Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is what I'd like to prove:

Let $R$ be a commutative, noetherian ring, and let $M$ and $N$ be finitely generated $R$-modules. Suppose $M\otimes_RN\cong R$. Does it follow that $M\cong N\cong R?$

Whether or not this is true, I'm wondering what conditions are necessary for something like it to be true, that is, what conditions $\star_R$, $\star_M$, and $\star_N$ make the following statement true:

Let $R$ be a ring satisfying $\star_R$, and let $M$ be a right $R$-module satisfying $\star_M$ and let $N$ be a left $R$-module satisfying $\star_N$. If $M\otimes_RN\cong R$, then $M\cong N\cong R$.

Also, what about higher ranks, that is, what can be said in the case $M\otimes_RN\cong\bigoplus_{i=1}^nR$? I'd appreciate proofs in the affirmative and/or counter examples. Thanks!

Edit: In light of Martin's and Qiaochu's responses, is $\text{Pic}(\text{Spec } R)$ trivial when $R$ is local?

share|cite|improve this question
You cannot prove this. Search for "Picard group" in google and in books. – Martin Brandenburg Apr 15 '13 at 17:49
up vote 4 down vote accepted

This is false. As Martin Brandenburg points out in the comments, the keyword is Picard group. For any commutative ring $R$, there is a group $\text{Pic}(\text{Spec } R)$ consisting of all modules which are invertible with respect to the tensor product, and your claim holds iff the Picard group is trivial (which it is generally not). If $R$ is a Dedekind domain, then this group can be identified with the ideal class group of $R$, so to exhibit a counterexample it suffices to exhibit a ring of integers in a number field which fails to have unique prime factorization.

For example, take $R = \mathbb{Z}[\sqrt{-5}]$. This fails to have unique prime factorization since $(1 + \sqrt{-5})(1 - \sqrt{-5}) = 2 \cdot 3$. Let $M$ be the ideal $(2, 1 + \sqrt{-5})$ regarded as an $R$-module. Since this is not a principal ideal, $M$ is not a free module. But as it turns out,

$$(2, 1 + \sqrt{-5}) \otimes (3, 1 - \sqrt{-5}) \cong R.$$

Edit: If $R$ is local, then $\text{Pic}(\text{Spec } R)$ is trivial. This follows from the observation that any element of the Picard group is necessarily (finitely generated and) projective (see, for example, MO) and that any projective module over a local ring is free.

share|cite|improve this answer
Thanks Qiaochu. Can you see my edit? – Jared Apr 16 '13 at 0:12
Awesome. Thanks so much. This is great. – Jared Apr 16 '13 at 0:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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