2010 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be an algebraic field extension and let be a ring with

Prove that is a field.

Proof.

Let . Since is algebraic over , it satisfies a relation of least positive degree

with and ; otherwise one could factor out and obtain a relation of smaller degree. Rearranging gives

The right side lies in because . Thus every nonzero element of is invertible in , so is a field.

Problem 2.

Show that a group of order is not simple.

Proof.

Since

Burnside's theorem implies that every group of this order is solvable. A nontrivial finite simple solvable group must be cyclic of prime order: its last nontrivial derived subgroup is a nontrivial abelian normal subgroup and hence must be the whole group. Since is composite, a group of this order cannot be simple.

Problem 3.

Determine the splitting field of over , and describe its Galois group abstractly and explicitly.

Proof.

In , , so the polynomial is . The element is one root; the other roots are obtained by multiplying it by fifth roots of unity. Thus the splitting field is the smallest finite extension containing all fifth roots of unity.

The multiplicative order of modulo is , since

Therefore the splitting field is

The polynomial is separable because its derivative is , which has no common root with it.

Every finite-field Galois group is cyclic, so

Explicitly, if is Frobenius, the four automorphisms are

Problem 4.

Let be the character of a -dimensional complex representation of a finite group . Prove that

for every , and that equality implies for some root of unity .

Proof.

Because is finite, the representation admits a -invariant Hermitian inner product, so every is unitary. Also has finite order, so is diagonalizable with eigenvalues that are roots of unity. Therefore

Equality in the triangle inequality occurs only when all the unit complex numbers have the same argument. Hence they are all equal to one root of unity . Since is diagonalizable, this gives

Problem 5.

Find the Galois group over of

Proof.

The rational-root test shows that the cubic has no rational root, so it is irreducible. For a cubic , the discriminant is

Here

which is not a square in . The Galois group of an irreducible cubic is exactly when its discriminant is a square, and is otherwise . Therefore the Galois group is

Problem 6.

Let . Prove that an ideal is maximal if and only if, for some coordinate ,

Proof.

Let be the standard coordinate idempotents. For any ideal , define

Each is an ideal of the field , hence is either or , and

The quotient is the product of the coordinates for which . It is a field exactly when precisely one is zero. In that case consists exactly of the tuples whose th coordinate vanishes. Since an ideal is maximal exactly when its quotient is a field, the result follows.

Problem 7.

Define ring, module, ring homomorphism, and module homomorphism.

Proof.

(a) A ring is an abelian group under addition with an associative multiplication distributing over addition; under the convention used here it has an identity.

(b) An -module is an abelian group with scalar multiplication by satisfying the distributive, associative, and identity axioms.

(d) An -module homomorphism is additive and satisfies

for every and .

Problem 8.

Let be a finite abelian group, let be its Sylow -subgroup, and suppose . Prove that

Proof.

Write

where is the direct sum of the primary components for primes other than . Since

it is enough to compute this quotient. The group is annihilated by , because its order is , so . Multiplication by is an automorphism of , because is relatively prime to , so . Therefore

as required.

Problem 9.

Give an inseparable field extension and compute its separable and inseparable degrees.

Proof.

Let

where is an indeterminate. The polynomial

is irreducible over because is not a th power in . Its derivative is zero, and in an algebraic closure it has only one distinct root. Hence is purely inseparable of degree . Therefore

Problem 10.

(a) How many similarity classes of rational matrices have characteristic polynomial

(b) Give one representative of each class.

Proof.

Factor over :

For each of the linear factors and , the multiplicity- primary part has either partition or . The irreducible factor occurs once. Hence there are

similarity classes.

Let denote the companion matrix of a monic polynomial . Representatives and minimal polynomials are: