I had my algebra qual yesterday. These are the problems along with my solutions. Hopefully they are correct

Some preliminary comments. First, I know my solution to 1(a) is incorrect. Second, I didn’t know how to prove problem 3. I literally wrote down garbage so I am expecting zero points for that one.

**1.** Let denote the additive group of rational numbers.

**a)** Prove that every finitely generated subgroup of is cyclic.

*Proof. *Let be a finitely generated subgroup of and let be the generators of . Pick . We want to show that can be generated by a single in . Then for some positive integer . But, . Thus, .

**b) **Prove that is not finitely generated.

*Proof.* Seeking a contradiction, suppose that is finitely generated. Note that is a subgroup of itself. Thus, is a finitely generated subgroup of and by part (a), this means that must be cyclic. This of course is a contradiction because is not cyclic.

**2. **List, up to isomorphism, all abelian groups of order .

*Solution.* The partitions of are , , and . The abelian groups associated with are:

The partitions of are and . The abelian groups associated with are:

The abelian groups associated with are:

By the fundamental theorem of finitely generated abelian groups, all abelian groups of order are isomorphic to one of the groups listed below and none of the groups is isomorphic to another.

**3. **Let be a vector space over a field and let be a linear operator such that . Prove that .

*Proof.* ðŸ˜¦

**4.** Let , be matrices with entries in a field and suppose that . Prove that if has an eigenspace of dimension one, then and share a common eigenvector.

*Proof.* Since has an eigenspace of dimension one, let be an eigenvalue of and let be the one eigenvector corresponding to . Then we have that and . Thus, which means that is an eigenvector of with corresponding eigenvalue . But is the only eigenvector corresponding to so . This shows that is an eigenvector of with corresponding eigenvalue .

**5. **Let be a principal ideal domain. Prove that every proper nontrivial prime ideal is maximal.

*Proof.* Let be a proper nontrivial prime ideal of . Let be an ideal of such that . Then for some in . Thus, . Since is prime, or .

Case 1: Suppose that . Then so .

Case 2: Suppose that . Then for some in . Thus, implies that and so is a unit. Thus, .

We have shown that or which means that is a maximal ideal of .

Each problem is worth 5 points, 15 is a passing score. Best case I think my score will be 2+5+0+5+5 which is 17 and a pass. Realistically, I think my score could be 1+5+0+4+5 which is 15 and a pass. But it could be 1+4+0+4+4 which is a 13 and not a pass. I feel very confidant about problems 2, 4, and 5; those should be five points each. But the qual committee can be very particular about grading. I also feel very confidant about 1(b); I just don’t know how many points that one is worth. All I can do now is wait…

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March 22, 2011 at 10:59 pm

j2kun1. (a)

Let be a subgroup generated by Let . Then we claim . This is relatively straightforward, because if then , so each can be obtained by a sum of , and then each is a sum of , establishing the claim.

3. For this one you should have read “Linear Algebra Done Right” by Sheldon Axler, because this is an exercise in one of the earlier chapters. (And I told ya’ll to read it!)

To show it suffices to show and . The second condition makes use of idempotency: if , then for some , and further , so , so cannot be in the kernel of , and these two subspaces are disjoint. The first then follows from the rank-nullity theorem, in that the dimensions of these two (disjoint) subspaces add up to the dimension of , and in particular if we append a basis for to a basis for , we will have linearly-independent vectors in an -dimensional space, and hence a basis for .

I hope none of my latex breaks, since I can’t preview the comment before I post it.

March 22, 2011 at 11:13 pm

j2kunYou made a mistake on problem 4. In particular, , and the eigenvalue is not . Since is an eigenvector of , you may only assume that is a multiple of , in the sense that the one-dimensional eigenspace of corresponding to is precisely . This follows since is again an eigenvector of . So you may write for some , which is enough to say that is an eigenvector of (corresponding to , which does not matter).

I hate to bear the bad news, but thats at least one point off for claiming a false statement.

March 23, 2011 at 12:17 am

RichardKendall mentioned to me after the exam that my solution to number four was incomplete. And it’s a really easy fix. Lets hope it’s only a one point deduction! Thanks for the insight on problem one. I’m really hoping that part b is worth 2 points.

As far as linear algebra done right (2nd edition), perhaps I’ll take a look at it in the future. I’m waiting for the third edition, when they really get it right.