2004 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) Maximal ideals of correspond to maximal ideals of containing , so they are

(b) The polynomial is irreducible over , so the quotient is a field. Its only maximal ideal is .

(c) Since over , the maximal ideals are

(d) We have . The irreducible factors give the two maximal ideals

Proof.

Over , the eigenvalues each have algebraic multiplicity . Hence

and

For each of and , the largest Jordan block has size , and the only partition of with largest part is . For , the exponent in the minimal polynomial is , so there are three size-one blocks. Thus there is exactly one complex conjugacy class.

A rational representative is

where denotes the companion matrix of the monic polynomial . The first two blocks have sizes and , so the total size is . Their characteristic polynomials multiply to the required polynomial, and their largest elementary divisors give the required minimal polynomial.

Proof.

The action of on the left cosets gives a homomorphism

Its kernel is normal. The action is nontrivial and transitive because , so the kernel is not all of . If were simple, the kernel would be trivial, and would embed in . Thus would divide .

The inequalities and divisibility force . The embedding would then be an isomorphism . Since , we have , but is not simple: is a nontrivial proper normal subgroup. This contradiction proves that is not simple.

Proof.

If , then

Every automorphism is of the form

Composition corresponds to multiplication of the subscripts modulo . The element has order modulo , so

generated by .

Proof.

The Euclidean algorithm, implemented by elementary integral column operations, reduces the unimodular row

to . Thus there is a matrix such that

Equivalently, the first row of is . Since is unimodular, . If it is , take . If it is , multiply on the left by

This leaves the first row unchanged and changes the determinant to . The resulting matrix is in and has the desired first row.

Proof.

(a) Each additive group is free abelian of rank . Explicit isomorphisms are

and

(b) The ring is an integral domain. The other two are not: in , and but in . Hence is not isomorphic to either one. Finally, has a nonzero nilpotent element, whereas has no nonzero nilpotents. Therefore those two rings are also not isomorphic.

Proof.

(a) As an -dimensional vector space over a field of elements, has

(b) Frobenius is injective and hence surjective on a finite field of characteristic . Thus finite fields are perfect, and all their algebraic extensions are separable.

(c) Every element of is a root of , and this polynomial has exactly the elements of as its roots. Its derivative is , so it is separable. Hence is the splitting field over of a separable polynomial and is Galois.

(d) The Frobenius automorphism

fixes . We have . If and , all elements of would be roots of , which has at most roots. Thus has order . Since , it follows that

Proof.

If is a root and , irreducibility makes the action of on the roots transitive, and

Because is the splitting field, the action is faithful, so the core of the stabilizer is trivial.

(a) Every subgroup of is normal. Therefore a core-free subgroup must be trivial. Hence and

(b) In , the core-free subgroups are the trivial subgroup and the order- reflection subgroups. They have indices and , respectively. All other nontrivial subgroups contain a nontrivial normal subgroup in their core. Thus

Both possibilities occur: a reflection fixed field has degree , while a primitive element with trivial stabilizer has degree .

Proof.

Let . The number of Sylow -subgroups satisfies

so or . If it is , is not simple.

Assume . The distinct Sylow -subgroups contribute

nonidentity elements. If were simple, the number of Sylow -subgroups would be greater than . It is congruent to modulo , so there are at least , contributing at least further nonidentity elements. Only three nonidentity elements remain. Every Sylow -subgroup has order and therefore has exactly three nonidentity elements, so all Sylow -subgroups must consist of the identity together with precisely those same three remaining elements. Hence the Sylow -subgroup is unique and normal, a contradiction. Thus cannot be simple.

Proof.

(a) If has finite order, its minimal polynomial divides and is a product of cyclotomic polynomials of total degree at most . A primitive th root can occur only if

which gives . A two-dimensional rational representation cannot combine a primitive root of degree with another eigenvalue, so no additional least common multiples occur. Hence the possible orders are exactly

For example,

has characteristic polynomial and order .

(b) Every positive integer occurs. For , rotation through angle ,

has order . The identity and give orders and . In particular, has order .

Proof.

For a rational prime ,

(a) Modulo , is a square because . Thus is reducible, so the quotient is not an integral domain. Explicitly,

but neither factor lies in . Hence is not prime.

(b) The nonzero squares modulo are

so is not a square modulo . Hence is irreducible over . The quotient is therefore a field, and is maximal, hence prime.