2011 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Show that is not contained in any field that is Galois over with

for any positive integer . You may use that the Galois group of over is .

Proof.

Suppose . Since is normal, it contains every conjugate of , hence it contains the splitting field

of . Because is Galois, restriction gives a surjection

But is not a quotient of any symmetric group. For , the only proper nontrivial normal subgroup of is , so its only nontrivial proper quotient is . For , the normal subgroups are , giving no quotient of order ; and for , the group order is too small. This contradiction proves the claim.

Problem 2.

Let be algebraically closed of characteristic , and write

where . How many th roots of unity are in ?

Proof.

In characteristic ,

Thus and have the same distinct roots. Since , the derivative

has no common root with , so is separable. It splits in the algebraically closed field and has exactly roots. Therefore there are exactly

distinct th roots of unity in .

Problem 3.

A commutative ring with is Boolean if for every .

(a) Find all Boolean integral domains.

(b) Prove that every prime ideal in a Boolean ring is maximal.

Proof.

(a) In a Boolean domain,

for every . Since there are no zero divisors, every element is or . Thus the only Boolean integral domain is , which is indeed Boolean.

(b) If is prime, then is an integral domain. It is also Boolean because the identity passes to quotients. By part (a),

which is a field. Hence is maximal.

Problem 4.

Determine the Galois closure of

Describe all elements of its Galois group by their actions on generators, and identify the group abstractly.

Proof.

Let

Then satisfies

The element is not a square in : writing it as would give and , which have no rational solution. Thus is irreducible and .

Since

the splitting field of , whose roots are , is

The field is real, so and .

Define

and

Then cycles the roots as

so has order ; has order , and . The eight automorphisms are

with their actions obtained from the displayed formulas. Hence

Problem 5.

Suppose is finite of odd order and has order . Show that .

Proof.

Conjugation defines a homomorphism

Since , its automorphism group has order . The image of the homomorphism has order dividing both , which is odd, and . Hence the image is trivial. Therefore every element of centralizes every element of , so .

Problem 6.

Classify all finite groups whose automorphism group is trivial.

Proof.

If , then every inner automorphism is trivial. Hence

so is abelian. Inversion is then an automorphism and must be the identity. Thus every element has order dividing , and

for some .

Its automorphism group is , which is trivial only for or . Therefore the only finite groups with trivial automorphism group are

Problem 7.

Let be irreducible, let be its splitting field, and suppose both and are roots. Show that .

Proof.

Because and have the same irreducible polynomial, the -embedding

extends to an -automorphism of the splitting field . Hence

for every positive integer .

The automorphism group of the finite extension is finite, so has finite order . Then

which gives . Therefore has positive characteristic.

Problem 8.

Let be the six-dimensional space of homogeneous quadratic polynomials. Let act by permuting the variables.

(a) Give the character table of .

(b) Find the character of this representation.

Proof.

(a) With classes represented by , the character table is

(b) Use the basis

The identity fixes all six basis vectors. A transposition fixes two of them, for example fixes and . A -cycle fixes none. Thus

respectively. Therefore

Problem 9.

Let be an complex matrix such that

and

Find its characteristic polynomial, minimal polynomial, Jordan form, and rational canonical form.

Proof.

For eigenvalue , the first kernel dimension shows there are two Jordan blocks, and the second shows their sizes are and . For eigenvalue , there are two blocks; the successive dimensions force sizes and . These blocks have total size , so there are no other eigenvalues.

Thus

and

The Jordan form is

Pairing elementary divisors from smallest to largest gives the invariant factors

Therefore the rational canonical form is

where denotes the companion matrix of .

Problem 10.

Determine whether each statement is true or false.

(a) If every finitely generated subgroup of is abelian, then is abelian.

(b) If all proper subgroups of are normal, then is abelian.

(d) If two complex matrices have the same minimal and characteristic polynomials, then they are similar.

(e) If lies in every maximal ideal of a commutative ring , then is a unit.

Proof.

(a) True. Any two elements lie in the finitely generated subgroup they generate, which is abelian, so they commute.

(b) False. Every subgroup of the quaternion group is normal, but is nonabelian.

(d) False. The nilpotent matrices with Jordan block partitions and both have characteristic polynomial and minimal polynomial , but are not similar.

(e) True. If were a nonunit, it would lie in a maximal ideal . Since , this would imply , a contradiction.