2001 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) Suppose and choose , so . Multiplying by gives a nonzero translation in . Conjugating it by gives translations by every vector in the -orbit of . This group acts transitively on the seven nonzero vectors, so contains every translation. Together with , these generate . Hence .

(b) The displayed formula defines an action on the eight-element set . If acts trivially, evaluating at gives , and then for every , so . Thus the action is faithful and yields an embedding .

(c) A nonzero translation partitions the eight vectors into four pairs , so as a permutation it is a product of four transpositions and is even. The group is generated by elementary transvections. A transvection fixes a hyperplane of four vectors and interchanges the remaining four in two pairs, so it too is even. Hence both the translation subgroup and act by even permutations, and therefore .

Proof.

(a) The center has order or . If it has order , then is abelian. If it has order , then has order and is cyclic. Whenever is cyclic, is abelian: if , all elements have the form and commute. Thus is abelian in either case.

(b) The group

has elements and is nonabelian; for example, the elementary matrices with nonzero and entries do not commute.

Proof.

(a) If , then

so , and

so . Both Sylow subgroups are normal, and therefore

(b) Write . There is always the cyclic group . A nonabelian group exists exactly when

because this is exactly when there is a nontrivial homomorphism

giving a nontrivial semidirect product . Thus there is exactly one group of order precisely when the smaller prime does not divide one less than the larger prime.

Proof.

(a) There is exactly one pair of nonreal roots. Complex conjugation fixes the real roots and interchanges that pair, so it acts as a transposition.

(b) Irreducibility makes the action on the roots transitive, so divides the group order. By Cauchy's theorem the group contains an element of order , necessarily a -cycle. Conjugating the transposition by powers of that cycle produces transpositions corresponding to the edges of a connected graph on the roots. Such transpositions generate . Hence the Galois group is .

(c) Modulo , the polynomial becomes , which is irreducible. For example, the finite-field criterion is verified by

Thus is irreducible over .

Its derivative is , with real critical points , where . Since ,

Together with the limits at , this gives exactly three real roots. Therefore

Proof.

Write

The conjugacy classes are

Thus representatives are . Moreover,

and

The standard representation is the symmetry action on the plane:

Since , there are four one-dimensional representations, obtained by independently assigning the values . The standard two-dimensional representation is irreducible and is the fifth. The degree sum

shows that the list is complete.

Proof.

(a) An antisymmetric matrix has zero diagonal and one free entry for every pair , so

(b) If , then

(c) Under the natural identification of antisymmetric bilinear forms with , is the operator induced by on the second exterior power. If the eigenvalues of over are , then the eigenvalues of are

counted with algebraic multiplicity.

If is diagonalizable, so is the induced operator, and therefore

counting multiplicities.

For an arbitrary matrix , the rank is not determined by its eigenvalues alone, so the rank request needs a diagonalizability hypothesis. For instance, and in dimension have the same eigenvalues, but while . In general the exact rank is

and its nullity depends on the Jordan structure of .

Proof.

(a) Let permute the standard basis:

These are permutation matrices. If , then fixes all three indices, so . Thus the kernel is trivial.

(b) The line

is fixed pointwise.

(c) Take

It is permutation-invariant and is generated by

Since and their dimensions add to ,

Proof.

The group is finite abelian. Let be its exponent, the least common multiple of the orders of its elements. Every element of is a root of , so

On the other hand, the exponent of a finite group divides its order, so . Hence .

For a finite abelian group, there is an element whose order equals the exponent: choose elements of maximal prime-power order in each primary component and multiply them. Therefore has an element of order , and that element generates .

Proof.

(a) Since , the maximal ideals are

(b) Since , the maximal ideals are

(c) The two factors in part (a) are coprime, so the Chinese remainder theorem gives

Proof.

(a) Take

It has no root in , so as a cubic it is irreducible. Therefore

and this field has elements; hence it is .

(b) The relation gives . Therefore

Using these coordinate vectors as columns, the matrix is

Proof.

(a) Since is squarefree and congruent to modulo ,

The minimal polynomial of is

(b) No. The field discriminant is , so ramifies. More explicitly, modulo ,

and

Thus is not prime.

(c) No. Since does not divide the discriminant , it is unramified. In fact

so splits: