The Stacks project

Lemma 15.83.10. 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 15.82.8 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 15.77.6 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 15.77.3. This implies that $K$ is $R$-perfect by Lemma 15.83.3 and the proof is complete. $\square$

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Your email address will not be published.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 0GHJ. Beware of the difference between the letter 'O' and the digit '0'.