On characterizing modules that don't annihilate any module under tensor product.

Question

Let $R$ be a commutative ring and $M$ an $R$-module. Then under what condition can we deduce that for any nonzero $R$-module $N$, $M\otimes_RN\neq0$?

Answer 1

Not necessarily. If $M$ is a flat $R$-module, then what you claim is true ($M \otimes_R N \neq 0$ whenever $N \neq 0$) if and only if $M$ is faithfully flat. Note that all projective modules are flat, but they don't have to be faithfully flat.

But the results are very nice when your ring is local. If $R$ is a local ring, then $M$ is free $\iff$ $M$ is projective. If $M$ is finitely generated, or if the unique maximal ideal of $R$ is nilpotent, then free/projective/flat/faithfully flat are all equivalent.

This last thing I mentioned is less well known, so let me give a proof. Suppose $M$ is flat, $(R,\mathfrak m)$ is local, and $\mathfrak m^N = 0$ for some $N$. One can show using the 'equational criterion for flatness' that a set of elements $x_1, ... , x_n \in M$ is linearly independent over $R$ if and only if the images $x_1 + \mathfrak m M, ... , x_n + \mathfrak m M$ in $M/\mathfrak m M$ are linearly independent over $R/\mathfrak m$.

It follows that a basis $x_i + \mathfrak m$ for $M/\mathfrak m M$ over $R/\mathfrak m$ lifts to a linearly independent set $x_i$ in $M$ over $R$. Letting $N = \sum\limits_i Rx_i$, you have $$M = N + \mathfrak m M = N + \mathfrak m(N + \mathfrak m M) = N + \mathfrak m^2 M$$ $$ = N + \mathfrak m^3 M = \cdots = N + \mathfrak m^N M = N + 0$$ so $M$ is free. Hence $M$ is also faithfully flat and projective.