2015 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) A proper ideal of a commutative ring is prime if implies or . Equivalently, is an integral domain.

(b) A proper ideal is maximal if the only ideals containing it are and . Equivalently, is a field.

(c) Take and

The quotient is a field, so is maximal and prime. The quotient has zero divisors and is not a field, so is neither prime nor maximal. Finally, is a domain but not a field, so is prime but not maximal. In a commutative ring with identity every maximal ideal is prime, so “maximal but not prime” cannot occur.

Proof.

If some had infinite order, the subgroups

would be infinitely many distinct subgroups. Hence every element of has finite order, so every cyclic subgroup of is finite.

By hypothesis there are only finitely many subgroups, say . Every element lies in the cyclic subgroup , which is one of these finitely many subgroups and is finite. Therefore

is a finite union of finite sets, and hence is finite.

Proof.

(a) Taking determinants gives

The left side is nonnegative because , so cannot be odd. Hence is even.

(b) The minimal polynomial of divides

which has distinct roots over . Therefore is diagonalizable over , and its eigenvalues are among and . Because is real, nonreal eigenvalues occur with equal algebraic multiplicities. Thus each occurs times, and is similar over to

up to permutation of the diagonal entries.

Proof.

The number of Sylow -subgroups satisfies

Thus . Let be the unique Sylow -subgroup; then .

The quotient has order . Its number of Sylow -subgroups is congruent to modulo and divides , so it has a unique, hence normal, subgroup of order . The inverse image is normal in , and

Proof.

Let and let

This is the splitting field of the separable polynomial , whose roots are , so is Galois. Eisenstein's criterion gives , and because the latter is real. Hence .

Define automorphisms

and

Then

The eight maps and , , are distinct and account for all automorphisms. Therefore

Proof.

Let be an ideal. If , it is principal. Otherwise, choose a nonzero polynomial of least degree. We claim that .

Certainly . For any , the division algorithm gives

Since , minimality of forces . Hence , so . Therefore , and is a PID.

Proof.

Take and , regarded as a -module. Suppose finitely many rationals

generated . Choose a positive integer divisible by all their denominators. Every integer linear combination of the then lies in . Choose a prime not dividing . Then , contradicting generation. Hence is not finitely generated as a -module.

Proof.

(a) Yes. Write . For every , normality gives . Conjugation cannot send the nonidentity element to , so . Thus commutes with every , and .

(b) No. In , the alternating subgroup

is normal and has order , but . Thus is not contained in the center.

Proof.

(a) A nonzero representation is irreducible if its only -invariant subspaces are and .

(b) Let generate and set . Since and the characteristic is ,

Thus is nilpotent and has nonzero kernel. Choose . Then , so the line is invariant under , and hence under all of . It is a nonzero proper subspace of , so the representation is reducible.

Proof.

(a) Choose the columns of an invertible matrix successively. With , the choices are

Therefore

(b) The determinant map is surjective and has kernel . Since ,

(c) In ,

Use the norm

There is no element of norm or : reducing or modulo would make congruent to or , neither of which is a quadratic residue modulo . It follows from multiplicativity of the norm that , , and are irreducible. The factors appearing in the two decompositions are not associates; in particular their norms rule out all comparisons except between and , and their quotient

does not lie in the ring. Thus has two inequivalent factorizations into irreducibles, and is not a UFD.