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 B = 1 and | G| = | A|| B|, we see that G A X B. Now groups of prime order p are isomorphic the cyclic group /p , while groups of order p2 are either isomorphic to /p2 or /p X /p . Therefore G is isomorphic to either /49 X /23 or /7 X /7 X /23 . 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 /49 X /23 and /7 X /7 X /23 .

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) 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,..., An of prime power order. If n > 1, then A1 A2 and A2 A1. Therefore n1. 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 B or B A are the cyclic groups of prime power order.

4.
(a)
Let x R/rad I and suppose xn = 0 where n > 0. Then we may write x = y +radl I where y R. Since xn = 0, we see that yn +rad I =rad I, which means that yn rad I. By definition of rad I we see that (yn)m = 0. Therefore ymn = 0, hence y rad I and we deduce that x = 0. This establishes that R/rad I has no nonzero nilpotent elements.
(b)
If x P1 P2 ... Pn, then x Pi for all i and hence xn P1P2... Pn. It follows that x rad (P1P2... Pn). Conversely suppose x rad (P1P2... Pn). Then x rad Pi for all i. This means that xm Pi for some m > 0 and since Pi is prime, we deduce that x Pi for all i as required.

(c)
If Pi is contained in every Pj, then P1 ... Pn = Pi and hence R/rad (P1... Pn) = R/Pi by (b). We deduce that R/rad (P1... Pn) is an integral domain.

Conversely suppose R/rad (P1... Pn) is an integral domain. Then by (b) we see that R/(P1 ... Pn) is also an integral domain. Suppose there does not exist an i such that Pi is contained in Pj for all j. Then for each i, we can choose xi Pi such that xiPj for some j (where j depends on i). Now set yi = xi + P1 ... Pn for i = 1,..., n. Then yi is a nonzero element of R/(P1 ... Pn) for all i, yet

y1... yn = x1... xn + P1 ... Pn = 0

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

5.
Obviously K() K(). Now

8 = [K() : K] = [K() : K()][K() : K]

which shows that [K() : K()] divides 8. Also satisfies the polynomial X3 - which shows that [K() : K()]3. Therefore [K() : K()] = 1 or 2. We need to eliminate the possibility that [K() : K()] = 2. If [K() : K()] = 2, then the polynomial X3 - could not be irreducible over K(), and it would follow that X3 - has a root in K(). But the roots of X3 - are , and and since K, it would follow that all the roots of X3 - are in K. In particular K(). 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 0S and hence 0 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 S = . 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 P. By maximality of P we have ai A and aj B for some i, j and hence ai + j P. This contradicts the fact that P T, and it follows that P is a prime ideal not containing a.
(b)
Let : RK denote the composition of the natural epimorphism RR/P followed by the natural monomorphism R/PK. If b S, then bP, hence the image of b in R/P is nonzero and we deduce that b is invertible in K. It follows that extends to a ring homomorphism : S-1RK.

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) S4/A4 /2 .

8.
First we find the Jordan canonical form of the matrix
 0 -2 1 3
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
 1 0 0 2
and we deduce that the Jordan canonical form of A is
 1 0 0 0 2 0 0 0 2
The matrices which commute with this canonical form are the matrices of the form
 p 0 0 0 a b 0 c d
where p, a, b, c, d are arbitrary complex numbers.

Peter Linnell
1998-05-15