Fall 1995 Algebra Prelim Solutions

1.

The order of G is 5³ $\cdot$ 7³. The number of Sylow 5-subgroups of G divides 7³ and is congruent to 1 modulo 5; the only possibility is 1. Therefore G has a normal Sylow 5-subgroup A. The number of Sylow 7-subgroups of G divides 5³ and is congruent to 1 modulo 7; the only possibility is 1. Therefore G has a normal Sylow 7-subgroup B. Since (| A|,| B|) = 1, we see that A $\cap$ B = 1. We next show that every element of A commutes with every element of B. Suppose a $\in$ A and b $\in$ B. Then aba^-1b^-1 = a(ba^-1b^-1) and since A $\lhd$ G, we see that ba^-1b^-1 $\in$ A and consequently aba^-1b^-1 $\in$ A. Similarly aba^-1b^-1 $\in$ B and we deduce that aba^-1b^-1 $\in$ A $\cap$ B = 1. Therefore aba^-1b^-1 = 1 and we conclude that ab = ba, in other words every element of A commutes with every element of B.

Since a group of prime power order has normal subgroups of order m for all m dividing the order of the group, we see that A has a normal subgroup H of order 25. From the previous paragraph, B centralizes H and so certainly normalizes H. Thus A and B normalize H, hence | A| and | B| divide the order of the normalizer of H in G and we conclude that H $\lhd$ G. This completes the solution.

2.

First we write G as a direct product of cyclic groups of prime power order: G $\cong$ $\mathbb {Z}$/4$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/9$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/9$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/5$\mathbb {Z}$ . Any subgroup of G is isomorphic to a product of subgroups, where one subgroup is taken from each factor. Thus H $\cong$ $\mathbb {Z}$/9$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/3$\mathbb {Z}$ or $\mathbb {Z}$/3$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/9$\mathbb {Z}$ . These last two groups are isomorphic, so we conclude that H $\cong$ $\mathbb {Z}$/3$\mathbb {Z}$$\mbox{$\times$}$$\mathbb {Z}$/9$\mathbb {Z}$ .

3.

(a)

Since D is a conjugacy class in f^-1(C), we may write D = {bdb^-1 | b $\in$ f^-1(C)} for some fixed d $\in$ D. Then

f (D) = {f (bdb^-1) = f (b)f (d )f (b)^-1 | b $$\in$$ f^-1(C)}.

Therefore f (D) = {cf (d )c^-1 | c $\in$ C} , and (a) follows.

(b)

Let D(g) denote the conjugacy class of g in f^-1(C). Since f (g) is centralized by C, the conjugacy class containing f (g) is precisely {f (g)} . Since f (D(g)) is by (a) the conjugacy class containing f (g), we see that | f (D(g))| = 1. Therefore all elements of D(g) are in the same coset of ker f and we conclude that | D(g)|$\le$| ker f| as required.

(c)

Let K denote the centralizer of g in f^-1(C). Then the order of the centralizer of g in G is at least | K|. Now the order of the conjugacy class of g in f^-1(C) is [f^-1(C) : K], and by (b) this order is at most | ker f|. Therefore [f^-1(C) : K]$\le$| ker f|, consequently

| K|$$\ge$$| f^-1(C)/ker f| = | C|

because f^-1(C)/ker f $\cong$ C. The result follows.

4.

(a)

If R has no prime elements, then R is a field and so certainly a PID. Therefore we may suppose that R has exactly one prime p (up to a multiple of a unit), and we need to prove that R is a PID. Let I be a nonzero ideal of R. Then each nonzero element of I can be written in the form upⁿ for some nonnegative integer n and some unit u, because p is the only prime (up to a multiple of a unit) of R. Let N be the smallest nonnegative integer such that up^N $\in$ I for some unit u. We now show that I = p^NR; clearly p^NR $\subseteq$ I. If x $\in$ I \ 0, then we may write x = vpⁿ for some unit v and some integer n$\ge$N. Thus x = p^Nvp^{n - N} which shows that x $\in$ p^NR, and the result follows.

(b)

Suppose every maximal ideal of R is principal. Then each maximal ideal of R is of the form pR where p is a prime of R. Suppose by way of contradiction that I is a nonprincipal ideal of R. Clearly 0$\ne$I$\ne$R. Choose a nonzero element x $\in$ I, and write x = p₁^d₁... p_n^d_n, where the p_i are nonassociate primes and the d_i are positive integers.

For each prime p, let e(p) denote the largest integer such that p^e(p)R $\supseteq$ I. If p is not an associate of one of the p_i, then e(p) = 0. Set y = p₁^e₁... p_n^e_n. We claim that I = yR.

First we show that I $\subseteq$ yR. If z $\in$ I, then by unique factorization we may write z = qp₁^f₁... p_n^f_n, where the f_i are nonnegative integers and q is a product of primes which are not associate to any of the p_i. Again using unique factorization, we must have f_i$\ge$e_i for all i and we deduce that z $\in$ yR.

Finally we show that yR $\subseteq$ I. Set J = {r $\in$ R | yr $\in$ I} (so J = y^-1I). Clearly J is an ideal of R and yJ = I. If J$\ne$R, then by Zorn's lemma J is contained in a maximal ideal of R, which we may assume is of the form pR where p is a prime of R. It would follow that ypR $\supseteq$ I, which contradicts the maximality of the e(p). Therefore J = R and we deduce that yR $\subseteq$ I. Thus I = yR and the proof is complete.

5.

Let $\pi$ : N $\oplus$ X$\to$N denote the projection onto N, so $\pi$(n, x) = n for all n $\in$ N and x $\in$ X, and let $\iota$ denote the identity map on N. Then ($\pi$f )i = $\pi$$\sigma$ = $\iota$ . This shows that i has a left inverse, consequently the sequence

0 $$\longrightarrow$$ N$$\overset {i}$$ $$\longrightarrow$$ M $$\longrightarrow$$ M/N $$\longrightarrow$$ 0

splits (the map M$\to$M/N above is of course the natural epimorphism). This shows that M $\cong$ N $\oplus$ M/N as required.

6.

Let G denote the Galois group of E over F. Since E is a Galois extension of F with [E : F] = pⁿ, we see that | G| = pⁿ. Since G is a p-group, it has a series G = G₀ $\supset$ G₁ $\supset$ ... $\supset$ G_{n - 1} $\supset$ G_n = 1, with G_i $\lhd$ G and | G_i/G_{i - 1}| = p for all i. Set K_i = {e $\in$ E | ge = e for all g $\in$ G_i} . Then by the Galois correspondence, K_i is normal over F and [K_i : K_{i - 1}] = [G_i : G_{i - 1}] = p for all i, as required.

7.

Suppose K is a finite field which is algebraically closed. Let n be a positive integer which is prime to the characteristic of K, and consider the polynomial Xⁿ - 1. The derivative of Xⁿ - 1 is nX^{n - 1} which is prime to Xⁿ - 1 in K[X], because n is prime to the characteristic of K. This tells us that the roots of Xⁿ - 1 in a splitting field for K are distinct. If K is algebraically closed, then all these roots would be in K, and we would deduce that | K|$\ge$n. Since n can be arbitrarily large, this would contradict the assumption that K is finite, and the result is proven.