2006 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) A proper ideal of a commutative ring is prime if implies or . Equivalently, is an integral domain.

(b) If with , a Sylow -subgroup of is a subgroup of order .

(c) The polynomial ring over a field is a UFD, but it is not a PID: the ideal is not principal.

(d) In , the zero ideal is prime because is an integral domain, but it is not maximal because is not a field.

Proof.

Let and . The three roots are , so

Eisenstein at shows that is irreducible, hence . The field is real while is not, so adjoining has degree . Therefore

Define

Then and . The six automorphisms are

so

Proof.

Choose an eigenvalue of and let

This is a nonzero subspace. If , then

so . Thus is invariant under . The restriction is an operator on a nonzero finite-dimensional complex vector space, so it has an eigenvector . Then and for some , making a common eigenvector.

Proof.

Let have order . Then . Since is finite, has finite order, so has finite order, say . The order of the image of in divides the order of , so . Therefore the element

has order in .

Proof.

(a) A matrix commuting with every invertible diagonal and elementary matrix must be scalar. Hence the center of consists of with . Since ,

(b) The group acts by fractional linear transformations on the projective line

which has four points. This gives a homomorphism

If a projective transformation fixes all four points, in particular it fixes . A fractional linear transformation fixing these three distinct points is the identity; hence the action is faithful.

Moreover,

Thus its faithful image is a subgroup of of order , so it is all of .

Proof.

Let . Completing the square gives

There are three cases.

If , the polynomial has two distinct real roots . The corresponding linear ideals are comaximal, and the Chinese remainder theorem gives

If , putting gives

the ring of dual numbers.

If , put . The class

satisfies , and hence

Proof.

Let be a Sylow -subgroup of . Since , its order is for some . It remains to show that a finite -group of order has a subgroup of every order , .

Proceed by induction on . A nontrivial finite -group has nontrivial center, so choose a central subgroup of order . The quotient has order . If , induction gives a subgroup of order . Its inverse image under is a subgroup of of order . Taking proves the result.

Proof.

Define a map of sets

It is well-defined. It is also injective: if and , then , so . The target is finite, and therefore

Proof.

Every complex matrix has a Jordan canonical form. An invertible one has no zero eigenvalue. Therefore every conjugacy class has exactly one representative of one of the following types, up to interchanging and in the first type:

or

The first family includes the scalar matrices when . Two diagonal representatives are conjugate exactly when they have the same unordered pair of eigenvalues; two Jordan-block representatives are conjugate exactly when their eigenvalues agree. No matrix in the second family is conjugate to one in the first because the former is not diagonalizable.

Proof.

Put . Since and is finite, the structure theorem over the PID gives

Taking base- logarithms of the three cardinalities gives

and

It follows that , , and . Therefore