2023 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

For each action of , identify the indicated stabilizer.

(a) The natural action on ; stabilize .

(b) Left multiplication on ; stabilize .

(c) Conjugation on ; stabilize .

Proof.

(a) The stabilizer consists of the permutations of that fix :

(b) The left-regular action is free, so

(c) The stabilizer is the centralizer of . Its elements may independently interchange and , so

Problem 2.

Let be a group whose automorphism group is cyclic. Prove that is abelian.

Proof.

The inner automorphisms form a subgroup

Every subgroup of a cyclic group is cyclic, so is cyclic. Since

the quotient is cyclic. A group whose quotient by its center is cyclic is abelian. Indeed, if is generated by , then every element is with , and any two such elements commute. Hence is abelian.

Problem 3.

Let be a nonzero finite commutative ring with identity. Prove that every prime ideal of is maximal.

Proof.

Let be prime. Then is a finite integral domain. Every finite integral domain is a field: for , multiplication by is injective and hence, on a finite set, surjective, so for some . Therefore is a field, which means is maximal.

Problem 4.

All ring homomorphisms are required to preserve identity.

(a) Describe all ring homomorphisms .

(b) Describe all ring homomorphisms .

Proof.

(a) There is exactly one. A unital homomorphism must send to and therefore sends

(b) Let and . They are idempotents, , and . The only idempotents in are and . Thus a unital homomorphism must send exactly one of to and the other to . This gives precisely the two coordinate projections

Problem 5.

Let be a ring and an -module. Prove that is finitely generated if and only if it is isomorphic to a quotient of a finite-rank free -module.

Proof.

If is generated by , define

This map is surjective, so the first isomorphism theorem gives

Conversely, if , then the images of the standard basis vectors of generate the quotient. Hence is finitely generated.

Problem 6.

Does there exist a integer matrix of order ?

Proof.

No. Suppose . Over characteristic zero, has no repeated roots, so is diagonalizable over , and the order of is the least common multiple of the orders of its eigenvalues. If has order , at least one eigenvalue must be a primitive th root of unity. Its minimal polynomial over is , whose degree is

Therefore the minimal polynomial of has degree at least . But the minimal polynomial of a matrix has degree at most , a contradiction.

Problem 7.

Prove that is irreducible over .

Proof.

Let and let be its ring of integers. The rational prime is unramified in the cyclotomic extension , because only the prime ramifies. Thus, for any prime ideal of lying over ,

The polynomial is therefore Eisenstein at : its constant term lies in but not in , and all other nonleading coefficients lie in . Eisenstein's criterion in the discrete valuation ring proves that is irreducible over .

Problem 8.

Find the cardinalities of the following subsets of :

(a) .

(b) .

Proof.

(a) The multiplicative group has order

Since , exponentiation by is an automorphism of this cyclic group. Every nonzero element is therefore a fifth power, and . The cardinality is .

(b) We have

These three roots are distinct in characteristic , so the set is and has cardinality .

Problem 9.

Suppose is Galois and . Can be the splitting field of a degree- polynomial over ?

Proof.

No. If a degree- polynomial had splitting field , the Galois group would act faithfully on its distinct roots. Thus would embed in for some , and hence in .

An order- subgroup of would be a Sylow -subgroup. Every Sylow -subgroup of is isomorphic to the dihedral group , not to . For example, contains no copy of because has six elements of order , whereas a Sylow -subgroup of has only two. Hence the required faithful action cannot exist.

Problem 10.

Let be prime, let be the splitting field over of an irreducible polynomial of degree , and let be distinct roots such that

Prove that .

Proof.

Put . Since is irreducible of degree ,

The assignment defines a -embedding of into an algebraic closure. Because generates the same field, its image is , so this is a nonidentity automorphism of . Hence

The automorphism group order divides the degree, so it has order . Therefore is Galois. It contains every conjugate of , so it contains every root of and must equal the splitting field . Thus , and every group of prime order is cyclic.