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When I read Thang Le's paper the coloured Jones polynomial and the A-polynomial of knots, it says in page 21 that:

Since $R=\mathbb{C}[t^{\pm1}]$ is a PID, and $C$ is free over $R$. So if we tensor the exact sequence

$0\to A\to B\to C \to 0 $ , (where $A$ is the kernal of $B\to C$)

with any $R$-module, in particular $\mathbb{C}$, we have the exact sequence

$0\to \mathbb{C}\otimes_{R}A\to \mathbb{C} \otimes_{R}B\to \mathbb{C}\otimes_{R} C $.

Here $A, B$ and $C$ are $R$-modules.

I failed to find any theorem saying something similar above. Could anyone help me to find a proof of this? I have tried myself but failed.

Remark: In the Thang Le's paper, $C=\mathcal{S}(X), A=\mathcal{P}$ and $B=\mathcal{T}^{\sigma}$. I suppose the detailed settings of $\mathcal{P}$ and $\mathcal{T}^{\sigma}$ are not important.

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If $C$ is free, then it is in particular projective, so the short exact sequence $0\to A\to B\to C\to 0$ splits. As a consequence of this, tensoring it with anything will result in another exact sequence.

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    @Mariano: well, for instance I first learned abut Tor in an algebraic topology class, not an algebra class. Of course it is "really" a pure algebra concept...2011-11-10