Spring 1995 Algebra Prelim Solutions

1.

Let G be a group of order 1127 = 7²*23. The number of Sylow 23-subgroups divides 49 and is congruent to 1 modulo 23. This means that G has exactly one Sylow 23-subgroup and therefore G has a normal Sylow 23-subgroup A. Also the number of Sylow 7-subgroups divides 23 and is congruent to 1 modulo 23. Therefore the number of Sylow 7-subgroups is 1 and we deduce that G has a normal Sylow 7-subgroup B. Since G has normal subgroups A, B such that A $\cap$ B = 1 and | G| = | A|| B|, we see that G $\cong$ A X B. Now groups of prime order p are isomorphic the cyclic group $\mathbb {Z}$/p$\mathbb {Z}$ , while groups of order p² are either isomorphic to $\mathbb {Z}$/p²$\mathbb {Z}$ or $\mathbb {Z}$/p$\mathbb {Z}$ X $\mathbb {Z}$/p$\mathbb {Z}$ . Therefore G is isomorphic to either $\mathbb {Z}$/49$\mathbb {Z}$ X $\mathbb {Z}$/23$\mathbb {Z}$ or $\mathbb {Z}$/7$\mathbb {Z}$ X $\mathbb {Z}$/7$\mathbb {Z}$ X $\mathbb {Z}$/23$\mathbb {Z}$ . In particular G is abelian and by the fundamental structure theorem for finitely generated abelian groups, these last two groups are not isomorphic. Therefore up to isomorphism there are two groups of order 1127, namely $\mathbb {Z}$/49$\mathbb {Z}$ X $\mathbb {Z}$/23$\mathbb {Z}$ and $\mathbb {Z}$/7$\mathbb {Z}$ X $\mathbb {Z}$/7$\mathbb {Z}$ X $\mathbb {Z}$/23$\mathbb {Z}$ .

2.

We shall prove the result by induction on | G|, the result being obviously true if | G| = 1. Also if G is abelian, then there is nothing to prove, so we may assume that G is not abelian. Since G is nilpotent and not 1, its center Z is not 1. By induction the result is true for G/Z. Note that if G/Z is cyclic, then G is abelian which is not the case. Therefore G/Z is noncyclic. By induction, G/Z has a normal subgroup H/Z such that (G/Z)/(H/Z) is a noncyclic abelian group. But (G/Z)/(H/Z) $\cong$ G/H and the result follows.

3.

Let G be a finitely generated abelian group with the given property. Then by the structure theorem, G is isomorphic to a direct product of nontrivial groups A₁, A₂,..., A_n of prime power order. If n > 1, then A₁ $\nsubseteq$ A₂ and A₂ $\nsubseteq$ A₁. Therefore n$\le$1. This means that G is cyclic of prime power order. Conversely if G is cyclic of prime power order, it has the given property, because then G has exactly one subgroup of each order dividing | G| and it follows that G has the property as stated in the problem. We conclude that the finitely generated abelian groups with the property that for all subgroups A, B, either A $\subseteq$ B or B $\subseteq$ A are the cyclic groups of prime power order.

4.

(a)

Let x $\in$ R/rad I and suppose xⁿ = 0 where n > 0. Then we may write x = y +radl I where y $\in$ R. Since xⁿ = 0, we see that yⁿ +rad I =rad I, which means that yⁿ $\in$ rad I. By definition of rad I we see that (yⁿ)^m = 0. Therefore y^mn = 0, hence y $\in$ rad I and we deduce that x = 0. This establishes that R/rad I has no nonzero nilpotent elements.

(b)

If x $\in$ P₁ $\cap$ P₂ $\cap$ ... $\cap$ P_n, then x $\in$ P_i for all i and hence xⁿ $\in$ P₁P₂... P_n. It follows that x $\in$ rad (P₁P₂... P_n). Conversely suppose x $\in$ rad (P₁P₂... P_n). Then x $\in$ rad P_i for all i. This means that x^m $\in$ P_i for some m > 0 and since P_i is prime, we deduce that x $\in$ P_i for all i as required.

(c)

If P_i is contained in every P_j, then P₁ $\cap$ ... $\cap$ P_n = P_i and hence R/rad (P₁... P_n) = R/P_i by (b). We deduce that R/rad (P₁... P_n) is an integral domain.

