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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.

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    This is false. Let $k[X] \to k[X]$ be the map that evaluates a polynomial at $0$. The Koszul complex for $X$ and the one for $0$ are not isomorphic, obviously. – 2017-02-27
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    @PedroTamaroff: No, you're using the wrong definition: $K(x,S)$ here means the Koszul complex where you consider $S$ as an $R$-module. So you would be considering $k[X]$ as a $k[X]$-module via the evaulation at $0$ map and letting $X$ act on it via this module structure (i.e., by $0$). – 2017-02-27
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    when we consider 2nd k[X] as a 1st K[X]-algebra, then $X.f=0, \forall f\in k[X] $ – 2017-02-27
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    @EricWofsey Alright, it is hard to gather that from the post... – 2017-02-27

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