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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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1 Answer 1

up vote 2 down vote accepted

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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But somehow the paper says if we tensor with any $R$-module, we still get the exact sequence. We are not tensoring the sequence with $C$. –  Junyu Nov 10 '11 at 5:19
2  
@Junyu, that is irrelevant, really: as I wrote, the exact sequence splits because $C$ is projective, so tensoring it with anything will result in a new exact sequence. –  Mariano Suárez-Alvarez Nov 10 '11 at 5:20
    
Mariano's answer is correct. If you're topologically inclined, you might want to think of it terms of the vanishing of $\operatorname{Tor}^1(C,M)$ for any $R$-module $M$. This shows that even flatness of $C$ would be enough. Since we're over a PID, this means that $C$ being torsionfree is enough. Otherwise it is not so clear why $R$ being a PID is relevant, although perhaps that is used in order to deduce that your module $C$ is free (a finitely generated module over a PID is free iff it is torsionfree). –  Pete L. Clark Nov 10 '11 at 5:34
    
@Pete: $Tor^1$ is topological now? :) –  Mariano Suárez-Alvarez Nov 10 '11 at 5:38
    
@PeteL.Clark: Thanks for your help. I understand Mariano's answer now. –  Junyu Nov 10 '11 at 5:56

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