2005 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be the multiplicative group of nonzero complex numbers, and let be its subgroup of th roots of unity. Give an explicit isomorphism

Proof.

Define

If , then , so and ; hence the map is well-defined. It is a homomorphism and is injective because exactly when . It is surjective because every nonzero complex number has an th root. Thus is an isomorphism.

Problem 2.

Suppose is a group of order and is a field. Prove that is isomorphic to a subgroup of .

Proof.

Let be the -dimensional -vector space with basis . For , define

and extend linearly. Each permutes the basis, so , and

If is the identity, then , so . Thus is injective, giving the required embedding. This is the left regular representation.

Problem 3.

Let denote the field with elements. Decide whether each ring is a field.

(a) .

(b) .

Proof.

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

(a) At the two elements of ,

Thus the cubic has no root and is irreducible. The quotient is a field, in fact .

(b) In ,

Thus divides the polynomial, so the quotient is not a field.

Problem 4.

Let .

(a) Show that is not a UFD.

(b) Factor the principal ideal as a product of prime ideals of .

Proof.

The norm is

In ,

There is no element of norm or , as is immediate from or . Hence and are irreducible. The elements have norm and cannot factor into nonunits, since such a factorization would require a factor of norm or . Thus they too are irreducible. The two factorizations are not equivalent up to units and order; the only units are . Therefore is not a UFD.

For the ideal factorization, put

The quotients by each of these ideals are isomorphic to or , so the ideals are maximal and hence prime. The prime ramifies and splits:

These identities can be checked by multiplying the displayed ideals, or by factoring modulo and . Consequently,

Problem 5.

Classify the groups of order up to isomorphism.

Proof.

There are exactly five isomorphism classes. The two abelian groups are

The three nonabelian groups are

and the dicyclic group

equivalently the nontrivial semidirect product in which a generator of acts by inversion.

To see completeness, Sylow's theorem gives or . If , conjugation on the four Sylow -subgroups identifies the group with . If , the normal subgroup is acted on by a Sylow -subgroup of order . The Sylow -subgroup is either or , and the action on is either trivial or has image . These possibilities yield precisely the four groups in the list other than (with the trivial action giving and the trivial action giving ). The listed groups are distinguished by commutativity, their numbers of elements of order , and whether their Sylow -subgroup is normal.

Problem 6.

Let be a matrix over with characteristic polynomial

and minimal polynomial

(a) Find and .

(b) How many distinct conjugacy classes of such matrices are there under conjugation by ?

Proof.

The eigenvalues are with algebraic multiplicity and with algebraic multiplicity . Hence

The -primary component must be one block . On the -primary component, the largest nilpotent block has size , and the partitions of with largest part exactly are

Thus there are exactly two conjugacy classes, represented by

and

Either displayed block-diagonal matrix answers part (c).

Problem 7.

Suppose is prime and is a field extension of degree .

(a) Prove that if , then is separable.

(b) Prove that if , then is separable.

Proof.

(a) Every algebraic extension of a characteristic-zero field is separable. Indeed, an irreducible polynomial in characteristic zero cannot have zero derivative, so it has no repeated roots. Thus is separable.

(b) Every finite field is perfect because its Frobenius map is injective and hence surjective. Therefore every finite algebraic extension of , including , is separable.

The element is not a th power in , so is irreducible and . Its derivative is zero, and over an algebraic closure

Thus is purely inseparable.

Problem 8.

Let be the splitting field over of .

(a) Find .

(b) Describe both abstractly and as a set of automorphisms.

Proof.

Let . Then

Every automorphism is

Thus

The complete subgroup/fixed-field correspondence is

and

Indeed, is complex conjugation, fixes , and fixes .

Problem 9.

Let have degree . Consider:

(i) has no roots in ;

(ii) modulo , , where are irreducible of degrees ;

(iii) modulo , , where are irreducible of degrees .

For each assertion, prove it or give a counterexample:

(a) (i) implies that is irreducible over .

(b) (ii) implies irreducibility.

(d) (i) and (ii) imply irreducibility.

(e) (i) and (iii) imply irreducibility.

(f) (ii) and (iii) imply irreducibility.

Proof.

(a) False. A product of an irreducible quadratic and an irreducible cubic over is reducible but can have no rational root. For example,

has no rational root but is reducible.

(b) False. The factorization pattern is itself compatible with a rational factorization of degrees and . The same example works: modulo , both and are irreducible, while their product is reducible over .

(c) False. Choose any monic irreducible quartic , lift it to a monic , and take . Then is reducible over and has the required factorization modulo .

(d) False. Again,

has no rational root, and its two factors remain irreducible modulo , but is reducible over .

(e) True. A proper rational factorization of a degree- polynomial compatible with the squarefree factorization pattern modulo must have factor degrees and . It would therefore give a rational linear factor, contrary to (i). Hence is irreducible.

(f) True. A proper rational factor must have degree or by (ii), but must have degree or by (iii). No proper degree satisfies both restrictions, so is irreducible.

Problem 10.

Determine whether each statement is true or false, with justification.

(a) and are isomorphic.

(b) .

(d) For every commutative ring , every subring of is an ideal of .

(e) For every commutative ring , every ideal of is a subring of .

(f) For every commutative ring with identity, every prime ideal is maximal.

Proof.

(a) True. Decomposing into primary cyclic factors gives

(b) False. The units are , and every nonidentity unit has order . Thus

not .

(d) False. The subring is not an ideal of .

(e) True under the usual convention that a subring need not contain the identity of . An ideal is an additive subgroup, and if , then because and .

(f) False. The ideal is prime in but is not maximal.