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 $\in$ G and x is an element of order two, then we define g $\cdot$ x = gxg^-1. This action yields a homomorphism $\theta$ : G$\to$S₃. Suppose ker $\theta$ = G. Then gxg^-1 = x for all g $\in$ 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 $\theta$$\ne$G, then since G is simple we must have ker $\theta$ = 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 : F$\to$S₃ 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 $\infty$ , 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 $\mathbb {Z}$/4$\mathbb {Z}$ . Thus $\bar{0}$ is the identity of $\mathbb {Z}$/4$\mathbb {Z}$ under the operation of addition. Define h : F$\to$$\mathbb {Z}$/4$\mathbb {Z}$ by h(x) = $\bar{0}$ and h(y) = $\bar{1}$ . Since h(x⁶) = 6*$\bar{0}$ = $\bar{0}$ , h(y⁴) = 4*$\bar{0}$ , and

h(yxy^-1) = $$\bar{1}$$ + $$\bar{0}$$ - $$\bar{1}$$ = $$\bar{0}$$ = h(x^-1),

we see that h induces a homomorphism from G to $\mathbb {Z}$/4$\mathbb {Z}$ , which has $\bar{1}$ in its image. Since $\bar{1}$ 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 0$\ne$s $\in$ R such that s is not a unit, equivalently sR$\ne$R. Define f : R$\to$R by f (r) = sr. Then f is an R-module homomorphism which is injective, because R is a PID and s$\ne$ 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 $\cong$ R $\oplus$ N for some R-submodule N of M. Then M $\cong$ sR $\oplus$ N and since sR $\oplus$ N is a proper submodule of R $\oplus$ N, we have proven that M is isomorphic to a proper submodule of itself.

4.

(i)

We will write mappings on the left. Let $\beta$ : B$\to$B $\oplus$ C, $\gamma$ : C$\to$B $\oplus$ C denote the natural injections (so $\beta$b = (b, 0)), and let $\pi$ : B $\oplus$ C$\to$B, $\psi$ : B $\oplus$ C$\to$C denote the natural epimorphisms (so $\pi$(b, c) = b). Define

$$\theta$$ : Hom_R(A, B $$\oplus$$ C)$$\to$$Hom_R(A, B) $$\oplus$$ Hom_R(A, C)

by $\theta$(f )= ($\pi$f,$\psi$f ), and

$$\phi$$ : Hom_R(A, B) $$\oplus$$ Hom_R(A, C)$$\to$$Hom_R(A, B $$\oplus$$ C)

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

$$\theta$$$$\phi$$(f, g) = $$\theta$$($$\beta$$f + $$\gamma$$g) = ($$\pi$$($$\beta$$f + $$\gamma$$g),$$\psi$$($$\beta$$f + $$\gamma$$g)) = (f, g)

because $\pi$$\gamma$ , $\psi$$\beta$ are the zero maps, and $\pi$$\beta$ , $\psi$$\gamma$ are the identity maps. Therefore $\theta$$\phi$ is the identity map. Also

$$\phi$$$$\theta$$(h) = $$\phi$$($$\pi$$h,$$\psi$$h) = $$\beta$$$$\pi$$h + $$\gamma$$$$\psi$$h = ($$\beta$$$$\pi$$ + $$\gamma$$$$\psi$$)h = h

because $\beta$$\pi$ + $\gamma$$\psi$ is the identity map. Thus $\phi$$\theta$ is the identity map and (i) is proven.

(ii)

Write Hom_R(A, A) = X. If Hom_R(A, A $\oplus$ A) $\cong$ $\mathbb {Z}$ , then by the first part we would have X $\oplus$ X $\cong$ $\mathbb {Z}$ . Thus X$\ne$ 0, and we see that $\mathbb {Z}$ 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 $\in$ R | r/1 $\in$ S^-1I} (where we view S^-1R as elements of the form r/s where r $\in$ R and s $\in$ S). Then J is an ideal of R and since R is Noetherian, there exist elements x₁,..., x_n which generate J as an ideal, which means J = x₁R + ... + x_nR. We claim that I is generated by {x₁/1,..., x_n/1} . Indeed if r/s $\in$ I, then r = r₁x₁ + ... + r_nx_n for some r_i $\in$ R, and hence r/s = r₁/s x₁ + ... + r_n/s x_n. 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]]$\to$R[[X]] extends to a homomorphism S^-1R[[X]]$\to$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 $\mathbb {Q}$[Y]. Since Y $\mapsto$ Y + 1 induces an automorphism of $\mathbb {Q}$[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).