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 10.168.1 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
\[ K = K_{A/R} \otimes _ A^\mathbf {L} K_{B/A} \cong \varphi ^!(R) \otimes _ A^\mathbf {L} \psi ^!(A) \cong \psi ^!(\varphi ^!(R)) \cong (\psi \circ \varphi )^!(R) \]
by Lemmas 47.24.10 and 47.24.7. Thus $K$ is a relative dualizing complex for $R \to B$. $\square$
Comments (0)