Is it true that the tensor product of any two nonzero $R$-modules is nonzero if and only if $R$ is a local ring with a nilpotent maximal ideal? If $R$ is a commutative ring and $I$ is a nilpotent ideal of $R$, then for any $R$-module $M$, if $IM=M$ (or equivalently, $M/IM=0$), then $M$ must necessarily be the zero module.
Now, my question is: Is it true that for a nonzero commutative ring $R$, the tensor product of any two nonzero $R$-modules is nonzero if and only if $R$ is a local ring with a nilpotent maximal ideal?
One direction can easily be proven using the above fact: If $R$ is a local ring with a nilpotent maximal ideal $\mathbf{m}$ and $M$ and $N$ are two nonzero $R$-modules, then $M/\mathbf{m}M$ and $N/\mathbf{m}N$ are nonzero vector spaces over the field $R/\mathbf{m}$ (by the above fact), hence $(M/\mathbf{m}M) \otimes_{R/\mathbf{m}} (N/\mathbf{m}N) \cong (M \otimes_{R} N)/\mathbf{m}(M \otimes_{R} N)$ is nonzero, hence $M \otimes_{R} N$ is nonzero.
Conversely, if the tensor product of any two nonzero $R$-modules is nonzero, then the sum of any two proper ideals of $R$ is proper (because $(R/I) \otimes_{R} (R/J) \cong R/(I+J)$), hence $R$ must be local. What could not be easily proven is that the unique maximal ideal of $R$ is nilpotent.
 A: Here is a counterexample.  Let $k$ be a field and let $R=k[x_1,x_2,\dots]/I$ where $I$ is generated by $x_n^{n+1}$ for each $n$ and $x_ix_j$ for each $i\neq j$.  This is a local ring with maximal ideal $\mathbf{m}=(x_1,x_2,\dots)$ which is not nilpotent.  I claim though that $R$ has the property that $M=\mathbf{m}M$ implies $M=0$ for any $R$-module $M$, and so the tensor product of two nonzero $R$-modules is nonzero.
To prove this, observe that $x_n\mathbf{m}^n=0$ for each $n$.  So, if $M=\mathbf{m}M$, then $x_nM=x_n\mathbf{m}^nM=0$.  Since $n$ is arbitrary, this means $\mathbf{m}M=0$ and hence $M=0$.
Here is a characterization of such rings that does work.
Theorem: Let $R$ be a nonzero commutative ring.  Then the following are equivalent.


*

*The tensor product of any two nonzero $R$-modules is nonzero.

*$R$ is a local ring and for any $R$-module $M$, $M=\mathbf{m}M$ implies $M=0$ where $\mathbf{m}$ is the maximal ideal.

*$R$ is a local ring and for any sequence $(x_n)$ of elements of the maximal ideal, there is some $N\in\mathbb{N}$ such that $\prod_{n=1}^Nx_n=0$.


Proof:  Your arguments show that (2) implies (1), and that (1) implies $R$ is local.  If $R$ is local and $M=\mathbf{m}M$ for some nonzero $R$-module $M$, then $M\otimes R/\mathbf{m}=0$ so $R$ does not satisfy (1).  Thus (1) and (2) are equivalent.
To show (2) implies (3), suppose $(R,\mathbf{m})$ is local but does not satisfy (3), so there is a sequence $(x_n)$ of elements of $\mathbf{m}$ such that $\prod_{n=1}^Nx_n\neq 0$ for all $N$.  Consider the module $M$ generated by elements $a_0,a_1,a_2,\dots$ with relations $x_na_n=a_{n-1}$.  Clearly this $M$ satisfies $M=\mathbf{m}M$, since each $x_n$ is in $\mathbf{m}$.  On the other hand, if you take only the generators $a_0,\dots,a_N$ and relations $x_na_n=a_{n-1}$ for $n\leq N$, you simply get a free module generated by $a_N$ in which $a_0=\prod_{n=1}^N x_n \cdot a_N$ is nonzero.  It follows that $a_0$ is nonzero in $M$, and so $M$ is nonzero.  Thus $R$ does not satisfy (2).
Finally, suppose $(R,\mathbf{m})$ is local but does not satisfy (2); let $M$ be a nonzero $R$-module such that $\mathbf{m}M=M$.  Suppose $r\in R$ is such that $rM\neq 0$.  Then $r\mathbf{m}M\neq 0$, and in particular there is some $x_1\in\mathbf{m}$ such that $rx_1M\neq 0$.  Replacing $r$ with $rx_1$ and iterating (starting from $r=1$), we can construct a sequence $(x_n)$ of elements of $\mathbf{m}$ such that $\prod_{n=1}^N x_n\cdot M\neq 0$ for each $N$.  In particular, $\prod_{n=1}^N x_n\neq 0$ for each $N$, so $R$ does not satisfy (3).
A: If $R$ is not local, take tow distinct maximal ideals $M,N$, $R/M\otimes_R R/N=0$. So, the assumption implies $R$ must be local, with maximal ideal $M$ say. If $M$ does not consist of nilpotent elements, let $f\in M$ be on such. Then, $R/f\neq 0\neq R_f$ and $R/f\otimes_R R_f=0$.
