Lemma 47.27.6. Let $R \to A \to B$ be a ring maps which are flat and of finite presentation. Let $K_{A/R}$ and $K_{B/A}$ be relative dualizing complexes for $R \to A$ and $A \to B$. Then $K = K_{A/R} \otimes _ A^\mathbf {L} K_{B/A}$ is a relative dualizing complex for $R \to B$.

**Proof.**
We will use reduction to the Noetherian case. Namely, 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 $A = A_0 \otimes _{R_0} R$. After increasing $R_0$ and correspondingly replacing $A_0$ we may assume there is a flat finite type ring map $A_0 \to B_0$ such that $B = B_0 \otimes _{R_0} R$ (use the same lemma). If we prove the lemma for $R_0 \to A_0 \to B_0$, then the lemma follows by Lemmas 47.27.2, 47.27.3, and 47.27.4. This reduces us to the situation discussed in the next paragraph.

Assume $R$ is Noetherian and denote $\varphi : R \to A$ and $\psi : A \to B$ the given ring maps. Then $K_{A/R} \cong \varphi ^!(R)$ and $K_{B/A} \cong \psi ^!(A)$, see references given above. Then

by Lemmas 47.24.10 and 47.24.7. Thus $K$ is a relative dualizing complex for $R \to B$. $\square$

## 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Your email address will not be published.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)