2021 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be linear. If are eigenvectors belonging to distinct eigenvalues, prove that is linearly independent.

Proof.

Suppose , where , , and . Applying gives

Subtracting yields

Since and , we get . The original relation then gives because . Thus the two vectors are linearly independent.

Problem 2.

(a) For an odd prime , prove that there are exactly two groups of order up to isomorphism.

(b) Find the order of a Sylow -subgroup of .

Proof.

(a) If , then the number of Sylow -subgroups divides and is congruent to modulo , so the Sylow subgroup is normal. By Cauchy's theorem there is a subgroup , and . The action is either trivial or has image the unique element of order , namely inversion. The trivial action gives ; inversion gives the dihedral group of order . These are not isomorphic because one is abelian and the other is not.

(b) We have

The second product is not divisible by . Hence a Sylow -subgroup has order

The upper unitriangular matrices form a subgroup of exactly this order.

Problem 3.

Let . If and are abelian, prove that is abelian.

Proof.

Since is abelian, every commutator of elements of lies in . Similarly every commutator lies in . Therefore

A quotient is abelian exactly when . Taking proves that is abelian.

Problem 4.

(a) Let act transitively on , where is prime. Prove that contains a -cycle.

(b) Give a transitive subgroup of containing no -cycle.

Proof.

(a) By orbit-stabilizer,

so divides . Cauchy's theorem gives an element of order in . A nonidentity permutation of order in must consist of one -cycle, so it is a -cycle.

(b) The Klein four group

acts transitively on four letters. All its nonidentity elements have order , so it contains no -cycle.

Problem 5.

For every , prove that the principal ideal is not maximal in .

Proof.

The zero ideal is not maximal because is not a field, and the unit ideal is not a proper ideal. Suppose first that is a nonconstant nonunit. Choose a prime not dividing its leading coefficient. Then the reduction is nonconstant. The ideal

is proper because its quotient is , and the first inclusion is strict because a nonzero constant cannot be a multiple of the nonconstant polynomial .

If is a nonunit constant, choose a prime dividing . Then

The maximal ideal strictly contains , since . Thus no principal ideal is maximal.

Problem 6.

Let be a field with elements. Prove that is not a square in .

Proof.

The prime subfield is . The nonzero squares in are

so is not a square there. If had a square root in , then would contain

But the subfields of are with , and . Hence is not a subfield of , a contradiction.

Problem 7.

Let be prime and .

(a) For which does contain a Galois extension with Galois group ?

(b) Prove that every such is contained in .

Proof.

(a) The Galois group of is cyclic of order . It has a quotient of order , and hence a corresponding degree- Galois subfield, exactly when

Thus the answer is precisely the primes .

(b) Complex conjugation restricts to an automorphism of whose order divides . But has order and therefore has no nonidentity element of order . Conjugation must act trivially on , so every element of is real and .

Problem 8.

Let be an integral domain and an -module. Define

(a) Prove that is a submodule.

(b) Prove that is torsion-free.

Proof.

(a) If and , with , then

Since is a domain, , so is torsion. Also, if , then , so is torsion. Additive inverses are immediate. Hence is a submodule.

(b) Suppose and

Then is torsion, so for some . Since , this means is torsion. Therefore , proving that the quotient is torsion-free.

Problem 9.

For each item, give an example or prove none exists.

(a) A simple group with a subgroup that is not simple.

(b) A nonabelian group for which is a homomorphism.

(c) A real matrix satisfying .

(d) A field and a polynomial whose splitting field is not Galois.

Proof.

(a) Take and

The group is simple, while the nontrivial abelian group is not simple.

(b) No such group exists. If squaring is a homomorphism, then

Thus , and cancellation gives for all . Hence the group is abelian.

(c) One example is , since .

(d) In characteristic , take the base field and

Its splitting field is . This is a nontrivial purely inseparable extension, so it is not separable and therefore is not Galois.