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Let $M$ be finitely generated Module over the Polynomial ring $R=k[x_{1},..,x_{n}]$, then there is a free resolution of M of the form $$0\rightarrow F_{n}\rightarrow ...\rightarrow F_{0}\rightarrow M\rightarrow 0$$ We can write the free modules in the form $F_{i}=\bigoplus_{j\in \mathbb{Z} } R(-j)^{\beta_{i,j}}$, these numbers $\beta_{i,j}$ are called betti numbers, my question is how to show that $$\beta_{i,j}= \dim_{k}\; \text{Tor}_{i}^{R}(M,k)_{j}$$ .

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What is $S$?${}$ – Rasmus May 30 '12 at 11:46
sorry, it is R. – kiranovalobas May 30 '12 at 14:56
up vote 2 down vote accepted

Here is a hint. In order to compute the Tor groups you're interested in, you have to apply $-\otimes_R k$ to your resolution then take homology. The terms of your resolution are sums of (shifts of) free modules, so because $R\otimes_R k=k$, you understand all the terms in $F_* \otimes _R k$. If your resolution is such that the images of the differentials $F_i \to F_{i+1}$ are contained in $(x_1,\ldots,x_n)F_{i+1}$, then the induced differential on $F_* \otimes _R k$ is zero. That makes it easy to calculate homology, and therefore Tor...

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