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I know tensor product is right exact, but I can't figure out why it's exact when it is acted on a split short exact sequence. In addition, can you give an example that tensor product acts on a short exact sequence but the result sequence is not exact? Thank you.

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The point is that in contrast to a short exact sequence, a split short exact sequence can be viewed as a certain kind of diagram with additive commutativity relations:

Definition. A sequence $A\xrightarrow{i} B\xrightarrow{\pi} C$ is split short exact if there exist $B\xrightarrow{r} A$, $C\xrightarrow{\sigma} B$ such that $$ri=\text{id}_A,\quad \pi\sigma=\text{id}_C\quad\text{and} \quad ir+\sigma\pi=\text{id}_A.$$

Now, functors preserve composition and identities by definition, and additive functors also preserve addition. Hence, split short exact sequences are preserved under any additive functors - the tensor product $X\otimes_R -$ is one such.

It is always helpful to check whether a definition can be formulated in such a purely diagrammatic way, as in the latter case it'll likely be stable under application of certain functors.

For example, slightly generalizing the above, you can contrast the notion of acyclicity and contractibility: Again, contractibility can be defined in terms of diagrams with additive relations and hence is preserved under any additive functor. For acyclicity, this is not the case, but stronger, more 'constructive' variants of acyclicity have been developed (absolute acyclicity, coacyclicity and contraacyclicity, for example - see this article if you're interested).

Concerning your second question, try to think of a simple short exact sequence over $k[x]/(x^2)$.

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  • $\begingroup$ shouldn't it be $ir+\sigma\pi=\text{id}_B$? $\endgroup$ – roi_saumon Jun 3 at 15:13

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