2022 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

If has finite index in , prove that contains a subgroup that is normal in and has finite index in .

Proof.

Let act by left multiplication on the finite set . The resulting homomorphism

has kernel normal in . Every element of fixes the coset , so . Moreover, has finite index in because the image of is finite. Therefore

Problem 2.

Prove that there is no simple group of order .

Proof.

Since , Sylow's theorems give

Thus or . The first case gives a normal Sylow subgroup. Suppose . Distinct subgroups of order intersect trivially, so their nonidentity elements number

Only elements, including the identity, remain. A Sylow -subgroup has order , so it consists of exactly those remaining elements. Hence it is the unique Sylow -subgroup and is normal. In either case the group has a nontrivial proper normal subgroup, so it is not simple.

Problem 3.

If , prove that is isomorphic to a subgroup of .

Proof.

Define

This is a homomorphism, and

The first isomorphism theorem gives

and the image is a subgroup of .

Problem 4.

Let be a field. For every nonzero ideal , prove that

where the ideals of each form a finite chain.

Proof.

Because is a PID, write

where the are pairwise nonassociate irreducibles. The ideals are pairwise comaximal, so the Chinese remainder theorem gives

Set . Ideals of correspond to ideals of containing . Since is a PID, these are exactly

Their images in form the finite chain

and there are no other ideals.

Problem 5.

For each item, give an example or prove none exists.

(a) A prime ideal in a finite ring that is not maximal.

(b) A nonzero prime ideal in an integral domain that is not maximal.

Proof.

(a) No such example exists. If is prime in a finite commutative ring , then is a finite integral domain, hence a field. Therefore is maximal.

(b) Take and . Then

is an integral domain, so is prime. It is not maximal because is not a field. Also .

Problem 6.

Let and be finite abelian groups of orders and . If , prove that

Proof.

Choose a prime dividing both and . By the structure theorem for finite abelian groups, both and have quotients isomorphic to . Thus there are surjections

Right exactness of tensor products produces a surjection

The target is nonzero, so is nonzero.

Problem 7.

Classify all -modules having elements.

Proof.

The Gaussian integers form a PID. Since , the only Gaussian prime that can occur in a module of cardinality a power of is

whose norm is . The structure theorem for finite modules over a PID says that every such module is a direct sum

Since , cardinality requires . The five partitions of give exactly five isomorphism classes:

and

Problem 8.

Suppose an irreducible degree- polynomial over has a cyclic Galois group. Prove that this group has order .

Proof.

Let be the splitting field, let , and let be a root. Irreducibility gives

The subgroup

therefore has index in . Because is cyclic, is normal. The Galois correspondence then shows that is Galois. Consequently it contains all conjugates of , hence all roots of the polynomial. It is therefore already the splitting field . Thus

Problem 9.

Let , where is prime and . If is irreducible over , prove that its splitting field has degree over .

Proof.

Because , the cyclic group contains all th roots of unity. Let be a root of . Irreducibility gives

Every root of has the form , where . Since every such already lies in , all roots lie in . Thus this field is the splitting field and has degree .

Problem 10.

Suppose . If is a root of a quadratic polynomial in , prove that .

Proof.

The minimal polynomial of over divides a quadratic, so

On the other hand, the tower law gives

so divides . The only positive integer dividing and at most is . Hence , so .