2007 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be the field of rational numbers. Find a field such that , the dihedral group with elements. Prove your answer.

Proof.

Let

the splitting field over of . Put . By Eisenstein's criterion at , is irreducible over , so . Since , it does not contain ; hence .

Define automorphisms by

and

Then , and

with the same equality on . Thus . The eight automorphisms

are distinct. Since , they form the full Galois group. Therefore

Problem 2.

Let denote the finite field of elements. Show that

and

Proof.

An invertible matrix is obtained by choosing an ordered basis of . The first column has choices; after independent columns have been chosen, the next column has choices. Hence

The determinant map is surjective and has kernel . Therefore

Factoring from the th factor and rearranging gives

The center of consists of the scalar matrices satisfying . Since is cyclic of order , this center has elements. As , division by this number gives the second formula.

Problem 3.

Let be an odd positive integer. Show that if is an integer such that divides , then

Proof.

Write with the odd primes. Since , we have . In particular, , and the residue class of in has order exactly : its square is , while its fourth power is . Lagrange's theorem therefore gives

so every . Consequently every prime power , and hence .

Problem 4.

Let be an matrix with entries in , with minimal polynomial

(a) What is the characteristic polynomial of ?

(b) What are the trace and determinant of ?

(c) How many conjugacy classes are there of matrices in with this minimal polynomial? Write down one matrix from each conjugacy class.

Proof.

The polynomial is irreducible over , and it is relatively prime to . If the characteristic polynomial is

the minimal polynomial forces and , while the dimension gives . The only possibility is , . Thus

The coefficient of in this monic polynomial is . Since that coefficient equals ,

Also, because the dimension is even, the constant term is , so

For the -primary part, the dimension is and the minimal polynomial is , so it consists of the single companion block . The -primary part has dimension , and its largest Jordan block has size exactly . The partitions of whose largest part is are

Hence there are exactly two conjugacy classes. Representatives are

and

where

Problem 5.

Prove that is a Euclidean domain with respect to the norm

Proof.

The norm is multiplicative because

Let with , and write

Choose integers such that and , and put and . Then

Thus with either or . This is the Euclidean division property, so is Euclidean.

Problem 6.

Prove that no group of order is simple.

Proof.

Let have order . By Sylow's theorems, the number of Sylow -subgroups satisfies

so or . Similarly,

so or . If either number is , the corresponding Sylow subgroup is normal and is not simple.

If both are larger than , the distinct Sylow -subgroups contribute nonidentity elements, and the distinct Sylow -subgroups contribute nonidentity elements. Distinct subgroups of prime order intersect only in the identity, so this would give at least nonidentity elements in a group having only . This is impossible. Therefore is not simple.

Problem 7.

Let be a finite field and let be a finite extension of . Show that both the norm map and the trace map from to are surjective. Is the same statement true if and are number fields?

Proof.

Write and . The multiplicative group is cyclic of order , and

If generates , then has order , so it generates . Together with , this proves that the norm is surjective.

The trace is an -linear map . Finite fields are perfect, so is separable, and the trace pairing is nondegenerate. In particular, the trace map is not the zero map. A nonzero linear map into the one-dimensional -space is surjective.

For number fields, the trace is still surjective as a map of fields: since the characteristic is zero,

so the -linear trace map is nonzero and therefore onto. The norm need not be onto. For example, for and ,

is nonnegative under the real embedding, so is not a norm. Thus the corresponding assertion for both maps is false for number fields.

Problem 8.

Let be a commutative ring with identity, and let be matrices over .

(a) Assume either or is invertible. Show that the characteristic polynomials of and are equal.

(b) For arbitrary and , show that the characteristic polynomials of and are equal.

Proof.

If is invertible, then

so and are similar. If is invertible, the analogous identity is . This proves (a).

For (b), work first in the polynomial ring . The block matrices

give, by block elimination,

Equivalently, one may use the general identity over any commutative ring. Both sides are polynomials in . Multiplying by and substituting yields

as polynomials in . Hence the characteristic polynomials are equal without any invertibility assumption.

Problem 9.

Let be a finite cyclic -group and let be a representation on a finite-dimensional vector space over a field of characteristic . Assume that is irreducible. Prove that is trivial.

Proof.

Let generate , say , and put . Then

In characteristic ,

so . Thus is nilpotent and has nonzero kernel. The fixed-point space

is therefore nonzero. Because is abelian, is a -invariant subspace. Irreducibility gives . Hence , and since generates , every element of acts trivially.

Problem 10.

Let be a positive integer. Prove that

is irreducible over .

Proof.

Set and translate the polynomial by :

Its constant term is , which is divisible by but not by . For , every binomial coefficient is even because is a power of . Thus every nonleading coefficient of is even; adding preserves divisibility by of every nonleading coefficient. The leading coefficient is .

Therefore is Eisenstein at , so it is irreducible over . Translation by is an automorphism of , so the original polynomial is also irreducible.