2022 Fall Qualifying Exam in Algebra

Proof.

Strictly speaking, the formula defines a right action if it is written as . Equivalently, one may obtain a left action by defining . The stabilizers and orbits occurring in the problem are the same under either convention, so we use the stated composition formula.

(a) For ,

Thus stabilizes precisely when

for every . This is equivalent to saying that commutes with every element of . Therefore,

(b) Suppose first that every automorphism of is inner. If , then for some . Hence

so lies in the orbit of . Thus every element of belongs to the same orbit, and the action is transitive.

Conversely, suppose the action on is transitive. Given , transitivity implies that lies in the orbit of . Therefore, for some ,

Thus is inner. Since was arbitrary, every automorphism of is inner.

Proof.

Suppose first that is cyclic of order . If , then Lagrange's theorem shows that has no elements of order . If , a cyclic group has a unique subgroup of order . That subgroup is cyclic, and all of its nonidentity elements have order . Therefore has exactly elements of order .

Conversely, suppose that for every prime , the group has either no elements of order or exactly such elements. Decompose into its primary components:

where is the Sylow -subgroup of . By the classification theorem for finite abelian groups,

for some integers . The elements annihilated by form the subgroup

This subgroup has elements. Its nonzero elements are exactly the elements of order , so has elements of order . By hypothesis,

which implies . Hence every primary component is cyclic.

The orders of the different primary components are relatively prime, so their direct product is cyclic. Therefore is cyclic.

Proof.

An invertible matrix is determined by an ordered basis of . The first column may be any nonzero vector, the second may be any vector outside the span of the first, and the third may be any vector outside the span of the first two. Hence

Factoring out powers of gives

None of , , and is divisible by . Therefore the largest power of dividing is , so every Sylow -subgroup has order .

Consider the subgroup

The product and inverse of upper unitriangular matrices are again upper unitriangular, so is a subgroup of . There are independent choices for each of , and therefore

Thus is a Sylow -subgroup of .

Proof.

Let

be the quotient map followed by a fixed isomorphism. By the correspondence theorem, ideals of containing correspond bijectively to ideals of through inverse image under . This correspondence preserves maximal ideals.

For each prime number , the ideal is maximal in because

is a field. Therefore

is a maximal ideal of containing . If , then , so the correspondence theorem gives . Since there are infinitely many primes, there are infinitely many maximal ideals of containing .

Proof.

Let be a representative of the unique associate class of irreducible elements.

(a) Since is a UFD, every nonzero nonunit is a product of irreducibles. Every irreducible is associated to , so every nonzero element of has the form

for a unit and an integer .

Let be a nonzero ideal of . Consider the set

This is a nonempty set of nonnegative integers, so it has a least element . Multiplying by a unit inverse shows that , and hence

If is nonzero, write . The minimality of implies . Therefore

Thus , and consequently . The zero ideal is also principal, so every ideal of is principal. Hence is a PID.

(b) In a PID, every nonzero prime ideal is generated by an irreducible element and is maximal. Since every irreducible is associated to , the only nonzero prime ideal is . It is therefore the unique maximal ideal.

Because is a UFD, it is an integral domain, so is prime. It is not maximal because is not a field: the element is a nonzero nonunit. Thus is the unique nonmaximal prime ideal.

Proof.

We prove the result by induction on . Since , it is enough to prove that every submodule of is free of rank at most .

If , the result is immediate. If , then is an ideal of . Since is a PID, either or for some nonzero . Thus is free of rank or .

Assume the result holds for submodules of , and let . Let

be projection onto the last coordinate. Its image is an ideal of , so

for some .

If , then , and the induction hypothesis applies.

Suppose . Choose such that . If , then for some . Therefore

so . Hence

This sum is direct. Indeed, if , then

Since a PID is an integral domain and , we have . Thus

By the induction hypothesis, is free of rank at most , while is free of rank . Therefore is free of finite rank at most .

Proof.

We regard as an -module through the quotient map .

(a) If , then is even, so is an integer. Thus is well-defined. Its additivity in each variable follows immediately from the additivity of the constant coefficient of the first input and the coefficient of in the second input.

It remains to prove the balancing identity. Let

belong to , and let

The constant coefficient of is , so

The coefficient of in is . Hence

Because , its constant coefficient is even. Therefore

and consequently

Thus is -balanced.

(b) By the universal property of the tensor product, induces an -module homomorphism

satisfying . We have

and

Therefore

so is nonzero.

(c) Set

Using the balancing relation , we obtain

Similarly,

Thus both and annihilate .

(d) Consider the surjective -module homomorphism

Its kernel is . Part (c) gives

Conversely, suppose . Applying from part (b), we get

Since in the -module , this equality says precisely that the image of in is zero. Thus . Therefore

The first isomorphism theorem now gives

Proof.

Clearly,

The element satisfies the polynomial

Therefore

If , then this degree cannot be , so it must equal . The tower law would then give

where the product on the right is divisible by . This contradicts the assumption that is odd. Hence , and therefore

Proof.

Since , we may identify with . Its multiplicative group is cyclic of order

Thus divides . In particular, the assumption actually implies , since no divisor of lies strictly between and .

Let

Every subfield of has order for some divisor of . Hence a proper subfield has order , , or . Its multiplicative group therefore has order , , or .

If were a proper subfield of , then the order of would divide one of , , or . This is impossible because . Therefore is not proper, and

Proof.

By the Galois correspondence, the conjugate subgroups

have fixed fields

and these fields are isomorphic to over .

Conversely, suppose is -isomorphic to . A -embedding extends to an automorphism of an algebraic closure of . Since is normal, the extension sends to itself and therefore restricts to an element . It follows that

Thus the desired fields correspond exactly to the conjugates of in .

There are

different -cycles in . Every cyclic subgroup of order contains exactly four -cycles, namely its four nonidentity elements. Moreover, two distinct subgroups of order have trivial intersection, since any nonidentity element would generate both subgroups. Therefore the number of cyclic subgroups generated by -cycles is

All such subgroups are conjugate in . Hence there are exactly

subfields of that are isomorphic to over .