Consider the polynomial ring $S = C[x\_1,\ldots,x\_n]$. Let
$$
‎‏0 ‎‎\rightarrow‎‎‏ ‎E\_{n-1} ‎‎\rightarrow ‎\cdots ‎‎‎\rightarrow ‎E\_1 ‎‎‎\rightarrow ‎E\_0 \rightarrow I \rightarrow 0‎ ‎,‎
$$
be a minimal free resolution of an ideal $I$ of $S$, where
$$
E\_p = \bigoplus \_j S(-a\_{pj}) .
$$
The regularity of $I$ is $\max\lbrace a\_{pj} - p \rbrace $. Let $m = (x\_1,x\_2,\ldots,x\_n)$ be a maximal ideal of the ring $S = C[x\_1,\ldots,x\_n]$. Why is the regularity of $I$ the largest $q$ for which $\operatorname{Tor}^S\_p(I,S/\mathfrak{m})\_{p+q} \neq 0$ for some $p$?



Castelnuovo Mumford Regularity
By the construction of derived functors, $\mathrm{Tor}(I,A)$ is the homology of the complex 
$$0 \to E\_{n-1} \otimes A \to \cdots \to E\_1 \otimes A \to E\_0 \otimes A.$$
(All tensor products are $\otimes\_S$.) In the particular case that $A = S/\mathfrak{m} =:k$, we have
$E\_p \otimes k= \oplus\_j k(-a\_{pj})$, so $\mathrm{Tor}\_p(I, k)$ is the homology of the complex of $k$-vector spaces defined by taking the matrices in your original complex and reducing all of their elements modulo $\mathfrak{m}$. The assumption that your complex is minimal is equivalent to saying that all of the maps are $0$ modulo $m$, so you are just taking the homology of a complex where all maps are zero, and you get
$$\mathrm{Tor}\_p(I, k) \cong \oplus\_jk(-a\_{pj})$$
as a graded $k$-vector space. Now use the definition of CM regularity.


