2019 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

No. In a nontrivial semidirect decomposition , the orders of and multiply to and their intersection is trivial. Both factors have even order unless one is trivial, so each contains an element of order . But has a unique element of order , namely . Therefore would lie in both and , contradicting . Thus only a trivial decomposition involving the identity group is possible.

Proof.

Sylow's theorems give

Thus or . In the second case, divides . Since , it must divide , so . The only consecutive primes with the smaller one prime are , . Hence and .

The four Sylow -subgroups contribute eight nonidentity elements, leaving exactly three nonidentity elements. Any Sylow -subgroup has order and must consist of the identity and those three elements, so it is unique and normal. Call it . It cannot be cyclic: if , then the conjugation action of a subgroup on would be trivial because , making the Sylow -subgroup normal. Therefore . The nontrivial action of cyclically permutes the three nonidentity elements of , and the resulting semidirect product is . Thus .

Proof.

(a) If and , every term of vanishes, so is nilpotent. Also , so is an ideal.

Every prime ideal contains every nilpotent: if , primality applied repeatedly gives . Hence is contained in the intersection of the prime ideals. Conversely, if is not nilpotent, the multiplicative set does not contain zero. An ideal maximal among those disjoint from this set is prime, and it does not contain . Thus is not in the intersection, proving equality.

(b) In , the matrices

are nilpotent, but

has square . Hence the nilpotent elements are not closed under addition.

Proof.

The norm is

Thus . Since is prime in , multiplicativity of the norm shows that any factorization of has a unit factor. Hence it is irreducible. The ring is Euclidean and therefore a PID, so irreducible elements are prime.

In the quotient, . The relation becomes , hence . Define

This is surjective and its kernel is . Therefore

Proof.

Distinct maximal ideals are comaximal, so the Chinese remainder theorem gives

where are fields.

An element is a unit exactly when both coordinates are nonzero. If at least one coordinate is zero, it is a zero divisor. This proves (a). The element is a unit, while and are distinct nonzero zero divisors, proving (b).

If both fields are finite, the whole product is finite. If either field is infinite, varying a nonzero coordinate gives infinitely many units and varying one coordinate along an axis gives infinitely many zero divisors. This proves (c).

Proof.

Let . It satisfies a polynomial over with finitely many coefficients . Since each is algebraic over , the extension

is finite over . The element is algebraic over , so is finite. Therefore is finite by the tower law, and is algebraic over . Since was arbitrary, is algebraic.

Proof.

Let

The roots are , and . Hence and the splitting field is . The quartic is irreducible over : a factorization into rational quadratics would force either or to be a rational square. Thus .

The automorphisms may independently apply

and

Both have order and they commute. These four automorphisms account for the full degree, so

Proof.

If , then would be a quadratic subfield. A cyclic group of order has a unique subgroup of order , so has a unique quadratic subfield.

Complex conjugation restricts to the unique element of order in . Its fixed field is also quadratic and is contained in . Uniqueness would force this real field to equal , which is impossible. Hence .

Proof.

Over ,

and the quartic factor is irreducible because the order of modulo is . The product is squarefree. Since , its minimal polynomial contains the quartic factor. Thus its minimal polynomial is either the quartic, of degree , or all of , of degree .

The corresponding modules are semisimple. Below dimension , a nontrivial module consists of one four-dimensional irreducible summand together with a uniquely determined number of trivial one-dimensional summands, so its similarity class is unique in each dimension. In dimension , there are two possibilities:

where is the irreducible four-dimensional module. Their minimal polynomials are respectively the quartic and times the quartic, so they are not similar. The smallest dimension is therefore .

Proof.

Put in Jordan normal form:

For each Jordan block, conjugation by the permutation matrix that reverses the basis order carries the block to its transpose. Therefore is similar to . Transposing the displayed similarity gives

Thus is similar to , which is similar to , which is similar to . Hence and are similar.