2012 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be a commutative ring. If are nilpotent, prove that is nilpotent. Does this remain true without commutativity?

Proof.

Suppose and . By the binomial theorem, every term in

contains either at least factors of or at least factors of . Hence every term is zero, so is nilpotent.

The assertion fails in a noncommutative ring. In , let

Both square to zero, but

has square and is not nilpotent.

Problem 2.

(a) How many Sylow -subgroups does have?

(b) Find a Sylow -subgroup of and describe it abstractly and explicitly.

Proof.

(a) There are different -cycles. Each subgroup of order has four nonidentity elements, all -cycles. Thus the number of Sylow -subgroups is

(b) Since , a Sylow -subgroup has order . Let

Then and . Hence

is a subgroup of order , fixing the letter . Abstractly, , and it is a Sylow -subgroup.

Problem 3.

For a -module , consider:

every ascending chain of submodules eventually stabilizes;

every submodule of is finitely generated.

Does either property imply the other?

Proof.

The two properties are equivalent, for modules over any ring.

Assume , and let be a submodule. If were not finitely generated, choose , then , and inductively

This produces a strictly increasing chain of submodules, contradicting . Thus implies .

Conversely, assume and let

be an ascending chain. Its union is a submodule, hence is generated by finitely many elements. All these generators lie in some common , so . Therefore for every , and the chain stabilizes. Thus implies .

Problem 4.

Show that all squares in a group lie in every subgroup of index . Must all cubes lie in every subgroup of index ?

Proof.

Every subgroup of index is normal, and has order . Therefore

for every , so .

The analogous assertion for cubes is false because a subgroup of index need not be normal. In , let

which has index , and let . Then

Problem 5.

Suppose is Galois and

Show that there is an irreducible quartic whose splitting field is .

Proof.

Let be a point stabilizer in , and let . Then

By the primitive element theorem, for some . Its minimal polynomial is irreducible of degree .

The splitting field of inside is the fixed field of the core

The intersection of all four point stabilizers is trivial. Hence the core is trivial, and the splitting field is .

Problem 6.

How many conjugacy classes are there in ?

Proof.

Conjugacy classes are classified by rational canonical form. Because the matrix is invertible, the polynomial cannot occur among its elementary divisors. The relevant monic irreducibles over are

The possibilities of total dimension are:

three -primary classes, corresponding to the partitions , , and ;
one class with elementary divisors and ;
one class for each of the two irreducible cubics.

Thus the total number of conjugacy classes is

Problem 7.

Let be finite, a field, and a representation.

(a) Must ?

(b) Must ?

Proof.

(a) Yes. The composite

is a homomorphism into an abelian group, so it annihilates the commutator subgroup. Hence every element of has determinant .

(b) No. Let , , and take the one-dimensional representation sending the generator to . Since but , the image of the center is not contained in .

Problem 8.

Determine the Galois group of the splitting field of over , , and .

Proof.

Over ,

The splitting field is already , so the Galois group is trivial.

Over , nonzero cubes are only and , so has no solution. Thus the cubic is irreducible. Its splitting field is , and the Galois group is cyclic of order , generated by Frobenius.

Over , , and

The quadratic factor has discriminant

which is not a square modulo . Hence the splitting field is , and its Galois group is cyclic of order .

Problem 9.

Answer each question briefly.

(a) If a group has elements of orders and , must it have an element of order ?

(b) If is cyclic Galois and , is cyclic Galois?

(d) What is the largest element order in , the symmetry group of a -gon?

(e) Find for finite-dimensional -vector spaces.

Proof.

(a) False. The group has elements of orders and but none of order .

(b) True. Under Galois correspondence, corresponds to a subgroup of the cyclic group . Every subgroup of a cyclic group is normal, so is Galois, and the quotient Galois group is cyclic.

(c) True. After labeling its elements, every group of order embeds in by its regular action and is determined by the permutations representing left multiplication. There are at most such lists, so certainly at most that many isomorphism classes.

(d) The rotations form a cyclic subgroup of order , and every reflection has order . Thus the maximum order is .

(e) If and , then the tensors of basis vectors form a basis, so

Problem 10.

For each item, give an example or explain why none exists.

(a) A quadratic field extension that is not separable.

(b) A nonabelian group all of whose proper subgroups are cyclic.

(d) A nonabelian group with trivial automorphism group.

(e) An element of order in .

Proof.

(a) Let and

The polynomial is irreducible and has derivative zero, so is an inseparable quadratic extension.

(b) The quaternion group is nonabelian, and every proper subgroup is cyclic.

(c) The algebraic closure is infinite. Every nonzero element belongs to some finite subfield and therefore has finite multiplicative order.

(d) No such group exists. If is trivial, then every inner automorphism is trivial, so . Hence is abelian.

(e) The class

has order in .