Equivalent definitions of invertible modules

Question

Let $R$ be commutative unital ring, and $M$ an $R$-module. $M$ is called invertible (a.k.a. projective module of rank one), if it is finitely generated, and $M_{\mathfrak{p}} \cong R_{\mathfrak{p}}$ for every $\mathfrak{p} \in \operatorname{Spec} R$. Equivalently, there are $a_1, \dotsc, a_n \in R$ such that $(a_1, \dotsc, a_n) = R$ and $M_{a_i} \cong R_{a_i}$ for every $i$. There is a third equivalent definition: $M$ is finitely generated, and there is an $R$-module $N$ such that $M \otimes_R N \cong R$. See e.g. Bourbaki: Commutative Algebra, II, 5.4, Theorem 3, or Proposition 19.8 in Pete Clark's note on Commutative Algebra. Both of these references assume that $M$ is finitely generated in the third definition.

My question: is the finitely generatedness really necessary in the last definition? It seems to me that if $M \otimes_R N \cong R$, then $M$ and $N$ are automatically finitely generated. This would make the last definition really simple.

Remark: I have a proof in mind, so as a second question: is the following argument correct?

If $\varphi \colon M \otimes N \to R$ is an isomorphism, then $\varphi^{-1}(1) = \sum_{i=1}^s x_i \otimes y_i$ for some $x_i \in M$ and $y_i \in N$. Then let $M'$ be the submodule of $M$ generated by $x_1, \dotsc, x_s$. The composition $M' \otimes N \xrightarrow{\sigma} M \otimes N \xrightarrow{\varphi} R$ is surjective (because $\sum_{i=1}^r x_i \otimes y_i \in M' \otimes N$ goes to $1 \in R$), and $\varphi$ is an isomorphism, so $\sigma$ is also surjective. Let $M'' = M/M'$, so $0 \to M' \to M \to M'' \to 0$ is exact. Then $M' \otimes N \xrightarrow{\sigma} M \otimes N \to M'' \otimes N \to 0$ is also exact. However $\sigma$ is surjective, thus $M'' \otimes N = 0$. But then $$ 0 = (M'' \otimes N) \otimes M \cong M'' \otimes (N \otimes M) \cong M'' \otimes R \cong M'', $$ therefore $M = M'$. So $M$ is indeed finitely generated.

Answer 1

I will happily incorporate this strengthening of the statement of Proposition 19.8 of my notes. Thanks for this, and in the future please feel free to contact me directly (as well).

As an aside, I well remember Rota's criticism of commutative algebra texts for their "hygienic theorems": see this previous MO answer. In many places in my notes if I state a result which has an implication without a converse implication or a converse only under additional hypotheses, I follow it with an exercise, remark or reference about whether/why the converse is not true. I would like to do this all the time -- but I am not a real expert in the subject and I read the same textbooks as everyone else, so often (as here) I do not know enough to do so. That's why it's great to get responses like this from other mathematicians.

Answer 2

Here are two categorical comments.

The first is that $M$ is an invertible module in the sense of having an inverse with respect to the tensor product (and why would you pick any other definition? "Invertible" is right there in the name!) iff the functor

$$\text{Mod}(R) \ni N \mapsto N \otimes_R M \in \text{Mod}(R)$$

is an autoequivalence. This functor sends the unit module $R$ to $M$, from which it follows that invertible modules inherit any categorical properties that the unit possesses. Finite presentation turns out to be such a categorical property (namely, it is the categorical property of compactness), as is projectivity, and since both properties hold of $R$ they necessarily hold of any invertible module. Finite presentation and projectivity together are equivalent to another categorical condition called tininess.

This is a completely general observation valid in any monoidal category: invertible objects always inherit any categorical properties the unit possesses.

The second is that we can do just as well with a natural weakening of invertibility, namely dualizability. It's straightforward to show, with a proof very similar to the proof in the OP (that doesn't pass through compactness or tininess), that the dualizable modules are precisely the finitely presented projective modules; see here for details.

Answer 3

This is more appropriate in a comment than answer. The proof can be simplified a little bit by noticing once we have $M'\otimes N\rightarrow M\otimes N$ to be surjective, then tesnoring with $M$ gives a surjection $M'=M'\otimes N\times M\rightarrow M\otimes N\otimes M=I$. So $M=M'$ and $M$ is finitely generated.