Spring 1994 Algebra Prelim Solutions

1.
Suppose G is a simple group with exactly three elements of order two. Consider the conjugation action of G on the three elements of order two: specifically if g G and x is an element of order two, then we define g x = gxg-1. This action yields a homomorphism : GS3. Suppose ker = G. Then gxg-1 = x for all g G and for all elements x of order two. Thus if x is an element of order two, we see that x is in the center of G and hence G is not simple. On the other hand if ker G, then since G is simple we must have ker = 1 and it follows that G is isomorphic to a subgroup of S3. The only subgroup of S3 which has exactly three elements of order two is S3 itself. But S3 is not simple (because A3 is a nontrivial normal subgroup), hence G is not simple and the result is proven.

2.
Let F denote the free group on generators x, y, and define a homomorphism f : FS3 by f (x) = (123) and f (y) = (12). Since f (x6) = (123)6 = e, f (y4) = (12)4 = e, and f (yxy-1) = (213) = x-1, we see that f induces a homomorphism from G to S3. This homomorphism is onto because its image contains f (x) = (123) and f (y) = (12), and the elements (123), (12) generate G. Thus G has a homomorphic image isomorphic to S3.

We prove that G is not isomorphic to S3 by showing it has an element whose order is a multiple of 4 or , which will establish the result because the orders of elements in S3 are 1,2 and 3. We shall use bars to denote the image of a number in /4 . Thus is the identity of /4 under the operation of addition. Define h : F/4 by h(x) = and h(y) = . Since h(x6) = 6* = , h(y4) = 4* , and

h(yxy-1) = + - = = h(x-1),

we see that h induces a homomorphism from G to /4 , which has in its image. Since has order 4, it follows that G has an element whose order is either a multiple of 4 or infinity. This completes the proof.

3.
(i)
Since R is not a field, we may choose 0s R such that s is not a unit, equivalently sRR. Define f : RR by f (r) = sr. Then f is an R-module homomorphism which is injective, because R is a PID and s 0, and is not onto because s is not a unit. This proves that R is isomorphic to the proper submodule sR of R.
(ii)
Using the fundamental structure theorem for finitely generated modules over a PID, we may write M as a direct sum of cyclic R-modules. Since M is not a torsion module, at least one of these summands must be R; in other words we may write M R N for some R-submodule N of M. Then M sR N and since sR N is a proper submodule of R N, we have proven that M is isomorphic to a proper submodule of itself.

4.
(i)
We will write mappings on the left. Let : BB C, : CB C denote the natural injections (so b = (b, 0)), and let : B CB, : B CC denote the natural epimorphisms (so (b, c) = b). Define

: HomR(A, B C)HomR(A, B) HomR(A, C)

by (f )= (f,f ), and

: HomR(A, B) HomR(A, C)HomR(A, B C)

by (f, g) = f + g. It is easily checked that and are R-module homomorphisms, so will suffice to prove that and are the identity maps. We have

(f, g) = (f + g) = ((f + g),(f + g)) = (f, g)

because , are the zero maps, and , are the identity maps. Therefore is the identity map. Also

(h) = (h,h) = h + h = ( + )h = h

because + is the identity map. Thus is the identity map and (i) is proven.
(ii)
Write HomR(A, A) = X. If HomR(A, A A) , then by the first part we would have X X . Thus X 0, and we see that is the direct sum of two nonzero groups. This is not possible and the result follows.

5.
(i)
Let I be an ideal of S-1R. We need to prove that I is finitely generated. Let J = {r R | r/1 S-1I} (where we view S-1R as elements of the form r/s where r R and s S). Then J is an ideal of R and since R is Noetherian, there exist elements x1,..., xn which generate J as an ideal, which means J = x1R + ... + xnR. We claim that I is generated by {x1/1,..., xn/1} . Indeed if r/s I, then r = r1x1 + ... + rnxn for some ri R, and hence r/s = r1/s x1 + ... + rn/s xn. This proves (i).
(ii)
Let S be the multiplicative subset {1, X, X2,...} . Then every element of S is invertible in R[[X, X-1]] and hence the identity map R[[X]]R[[X]] extends to a homomorphism S-1R[[X]]R[[X, X-1]]. It is easily checked that this map is an isomorphism. Since R[[X]] is Noetherian, it follows from (i) that S-1R[[X]] is Noetherian and hence R[[X, X-1]] is Noetherian as required.

6.
(i)
Set Y = X - 1. Then
 X4 + X3 + X2 + X + 1 = (X5 - 1)/(X - 1) = ((Y + 1)5 - 1)/Y = (Y5 + 5Y4 + 10Y3 + 10Y2 + 5Y)/Y = Y4 + 5Y3 + 10Y2 + 10Y + 5.
Applying Eisenstein's criterion for the prime 5, we see that Y4 + 5Y3 + 10Y2 + 10Y + 5 is irreducible in [Y]. Since Y Y + 1 induces an automorphism of [Y], we deduce that X4 + X3 + X2 + X + 1 is irreducible.
(ii)
Let c(X) denote the characteristic polynomial of A, and let m(X) denote the minimum polynomial of A. Since A5 = I, we see that m(X) divides X5 - 1, and since 1 is not an eigenvalue of A, we see that X - 1 does not divide m(X). Therefore m(X) divides X4 + X3 + X2 + X + 1 and using (i), we deduce that the only irreducible factor of m(X) is X4 + X3 + X2 + X + 1. It follows that the only irreducible factor of c(X) is X4 + X3 + X2 + X + 1, which shows that the degree of c(X) is a multiple of 4. This completes the proof, because n is the degree of c(X).

Peter Linnell
1998-07-09