Lemma 15.82.10. Let $R \to A$ be a ring map which is flat and of finite presentation. Let $K \in D(A)$ be pseudo-coherent. The following are equivalent

1. $K$ is $R$-perfect, and

2. $K$ is bounded below and for every prime ideal $\mathfrak p \subset R$ the object $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ is bounded below.

Proof. Observe that (1) implies (2) as an $R$-perfect complex has bounded tor dimension as a complex of $R$-modules by definition. Let us prove the other implication.

Write $A = R[x_1, \ldots , x_ d]/I$. Denote $L$ in $D(R[x_1, \ldots , x_ d])$ the restriction of $K$. By Lemma 15.81.8 we see that $L$ is pseudo-coherent. Since $L$ and $K$ have the same image in $D(R)$ we see that $L$ is $R$-perfect if and only if $K$ is $R$-perfect. Also $L \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ and $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ are the same objects of $D(\kappa (\mathfrak p))$. This reduces us to the case $A = R[x_1, \ldots , x_ d]$.

Say $A = R[x_1, \ldots , x_ d]$ and $K$ satisfies (2). Let $\mathfrak q \subset A$ be a prime lying over a prime $\mathfrak p \subset R$. By Lemma 15.76.6 applied to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$. Since $K$ is bounded below, we see that $K$ is perfect in $D(A)$ by Lemma 15.76.3. This implies that $K$ is $R$-perfect by Lemma 15.82.3 and the proof is complete. $\square$

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