2013 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

If , then has only two distinct prime divisors. Burnside's theorem implies that is solvable.

Suppose . Then is square-free. A standard square-free-order theorem states that every finite group of square-free order is metacyclic: it has a cyclic normal subgroup with cyclic quotient. Briefly, its Sylow subgroups are cyclic, and the Sylow congruences, applied successively from the largest prime downward, produce a cyclic normal Hall subgroup; the remaining cyclic Sylow factors act on it by conjugation. Hence there is a normal cyclic subgroup such that is cyclic.

Both and are abelian, so

and . Thus is solvable. This covers both cases.

Proof.

(a) The ideal is prime in because is a domain, but it is not maximal because is not a field.

(b) Take

Both additive groups are isomorphic to . But is a field, whereas has nonzero zero divisors, so the rings are not isomorphic.

(c) For any field , the polynomial ring is a UFD by Gauss's lemma. It is not a PID because the ideal is not principal.

Proof.

(a) An element of order must contain a -cycle, so it requires at least seven letters. Conversely, a -cycle generates such a subgroup. Thus the answer is

(b) A permutation of order must have disjoint cycle lengths whose least common multiple is . The smallest possible support is a disjoint -cycle and -cycle, using nine letters. Their product has order . No permutation on fewer than nine letters can contain both factors and in its order. Thus the answer is

Proof.

(a) Let and define

These are ideals. If , multiplication by the idempotents and shows that . Thus , and the reverse inclusion follows by addition. Hence .

(b) If is prime, then

so contains or . It follows that every prime ideal has one of the forms

where is prime in or is prime in . Conversely, these ideals are prime because the corresponding quotients are and . Similarly, the maximal ideals are exactly

with maximal in the respective factors.

Proof.

(a) Irreducibility makes a transitive subgroup of , so it is or . The two nonreal roots are interchanged by complex conjugation, while the real root is fixed. Thus contains a transposition and cannot be . Therefore .

(b) Put

Its conjugates are for , and a calculation from

gives its minimal polynomial

This cubic has no rational root and is therefore irreducible. The field is the fixed field of complex conjugation in , so it is the unique cubic subextension, and the roots of generate it.

Proof.

The splitting field is the smallest whose multiplicative group contains the th roots of unity. Thus is the multiplicative order of modulo . We have

so

Therefore

and .

The subfields of are exactly for positive divisors of . Since the divisors are

the field has six subfields, counting and itself.

Proof.

Factor

The operator is injective and hence invertible, because is finite-dimensional. From

we conclude that .

The cyclotomic polynomial is irreducible over . Therefore the action of makes a vector space over

which has degree over . Hence

so .

Proof.

Distinct maximal ideals are comaximal, so choose

For a pure tensor , balancedness gives

Pure tensors generate the tensor product, so the entire tensor product is zero.

Proof.

Let , so . The number of Sylow -subgroups satisfies

Since , this forces . Let be the unique Sylow -subgroup. Then .

Set . The Galois correspondence gives

Because is normal, is Galois.

Proof.

Let generate and suppose its order is . Put . Then

In characteristic ,

Thus is nilpotent, so its kernel is nonzero. The subspace

is invariant under and hence under the cyclic group . Irreducibility forces it to be all of . Therefore , and since generates , the representation is trivial.