Lemma 15.81.11. Let $R$ be a ring, $f \in R$ an element, $R_ f \to A$ is a finite type ring map, $g \in A$, and $K^\bullet $ a complex of $A$-modules. If $K^\bullet $ is $m$-pseudo-coherent (resp. pseudo-coherent) relative to $R_ f$, then $K^\bullet \otimes _ A A_ g$ is $m$-pseudo-coherent (resp. pseudo-coherent) relative to $R$.
Proof. First we show that $K^\bullet $ is $m$-pseudo-coherent relative to $R$. Namely, suppose $R_ f[x_1, \ldots , x_ n] \to A$ is surjective. Write $R_ f = R[x_0]/(fx_0 - 1)$. Then $R[x_0, x_1, \ldots , x_ n] \to A$ is surjective, and $R_ f[x_1, \ldots , x_ n]$ is pseudo-coherent as an $R[x_0, \ldots , x_ n]$-module. Hence by Lemma 15.64.11 we see that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R[x_0, x_1, \ldots , x_ n]$-modules.
Choose an element $g' \in R[x_0, x_1, \ldots , x_ n]$ which maps to $g \in A$. By Lemma 15.64.12 we see that
\begin{align*} K^\bullet \otimes _{R[x_0, x_1, \ldots , x_ n]}^{\mathbf{L}} R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}] & = K^\bullet \otimes _{R[x_0, x_1, \ldots , x_ n]} R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}] \\ & = K^\bullet \otimes _ A A_ f \end{align*}
is $m$-pseudo-coherent as a complex of $R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}]$-modules. write
\[ R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}] = R[x_0, \ldots , x_ n, x_{n + 1}]/(x_{n + 1}g' - 1). \]
As $R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}]$ is pseudo-coherent as a $R[x_0, \ldots , x_ n, x_{n + 1}]$-module we conclude (see Lemma 15.64.11) that $K^\bullet \otimes _ A A_ g$ is $m$-pseudo-coherent as a complex of $R[x_0, \ldots , x_ n, x_{n + 1}]$-modules as desired. $\square$
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