2007 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

Use the presentation

The conjugacy classes are

Thus there are five irreducible complex characters. The abelianization is , giving four one-dimensional characters. Together with the two-dimensional geometric representation, the complete character table is

Their squared degrees sum to , so the list is complete.

Proof.

Let be a finite field of characteristic . The Frobenius map

is injective because implies , hence . Since is finite, is therefore surjective.

An irreducible polynomial over a field of characteristic can be inseparable only if its derivative is zero, in which case it has the form

Because Frobenius is surjective, write each . Then

contradicting irreducibility unless has degree . Thus every irreducible polynomial over is separable, so is perfect and every finite extension of finite fields is separable.

Proof.

Put , where . Then

which is cyclic of order . A cyclic group of even order has a unique subgroup of index , namely its subgroup of squares. By the Galois correspondence, therefore has a unique intermediate field of degree over . More explicitly, it is

For (b), take the maximal real subfield of the seventh cyclotomic field:

The group is cyclic of order , and complex conjugation is its unique subgroup of order . Its fixed field is , so

Thus is a cyclic Galois extension of degree .

Proof.

Let . The number of Sylow -subgroups satisfies

Among the divisors of , only is congruent to modulo . Hence . The unique Sylow -subgroup is normal, so is not simple.

Proof.

(a) Since , the Chinese remainder theorem gives

(b) These rings are not isomorphic. The ring is an integral domain, whereas the class in is nonzero and satisfies . Thus the second ring has a nonzero nilpotent element and is not a domain.

(c) The discriminant of is , so the polynomial is irreducible over . Completing the square gives

If denotes the class of , then

satisfies . Hence the map

is an isomorphism of real algebras.

Proof.

Since

we have . The hypothesis says that is injective, hence invertible because is finite-dimensional. Therefore

The polynomial is irreducible over . Consequently becomes a vector space over the field

with acting as . This field has degree over . If is the dimension of over , then

Thus .

Proof.

Consider the descending chain

If and , then applying repeatedly shows

for every , contradicting nilpotence. Thus, until the chain reaches , every inclusion is strict. Each strict inclusion lowers dimension by at least one. Starting from dimension , after at most such decreases the space is zero. Hence , or .

Proof.

Relative to the basis , the matrix of is a permutation matrix. Its diagonal entry indexed by is exactly when , and otherwise it is . Therefore its trace is

the number of fixed points.

The inner product with the trivial character is

By Burnside's orbit-counting lemma, this average equals the number of -orbits in . Equivalently, it is the dimension of the invariant subspace , whose basis consists of the sums of the elements in each orbit.

Proof.

The roots of the first factor are , and those of the second are . Hence

Neither lies in nor lies in , so . Independently changing the signs of and gives four automorphisms. Thus

The three subgroups of order fix, respectively,

Together with the fields fixed by the trivial subgroup and the whole group, the complete intermediate-field lattice is

Proof.

Let have order . By Sylow's theorems, the number of Sylow -subgroups satisfies and , hence . Let be this normal subgroup, and let be a Sylow -subgroup. Then

There are only two homomorphisms .

If is trivial, then

If is nontrivial, then , and

These groups are not isomorphic because is abelian and is not. Hence these are exactly the two isomorphism classes.