2

If $R$ is a commutative ring with $1\ne 0$ and $R^m \cong R^n$ as direct sums of $R$-modules, then $m=n$.

There is a hint to this problem (I don't know how to use it): Reduce this to the case of a field.

I noticed that there was a similar question. But as a beginner of commutative algebra, I don't know if my question is closely related to that one. Thanks in advance!

Community
  • 1
No One
  • 7,939
  • Localize at an arbitrary prime. – anomaly Sep 09 '15 at 15:23
  • Closely related: http://math.stackexchange.com/questions/1359326/prove-that-mathbb-zn-is-not-isomorphic-to-mathbb-zm-for-m-neq-n/1359343#1359343 – user26857 Sep 09 '15 at 21:38
  • This is not a duplicate of the other two. The one from AM assumes they're isomorphic as rings, this one assumes they're isomorphic as $R$-modules, which makes for a very simple proof. – Robert Cardona Nov 05 '15 at 22:39
  • @RobertCardona: I don't see anything in this question or either of the duplicates saying that they are isomorphic as rings, as opposed to as modules. – Eric Wofsey Nov 05 '15 at 22:59
  • @EricWofsey the first one is referencing AM 2.11 which assumes they are isomorphic as rings (which is what makes the problem challenging); you're right about the second one. Let me remove my vote to reopen! My bad! – Robert Cardona Nov 05 '15 at 23:03
  • @RobertCardona: Oh, huh, the original version of the first linked duplicate was asking about rings rather than modules, but it has since been edited to talk about modules instead. – Eric Wofsey Nov 05 '15 at 23:08

1 Answers1

6

Asume $R^m \cong R^n$ as $R$-modules. $\mathfrak{m} \subset R$ be a maximal ideal, $R/\mathfrak{m}$ the residue field.

Tensoring with $R/\mathfrak{m}$, we get $R^m \otimes R/\mathfrak{m} \cong R^n \otimes R/\mathfrak{m}$, hence $R^m/\mathfrak{m} R^m \cong R^n/\mathfrak{m} R^n$. Both of these are finite dimensional vector spaces over $R/\mathfrak{m}$ of dimension $m$ and $n$ respectively.

Hence, by classification of finite dimensional vector spaces, $m = n$.

Balarka Sen
  • 13,910
  • @downvoter Care to explain the downvote? – Balarka Sen Sep 16 '15 at 10:41
  • @user26857 I thought the point of voting system was to evaluate answers based on it's quality? Besides, when I posted the answer, the question was not yet marked as a duplicate, so I did not know (although suspected) that this proof had appeared before in MSE. Otherwise I would not have answered it and would have voted to close as duplicate, of course. – Balarka Sen Oct 29 '15 at 08:26