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I have a basic question about exact sequences. I want to show that if I have that whenever $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ is exact then $0 \to A \otimes N \to B \otimes N \to C \otimes N \to 0$ is then $N$ is flat.

So let $\dots \to A \xrightarrow{f} B \xrightarrow{g} C \to D \dots$ be an exact sequence. Now I want to split it into short exact sequences so I can apply what I have: $0 \to A / \operatorname{Ker}{f} \to B \to \operatorname{Im}{g} \to 0$ is exact.

From this I get that $ A / \operatorname{Ker}{f} \otimes N \to B\otimes N \to \operatorname{Im}{g}\otimes N \to 0$ is exact since $- \otimes N$ is right exact. But how do I get from that to $0 \to A \otimes N \to B\otimes N \to C \otimes N \to 0$? I think I can't. So how do I split the long exact sequence in a better way?

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I agree with Serkan's answer, but it's worth noting that in your example, $\mbox{Ker}(f) = 0$ and $\mbox{Im}(g) = C,$ just by exactness. So the equivalence you desired actually holds. – rotskoff Jun 11 '12 at 8:23
up vote 2 down vote accepted

The point is that you need to show that $\ldots \rightarrow A\otimes N \rightarrow B \otimes N \rightarrow C \otimes N \rightarrow D\otimes N \rightarrow \ldots$ is exact at $B \otimes N$, not that $0 \rightarrow A\otimes N \rightarrow B \otimes N \rightarrow C \otimes N \rightarrow 0$ is an exact sequence. Now, the sequence which you have shown to be exact gives the desired result.

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Yes, Serkan is right. Actually, I am typing the something just a moment ago. Anyway, it does not need now. – Joy-Joy Jun 11 '12 at 8:18
How do I get from $0 \to A / \operatorname{Ker}{f} \otimes N \to B\otimes N \to \operatorname{Im}{g}\otimes N \to 0$ is exact to $\dots \to A \otimes N \to B\otimes N \to C \otimes N \to \dots$? – Rudy the Reindeer Jun 11 '12 at 8:21
I think you mean it the other way around. I don't understand how I get exactness at $B \otimes N$ in $\dots \to A \otimes N \to B \otimes N \to C \otimes N \dots$ from the fact that the following is exact: $0 \to A / ker(f) \otimes N \to B \otimes N \to im(g) \otimes N \to 0$. – Rudy the Reindeer Jun 11 '12 at 8:40
Thank you very much. Now it's clear. – Rudy the Reindeer Jun 11 '12 at 8:52

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