2021 Spring Algebra

Problem 1.

(a) Let be a group of order . Prove that is isomorphic to a subgroup of .

(b) Is every group of order 6 necessarily isomorphic to a subgroup of ? Explain

Proof.

(a) Given any group action of on a set , we have an injective group homomorphism . If we consider the group action of acting on itself by left multiplication, , we deduce that there is an injective group homomorphism from to , and the image of under this map is a subgroup of isomorphic to .

(b) Consider the cyclic group of order 6. We claim that no subgroup of is isomorphic to this group. (There is a subgroup isomorphic to , so this is the only possible example.) The only elements in of order 6 are either 6-cycles or 3,2-cycles, and neither such element is even.

Problem 2.

Let denote a finite group with identity element . Assume is a fixed integer for which for all .

(a) Prove that such an always exists.

(b) Define

Prove that the index of in is equal to the order of .

Proof.

(a) We may certainly assume is not trivial, otherwise any value of will work. In this case, is a mulitple of the order of every element in , so that

for all .

(b) Let be the function defined by . It follows from the choice of that this is a group homomorphism. It is clear from the choice of that it has kernel and image . From the First Isomorphism Theorem we have

from which it follows that

Problem 3.

(a) Define what it means for an integral domain to be a Euclidean domain.

(b) Prove that is a Euclidean domain.

Proof.

(a) The integral domain is Euclidean if there exists a function satisfying the following property:

(b) We have a norm function on defined by . We note that this norm is multiplicative. We show that if with , then there exist with

where or . Then main idea is to divide in . We have

Let be the nearest integer to and be the nearest integer to . So

where . Let

Note that

This completes the proof.

Problem 4.

Does there exist a ring (commutative, with unity) such that its underlying additive group is isomorphic to ? Prove your answer.

Proof.

The answer is no. Notice that is a torsion abelian group with elements of every possible finite order. If a ring has characteristic 0, then is a non-torsion element. On the other hand, if has characteristic , then every element of is -torsion because

In either case, the characteristic of prohibits its underlying additive group from being isomorphic to .

Problem 5.

Let be a commutative ring.

(a) Assume that every ideal of is finitely generated. Prove that, for every countably infinite chain of ideals in , there exists an integer such that for all .

(b) Give an example of a chain of ideals in a PID which contains 5 distinct ideals:

Proof.

(a) First suppose that every ideal of is finitely generated, and consider a chain of ideals . Then union is also an ideal of , and thus has the form for a finite list of generators . Because each , there exists such that . Then for each we have

So for all .

(b) If we take , then we may take the following chain of ideals as an example:

Problem 6.

Assume is algebraic over and assume . Prove that there exists a polynomial such that .

Proof.

Fix a polynomial of minimal degree such that . Writing , we have

Because has minimal degree, we must have , otherwise we could divide by the lowest power of in the polynomial above to reduce the degree. Multiplying by and rearranging, we have

Thus is a suitable polynomial.

Problem 7.

Let denote a splitting field of over

(a) Determine the degree of . Briefly justify your answer.

(b) Write out the elements in . (To receive full credit, explain how you know that the elements you write down are actually automorphisms.)

(c) Write out the intermediate fields between and , and for each intermediate field, list the subgroup of to which it corresponds. (No justification is necessary for this part.)

Proof.

(a) The four roots of this polynoial are , . It's clear that the tower consists of two extensions of degree 2 (because the middle field is real and the top field is not real, we know that the second field extension is not degree 1). Thus the degree is 4.

(b) These correspond to the four possibilities and . Because automorphisms must send roots of a polynomial to other roots of a polynomial, these are the only possibilities, and because we know the extension is Galois (being a splitting field over a characteristic zero field) and of degree 4, we know there are exactly 4 automorphisms, and hence each of these possibilities is indeed an automorphism.

corresponds to
corresponds to
corresponds to
corresponds to
corresponds to

Problem 8.

Let be a prime and be a positive integer satisfying . Let be a Galois extension where where is a positive integer. Prove that there is a subfield of containing with such that is Galois.

Proof.

Let and let be a subfield of containing . Let be the subgroup of corresponding to under the Galois correspondence. Then is Galois if and only if is normal in . Therefore, we need only show that contains a proper non-trivial normal subgroup. That is, is not simple.

Let be the number of Sylow -subgroups of . By the Sylow Theorems, and . Since , . Since has a unique Sylow -subgroup, this subgroup in normal in . The fixed field of this subgroup satisfies and Galois.

Problem 9.

Determine the number of isomorphism classes of -modules of order 9. Include a full justification of your claim.

Proof.

An -module is a vector space over together with a linear transformation . (The linear transformation comes from the action of on .) The order of the -module is . So in our case, is a 2-dimensional vector space over .

The classification of modules over a PID applied to the case of (which is a PID) says

where each is a monic polynomial of degree at least 1 and for . These are the invariant factors of the -module. Moreover, two such modules are isomorphic if and only if they have the same factors.

It is clear that . For example, note that is a basis for this quotient vector space. Therefore, we need only list the possible invariant factors for which . There are two possibilities:

and is a monic polynomial of degree 2. There are 9 such polynomials. (You do not have to list them.)
. In this case and are linear polynomials. Since they must be the same linear polynomial. There are 3 possibilities since there are 3 monic linear polynomials in .

Thus there are 12 such -modules up to isomorphism.

Problem 10.

Let be positive integers and let . Prove that

Proof.

Solution 1:

The tensor product is a -module, a finite abelian group. Since 1 is a generator for and 1 is a generator for , we see that is a generating set for . So this tensor product is a cyclic group.

We now show that the order of this group divides . We have

and also

Therefore the order of this cyclic group divides and divides , so it divides .

To show that we show that there is a surjective -module homomorphism from to . We know that -module homomorphisms from to correspond to -bilinear maps

Therefore, we need only construct such a map. Let . We need to check that and that . Checking each claim is straightforward. For example,

Solution 2:

If is an -module, we always have , so it suffices to show that the principal ideal in is equal to the principal ideal . Since , the inclusion is clear. For the converse direction, notice that for some integers , so .