2000 Winter Qualifying Exam in Algebra (AI-generated)

Proof.

Count the set of pairs satisfying . Counting first by gives the first sum, while counting first by gives the second. This proves (a).

For each orbit and , orbit-stabilizer gives . Therefore

Summing over all orbits and using (a) proves Burnside's formula.

Proof.

Put in rational canonical form. If its companion blocks have sizes , a companion block of size has possible nonzero entries only in its subdiagonal positions and its last-column positions. Thus the whole block-diagonal matrix has at most

possibly nonzero entries. It consequently has at least

zero entries. For ,

and the assertion is immediate for .

Proof.

The operator is nilpotent because , but it is not the zero operator because . If a nilpotent operator were diagonalizable, all its eigenvalues would be , so its diagonal form would be the zero matrix and the operator itself would be zero. This contradiction proves that is not diagonalizable.

Proof.

Conjugation gives a homomorphism

Since , its automorphism group has order . The order of the image divides both and . Any prime divisor of a nontrivial image would therefore divide and be smaller than , contradicting the minimality of . Hence the image is trivial. Every element of centralizes , so .

Proof.

Discreteness implies that there is a shortest nonzero vector . The subgroup is cyclic: if it contained an element not in , subtracting a suitable integer multiple of would produce a shorter nonzero vector on the same line.

Because is not cyclic, choose an element outside . Orthogonally project onto a line perpendicular to . Discreteness implies that the nonzero projected lengths have a positive minimum; choose attaining it. For any , subtract an integer multiple of so that the remaining projection has absolute value smaller than that of . Minimality forces that projection to be zero. The remainder lies in . Thus

for integers . Since are linearly independent over , this expression is unique. Hence

Proof.

If had index , the action on the five left cosets would give a homomorphism

The action is transitive and hence nontrivial. Since is simple, its kernel would be trivial. This would embed a group of order

into , which has order , an impossibility. Therefore no such subgroup exists.

Proof.

All complex eigenvalues of a real orthogonal matrix have absolute value . Nonreal eigenvalues occur in conjugate pairs, each pair having product . The real eigenvalues are or . Since the product of all eigenvalues is , an odd number of the real eigenvalues must equal . In particular, is an eigenvalue.

Without the determinant hypothesis, the rotation

is orthogonal with determinant and eigenvalues , so it has no eigenvalue .

Proof.

Every representation of a compact group is completely reducible. Let be the unique irreducible -representation of dimension . Decompositions of a -dimensional representation correspond to partitions of , giving exactly

and

Uniqueness of irreducible decomposition makes them pairwise nonisomorphic.

Proof.

Using ,

gives two inequivalent factorizations into irreducibles. The needed irreducibility follows because the norm equations and have no solutions. Thus is not a UFD.

Put

Then

and hence

The displayed ideals are prime because their quotients are or .

Proof.

The relation matrix

has rank , the gcd of its entries is , and the gcd of its minors is . Its Smith normal form is therefore

Consequently the group is

Proof.

For ,

where the sum is in .

For , the number of zeros satisfies

in . The constant sum is zero. Every monomial in has total degree , so at least one variable exponent is less than . Summing in that variable makes the monomial contribution zero. Thus in , meaning . Since the zero vector is one solution, cannot equal , and therefore a nonzero solution exists.

Proof.

By Hilbert's basis theorem, each is Noetherian (over its given Noetherian coefficient ring, in particular over a field). Every ideal of the product has the form

where is an ideal of : multiply elements of by the idempotents and to separate their components. If and , then is generated by

Thus every ideal is finitely generated, so the product is Noetherian.

Proof.

Let and . The splitting field is

Eisenstein at gives , while . Their intersection has degree dividing the coprime integers and , hence is . Therefore

The automorphisms are

with and . Consequently,

where the second factor acts faithfully by multiplication on .

Proof.

Let , a power of , and translate by :

Every interior binomial coefficient is even because is a power of . Hence every nonleading coefficient of is divisible by . Its constant term is

which is not divisible by . Thus is Eisenstein at . It is irreducible, and translation preserves irreducibility, so the original polynomial is irreducible.