2006 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

(a) Give an example of an infinite group in which every element has finite order.

(b) Prove that

has no multiple roots in .

Proof.

(a) One example is the additive group

It is infinite, but every element has finite order dividing .

(b) We have

and therefore

If were a multiple root, then , so the displayed identity would give , hence . But , a contradiction. Thus has no multiple root.

In particular, the order of every subgroup of , and hence the order of every element of , divides .

Problem 2.

Let be the splitting field of over .

(a) Find .

(b) Describe both as an abstract group and as a set of automorphisms.

Proof.

Let . The roots are , so

Eisenstein's criterion at shows that is irreducible, hence . Since is real, it does not contain , and therefore

Every automorphism is determined by sending to any of its four roots and to either or . Define

Then and . The eight automorphisms are

Thus

Problem 3.

For which primes can one find a nonzero homomorphism

Proof.

Any nonzero ring homomorphism to the field sends to . It is therefore determined by the image of , which must satisfy

For an odd prime , this is possible exactly when is a quadratic residue modulo , equivalently when

For , the element satisfies in . Hence such a homomorphism exists exactly for

Explicitly, whenever , the map is .

Problem 4.

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

(b) How many groups of order are there, up to isomorphism?

Proof.

Since , Sylow's theorems give

so . Also

The only divisors of are and , and , so . Thus both Sylow subgroups and are normal. Their intersection is trivial and, for , , the commutator lies in both and , hence is . Therefore

In particular, every such group is abelian, and there is exactly one isomorphism class.

Problem 5.

Let be submodules of a module and suppose that . Prove that there exist natural homomorphisms

such that .

Proof.

Define

These maps are well-defined because for and for . Their kernels are

and

Thus the kernels are naturally isomorphic; indeed, both are canonically the same quotient module.

Problem 6.

Determine, as a direct product of cyclic groups, the group of units of

Proof.

Over ,

The discriminant of is in , and is not a square modulo . Thus the quadratic factor is irreducible. By the Chinese remainder theorem,

Taking unit groups gives

The multiplicative group of every finite field is cyclic, so

Problem 7.

Suppose that is a matrix over , is not diagonalizable, and satisfies and .

(a) List all possibilities for the characteristic polynomial of .

(b) List all possibilities for the minimal polynomial of .

Proof.

Because is not diagonalizable, its characteristic polynomial has a repeated root. Let the repeated eigenvalue be and the remaining eigenvalue be . The trace and determinant conditions give

Eliminating yields

Thus either all three eigenvalues are , or the repeated eigenvalue is and the third is . The possible characteristic polynomials are therefore

For , non-diagonalizability permits the minimal polynomials

For the other characteristic polynomial, both eigenvalues must occur and non-diagonalizability forces a size-two block for , so the minimal polynomial is

Accordingly, the possible Jordan forms are

Problem 8.

Let be a prime power and a positive integer.

(a) Prove that is an automorphism of that fixes .

(b) Prove that generates .

Proof.

In characteristic , raising to the th power is a field homomorphism because is a power of . It is injective, hence bijective on the finite field . Every satisfies , so fixes .

For , the fixed points of satisfy . If , this equation has at most roots, so is not the identity. On the other hand, every element of satisfies , so . Thus has order .

The extension has degree and is Galois, so its Galois group has order . Since already has elements,

Problem 9.

For each statement, answer true or false and justify the answer.

(a) Every Euclidean domain is a principal ideal domain.

(b) For every commutative ring with identity, every subring of is an ideal of .

(d) If is a group, , and , then .

Proof.

(a) True. If is an ideal, choose a nonzero of least Euclidean value. Division of any by gives with either or the Euclidean value of smaller than that of . Since , minimality forces . Thus .

(b) False. The subring is not an ideal: but .

(d) False. Let , let

and let . Since is abelian, , and . But conjugation in sends to the other double transpositions, so is not normal in .