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.
Best Answer
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.