2009 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

Write

(a) The quotient has order . Any odd-order subgroup maps trivially to this quotient, so every -subgroup lies in the cyclic rotation group . A cyclic group has a unique subgroup of each possible order. Thus its Sylow -subgroup is cyclic and characteristic in , hence normal in .

(b) Since , the rotation has order . For , put

Each is dihedral of order , so it is a Sylow -subgroup. Every Sylow -subgroup contains the unique Sylow -subgroup of the rotation group and some reflection . Two such groups coincide exactly when their indices are congruent modulo . Therefore the complete list is

and there are of them.

Proof.

The inner automorphism group

is a subgroup of the cyclic group , so is cyclic. If is generated by , every element of has the form with . Any two such elements commute. Hence is abelian.

Proof.

(a) The polynomial splits over because , so

On the other hand, has no root because is not a square modulo . Thus

is a field. The rings are not isomorphic.

(b) We have

Each factor is a field, so the four ideals are

where . Under the product decomposition, the last two are and , respectively.

Proof.

Let and let be a primitive fifth root of unity. The splitting field is

Eisenstein's criterion gives , while . Their intersection has degree dividing both and , so it is . Hence .

Every automorphism has the form

with and , and all choices occur. Composition gives

which agrees with multiplication of the displayed matrices. Thus the correspondence is an isomorphism.

Proof.

For any two square matrices over a commutative field,

Taking traces of the proposed equation would give

This contradicts the assumption that the characteristic of does not divide .

Proof.

Over a splitting field, is diagonalizable because has distinct roots in characteristic zero. Every eigenvalue has order dividing . If an eigenvalue had order dividing , its eigenvector would be fixed by ; if its order divided , that vector would be fixed by . Hence every eigenvalue has order exactly .

Therefore the minimal polynomial and characteristic polynomial of are products of copies of the irreducible cyclotomic polynomial , whose degree is

Equivalently, is a vector space over . Thus

Proof.

Write , where is the Sylow -subgroup and is prime to . Since

we compute the quotient. The group is annihilated by , while multiplication by is an automorphism of . Hence

Proof.

(a) One-dimensional characters factor through the abelianization. Since

there are exactly two: the trivial and sign characters.

(b) The group has five conjugacy classes, corresponding to the five partitions of , so it has five irreducible complex representations. Thus three degrees remain. The sum of the squares of all irreducible degrees is . After the two linear characters, the remaining squares sum to

Each remaining degree is at least , and the only way to write as a sum of three squares of integers at least is

Hence the remaining degrees are .

Proof.

(a) If were not a unit, it would lie in the unique maximal ideal . Then

a contradiction. Thus is a unit.

(b) As printed, the assertion is false because satisfies . The intended assertion is true for a proper ideal . In that case . Since is Noetherian, is finitely generated, and

implies , while always . Hence . Nakayama's lemma gives .

Proof.

If is even, Frobenius is an automorphism of , so every is itself a square and

Suppose is odd. The set of squares, including zero, has size . Let

The translate has the same size. Since

the two subsets intersect. Thus for some , and hence