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 $$ri=\text{id}_A,\quad \pi\sigma=\text{id}_C\quad\text{and} \quad ir+\sigma\pi=\text{id}_B.$$
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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3$\begingroup$ shouldn't it be $ir+\sigma\pi=\text{id}_B$? $\endgroup$ Commented Jun 3, 2019 at 15:13
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$\begingroup$ I just want to say this has to be the most elegant proof in the world :) $\endgroup$– DebugCommented Apr 3, 2023 at 5:02
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1$\begingroup$ @DLeftAdjointtoU I am happy you like it :-) $\endgroup$– HannoCommented Apr 3, 2023 at 5:03
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$\begingroup$ @Hanno I'm learning about Flat Modules now out of Lang / Algebra. This is exciting stuff. A lot of beautiful proofs include a good diagram chase. Lang's construction of the tensor product using categorical language is amazing. I recently came to grips with the notion of "factoring through" something. "Since $f$ is multilinear, the induced map $M \to G$ takes on the value $0$ on $N$. Hence by the universal property of factor modules, it can be factored through $M/N$ and and we have [an arrow that makes this triangle commute]." Now, this proof step alway confused me until yesterday. =) $\endgroup$– DebugCommented Apr 3, 2023 at 5:08