Lemma 47.27.5. Let $R \to A$ be a flat ring map of finite presentation. Let $K$ be a relative dualizing complex. Then $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K)$ is an isomorphism.

**Proof.**
By Algebra, Lemma 10.166.1 there exists a finite type $\mathbf{Z}$ subalgebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ such that $A = A_0 \otimes _{R_0} R$. By Lemmas 47.27.2, 47.27.3, and 47.27.4 there exists a relative dualizing complex $K_0 \in D(A_0)$ and its derived base change is $K$. This reduces us to the situation discussed in the next paragraph.

Assume $R$ Noetherian and let $K$ be a relative dualizing complex for $R \to A$. Given a ring map $R \to R'$ set $A' = A \otimes _ R R'$ and $K' = K \otimes _ A^\mathbf {L} A'$. To finish the proof we show $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K', K') = A'$. By Lemma 47.25.2 we know this is true whenever $R'$ is Noetherian. Since a general $R'$ is a filtered colimit of Noetherian $R$-algebras, we find the result holds by More on Algebra, Lemma 15.78.7. $\square$

## 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Your email address will not be published.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.

## Comments (0)