Lemma 15.81.13. Let $R \to A \to B$ be finite type ring maps. Let $m \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $A$-modules. Assume $B$ as a $B$-module is pseudo-coherent relative to $A$. If $K^\bullet $ is $m$-pseudo-coherent (resp. pseudo-coherent) relative to $R$, then $K^\bullet \otimes _ A^{\mathbf{L}} B$ is $m$-pseudo-coherent (resp. pseudo-coherent) relative to $R$.

**Proof.**
Choose a surjection $A[y_1, \ldots , y_ m] \to B$. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Combined we get a surjection $R[x_1, \ldots , x_ n, y_1, \ldots y_ m] \to B$. Choose a resolution $E^\bullet \to B$ of $B$ by a complex of finite free $A[y_1, \ldots , y_ n]$-modules (which is possible by our assumption on the ring map $A \to B$). We may assume that $K^\bullet $ is a bounded above complex of flat $A$-modules. Then

in $D(A[y_1, \ldots , y_ m])$. The quasi-isomorphism $\cong $ comes from an application of Lemma 15.59.7. Thus we have to show that $\text{Tot}(K^\bullet \otimes _ A E^\bullet )$ is $m$-pseudo-coherent as a complex of $R[x_1, \ldots , x_ n, y_1, \ldots y_ m]$-modules. Note that $\text{Tot}(K^\bullet \otimes _ A E^\bullet )$ has a filtration by subcomplexes with successive quotients the complexes $K^\bullet \otimes _ A E^ i[-i]$. Note that for $i \ll 0$ the complexes $K^\bullet \otimes _ A E^ i[-i]$ have zero cohomology in degrees $\leq m$ and hence are $m$-pseudo-coherent (over any ring). Hence, applying Lemma 15.81.6 and induction, it suffices to show that $K^\bullet \otimes _ A E^ i[-i]$ is pseudo-coherent relative to $R$ for all $i$. Note that $E^ i = 0$ for $i > 0$. Since also $E^ i$ is finite free this reduces to proving that $K^\bullet \otimes _ A A[y_1, \ldots , y_ m]$ is $m$-pseudo-coherent relative to $R$ which follows from Lemma 15.81.12 for instance. $\square$

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