Understanding How We Get Koszul Complexes From Regular Sequences Let $ R $ be a ring, and $ M $ be an $ R $-module. We say that a sequence $ x_{1},\dots,x_{r} $ of elements of $ R $ is regular if:
1) $ (f_{1},\dots,f_{r})M \neq M, $ and
2) $ f_{i} $ is a non-zero divisor on $ M/(f_{1},\dots, f_{i-1})M $ for each $ i = 1,\dots,r. $
Apparently, it is the case that whenever we have a regular sequence $ f_{1},\dots,f_{r} \subset R, $ there is an induced Koszul complex which is a free resolution of $ R/(f_{1},\dots,f_{r}). $
I'm not quite sure what the form of this complex is, or indeed how to construct it. 
 A: If $x_1,\dots,x_r$ is a sequence of elements of $R$, define for each $1\le i \le r$ the Koszul complex $\mathcal K_\bullet^{(i)}(x_i; R)$ by 
$$0 \to R \xrightarrow{\cdot x_i} R \to 0,$$
with the right hand copy of $R$ in degree zero, i.e., the right hand copy of $R$ is $\mathcal K_0^{(i)}(x_i; R)$, and the left hand copy in degree 1. Let $d^{(i)}$ be the differential map on this sequence.
The Koszul complex $\mathcal K_\bullet(x_1,\dots,x_r; R)$ is defined in degree $n$ by
$$\bigoplus_{i_1+\dots+i_r = n} \mathcal K_{i_1}^{(1)}(x_1; R)\otimes_R \dots \otimes_R \mathcal K_{i_r}^{(r)}(x_r; R)$$
with differential in degree-$n$ given by the direct sum of the differentials $d_{i_1,\dots,i_r}:\mathcal K_{i_1}^{(1)}(x_1; R)\otimes_R \dots \otimes_R \mathcal K_{i_r}^{(r)}(x_r; R) \to \mathcal K_{n-1}(x_1,\dots,x_r; R)$, with $i_1+ \dots + i_r = n$, defined on simple tensors by
$$d_{i_1,\dots,i_r}(u_{i_1}^{(1)}\otimes \dots \otimes u_{i_r}^{(r)}) = \sum_{j=1}^r (-1)^{i_1+\dots+i_{j-1}}\; u_{i_1}^{(1)} \otimes \dots \otimes u_{i_{j-1}}^{(j-1)}\otimes d^{(j)}\left(u_{i_j}^{(j)}\right) \otimes u_{i_{j+1}}^{(j+1)} \otimes\dots \otimes u_{i_r}^{(r)}.$$
It is a theorem that whenever $x_1, \dots, x_r$ satisfy


*

*$x_i$ is a nonzerodivisor on $R/(x_1,\dots,x_{i-1})$ for $1\le i \le r$,


then $\mathcal K_\bullet(x_1,\dots,x_r; R)$ is a free resolution of $R/(x_1,\dots,x_r)$. 
