January 1996 Algebra Prelim Solutions

1.

Since f is an epimorphism from G to H, the fundamental isomorphism theorem tells us that G/kerf $\cong$ H, so in particular | G/ker f| = | H|. Therefore | G|/| ker f| = | H|, hence | ker f| = | G|/| H| and we deduce that | kerf| = 3³ $\cdot$ 11². Using the fundamental theorem for finitely generated abelian groups, we see that there are up to isomorphism six abelian groups of order | kerf| = 3³ $\cdot$ 11², namely

$$\mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z}$$

$$\mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z} \mathbb {Z}$$

2.

(i) Since A $\cap$ B is a subgroup of G whose order divides | A| = p⁴ and | B| = q⁵, we see that | A $\cap$ B| = 1 and hence A $\cap$ B = 1. Next if a $\in$ A and b $\in$ B, then a^-1b^-1ab = (a^-1b^-1a)b $\in$ B, because B $\lhd$ G. Similarly a^-1b^-1ab $\in$ A and we deduce that a^-1b^-1ab $\in$ A $\cap$ B = 1. Therefore a^-1b^-1ab = 1, consequently ab = ba for all a $\in$ A and b $\in$ B. We can now define a map $\theta$ : A X B - > G by $\theta$(a, b) = ab. Then

$$\theta$$((a₁, b₁)(a₂, b₂)) = $$\theta$$(a₁a₂, b₁b₂) = a₁a₂b₁b₂ = (a₁b₁)(a₂b₂)

(because a₂ commutes with b₁), hence $\theta$((a₁, b₁)(a₂, b₂)) = $\theta$(a₁, b₁)$\theta$(a₂, b₂) and we deduce that $\theta$ is a homomorphism. If (a, b) $\in$ ker $\theta$ , then ab = 1 and so a = b^-1. Thus a = b^-1 $\in$ A $\cap$ B = 1 and we deduce that a = b = 1. Therefore ker $\theta$ = 1 and so $\theta$ is a monomorphism. Since | G| = | A| X | B| we conclude that $\theta$ is also onto, consequently $\theta$ is an isomorphism and the result follows.

(ii) Since A is a p-group, it has a normal subgroup P of order p. Similarly B has a normal subgroup Q of order q. Since P X Q is a normal subgroup of A X B of order pq, we see that $\theta$(P X Q) is a normal subgroup of G of order pq, and so we may set N = P X Q to satisfy the requirements of the problem.

3.

Let G be a simple group of order 2² $\cdot$ 3 $\cdot$ 11². The number of Sylow 11-subgroups is congruent to 1 modulo 11 and divides 12, so the possibilities are 1 and 12. If there is 1 Sylow 11-subgroup, then it would have to be normal, which is not possible because G is simple. Therefore there are 12 Sylow 11-subgroups. If N is the normalizer of a Sylow 11-subgroup, then [G : N] is the number of Sylow 11-subgroups, so [G : N] = 12. By considering the permutation representation of G on the left cosets of N in G and using the fact that G is simple, we see that there is a monomorphism of G into A₁₂, the alternating group of degree 12. This means that G is isomorphic to a subgroup of A₁₂. This is not possible because 121 divides | G|, but 121 does not divide | A₁₂|. We now have a contradiction and we deduce that no such G exists, as required.

4.

Since ($\sqrt{2}$ + $\sqrt{3}$)³ - 9($\sqrt{2}$ + $\sqrt{3}$) = 2$\sqrt{2}$ , we see that $\sqrt{2}$ $\in$ $\mathbb {Q}$[$\sqrt{2}$ + $\sqrt{3}$] and we deduce that $\mathbb {Q}$[$\sqrt{2}$ + $\sqrt{3}$] = $\mathbb {Q}$[$\sqrt{2}$,$\sqrt{3}$]. Now [$\mathbb {Q}$[$\sqrt{2}$] : $\mathbb {Q}$] = 2, and [$\mathbb {Q}$[$\sqrt{2}$,$\sqrt{3}$] : $\mathbb {Q}$[$\sqrt{2}$]] = 1 or 2, because $\sqrt{3}$ satisfies x² - 3, a degree 2 polynomial over $\mathbb {Q}$ . Therefore [$\mathbb {Q}$[$\sqrt{2}$ + $\sqrt{3}$] : $\mathbb {Q}$] = 4 or 2, depending on whether or not $\sqrt{3}$ $\in$ $\mathbb {Q}$[$\sqrt{2}$].

Suppose $\sqrt{3}$ $\in$ $\mathbb {Q}$[$\sqrt{2}$]. Then we may write $\sqrt{3}$ = a + b$\sqrt{2}$ where a, b $\in$ $\mathbb {Q}$ . Clearly a, b$\ne$ 0. Squaring we obtain 3 = a² + 2ab$\sqrt{2}$ + 2b² and we deduce that $\sqrt{2}$ is rational, which is not so. Therefore $\sqrt{3}$$\notin$$\mathbb {Q}$[$\sqrt{2}$] and consequently $\mathbb {Q}$[$\sqrt{2}$ + $\sqrt{3}$] = 4. Note we also have that {1,$\sqrt{2}$} is a $\mathbb {Q}$ -basis for $\mathbb {Q}$[$\sqrt{2}$], and {1,$\sqrt{3}$} is a $\mathbb {Q}$[$\sqrt{2}$]-basis for $\mathbb {Q}$[$\sqrt{2}$,$\sqrt{3}$]. Recall that if e_i is an F-basis for E over F and f_j is an E-basis for K, then e_if_j is an F-basis for K. It follows that {1,$\sqrt{2}$,$\sqrt{3}$,$\sqrt{6}$} is a $\mathbb {Q}$ -basis for $\mathbb {Q}$[$\sqrt{2}$,$\sqrt{3}$].

