2024 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

Suppose first that is abelian. Then is abelian, so every subgroup, including , is normal.

Conversely, suppose is normal. For ,

must lie in . Hence . Taking gives for every . Thus every two elements commute, and is abelian.

Proof.

Let be the number of Sylow -subgroups. Then

Therefore is either or . If , the unique Sylow -subgroup is a nontrivial proper normal subgroup.

Suppose . Conjugation gives a nontrivial homomorphism

If were simple, its kernel would be trivial, so would embed in . But the highest power of dividing is , whereas with divides . Lagrange's theorem makes such an embedding impossible. Hence is not simple.

Proof.

Because a PID is an integral domain, and therefore are integral domains. Let . Since is a PID,

for some . The polynomial divides both the nonzero constant and the polynomial . Since degrees add in a polynomial ring over a domain, must be constant, so .

Because , we have for some . Comparing the coefficients of gives for the leading coefficient of . Thus is a unit. Hence , so there exist such that

Setting gives , so is a unit. Every nonzero element of is a unit, and therefore is a field.

Proof.

(a) Consider the composite quotient map

It is surjective. Its kernel consists precisely of those whose class modulo belongs to the ideal generated by , which is equivalent to . The first isomorphism theorem gives the desired isomorphism.

(b) By part (a),

But

in . The quotient therefore has zero divisors and is not a field. Consequently is not maximal.

Proof.

Assume first that every ideal is finitely generated. Given an ascending chain

the union is an ideal. Write . All finitely many generators lie in some , so

Thus for all .

Conversely, assume ACC and let be an ideal. If were not finitely generated, choose , and after choosing , choose

Then

would be an ascending chain that never stabilizes, contradicting ACC. Therefore every ideal is finitely generated.

Proof.

Suppose and satisfy , with . In the quotient,

Since is torsion-free, , so . Now the equality holds inside the torsion-free module . Since , it follows that . Thus is torsion-free.

Proof.

(a) A characteristic polynomial with distinct roots splits over , so is diagonalizable. Write

Each is a root of unity. If is the least common multiple of their orders, then for every , and hence .

(b) No. For , the matrix satisfies the hypotheses, but .

(c) No. Consider

Its only eigenvalue is the root of unity , but its characteristic polynomial is not separable. For every ,

Proof.

Suppose instead that is odd. Choose . By the tower law,

Thus the degree of the minimal polynomial of over , namely , is an odd integer greater than . By hypothesis, this minimal polynomial has a root in . An irreducible polynomial over having a root in must be linear, a contradiction. Therefore is even.

Proof.

Let , so . A group of order is either cyclic or isomorphic to . If were cyclic of order , it would have a unique subgroup of order and hence only one strict intermediate field. The hypothesis therefore forces

The proper nontrivial subgroups of this group are exactly the one-dimensional subspaces of the two-dimensional vector space . Their number is

By the Galois correspondence, there are exactly six strict intermediate fields.

Proof.

Put . The element satisfies

over . Since , this shows

On the other hand,

Thus fixes . But it does not fix , because . Hence , so . Combining both inequalities gives .