The Stacks project

Lemma 47.26.1. Let $A \to B$ be a faithfully flat map of Noetherian rings. If $K \in D(A)$ and $K \otimes _ A^\mathbf {L} B$ is a dualizing complex for $B$, then $K$ is a dualizing complex for $A$.

Proof. Since $A \to B$ is flat we have $H^ i(K) \otimes _ A B = H^ i(K \otimes _ A^\mathbf {L} B)$. Since $K \otimes _ A^\mathbf {L} B$ is in $D^ b_{\textit{Coh}}(B)$ we first find that $K$ is in $D^ b(A)$ and then we see that $H^ i(K)$ is a finite $A$-module by Algebra, Lemma 10.83.2. Let $M$ be a finite $A$-module. Then

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(M, K) \otimes _ A B = R\mathop{\mathrm{Hom}}\nolimits _ B(M \otimes _ A B, K \otimes _ A^\mathbf {L} B) \]

by More on Algebra, Lemma 15.99.2. Since $K \otimes _ A^\mathbf {L} B$ has finite injective dimension, say injective-amplitude in $[a, b]$, we see that the right hand side has vanishing cohomology in degrees $> b$. Since $A \to B$ is faithfully flat, we find that $R\mathop{\mathrm{Hom}}\nolimits _ A(M, K)$ has vanishing cohomology in degrees $> b$. Thus $K$ has finite injective dimension by More on Algebra, Lemma 15.69.2. To finish the proof we have to show that the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K)$ is an isomorphism. For this we again use More on Algebra, Lemma 15.99.2 and the fact that $B \to R\mathop{\mathrm{Hom}}\nolimits _ B(K \otimes _ A^\mathbf {L} B, K \otimes _ A^\mathbf {L} B)$ is an isomorphism. $\square$

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Your email address will not be published.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 0E4A. Beware of the difference between the letter 'O' and the digit '0'.