2013 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) The first column of an invertible matrix can be any nonzero vector, giving choices. The second column can be any vector outside the span of the first, giving choices. This proves the formula.

(b) The subgroup

has order . Since the highest power of dividing

is , this is a Sylow -subgroup.

(c) The normalizer of is the group of invertible upper-triangular matrices:

which has order . Therefore the number of conjugates of , and hence the number of Sylow -subgroups, is

Proof.

(a) As an abelian group,

Every automorphism preserves the primary components. On the -primary component it is an arbitrary element of , and on the -primary component it is an arbitrary automorphism of . Thus

and the number of automorphisms is

(b) By the Chinese remainder theorem,

as rings. A ring automorphism may permute the two isomorphic factors, while it must fix the unique factor. Each prime field has only the identity automorphism. Hence the ring has exactly two automorphisms: the identity and the transposition of the two factors.

Proof.

Suppose were not a unit. Then the proper ideal would be contained in some maximal ideal . By hypothesis, as well. Therefore

contradicting that is proper. Hence is a unit.

Proof.

(a) If is nilpotent, all its eigenvalues in an algebraic closure are zero, and therefore every has trace zero.

Conversely, let be the eigenvalues in an algebraic closure, counted with multiplicity. The hypotheses say that the power sums

vanish for . Newton's identities express the elementary symmetric functions recursively by

Because the characteristic is zero, each is invertible in . Inductively, . Hence the characteristic polynomial of is , and Cayley--Hamilton gives .

(b) Over , take the identity matrix . For every ,

in , but is not nilpotent.

Proof.

The set is a finite subgroup of , of order at most . We use the standard fact that every finite subgroup of the multiplicative group of a field is cyclic.

To prove the fact, let be the exponent of the finite abelian group . Every element of is a root of , so

On the other hand, , and the structure theorem for finite abelian groups supplies an element whose order is . Therefore , and that element generates . Applying this to proves the claim.

Proof.

(a) Direct substitution gives

Thus has no root in . A quadratic is reducible if and only if it has a root, so is irreducible.

(b) Every monic irreducible polynomial over whose degree divides divides . Since is irreducible of degree , it follows that

(c) The elements of consist of the five elements of and elements of degree exactly . Each monic irreducible quadratic contributes its two conjugate roots, so there are

monic irreducible quadratics. If scalar multiples are counted as distinct polynomials, each monic one has four nonzero scalar multiples, so there are irreducible quadratic polynomials in total.

Proof.

(a) The splitting field is the cyclotomic field

Therefore

(b) Its Galois group is

Every subgroup is the product of its -primary and -primary parts. The group has five subgroups, and has two. Hence the Galois group has

subgroups. By the Galois correspondence, has ten subfields, including and .

Proof.

Choose integers such that . For every pure tensor ,

Since pure tensors generate the tensor product, the tensor product is zero.

Proof.

Let . The polynomial is irreducible over by Eisenstein's criterion at , so the extension has degree .

(a) Its roots are . The real field does not contain , so the polynomial does not split there. Thus the extension is not normal and hence not Galois.

(b) Its Galois closure is . Let

and let be complex conjugation, so and . Then

Moreover, . Intermediate fields inside correspond to subgroups of containing . These are

Their fixed fields are respectively

Thus the complete list is

Proof.

(a) In the basis , the matrix of is a permutation matrix. Its diagonal entry corresponding to is exactly when , and is otherwise . Therefore its trace is

(b) Since the trivial character is identically ,

By Burnside's orbit-counting lemma, this average is exactly the number of -orbits in .