2017 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Suppose has eigenvalues and , and no others. Determine all possible pairs of characteristic and minimal polynomials, and give an example for each.

Proof.

Both eigenvalues must occur. Thus the algebraic multiplicities are either or . The possible pairs and examples are

These exhaust the possibilities because the exponent of each factor in the minimal polynomial is the size of the largest Jordan block for that eigenvalue.

Problem 2.

Prove that no group of order is simple.

Proof.

Since , the number of Sylow -subgroups satisfies

Thus or . If , the Sylow -subgroup is normal.

If , conjugation on the six Sylow -subgroups gives a nontrivial homomorphism

Were simple, its kernel would be trivial, so would embed in . Then would divide by Lagrange's theorem, which is false because . Therefore is not simple.

Problem 3.

Suppose is abelian and . Decide whether

Proof.

The statement is false. Take

Let

and let

Both subgroups have index , and hence

But and are not isomorphic.

Problem 4.

Determine, up to isomorphism, all -modules of order .

Proof.

Such a module is a two-dimensional -vector space, and multiplication by is an arbitrary linear operator. Module isomorphism is therefore the same as similarity of the corresponding matrices over .

The rational canonical forms give exactly six possibilities:

They correspond respectively to the zero operator, the identity, the two nontrivial Jordan blocks, an operator with distinct eigenvalues , and the companion matrix of the unique irreducible quadratic over .

Problem 5.

Suppose has characteristic , is the splitting field of an irreducible , and is abelian. If is a root of , prove that .

Proof.

Let and let

Because is irreducible in characteristic , it is separable, and acts transitively on its roots. If is another root and , then, using commutativity of ,

Thus every element of fixes every root of . Since those roots generate the splitting field , every element of fixes all of . Hence , and the Galois correspondence gives .

Problem 6.

(a) For , find the Galois group of its splitting field.

(b) Do the same over .

Proof.

The splitting field over is the smallest containing all st roots of unity, so is the multiplicative order of modulo .

(a) Since

and no smaller positive power of is modulo , the splitting field is . Therefore its Galois group is cyclic of order , generated by .

(b) Since

while and , the splitting field is . Its Galois group is cyclic of order , generated by .

Problem 7.

Which of the following ideals of are prime, and which are maximal?

Proof.

We examine the quotient rings.

For ,

which is a domain but not a field. Thus the ideal is prime but not maximal.

For ,

The nonzero classes of and multiply to zero, so the ideal is neither prime nor maximal.

For ,

which is a domain but not a field. Thus the ideal is prime but not maximal.

For , the quotient is

The polynomial has no root in , so it is irreducible. The quotient is a field; hence the ideal is maximal and prime.

For , the quotient is

But over , so the quotient is not a domain. The ideal is neither prime nor maximal.

Problem 8.

If is a finite normal subgroup of a group , , and has an element of order , prove that has an element of order .

Proof.

Let have order . Then . Since is finite, has finite order, so has finite order, say . The order of the image divides the order of , so . In the cyclic group of order , the element

has order exactly . Thus contains an element of order .

Problem 9.

Determine whether each statement is true or false, with justification.

(a) Every commutative ring with identity having exactly elements has zero divisors.

(b) For every prime , there is a nonzero ring homomorphism .

(d) If is a subfield of and is isomorphic to as fields, then .

(e) For every integral domain and every -module , the torsion elements form a submodule.

Proof.

(a) True. A finite commutative ring with identity and no zero divisors is a field. A finite field has prime-power order, but

is not a prime power.

(b) False. Any nonzero homomorphism to a field sends to . The image of would have to satisfy . For , no such element exists in .

is abelian because every element of commutes with , and it properly contains .

(d) False. For example,

but the fields are isomorphic by sending the transcendental element to .

(e) True. If and for nonzero , then

Since is a domain, , so is torsion. Also, if is torsion and , the same nonzero element annihilating annihilates . Thus the torsion elements form a submodule.