Question on minimal free resolution 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 : 


*

*Why the maps of $F_{\bullet}\otimes_{R}k$ being zero maps if $F_{\bullet}$ is a minimal free resolution

*Why $\text{Tor}_{i}(M,k)=F_{i}\otimes k$

 A: 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 $\text{image}(\phi_{i+1}\otimes k) = 0$.
