2020 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

For , prove that there are involutions such that has order .

Proof.

Let and regard the first letters as . Let be the reflection . Then

so is also an involution and . To ensure both permutations have order exactly even in the degenerate small case, let , which is disjoint from , and put

Both have order , and

which has order .

Problem 2.

Give a nonabelian semidirect product of two abelian groups.

Proof.

The symmetric group is

where the nonidentity element of acts on by inversion. Both factors are abelian. The product is not abelian because, for generators of and of ,

Problem 3.

Let

(a) Prove that is a PID.

(b) Prove that has only one irreducible element up to associates.

Proof.

This is the localization of the PID at the multiplicative set of polynomials not divisible by . Every nonzero polynomial can be written uniquely as

The element is a unit in . Thus every nonzero element of is a unit times .

If is an ideal, choose an element of with least exponent of . Multiplying by a unit gives , and every element of is divisible by . Hence , proving (a).

An element is irreducible exactly when . Therefore every irreducible is associated to , proving (b).

Problem 4.

Let , where is prime and is nonconstant. Prove that is maximal if and only if is irreducible modulo .

Proof.

Successive quotienting gives

The ideal is maximal exactly when this quotient is a field. Since is a PID, is a field exactly when is irreducible. This proves the equivalence.

Problem 5.

Let have characteristic . For which is

separable?

Proof.

Let . Its derivative is

If , then , so is separable. Suppose . A common root of and cannot be zero. From we get , while gives

These equations are compatible exactly when . Hence, for , the polynomial is separable exactly when

For , the polynomial is zero and is not separable. Thus the complete answer is: all with .

Problem 6.

For which prime powers is the additive group of cyclic?

Proof.

Write . The additive group is an -dimensional vector space over , so

This group is cyclic exactly when . Therefore the additive group is cyclic precisely when is prime.

Problem 7.

Let be a field extension of degree , and let be the smallest field containing that is Galois over . Prove that

Proof.

Because such a finite Galois closure exists, is separable. There are at most five -embeddings of into an algebraic closure. The Galois group acts faithfully on the five conjugate embeddings, and therefore embeds in . Hence

Problem 8.

Give an injective homomorphism of abelian groups and an abelian group such that

is not injective.

Proof.

Take the inclusion

and let . Identifying , the inclusion corresponds to multiplication by . After tensoring with , the induced map is

which is the zero map. Its domain is nonzero, so it is not injective.

Problem 9.

For , prove that the following are equivalent:

(a) The only complex eigenvalue of is .

(b) for some .

(c) .

Proof.

If (a) holds, the characteristic polynomial is . Cayley-Hamilton gives , so (a) implies (c). Clearly (c) implies (b). Finally, if and with , then

so . Thus (b) implies (a).

Problem 10.

Let be a linear operator on a finite-dimensional vector space over . If its characteristic polynomial is irreducible over , prove that its matrix is diagonalizable over .

Proof.

By Cayley-Hamilton, the minimal polynomial divides the characteristic polynomial. Since the characteristic polynomial is irreducible and nonconstant, the minimal polynomial must equal it. Every irreducible polynomial over a characteristic-zero field is separable, so this polynomial has distinct roots in . Hence the minimal polynomial of splits over into distinct linear factors. This is exactly the criterion for diagonalizability over .