# 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.

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The tensor is product symbol is \otimes. Are you taking that, or the direct product? –  Dylan Moreland Jan 21 '12 at 16:39
in my notes it says $\otimes_R$ maybe that means direct? –  Kyle Schlitt Jan 21 '12 at 16:40
oh i just read further, and my notes go on to say that if the resulting sequence is guaranteed to be exact, then the $R$-module $N$ is said to be flat. but he refers to the resulting sequence as "tensored" so the $\otimes_{R}$ must denote tensor product after all does this sound correct? –  Kyle Schlitt Jan 21 '12 at 16:43

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.
Of course Dylan is correct, but in fact, right exactness gives us the more general statement that $M\to N\to L\to0$ gets mapped to exact sequences under $\bullet\otimes_R A$. If tensor was exact, the tor functors wouldn't be very interesting. –  Alex Youcis Jan 21 '12 at 18:19
The sequence you constructed will not be exact because the composition of $\epsilon\times\mathrm{id}$ and $\sigma\times\mathrm{id}$ is $0\times\mathrm{id}$, which is not the zero map.