3
$\begingroup$

So, I'm a biologist at KCL, but I quite like mathematics and so am going through a book of exercises in algebra. Unfortunately, I've run into a problem in trying to answer some of the questions. I've been told that here might be a nice place to ask and so I was wondering if anyone would be able to help me (my mathematician friends also tend to be strict about notation, so apologies if it's not right). The questions lead you to showing if you have a principal ideal domain and a finitely generated module over it, then that module must be the distinct sum of a free module and its torsion. I can get to this point. However, it then asks if this free module is unique and why this may be false if it's not over a principal ideal domain. Here is where I get stuck.

Attempt at solution: I don't fully understand what it means by unique. I tried to generate some modules to see if I could see what was happening, but I think I'm falling down at understanding how to use the torsion. I tried things like the integers, but ended up with no torsion for any module I could think of. I the tried things like Z modulo nZ, but couldn't think of modules over them. Naturally, I looked online and I think this is known as the Structure Theorem For PIDs, hence the title, but couldn't find any concrete examples. I've also read that the ideal (x,2) over the integer polynomial ring provides a counter example, i.e. the second part, but I can't see how.

$\endgroup$
  • $\begingroup$ The rank of the free module is unique. For examples of modules over the integers it is easy to find ones with torsion because every abelian group is a module over the integers so you can just find an abelian group with torsion. For example $\mathbb{Z}/n\mathbb{Z}$ is a torsion $\mathbb{Z}$ module. $\endgroup$ – Seth Mar 23 '14 at 19:35
  • $\begingroup$ Thank you! Sorry to ask though, but what is the rank of a free module? I've never seen that term before. I'll see if I can find it in the algebra book. So, when it asks if the free module is unique, it means is the rank unique and the answer is yes? $\endgroup$ – Zigzag Mar 23 '14 at 19:37
  • $\begingroup$ All free modules $R$ modules are direct sums of copies of $R$. The rank of the free module is the cardinality of the number of copies. For a FGM over a PID the statement is that it is the sum of a free part and a torsion part (where the torsion part has a particularly nice form). There is a uniqueness statement which tells you that the rank of the free part is unique. $\endgroup$ – Seth Mar 23 '14 at 19:39
  • $\begingroup$ For examples of free modules where the "rank" is not well defined see en.wikipedia.org/wiki/Invariant_basis_number $\endgroup$ – Seth Mar 23 '14 at 19:39
  • $\begingroup$ So, over Z, it is saying that the free module is just some (x,y,z,...) such that each entry is an integer where the number of letters is its cardinality? $\endgroup$ – Zigzag Mar 23 '14 at 19:41
1
$\begingroup$

Let's focus on the example you gave: view the ideal $I=(x,2)$ inside of the ring $R=\Bbb Z [x]$ as an R module.

Obviously it has no torsion because it's in a domain. That being the case, the entire module would have to be free, in order to fit the decomposition description. We will see this is not possible, so that we have an example of a finitely generated module over a domain which doesn't decompose nicely.

Lets now suppose that it is a free module. (Checking uniqueness of the module or it's rank is premature because we don't even know if it's free yet.)

If $I$ were a direct sum of more than one copy of $R$, then $I$ would contain two nonzero ideals of $R$ that have intersection zero. But this is not possible in a domain! On the other hand, if $I$ were isomorphic to $R$, it would be a cyclic module: but this is also false! Thus $I$ does not decompose into a torsion and free part.

$\endgroup$
  • $\begingroup$ I'm just going to have a look at what a domain is, but that seems quite intuitive and I understand it in part. So thank you. $\endgroup$ – Zigzag Mar 23 '14 at 20:11
  • $\begingroup$ Domain means Integral Domain? $\endgroup$ – Zigzag Mar 23 '14 at 20:12
  • $\begingroup$ @zigzag I'll be waiting if you have questions. Yes, I mean integral domain. Let me know if any other breadcrumbs need more explanation, too. $\endgroup$ – rschwieb Mar 23 '14 at 20:13
  • $\begingroup$ Yeah, I think that makes sense. As I said, I'm a biologist and I'm doing it for self-interest so I don't know how precise it needs to be but it satisfies me so thank you. The only other thing I would like to know is what Lang means by unique? Does he mean the rank or does he mean the representation? $\endgroup$ – Zigzag Mar 23 '14 at 20:15
  • $\begingroup$ Or does he mean both? $\endgroup$ – Zigzag Mar 23 '14 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.