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Let $F\rightarrow G\rightarrow H$ be an exact sequence of free $R$-modules and $M$ be any $R$-module.

Is that true that $F\otimes_RM\rightarrow G\otimes_RM\rightarrow H\otimes_RM$ is exact?

If the answer is no, does that hold when all the modules above are finitely generated?

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Let $R=\mathbb{Z}$. Then $$0\rightarrow\mathbb{Z}\xrightarrow{\times2}\mathbb{Z}$$ is an exact sequence of free $R$-modules, but the tensor product with $M=\mathbb{Z}/2\mathbb{Z}$ is not exact.

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  • $\begingroup$ Thank you. That's a good counterexample. $\endgroup$
    – Hydrogen
    Jan 30 '21 at 17:46
  • $\begingroup$ I actually meet this in the following assumption: Let $\mathcal K_*$ be the Koszul Complex $\wedge^{n+1-i}V\otimes \mathcal O(-n-1+i)$ over $\mathbb P^n$ which is locally free and exact. Let $\mathcal F$ be any coherent sheaf. Then the author claims $K_*\otimes \mathcal F(m+1)$ is exact as well because terms of $K_*$ is locally free. Do you have any idea about this? $\endgroup$
    – Hydrogen
    Jan 30 '21 at 17:47
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    $\begingroup$ @Hydrogen In your question, you were only asking about a sequence that was exact at one place. If you have a bounded complex of free modules that is exact everywhere, then it is contractible, and tensoring with anything will preserve exactness. Exactness of a complex of sheaves on $\mathbb{P}^n$ is determined locally, so that should generalize to an exact complex of locally free sheaves, I think. $\endgroup$ Jan 30 '21 at 18:53
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    $\begingroup$ @Hydrogen A chain complex is contractible if it is chain homotopy equivalent to the zero complex. Applying any additive functor to a contractible complex gives a contractible (and therefore exact) complex. Most introductory texts on homological algebra will cover this. $\endgroup$ Jan 31 '21 at 11:04
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    $\begingroup$ @Hydrogen See this question, for example. $\endgroup$ Jan 31 '21 at 17:47

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