5.

Let P be a prime ideal of D and suppose P was not maximal. Then there would exist M $\lhd$ D such that M$\ne$D and M properly containing P. Since D is a PID, we may write P = pD and M = mD for some m, p $\in$ D with p$\ne$ 0. Then p = mx for some x $\in$ D because M contains P. Since P is a prime ideal, we must have m or x $\in$ P. We cannot have m $\in$ P because M properly contains P. Therefore we must have x $\in$ P and then we may write x = py for some y $\in$ D. This yields p = mpy and since D is a domain, we see that 1 = my. This shows that mD = D, which contradicts the fact that M is a proper ideal of D and the first part of the problem is proven.

Suppose now that f : D - > K is a ring epimorphism onto the integral domain K with kerf$\ne$ 0. Then D/kerf $\cong$ K, so ker f is a prime ideal because D/ker f is an integral domain. Using the first part of the problem, we see that ker f is a maximal ideal of D. Therefore D/ker f is a field and it follows that K is a field as required.

For the last part a counterexample is R = $\mathbb {Z}$/6$\mathbb {Z}$ . Let P be the prime ideal 2$\mathbb {Z}$/6$\mathbb {Z}$ , Q the prime ideal 3$\mathbb {Z}$/6$\mathbb {Z}$ , and S = R $\setminus$ P. Note that the ideals of R are precisely 0, R, P and Q, and the prime ideals of R are precisely P and Q. Then S^-1P = 0 because s = 6$\mathbb {Z}$ + 3 $\in$ S and sp = 0 for all p $\in$ P, and S^-1Q = R_P because Q $\cap$ S$\ne$ $\emptyset$ . Since all ideals of R_P are of the form S^-1I for some I $\lhd$ R, we see that 0 and R_P are the only ideals of R_P and it follows that R_P is a field. Similarly R_Q is a field. Since R is not an integral domain, we have now established that R is a counterexample.

6.

Suppose $\mathcal {P}$ is a prime ideal of R and R_{$\mathcal {P}$} has a nonzero nilpotent element. Then we may assume that R_{$\mathcal {P}$} has a nonzero element $\alpha$ such that $\alpha^{2}_{}$ = 0. If S = R $\setminus$ $\mathcal {P}$ , then we may write the nilpotent element as r/s where r $\in$ R and s $\in$ S. Since (r/s)² = 0, we see that r²t = 0 for some t $\in$ S, and rt$\ne$ 0 because r/s$\ne$ 0. Also (rt)² = r²tt = 0, so rt is a nonzero nilpotent element of R.

Suppose r is a nonzero nilpotent element of R. It remains to prove that R_{$\mathcal {P}$} has a nonzero nilpotent element for some prime ideal $\mathcal {P}$ of R. We may assume that r² = 0. Let I = {s $\in$ R | rs = 0} , an ideal of R which does not contain 1. By Zorn's lemma, there is a maximal ideal $\mathcal {P}$ of R containing I; of course $\mathcal {P}$ will also be a prime ideal. Then the image r/1 in R_{$\mathcal {P}$} is nonzero because rt$\ne$ 0 for all t $\in$ R $\setminus$ $\mathcal {P}$ . Since (r/1)² = r²/1² = 0/1² = 0, we see that r/1 is a nonzero nilpotent element of R_{$\mathcal {P}$}. This completes the proof.

7.

The submodule of M generated by A and B is A + B; this is the set {a + b | a $\in$ A and b $\in$ B} . Thus we need to prove that A $\oplus$ B $\cong$ A + B. We define a map $\theta$ : A $\oplus$ B - > M by $\theta$(a, b) = a + b. Clearly this is an R-module homomorphism of A $\oplus$ B onto A + B. If (a, b) $\in$ ker $\theta$ , then a + b = 0, consequently a = - b. This shows that a and - b are both in A $\cap$ B = 0. Therefore a = b = 0 and hence ker $\theta$ = 0. It follows that $\theta$ is an isomorphism and so A $\oplus$ B $\cong$ A + B as required.

8.

(i) Since there is a one-one correspondence between the subgroups of Gal(E/F) (the Galois group of E over F) and the proper subfields of E containing F, we need to show that S₆ has at least 35 proper subgroups. One way to do this is to note that S₆ has 144 5-cycles which gives 36 subgroups of order 5.

(ii) The subfield L required is Fix(A₆), the subset of E which is fixed pointwise by all elements of the alternating subgroup A₆ of S₆. Since A₆ is a normal subgroup of S₆, we see that L is a Galois extension of F, and that Gal (E/L) $\cong$ A₆. Since A₆ is a simple group, there is no subfield between E and L which is Galois over L.

(iii) The dimension of L over F is | S₆/A₆| = 2.