2012 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Recall that the exponent of a group is the least positive integer such that for every group element.

(a) Compute the exponent of .

(b) Compute the exponent of .

Proof.

(a) By the Chinese remainder theorem,

Its exponent is therefore

(b) The order of a permutation is the least common multiple of its cycle lengths. The exponent of is consequently

Each of the needed prime-power factors is realized by a cycle type in , so no smaller exponent works.

Problem 2.

Fix a prime . For positive integers , let be the number of nonzero ring homomorphisms

(a) Find .

(b) Find .

Proof.

A nonzero homomorphism between fields is injective. The field contains a copy of exactly when . When it does, that subfield is unique, and its embeddings are the Frobenius maps

Therefore

and

Problem 3.

Show that a group with exactly three elements of order is not simple.

Proof.

Let be the set of the three involutions. Conjugation preserves element order, so it defines a homomorphism

If were simple, then would be either or . If the kernel were , all three involutions would be central, and each of their order- subgroups would be a nontrivial proper normal subgroup, a contradiction.

Thus a simple would embed in . But the only simple subgroups of are cyclic of prime order, and none has exactly three involutions; itself is not simple. This contradiction proves that is not simple.

Problem 4.

List all ideals in

Proof.

Factor

Ideals of correspond to ideals of the PID containing the defining ideal, equivalently to monic divisors of . Hence the complete list is

The first and last correspond to the divisors and , respectively.

Problem 5.

Let

Show that is Galois and that

Proof.

Let . The polynomial is Eisenstein at , so

Put , so , and observe that

Consequently

Thus contains all roots

of . Conversely, can be recovered inside its splitting field because and , for example

Hence is the splitting field of the separable polynomial , so it is Galois.

The irreducible cubic has one real root and two nonreal roots. Its Galois group is therefore a transitive subgroup of containing complex conjugation, a transposition. It must be . Equivalently, its order is already , so

Problem 6.

Give and justify the complex character table of the quaternion group .

Proof.

The conjugacy classes of are

Since

there are four one-dimensional characters. The sum-of-squares formula then leaves one irreducible character of degree . The table is

The rows are pairwise orthonormal, and their squared degrees total

so these are all irreducible characters.

Problem 7.

Suppose

is an exact sequence of modules over a commutative ring . Show that if and are finitely generated, then is finitely generated.

Proof.

Let generate , and let generate . Choose with . We claim that

generate .

For , write

Then

By exactness, the expression in parentheses lies in and is therefore a linear combination of the . This proves the claim.

Problem 8.

Which of the following matrices are similar over ? Which are similar over ?

Proof.

The matrices and are similar over ; for example, conjugating by changes the superdiagonal entry from to . Both have characteristic polynomial and minimal polynomial .

The matrix has minimal polynomial , so it is not similar to or . The matrix has characteristic polynomial , so it is not similar to any of the first three. Thus the only nontrivial similarity is

This classification is the same over and over .

Problem 9.

Answer each item with justification.

(a) If both have finite order in a group, must have finite order?

(b) Does have exactly two subgroups of index ?

(d) If are nonzero modules over a commutative ring , must be nonzero?

(e) Are and isomorphic as fields?

Proof.

(a) False. In the infinite dihedral group, two reflections have order , but their product can be a rotation of infinite order.

(b) False. Index- subgroups are kernels of nonzero homomorphisms to . Since

there are three nonzero homomorphisms and three distinct index- subgroups.

with abelian factors.

(d) False. For example,

(e) No. A field isomorphism fixes . If with , squaring forces and then yields no rational solution to the remaining equation. Hence the two quadratic fields are not equal and therefore not isomorphic over .

Problem 10.

For each item, give an example or explain why none exists.

(a) A group in which the set of squares is not a subgroup.

(b) An element of order in some .

(d) A commutative ring with identity that is not a field and has exactly one prime ideal.

(e) A nonprincipal ideal in .

Proof.

(a) In , every -cycle is a square, while the nonidentity double transpositions are not squares. The product of suitable -cycles is a double transposition, so the set of squares is not closed under multiplication.

(b) Let be the companion matrix of the cyclotomic polynomial . It is an rational matrix whose eigenvalues are primitive fifteenth roots of unity. Hence has order .

(c) No such extension exists. The field is real closed, and its algebraic closure has degree . Every finite extension of therefore has degree or .

(d) The ring is not a field and has exactly one prime ideal, namely .

(e) Consider

The quotient is , so has index . If were principal, its index would be

This Diophantine equation has no integer solution. Hence is nonprincipal.