2020 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

A cube has four pairs of opposite vertices, equivalently four long body diagonals. Every rotational symmetry permutes these four diagonals, giving a homomorphism

If a rotation fixes all four body diagonals, it fixes at least three linearly independent directions. An orientation-preserving orthogonal transformation with this property is the identity, so is injective.

There are rotational symmetries of the cube: the image of a chosen face can be any of six faces, and after choosing that image there are four rotations determining the image of an adjacent vertex. Hence . The injective homomorphism is therefore an isomorphism.

Proof.

Let be a Sylow -subgroup and let be the number of its conjugates. Then

so . Conjugation on the Sylow -subgroups gives a homomorphism . If were simple, this nontrivial action would be faithful, so would divide and hence .

Legendre's formula gives

There are no further terms. But , so cannot divide . This contradiction proves that is not simple.

Proof.

No. If there were a surjection , its kernel would have order

Thus would have a normal subgroup of order . Its subgroups of order are generated by -cycles, and conjugation acts transitively on the eight -cycles. Hence no individual subgroup of order is normal. Therefore cannot be a quotient of .

Proof.

For define

Given with , write

Choose integers with and , and put . Then

For , multiplicativity of the norm gives unless . Thus the ring is Euclidean.

The element has norm . If it factored into two nonunits, their positive integer norms would both be at least , while their product would be , impossible. Hence is a noninteger irreducible.

Proof.

(a) If , then both and are idempotent. Thus

Subtracting gives .

(b) If , then . Consequently

and similarly . Thus , and additive inverses equal the elements themselves. Hence is a subgroup.

Since , inclusion-exclusion gives

As divides , its index is an integer smaller than , hence at most .

(c) If , then every element is idempotent. Otherwise . Since , choose . For ,

because . Thus , and already . The two cosets exhaust , so . In particular, is an additive subgroup and every element is idempotent.

Proof.

(a) By the structure theorem, a finitely generated abelian group is . If , comparing ranks gives . Tensoring the torsion subgroup of either module with the free rank-one part of the other would produce torsion in , so both torsion subgroups vanish. Hence , and is indeed invertible. It is the only isomorphism class.

(b) If , then associativity and commutativity give

Thus is invertible.

(c) Let be an isomorphism and write

Define by . The inverse identities for the tensor equivalence imply the dual-basis formula

Thus has a finite dual basis. By the dual-basis criterion, is finitely generated projective.

Proof.

(a) In characteristic ,

The roots of are the sixth roots of unity that are not roots of . They all lie in because , but not all lie in because . Hence the splitting field is .

(b) In characteristic , the roots of are the th roots of unity that are not th roots. Thus the splitting field is the smallest containing all th roots of unity, equivalently the smallest with

The order of modulo is , so the splitting field is .

Proof.

Put and . The three displayed roots are . Every automorphism of the splitting field permutes , and therefore permutes . Their elementary symmetric functions are fixed by the Galois group and lie in , proving (a).

For (b), the three transformed roots are distinct. Indeed,

If , then and , contradicting the transitivity of the -action on the roots of . The same applies to other pairs. Thus the -action on the transformed roots is the same faithful permutation action as on . Their splitting field is therefore the original splitting field, and its Galois group is .

Proof.

In characteristic ,

and for , . Hence and , so is -linear.

Every satisfies , so . The map is not the identity, since its fixed field is exactly , a proper subfield of . Therefore its minimal polynomial is

Since the vector-space dimension is , the characteristic polynomial is also

In characteristic , this is , and the conclusion remains valid.

Proof.

The required identity is equivalent to

Over , let be the companion matrix of

Then , is nonsingular because the constant term of is nonzero, and is not scalar. For , set

The scalar is also a root of in , since . Hence , so

Thus . The matrix is nonsingular and is not because of the companion block.