2021 Winter Qualifying Exam in Algebra (AI-generated)

Proof.

If is a Sylow -subgroup of , then is a Sylow -subgroup of because is normal. Hence

has order equal to the full -part of , so it is a Sylow -subgroup of .

Conversely, let be a Sylow -subgroup of . Choose a Sylow -subgroup of the inverse image and then a Sylow -subgroup of containing . Its image in is . Thus the map

is surjective. The cardinality of the target is therefore at most that of the source.

Proof.

No. If a finite group acts transitively on a set of size , orbit-stabilizer gives

But is not divisible by . Equivalently, has no subgroup of index . Hence no such transitive action exists.

Proof.

Relative primality means . The Chinese remainder theorem gives

For comaximal ideals, the intersection equals the product, so

Substitution gives the desired isomorphism.

Proof.

Assume toward a contradiction that is a UFD. A monic polynomial is primitive. Gauss's lemma for a UFD says that a factorization of a primitive polynomial in its fraction field can be rescaled to a factorization in by primitive polynomials. Because are monic, their leading coefficients are already . The rescaling units must therefore cancel without changing the monic factors, and Gauss's lemma implies

This contradicts the hypothesis that at least one factor is not in . Therefore cannot be a UFD.

Proof.

The maps restrict because and . The restricted is injective because the original map is injective.

If , choose with . Then

so the restricted is surjective.

Finally, suppose and . Exactness of the original sequence gives with . Using ,

Thus the kernel of the restricted equals the image of the restricted , proving exactness.

Proof.

If , the assertion is immediate. Otherwise write with . Since is a domain, multiplication by gives an -module isomorphism

Consequently

The module over itself is torsion-free because is an integral domain. Therefore has no nonzero torsion.

Proof.

Take

It has no root in , since and . A cubic over a field is reducible if and only if it has a root, so is irreducible. Therefore the quotient is a field, and as a three-dimensional vector space over it has

elements. The finite field of order is unique up to isomorphism, giving the result.

Proof.

Put and let , where is the group of th roots of unity. The extension

is cyclic of order dividing the odd integer . Let . Base change of the Galois extension shows that is Galois, and it is a subextension of . Hence

has odd order.

Complex conjugation acts on the cyclic radical group by inversion: an automorphism sending to is carried by conjugation to the automorphism sending to . Thus for . On the other hand, , so fixes pointwise. Since , this implies that conjugation by acts trivially on . Therefore every satisfies

An odd-order group has no nonidentity element of order , so is trivial. Hence , so . The standard radical-intersection lemma for a real pure extension of odd exponent gives

Indeed, the same conjugation-and-inversion argument applied to the normal closure shows that every element of the intersection is fixed by the radical group and hence lies in . Since lies in both fields, it follows that .

Proof.

Put in Jordan form. It is enough to construct a square root of each Jordan block

where and is nilpotent. Choose with . If the block has size , define

Because , this is a finite sum. The formal binomial identity , truncated modulo , gives

Taking the block diagonal sum of these square roots and conjugating back from Jordan form produces a matrix with .

Proof.

Dimension is impossible because every matrix is scalar. In dimension , take

Then is nonscalar and

Thus

in characteristic , and . Also , so is nonsingular. Therefore the smallest possible dimension is .