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

is $m$-pseudo-coherent as a complex of $R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}]$-modules. write

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)