1997 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

If is a finite normal subgroup of and contains an element of order , prove that contains an element of order .

Proof.

Let have order . Then , so has finite order because is finite. Hence has finite order , and the order of its image divides , so . The element then has order .

Problem 2.

Let be an odd prime and let be a finite simple group having exactly Sylow -subgroups. Prove that its Sylow -subgroups are abelian.

Proof.

Conjugation gives an action of on the set of its Sylow -subgroups. The kernel is normal. The action is nontrivial, and simplicity therefore makes it faithful. Thus every Sylow -subgroup embeds as a -subgroup of .

The largest power of dividing is , because

for odd . Hence . Every group of order or is abelian, so is abelian.

Problem 3.

Suppose . For fixed , prove that

is a homomorphism .

Proof.

The value lies in . The commutator identity gives

Because is central, this becomes

Therefore .

Problem 4.

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

Proof.

(a) The quotient is , an integral domain but not a field. Thus the ideal is prime but not maximal.

(b) The quotient is . The nonzero classes of and have product zero, so it is not a domain. The ideal is neither prime nor maximal.

(d) The quotient is

The quadratic has no root in , so the quotient is a field. The ideal is maximal and prime.

(e) Modulo ,

so the quotient has zero divisors. The ideal is neither prime nor maximal.

Problem 5.

Let be a commutative ring with identity. Suppose the coefficients of each of generate the unit ideal. Prove that the coefficients of also generate the unit ideal.

Proof.

Let be the ideal generated by the coefficients of . If , place it inside a maximal ideal . Reducing modulo , we obtain

in the polynomial ring over the field . Since that polynomial ring is a domain, either or . Thus all coefficients of , or all coefficients of , lie in , contradicting the assumption that each coefficient ideal is . Hence .

Problem 6.

How many elements of multiplicative order are in ?

Proof.

The multiplicative group is cyclic of order . Since , it has a unique subgroup of order . A cyclic group of order has

generators, so exactly six elements have order .

Problem 7.

Let be generated over by , with relations

Express as a direct sum of cyclic modules.

Proof.

The relation matrix is

It has rank . The gcd of its entries is , and its minors are , whose gcd is . Thus its Smith normal form is

Therefore

The cyclic summands have orders infinite and .

Problem 8.

Find the minimal polynomial over of a primitive complex twentieth root of unity.

Proof.

It is the twentieth cyclotomic polynomial. Since

we obtain

This polynomial is irreducible over by the cyclotomic-polynomial theorem.

Problem 9.

Let be the splitting field over of . Determine the number of subfields of .

Proof.

We have and

The -primary group has five subgroups: the trivial subgroup, three subgroups of order , and the whole group. The group has two subgroups. Because the primary factors have coprime orders, every subgroup is the product of a subgroup from each factor. Hence the Galois group has

subgroups. By the Galois correspondence, has ten subfields, including and .

Problem 10.

Let have characteristic zero and not contain a primitive th root of unity. Suppose is irreducible over . Show that the Galois group of its splitting field embeds in the affine linear group on .

Proof.

Let and let be a primitive th root of unity. The splitting field is

and its roots are , indexed by .

For , there are and such that

Therefore

Thus acts on the root labels by the affine transformation

The action on all roots is faithful because they generate , so this gives an embedding of the Galois group into the group of such affine transformations.