2011 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

Let . Then

which is cyclic of order . A quadratic subfield corresponds to a subgroup of index , and a cyclic group has exactly one such subgroup. Thus contains a unique quadratic extension of .

The index- subgroup consists of the squares in . The quadratic field is contained in the maximal real subfield exactly when complex conjugation belongs to this subgroup. Complex conjugation corresponds to , which is a square modulo exactly when

Thus the quadratic field is real precisely for . More explicitly, the unique field is

Proof.

Define

This is a ring homomorphism because the proposed image of is , and

in . It is surjective. Its kernel consists of elements with even.

The generators and lie in the kernel. Conversely, modulo their ideal we have and , so every element of the kernel vanishes. Therefore

The quotient is , a field, so the ideal is maximal.

Proof.

Let be an ideal of a Noetherian ring. If were not finitely generated, choose , and after choosing , choose

Then

would be an infinite strictly increasing chain, a contradiction. Hence every ideal is finitely generated.

Every ideal of is principal, so it is finitely generated. Equivalently, in any ascending chain of ideals, the union is an ideal ; once belongs to one term, that term equals the union and the chain stabilizes. Thus is Noetherian.

Proof.

The positive real sixth root of is . The roots are

Since

the splitting field is

The two quadratic fields and are distinct, so .

Every automorphism independently chooses the signs of and . Thus the four automorphisms are

and both and have order and commute. Therefore

Proof.

The number of Sylow -subgroups satisfies

The only possibility is . Let be the unique Sylow -subgroup, so and .

Conjugation gives a homomorphism

whose image has order dividing both and

These integers are relatively prime, so the image is trivial. Hence , and therefore divides .

Proof.

Maximal ideals of correspond to maximal ideals of containing . Such an is prime, so from we get either or .

If , then is maximal in , so

for some rational prime .

If , then is maximal in , so

where is a monic irreducible polynomial over .

Taking the images of these ideals in gives the complete list. The overlap appears in both families when and .

Proof.

(a) In characteristic , the binomial coefficients vanish for , so

Also and . Thus is a field homomorphism.

(b) Every field homomorphism is injective. If is finite, an injective self-map is surjective, so is an automorphism.

(c) Take . The element is not a th power in this rational function field, so Frobenius is not surjective and hence not an automorphism.

Proof.

(a) The exponent in the minimal polynomial says that the largest Jordan block for has size . Since the whole matrix has size , there is only one block:

(b) Write , where . Then

When , the second factor is invertible and commutes with . Consequently the kernels of powers of have the same dimensions as those of , so there is one Jordan block of size :

(c) Now , so moves each standard basis vector two positions along the Jordan chain. The odd- and even-indexed vectors form two chains of lengths

Thus

omitting the zero-size block when .

Proof.

(a) For classes , the table is

(b) Conjugation permutes the basis . The trace at is the number of group elements fixed by conjugation by , namely . These centralizer orders are , so

(c) Taking character inner products, using class sizes , gives multiplicities

Hence

Proof.

(a) False. Every finitely generated subgroup of the additive group is cyclic, but itself is not cyclic.

(b) True. Choose . Then is abelian and properly contains .

(c) True. In dimension , the characteristic polynomial and the largest invariant factor, which is the minimal polynomial, uniquely determine all invariant factors.

(d) False. Let , a nonreal root of . Then

but .

(e) False. The ring has the unique prime ideal but is not a field.