Suppose M is a finitely generated non-zero R-module, where R is a commutative unital ring. Show that the tensor product of M with itself is non-zero. Suppose M is a finitely generated non-zero R-module, where R is a commutative unital ring. Show that the tensor product of M with itself is non-zero.
I know one way to show this is to find an R-bilinear map which is nonzero, but am not sure how to find it.
 A: Not sure this is the best proof, but here goes. Note that $M\otimes_R M = 0$ if and only if $(M\otimes_R M)_{\mathfrak p} = 0$ for all primes $\mathfrak p\subset R$ (Atiyah-MacDonald 3.8). But
$$
(M\otimes_R M)_{\mathfrak p}\cong M_\mathfrak p\otimes_{R_{\mathfrak p}}M_{\mathfrak p}
$$
by Atiyah-MacDonald 3.7.
Hence we have reduced to the case where $R$ is a local ring. The statement now follows from the fact that tensor product is faithful for finitely-generated modules over local rings, e.g. exercise 3 in the second chapter of Atiyah-MacDonald. (This is a consequence of Nakayama's lemma, as far as I remember.)
Remark. So why doesn't the argument given here show that tensor products are faithful over any ring via reduction to the local case? Simply because a pair of distinct modules $M$ and $N$ might be supported on disjoint sets of primes. For instance, this is the case when $R = \mathbb Z$, $M = \mathbb Z/p$ and $N = \mathbb Z/q$ for distinct primes $p,q$. Then $\mathbb Z /p\otimes\mathbb Z/q = 0$ because, with $r$ ranging over primes, $(\mathbb Z/p)_{(r)} = 0$ for $r\neq p$ and similarly for $q$.
A: This is false. Let $R=\mathbb{C}[X]/(X^2)$. Let $x=\bar{X}$, so $x^2=0$, and let $M= R x$. Then $M$ is a finitely generated nonzero $R$-module, but $M\otimes M$ is generated by $x\otimes x=x^2\otimes 1=0\otimes 1=0$.
More generally, if $R$ is a commutative ring having a nonzero ideal $I$ satisfying $I^2=0$, then $I\otimes_R I=0$ and you get a counterexample.
