2010 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

Let have order . Sylow's theorems give

so the Sylow -subgroup is normal. By Schur--Zassenhaus, is a semidirect product

where has order , so or .

Since

the image of the action has order at most .

If , there are two actions up to isomorphism: the trivial action and the action in which a generator of acts by inversion. If , there are again two actions up to isomorphism: the trivial action and a nontrivial action onto the unique subgroup of order in . All nonzero homomorphisms from to are equivalent under .

Thus the four groups are

and

The first two are abelian, and the last two are nonabelian.

Proof.

(a) False. Conjugating by gives , which is not in the subgroup.

(b) True. If and , then , so the center is invariant under conjugation.

(c) False. In general,

need not equal . For example, this fails for suitable noncommuting elements of .

Proof.

(a) In ,

Use the norm

There are no elements of norm or . It follows by norm multiplicativity that , , and are irreducible. Their norms show that no factor from one decomposition is associate to a factor in the other. Thus factorization is not unique.

(b) Put

Each quotient is respectively or , so these ideals are maximal and prime. Reduction of modulo gives , while reduction modulo gives . Consequently

Therefore

Proof.

Irreducibility is unchanged by translation, so consider

Every intermediate binomial coefficient

is even. Thus every nonleading coefficient of is divisible by . Its constant term is

which is not divisible by . Hence is Eisenstein at , so it is irreducible over . Therefore is irreducible as well.

Proof.

The minimal polynomial of a nilpotent operator is for some positive integer . Its degree is at most , so . Therefore

Equivalently, in Jordan form every nilpotent block has size at most .

Proof.

Write

Its conjugacy classes are

There are four linear characters because the abelianization is , and one two-dimensional character from the geometric action on the square. The table is

The squared degrees sum to , so the table is complete.

Proof.

Over , the cubic has no root:

A cubic is reducible if and only if it has a root, so the polynomial is irreducible and the quotient is a field, namely .

Over ,

Thus the polynomial is reducible and its ideal is not maximal. The quotient is not a field.

Proof.

The Jordan form must be unchanged when every nonzero eigenvalue is replaced by its reciprocal. The complete list is

and the family

where parameters and represent the same class. The scalar and mixed-sign cases cover the self-reciprocal eigenvalues, while the last family covers a distinct reciprocal pair. A nontrivial Jordan block at or is similar to its inverse because changing the nonzero superdiagonal entry does not change its Jordan type. Hence every listed matrix works, and the Jordan classification shows that none are missing.

Proof.

Let . The roots are , so the splitting field is

The polynomial is Eisenstein at , giving , and is not in the real field . Thus .

Define

Then

so

The complete subgroup--fixed-field correspondence is

The subgroup lattice has at the top; its three order- subgroups are , , and . The first contains ; the second contains ; and the third contains . All order- subgroups contain . The subfield lattice is the reverse of these inclusions.

Proof.

Factor

The factor is irreducible over and occurs once. The -primary part has total multiplicity , whose possible partitions are , , and . Thus there are exactly three classes. With

representatives are

and

Their distinct invariant-factor data show that they are pairwise nonsimilar and exhaustive.