2003 Winter Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let denote the field with elements. Prove:

(a) is a field.

(b) is not a field.

Proof.

A cubic over a field is reducible exactly when it has a root.

(a) In ,

Thus has no root and is irreducible. Therefore its quotient is a field.

(b) In ,

Thus divides , so the quotient has zero divisors and is not a field.

Problem 2.

Let be a finite group. The exam defines a character of to be a homomorphism . Prove that the following are equivalent:

(a) Every element of is conjugate to its inverse.

(b) Every character of is real-valued.

Proof.

With the definition printed in the problem, only one implication is valid. If is conjugate to , then any homomorphism satisfies

Because is finite, is a root of unity, and its inverse is its complex conjugate. Hence is real, proving (a)(b).

The converse is false for one-dimensional characters. For example, is perfect, so every homomorphism is trivial and therefore real-valued. However, a -cycle in is not conjugate in to its inverse: the -class of -cycles splits into two -classes, and inversion interchanges them because is not a quadratic residue modulo .

The standard correct theorem uses all irreducible complex characters: every element is conjugate to its inverse if and only if every irreducible character is real-valued. Indeed, character values satisfy , and irreducible characters separate conjugacy classes. Thus the printed definition makes the requested equivalence incorrect.

Problem 3.

Let be prime, let have dimension over , and let satisfy . Find all possible rational canonical forms for and the characteristic polynomial of each.

Proof.

Over ,

where is irreducible of degree . The polynomial is squarefree, so is semisimple as a -module.

If , the dimension equation is

where is the number of blocks and the number of companion blocks. The only possibilities are

Thus the forms and characteristic polynomials are

If , then and the possibilities are

with characteristic polynomials , , and , respectively.

Problem 4.

Let be the group of upper triangular matrices over with all diagonal entries equal to . Let be a -group of order . Show that is isomorphic to a subgroup of .

Proof.

The left regular representation embeds in by permutation matrices. Now

The exact power of dividing this order is

On the other hand, an element of is determined freely by its entries above the diagonal, so

Hence is a Sylow -subgroup of . The embedded image of is a -subgroup, so it is contained in a conjugate of this Sylow subgroup. Conjugating the embedding places inside .

Problem 5.

Let be a vector space and linear.

(a) If , prove that is onto if and only if it is one-to-one.

(b) Give examples showing that both implications fail when .

Proof.

(a) Rank-nullity gives

Thus exactly when , which is exactly when is surjective.

(b) Let have countable basis . The right shift

is injective but not surjective because is not in its image. The left shift

is surjective but not injective. Thus both implications can fail.

Problem 6.

Let be a submodule of an -module . Prove that is a direct summand of if and only if there is an endomorphism such that and .

Proof.

If , define the projection

Then and .

Conversely, suppose and . For every ,

where and

Thus . If , write . Then

Therefore , so is a direct summand.

Problem 7.

Show that has exactly seven complex representations of dimension , up to isomorphism, and write them in terms of its irreducible representations. You may use that has exactly one irreducible representation of each positive dimension.

Proof.

Because is compact, every finite-dimensional complex representation is completely reducible. Let denote its unique irreducible representation of dimension . Isomorphism classes of -dimensional representations therefore correspond exactly to partitions of . The seven partitions give

and

These are pairwise nonisomorphic by uniqueness of irreducible decomposition, so there are exactly seven.

Problem 8.

Let .

(a) Show that is not a UFD.

(b) Factor as a product of prime ideals in .

Proof.

The norm is . The equality

gives two inequivalent factorizations into irreducibles. Indeed, the norm equations and have no solutions, proving the necessary irreducibility statements. Hence is not a UFD.

Set

The quotients by these ideals are finite fields, so they are prime. Factoring modulo and gives

Therefore

Problem 9.

Let have characteristic .

(a) For positive , prove

(b) Let , and let have total degree over . Prove that the number of zeros of in is divisible by .

Proof.

(a) Let generate . Then

If , this geometric sum is . If , it is in . These cases are equivalent to and , respectively.

(b) For , the expression is when and otherwise. Hence, in ,

The sum of the constant term is in . Expand into monomials. Every resulting monomial has total degree at most . Therefore in each monomial some variable has exponent strictly less than . Summing that monomial over all values of this variable gives ; this also holds when the exponent is , because then the sum is in . Thus

Since is an integer, this says precisely that .

Problem 10.

For , give examples of modules such that:

(a) is torsion-free and no linearly independent subset generates .

(b) is free and has a maximal linearly independent subset that does not generate .

Proof.

(a) Take as a -module. It is torsion-free. Any two nonzero rationals are -linearly dependent: if and are nonzero, suitable nonzero integer multiples of them agree. Thus an independent subset has at most one element, and the cyclic subgroup generated by one rational is never all of .

(b) Take and . This set is linearly independent. It is maximal independent because adding any nonzero integer produces a two-element dependent set, while adding destroys independence. Yet generates only , not .

Problem 11.

Let be the splitting field over of

Determine its Galois group and all intermediate fields explicitly.

Proof.

The roots of the first factor are , and the roots of the second are . Therefore

The two quadratic fields and are distinct, so . Independently changing the signs of and gives all four automorphisms. Hence

The complete list of intermediate fields is

The three quadratic fields correspond to the three subgroups of order .