Lemma 15.83.4. Let $R \to A$ be a flat ring map of finite presentation. Let $K \in D(A)$. The following are equivalent

$K$ is $R$-perfect, and

$K$ is isomorphic to a finite complex of $R$-flat, finitely presented $A$-modules.

Lemma 15.83.4. Let $R \to A$ be a flat ring map of finite presentation. Let $K \in D(A)$. The following are equivalent

$K$ is $R$-perfect, and

$K$ is isomorphic to a finite complex of $R$-flat, finitely presented $A$-modules.

**Proof.**
To prove (2) implies (1) it suffices by Lemma 15.83.2 to show that an $R$-flat, finitely presented $A$-module $M$ defines an $R$-perfect object of $D(A)$. Since $M$ has finite tor dimension over $R$, it suffices to show that $M$ is pseudo-coherent. By Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$-algebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ and a finite $A_0$-module $M_0$ flat over $R_0$ such that $A = A_0 \otimes _{R_0} R$ and $M = M_0 \otimes _{R_0} R$. By Lemma 15.64.17 we see that $M_0$ is pseudo-coherent $A_0$-module. Choose a resolution $P_0^\bullet \to M_0$ by finite free $A_0$-modules $P_0^ n$. Since $A_0$ is flat over $R_0$, this is a flat resolution. Since $M_0$ is flat over $R_0$ we find that $P^\bullet = P_0^\bullet \otimes _{R_0} R$ still resolves $M = M_0 \otimes _{R_0} R$. (You can use Lemma 15.61.2 to see this.) Hence $P^\bullet $ is a finite free resolution of $M$ over $A$ and we conclude that $M$ is pseudo-coherent.

Assume (1). We can represent $K$ by a bounded above complex $P^\bullet $ of finite free $A$-modules. Assume that $K$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. By Lemma 15.66.2 we see that $\tau _{\geq a}P^\bullet $ is a complex of $R$-flat, finitely presented $A$-modules representing $K$. $\square$

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