theorem for finitely generated abelian groups, we know that G is a direct product of nontrivial cyclic p-groups. Since {x $\in$ G | x^p = 1} has order p², we see that G is a direct product of exactly two nontrivial cyclic p-groups. It now follows that G $\cong$ $\mathbb {Z}$/p⁵$\mathbb {Z}$ X $\mathbb {Z}$/p$\mathbb {Z}$ or $\mathbb {Z}$/p⁴$\mathbb {Z}$ X $\mathbb {Z}$/p²$\mathbb {Z}$ or $\mathbb {Z}$/p³$\mathbb {Z}$ X $\mathbb {Z}$/p³$\mathbb {Z}$ (so there are three possible groups up to isomorphism).

6. Since K is a splitting over k, it can be written as k(a₁,..., a_n) where a₁,..., a_n are all the roots of some polynomial f $\in$ k[X]. If s $\in$ Gal(L/k), then sa_i also satisfies f, because s fixes all the coefficients of f, and so s permutes the a_i. It follows that sK = k(sa₁,...,sa_n) = K.

7. Let G be a simple group of order 280. The number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 40, so there are 1 or 8 Sylow 7-subgroups. There cannot be 1 Sylow 7-subgroup, because then the Sylow 7-subgroup would be normal which contradicts the hypothesis that G is simple. Therefore there are 8 Sylow 7-subgroups. Since two distinct Sylow 7-subgroups must have trivial intersection, we see that there are at least 8*6 = 48 elements of order 7. The number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 56. There cannot be 1 Sylow 5-subgroup, for then it would be normal which would contradict the hypothesis that G is simple. Therefore there are 56 Sylow 5-subgroups. Since two distinct Sylow 5-subgroups must intersect in the identity, we see that there are at least 56*4 = 224 elements of order 5. Finally since the Sylow 2-subgroup is not normal, there must be at least 9 elements whose order is a power of 2. We now count elements: we find that G has at least 48 + 224 + 9 = 281 elements, which is impossible because G has only 280 elements. Therefore no such G can exist and we deduce that there is no simple group of order 280.

8. Let K be a splitting field over F which contains E, let G = Gal(K/F), and let H = Gal(K/E). Since we are in characteristic zero, everything is separable and hence K is a Galois extension of F. Therefore by the fundamental theorem of Galois theory, we see that the number of fields between F and E is equal to the number of subgroups between G and H. Also [G : H] = n. By considering the permutation representation of G on the left cosets of H in G, we see that there is a normal subgroup N of G contained in H such that | G/N|$\le$n!. The number of subgroups between G and H is at most the number of subgroups between G and N, which is at most the number of subsets of G/N. Since the number of subsets of G/N is 2^{| G/N|}, we deduce that the number of subfields between F and E is at most 2^n!, as required.