2019 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Does contain a subgroup isomorphic to (a) or (b) ?

Proof.

(a) Yes. The permutations

fix and satisfy and . Hence they generate a subgroup isomorphic to .

(b) No. If embedded in , its action on five letters would decompose into orbits whose sizes divide . Faithfulness requires an orbit of size or ; size is impossible, so it would give a faithful embedding into . An order- subgroup of is Sylow and is isomorphic to , not . Therefore no such embedding exists.

Problem 2.

Let be finitely generated abelian, , and .

Prove that if is torsion-free, the isomorphism classes of and determine that of . Give a counterexample when has torsion.

Proof.

A finitely generated torsion-free abelian group is free, so . Free modules are projective, and therefore the exact sequence

splits. Hence

which is determined by the two isomorphism classes.

For a counterexample, take . Both

contain a subgroup isomorphic to with quotient isomorphic to , but and are not isomorphic.

Problem 3.

Define and . Prove that every finite -group is nilpotent.

Proof.

We use induction on . A nontrivial finite -group has nontrivial center. If , then . Otherwise, is a smaller -group and is nilpotent by induction. Suppose its lower central series reaches the identity after steps. Then

Consequently

Thus the lower central series of terminates, and is nilpotent.

Problem 4.

Let be commutative with identity.

(a) Let be multiplicatively closed and let be maximal among ideals disjoint from . Prove that is prime.

(b) If is the nilradical, prove the equivalence of:

has exactly one prime ideal.
Every element is nilpotent or a unit.
is a field.

Proof.

(a) Suppose but . By maximality, both and meet . Choose

with . Expanding shows that every term lies in , including , so . But , contradicting . Hence is prime.

(b) Suppose there is exactly one prime ideal . Since the nilradical is the intersection of all prime ideals, . It is also the unique maximal ideal. If is not nilpotent, then . Hence is contained in no maximal ideal, so and is a unit. This proves 1 implies 2.

If 2 holds, every nonzero class in is represented by a nonnilpotent element and is therefore a unit. Thus the quotient is a field, proving 2 implies 3.

If is a field, it has only one prime ideal, namely zero. Every prime ideal of contains , and the correspondence theorem then shows that is the unique prime ideal of . Thus 3 implies 1.

Problem 5.

Recall that is Euclidean.

(a) Prove that is finite for every nonzero ideal .

(b) Identify .

Proof.

(a) Since is a PID, write with . Multiplication by identifies as a sublattice of index

in the rank-two lattice . Therefore the quotient has elements and is finite.

(b) The norm of is , so the quotient has two elements. The map

has kernel . Hence

Problem 6.

Let be squarefree. Prove that the primitive th roots of unity form a basis of over .

Proof.

For a prime , the primitive th roots

form a basis of : the only rational linear relation among all roots is their sum, so deleting leaves independent elements.

Now write the odd part of the squarefree integer as . Cyclotomic fields of distinct prime conductors are linearly disjoint over ; this follows, for example, because their ramified-prime sets are disjoint. Therefore products of the displayed prime-level bases form a basis of

By the Chinese remainder theorem, these products are exactly the primitive th roots, each occurring once. If , then with odd and ; the primitive th roots are the negatives of the primitive th roots. Thus the conclusion also holds in the even squarefree case. Their number is , the field degree, so they form a basis.

Problem 7.

How many primitive elements does have over ?

Proof.

The subfields of are for divisors of . The only proper subfield is therefore . An element generates over exactly when it is not in the proper subfield. Hence the number is

Problem 8.

Let be a Galois algebraic extension with no proper intermediate fields. Prove that is prime.

Proof.

Choose . Since the extension is algebraic, is finite. The absence of proper intermediate fields forces , so is finite.

By the Galois correspondence, subgroups of correspond to intermediate fields. Thus has no nontrivial proper subgroup. If were composite, Cauchy's theorem would give a subgroup of prime order strictly between and . Therefore is prime. Since , the degree is prime.

Problem 9.

Let be a rational vector space with , where is prime. If , prove that .

Proof.

The minimal polynomial of divides

Over , the cyclotomic polynomial is irreducible and has degree . If , its minimal polynomial must contain , and would therefore have degree at least . But the degree of a minimal polynomial is at most , a contradiction. Hence .

Problem 10.

Let be nilpotent, with the same nilpotency index and satisfying

Prove that (i) nonsimilar examples exist for , while (ii) and are similar for .

Proof.

A nilpotent similarity class is determined by the partition of given by its Jordan block sizes. For a partition ,

and the nilpotency index is .

(i) In dimension , the distinct partitions

both have nilpotency index , rank , and square-rank . Their Jordan matrices are not similar. For every , append blocks of size to both partitions. For use the displayed pair, so examples exist for every .

(ii) Let be the number of blocks of size . The three given invariants determine

together with the largest for which . For , these data determine the uniquely: if the largest block is at most , solve successively for from the displayed equations; if it is at least , there is only one such block and the remaining total size is at most , where the same equations determine the remainder. Therefore the Jordan partitions agree, and and are similar.