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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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    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?2012-01-21

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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.

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    @AlexYoucis That's a good point. I think the two notions of right exactness are equivalent, but that isn't obvious (at least to me). I'll try to work out the diagram later on. Thanks for the comment!2012-01-21
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[I assume that you are interested in the direct product as you wrote.]

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.