2008 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Compute the following.

(a) If is cyclic of order , how many automorphisms does it have?

(b) How many homomorphisms are there from to ?

Proof.

(a) An automorphism of a cyclic group is determined by the image of a generator, which may be any generator. Thus the number is

(b) A homomorphism is uniquely determined by the arbitrary image of . Hence there are

homomorphisms.

Problem 2.

Show that if , where are odd primes not necessarily distinct, then is not simple.

Proof.

If , then the number of Sylow -subgroups divides and is congruent to modulo the odd prime . It is therefore , so the Sylow -subgroup is normal.

Now suppose . The number divides and is congruent to modulo . If , we are done. Otherwise . Distinct Sylow -subgroups intersect trivially, so they contribute

nonidentity elements. Thus only other nonidentity elements remain.

If the Sylow -subgroup were not normal, then , because . Its distinct conjugates would contribute at least

nonidentity elements. But for odd ,

a contradiction. Hence . In every case has a nontrivial proper normal subgroup.

Problem 3.

Factor and find its splitting field when the ground field is:

(a) ;

(b) ;

Proof.

(a) Over , is irreducible. Its roots are the primitive eighth roots of unity, and its splitting field is

of degree .

(b) In characteristic ,

It already splits over , though with a repeated root.

Both quadratics have nonreal roots, and the splitting field over is .

Problem 4.

In , determine whether each ideal is prime and whether it is maximal:

Proof.

For , the quotient is , a domain but not a field. Hence the ideal is prime but not maximal.

For , the quotient is

which has the nonzero zero divisors and . Hence the ideal is neither prime nor maximal.

For , the quotient is

a domain but not a field. Thus the ideal is prime but not maximal.

For , the quotient is

Since over , the quotient is not a domain. Hence the ideal is neither prime nor maximal.

Problem 5.

Let be the splitting field of over . Determine the number of fields with

Proof.

The splitting field is

Its Galois group is

which is cyclic of order

A cyclic group has exactly one subgroup for every divisor of its order. The positive divisors of are

so there are eight subgroups. By Galois correspondence, there are exactly eight intermediate fields, including both endpoints.

Problem 6.

Suppose is finite and is a subgroup containing every proper subgroup of .

(a) Prove that is a prime power.

(b) Prove that if is abelian, then is cyclic.

Proof.

(a) If two distinct primes divided , every Sylow subgroup would be proper and hence contained in . Thus would be divisible by the full prime-power part of for every prime divisor. It would follow that , impossible because is proper. Therefore only one prime divides , so is a prime power.

(b) In fact, the conclusion follows without assuming commutativity. Choose . If were proper, the hypothesis would force , contradicting . Hence , so is cyclic.

Problem 7.

Find all prime ideals of .

Proof.

Prime ideals in a product are exactly

where and are prime in the corresponding factors. The prime ideals of are and for rational primes . Hence the complete list is

where ranges over all rational primes.

Problem 8.

Let be the set of rational matrices whose characteristic polynomial is and whose minimal polynomial is .

(a) Show that all matrices in are similar.

(b) Give an example.

Proof.

Factor

and

The minimal polynomial is square-free. Therefore every primary component is semisimple. The characteristic polynomial forces two one-dimensional zero blocks and one block for each of , , and . Thus the rational canonical data are unique, proving (a).

(b) One representative is

(c) On the primary component, , while on the , , and primary components it is invertible. The component has dimension , so

Problem 9.

Show that the quaternion group is not a semidirect product of two proper subgroups.

Proof.

Every nontrivial subgroup of contains its unique element of order , namely . Thus any two nontrivial subgroups have nontrivial intersection.

In an internal semidirect product with proper, one factor is normal and one must have

This is impossible if both factors are nontrivial. If one factor is trivial, the other would have to be all of , contrary to properness. Hence no such semidirect decomposition exists.

Problem 10.

Let be algebraically closed. Find all monic separable polynomials whose zero set is closed under multiplication.

Proof.

Let be the finite set of roots. If is nonempty, it is a finite multiplicatively closed subset of a group. Repeated powers show that it contains and inverses, so it is a finite subgroup of . Every finite subgroup of a field's multiplicative group is cyclic, so

for some not divisible by .

The set may include or omit . Hence the nonconstant possibilities are

with and, in positive characteristic, . Conversely, all these polynomials are monic and separable and have multiplicatively closed root sets. If constant polynomials are included, is the additional case.