The Stacks project

Lemma 37.56.17. Let $f : X \to Y$ be a morphism of schemes locally of finite type over a base $S$. Let $m \in \mathbf{Z}$. Let $E$ be an object of $D(\mathcal{O}_ X)$. Assume $\mathcal{O}_ Y$ is pseudo-coherent relative to $S$1. Then the following are equivalent

  1. $E$ is $m$-pseudo-coherent relative to $Y$, and

  2. $E$ is $m$-pseudo-coherent relative to $S$.

Proof. The question is local on $X$, hence we may assume $X$, $Y$, and $S$ are affine. Arguing as in the proof of More on Algebra, Lemma 15.81.13 we can find a commutative diagram

\[ \xymatrix{ X \ar[r]_ i \ar[d]_ f & \mathbf{A}^ m_ Y \ar[r]_ j \ar[ld]^ p & \mathbf{A}^{n + m}_ S \ar[ld] \\ Y \ar[r] & \mathbf{A}^ n_ S } \]

The assumption that $\mathcal{O}_ Y$ is pseudo-coherent relative to $S$ implies that $\mathcal{O}_{\mathbf{A}^ m_ Y}$ is pseudo-coherent relative to $\mathbf{A}^ m_ S$ (by flat base change; this can be seen by using for example Lemma 37.56.15). This in turn implies that $j_*\mathcal{O}_{\mathbf{A}^ n_ Y}$ is pseudo-coherent as an $\mathcal{O}_{\mathbf{A}^{n + m}_ S}$-module. Then the equivalence of the lemma follows from Derived Categories of Schemes, Lemma 36.12.5. $\square$

[1] This means $Y \to S$ is pseudo-coherent, see Definition 37.57.2.

Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09UT. Beware of the difference between the letter 'O' and the digit '0'.