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Let $M$ be a finitely generated module over a polynomial ring $R$ over a field $k$. Let $F_{\bullet}$ be a minimal free resolution of $M$ : $$0\longrightarrow F_p \longrightarrow ....\longrightarrow F_1 \longrightarrow F_0\longrightarrow M$$

In one paper of M.Chardin, he claimed that the maps of $F_{\bullet}\otimes_{R}k$ being zero maps, $\text{Tor}_{i}(M,k)=H_{i}(F_{\bullet}\otimes k)=F_{i}\otimes k$. This claim also appears in the book "The Geometry of Syzygy" of D.Eisenbud, in the proof of proposition 1.7 on page 7.

My question is :

  1. Why the maps of $F_{\bullet}\otimes_{R}k$ being zero maps if $F_{\bullet}$ is a minimal free resolution
  2. Why $\text{Tor}_{i}(M,k)=F_{i}\otimes k$
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1 Answer 1

What's your definition of minimality? The usual one is that the image of each map $\phi_i : F_i \rightarrow F_{i-1}$ is contained in $\mathfrak{m} F_{i-1}$, where $\mathfrak{m} = (X_1, \dots, X_n)$ is the irrelevant ideal. By definition, the residue field $k = R / \mathfrak{m}$, so the map $\phi_i \otimes k : F_i \otimes k \rightarrow F_{i-1} \otimes k$ is the zero map. More concretely, the map $\phi_i \otimes k$ is obtained from the map $\phi_i$ by setting each of the variables $X_1, \dots, X_n$ to be zero, since the map $R \rightarrow k$ is given by $f(X_1,\dots,X_n) \mapsto f(0,\dots,0)$. Therefore, the fact that the image of $\phi_i$ is contained in $\mathfrak{m} F_{i-1}$ is equivalent to the map $\phi_i \otimes k$ being the zero map.

By definition, $\text{Tor}_i(M,k) = \text{ker}(\phi_i\otimes k) / \text{image}(\phi_{i+1}\otimes k)$. In other words, you can compute this Tor by taking a free resolution of $M$ and tensoring it with $k$. But since each $\phi_i\otimes k$ is the zero map, $\text{ker}(\phi_i\otimes k) = F_i \otimes k$ and $\text{image}(\phi_{i+1}\otimes k) = 0$.

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@navigator23: Thanks for correcting my slip-up! –  Michael Joyce Oct 20 '12 at 15:49
    
I don't mind at all! I'm glad you made the correction; unfortunately, I had went out for a run so I didn't have a chance to approve it until now. –  Michael Joyce Oct 20 '12 at 15:52

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