2024 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

(a) Prove that every group of order is cyclic.

(b) Is every group of order cyclic?

Proof.

(a) Let have order . Sylow's theorems give

so . Similarly,

so . The Sylow subgroups and are therefore normal. Their intersection is trivial, and elements of the two subgroups commute because their commutator lies in both. Hence

(b) No. Since the automorphism of has order , the semidirect product

has order and is nonabelian, hence is not cyclic.

Problem 2.

(a) If and has order , prove that for every .

(b) Prove that has no subgroup of order .

Proof.

(a) The subgroup has index , so it is normal and has order . Therefore

for every , which is equivalent to .

(b) Suppose had order . By part (a), would contain for every . Every -cycle is a square, since

Thus would contain all eight -cycles of , as well as the identity. This gives at least nine elements in a group of order , a contradiction.

Problem 3.

If has finite index in a group , prove that there is a normal subgroup such that

Proof.

Let act by left multiplication on the left cosets of . This action gives a homomorphism

Set . Then . Every element of fixes the coset , so . By the first isomorphism theorem,

Since the image is a subgroup of , its order is at most . This proof does not require to be finite.

Problem 4.

Is finitely generated as a -module?

Proof.

No. Suppose it were generated by finitely many elements. All those generators can be written with denominators dividing some fixed power . Every integer linear combination of them would then belong to

But and

This contradiction proves that is not finitely generated as a -module.

Problem 5.

Let be Noetherian, let be ideals, and suppose is the only prime ideal containing . Prove that

for some .

Proof.

Pass to . By the correspondence theorem, the only prime ideal of is . The nilradical is the intersection of all prime ideals, so

Thus every element of is nilpotent. Because is Noetherian, is finitely generated, say

For each , choose such that . If

then every monomial of total degree in the contains some and is zero. Hence , which is equivalent to .

Problem 6.

Let be a field and let satisfy and . Prove that is an eigenvalue of .

Proof.

We have

If were not an eigenvalue, then would be invertible. Multiplying the displayed equation by its inverse would give , contrary to . Therefore is singular and is an eigenvalue. This argument also covers characteristic , where .

Problem 7.

For which positive integers is irreducible over ?

Proof.

The roots of are the nonidentity cube roots of unity, because

Thus the polynomial is reducible over exactly when contains an element of order . This multiplicative group is cyclic of order , so such an element exists exactly when

Since , this divisibility holds exactly when is even. Therefore is irreducible precisely for odd positive integers .

Problem 8.

Let be finite Galois. Suppose , , and

for some . Prove that has positive characteristic.

Proof.

Because the Galois group is finite, has finite order, say . Since fixes , induction gives

for every . Taking yields

Thus . Since , the field cannot have characteristic zero. Its characteristic is therefore a positive prime dividing .

Problem 9.

As printed, the problem states that has degree four over , and asks to prove that it is Galois and compute its Galois group.

Proof.

The stated degree assumption is inconsistent with the displayed element, because

For the positive square root,

and for the other choice it is . In either case,

which has degree , not , over . It is the splitting field of , hence is Galois, and

generated by . Thus the problem as printed appears to contain a typographical error; the conclusions above solve the displayed version.

Problem 10.

Give an example or explain that none exists.

(a) A UFD that is not a Euclidean domain.

(b) A positive integer such that contains two distinct quadratic extensions of .

Proof.

(a) Let be any field. The ring is a UFD by Gauss's lemma, applied successively to polynomial rings. It is not a PID because the ideal is not principal. Since every Euclidean domain is a PID, is not Euclidean.

(b) Take . We have

It contains the two distinct quadratic fields and . Therefore is an example.