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.

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).