Conversely suppose R/rad (P₁... P_n) is an integral domain. Then by (b) we see that R/(P₁ $\cap$ ... $\cap$ P_n) is also an integral domain. Suppose there does not exist an i such that P_i is contained in P_j for all j. Then for each i, we can choose x_i $\in$ P_i such that x_i$\notin$P_j for some j (where j depends on i). Now set y_i = x_i + P₁ $\cap$ ... $\cap$ P_n for i = 1,..., n. Then y_i is a nonzero element of R/(P₁ $\cap$ ... $\cap$ P_n) for all i, yet

y₁... y_n = x₁... x_n + P₁ $$\cap$$ ... $$\cap$$ P_n = 0

This shows that R/(P₁ $\cap$ ... $\cap$ P_n) is not an integral domain and we have a contradiction. This completes the proof.

5.

Obviously K($\alpha^{3}_{}$) $\subseteq$ K($\alpha$). Now

8 = [K($$\alpha$$) : K] = [K($$\alpha$$) : K($$\alpha^{3}_{}$$)][K($$\alpha^{3}_{}$$) : K]

which shows that [K($\alpha$) : K($\alpha^{3}_{}$)] divides 8. Also $\alpha$ satisfies the polynomial X³ - $\alpha^{3}_{}$ which shows that [K($\alpha$) : K($\alpha^{3}_{}$)]$\le$3. Therefore [K($\alpha$) : K($\alpha^{3}_{}$)] = 1 or 2. We need to eliminate the possibility that [K($\alpha$) : K($\alpha^{3}_{}$)] = 2. If [K($\alpha$) : K($\alpha^{3}_{}$)] = 2, then the polynomial X³ - $\alpha^{3}_{}$ could not be irreducible over K($\alpha^{3}_{}$), and it would follow that X³ - $\alpha^{3}_{}$ has a root in K($\alpha^{3}_{}$). But the roots of X³ - $\alpha^{3}_{}$ are $\alpha$,$\omega$$\alpha$ and $\omega^{2}_{}$$\alpha$ and since $\omega$ $\in$ K, it would follow that all the roots of X³ - $\alpha^{3}_{}$ are in K. In particular $\alpha$ $\in$ K($\alpha^{3}_{}$). This establishes the result.

6.

(a)

Let T denote the ideals of R which have trivial intersection with S. Since a is not nilpotent, we see that 0$\notin$S and hence 0 $\in$ T. Therefore T is nonempty. Moreover T is ordered by inclusion, and the union of a chain in T is still in T. It now follows from Zorn's lemma that T has maximal elements; let P be one of these maximal elements. Then P $\cap$ S = $\emptyset$ . We claim that P is prime. If P is not a prime ideal, then there exist ideals A, B strictly containing P such that AB $\subseteq$ P. By maximality of P we have aⁱ $\in$ A and a^j $\in$ B for some i, j and hence a^{i + j} $\in$ P. This contradicts the fact that P $\in$ T, and it follows that P is a prime ideal not containing a.

(b)

Let $\theta$ : R$\to$K denote the composition of the natural epimorphism R$\to$R/P followed by the natural monomorphism R/P$\to$K. If b $\in$ S, then b$\notin$P, hence the image of b in R/P is nonzero and we deduce that $\theta$b is invertible in K. It follows that $\theta$ extends to a ring homomorphism $\phi$ : S^-1R$\to$K.

7.

(a)

The proper subfields of F containing K are in a one-one correspondence with the proper subgroups of Gal(F/K). Therefore we need to show that S₄ has at least 9 proper subgroups. There are 6 elements of order 2 and 8 elements of order 3 in S₄. Since any two subgroups of order 2 or 3 intersect in the identity, we see that there are 6 subgroups of order 2 and 4 subgroups of order 3, and we have shown that Gal(F/K) has at least 10 proper subgroups. This finishes part (a).

(b)

The Galois extensions E of K in F correspond to the normal subgroups of Gal(F/K), so we need a nontrivial normal subgroup of Gal(F/K). The simplest one is the alternating subgroup A₄ of S₄. The corresponding subfield E of K is the elements of F fixed by A₄. Also Gal (E/K) $\cong$ S₄/A₄ $\cong$ $\mathbb {Z}$/2$\mathbb {Z}$ .

8.

First we find the Jordan canonical form of the matrix

$$\Biggl[ \Biggr]$$

The characteristic equation of this matrix is -x(3 - x) + 2 = 0, which has roots 1 and 2. Therefore the Jordan canonical form of this matrix is

$$\Biggl[ \Biggr]$$

and we deduce that the Jordan canonical form of A is

$$\Biggl[ \Biggr]$$

The matrices which commute with this canonical form are the matrices of the form

$$\Biggl[ \Biggr]$$

where p, a, b, c, d are arbitrary complex numbers.