Property of exact sequences? If I have a commutative ring $R$ and an exact sequence $0\to M'\to M\to M''\to 0$ where $\epsilon:M'\to M$ and $\sigma:M\to M''$ do I get an exact sequence $0\to M'\to M\to M''\to 0$ by means of $\epsilon \circ id:M'\times N\to M\times N$ and $\sigma\circ id:M\times N\to M''\times N$? By $f\circ g$ I mean the mapping $(x,y)$ to $(f(x),g(y))$ (not sure how to do the tensor product symbol). The reason I ask is that my notes are a scrambled scrawling of material I cannot make sense of. And It looks like I have a lemma without a conclusion here but this is what I Guess it is. Can anyone confirm? Thanks so much.

This is not, in general, true. Modules $N$ for which the resulting sequence is always exact were given the adjective flat by Serre, and this property has interesting geometric implications. The standard counterexample is tensoring the injection \\[ 0 \to \mathbf Z \stackrel{\times 2}{\longrightarrow} \mathbf Z \\] with $\mathbf Z/2\mathbf Z$. However, tensoring _is_ right exact, i.e. \\[ M' \otimes N \to M \otimes N \to M'' \otimes N \to 0 \\] must be an exact sequence.