Modules which are isomorphic to their tensor product. Suppose that we have a commutative ring $R$. I am interested in finding the (finitely generated and projective, if you want) $R$-modules $M,$ such that $M\cong M\otimes_R M$ as $R$-modules.
 I know that it is true for $M=R$ or when $M$ is generated by a finite  set of orthogonal  idempotents.
I want to know if whether or not those are all the modules which that property. Thanks.
 A: Here are some examples:


*

*Localizations or quotients of $R$.

*More generally any epimorphism of commutative rings with domain $R$.

*Free $R$-modules of infinite rank.

*$R \oplus \bigoplus_{i \in I} N$ for any $R$-module $N$ with $N \otimes N = 0$.


The finitely generated examples can be classified:
Claim: When $M$ is a finitely generated $R$-module with $M \otimes M \cong M$, then $M$ is cyclic, i.e. $M \cong R/I$ for some ideal $I \subseteq R$.
Proof: We may base change to $R/\mathrm{Ann}(M)$ and therefore assume that $\mathrm{Ann}(M)=0$. Then we have to show $M \cong R$. Since $M$ is finitely generated, we have $\mathrm{supp}(M)=V(\mathrm{Ann}(M))=\mathrm{Spec}(R)$. For every $\mathfrak{p} \in \mathrm{Spec}(R)$ therefore $M \otimes \kappa(\mathfrak{p})$ is $1$-dimensional and $M_{\mathfrak{p}}$ is generated by a single element (Nakayama). Since $\mathrm{Ann}(M_{\mathfrak{p}})=0$, this implies that $M_{\mathfrak{p}}$ is free of rank $1$. Hence $M$ is locally free of rank $1$, hence invertible, and $M \otimes M \cong M$ implies $M \cong R$. $\square$
Of course, $R$ is the only finitely generated projective example (when $R$ is a domain).
