Let $R$ and $S$ be a commutative rings and $M$ be an $R$-module. Let $f:R\to S$ be a ring homomorphism.
For $x \in R$, define koszul complex $ K(x,R)$ to be
$$
0\longrightarrow R \stackrel{x}{\longrightarrow} R \longrightarrow 0.
$$
For a sequence $ x =x_{1},\cdots ,x_{n} ,$ the koszul complex
$K(x,R)$ is defined to be the chain complex $K(x_{1},R)\otimes_{R}
\cdots \otimes_{R} K(x_{n},R) $.
define $K(x,M) = K(x,R)\otimes_{R} M$
Let $x\neq 0$. I want to know if there is an $R$-isomorphism
$$K(x,S)\cong K(f(x),S)?$$
(For the left side, we consider S as an R-module via f.)
My try For the case n=1:
$\require{AMScd}$
\begin{CD}
0 @>>> S @>x>> S @>>> 0\\
@. @V id V V @VV id V\\
0 @>>> S @>f(x)>> S @>>> 0
\end{CD}
(Notes: 1-for an element $s\in S$, one has $x.s:=f(x)s$. 2-In the top row, in fact, I should write $R\otimes S$ instead of $S$.)
But I can not give a general proof. I think it is true, because matrices representing homomorphisms has $x_i$s as entries.