Composition and relative dualizing complex

The case of algebra. The case of schemes is true as well but it would
take more effort to prove it

dualizing.tex

@@ -4774,6 +4774,45 @@ \section{Relative dualizing complexes}
 More on Algebra, Lemma \ref{more-algebra-lemma-colimit-relatively-perfect}.
 \end{proof}
 
+\begin{lemma}
+\label{lemma-relative-dualizing-composition}
+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 $R \to B$.
+\end{lemma}
+
+\begin{proof}
+We will use reduction to the Noetherian case.
+Namely, by Algebra, Lemma
+\ref{algebra-lemma-flat-finite-presentation-limit-flat}
+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 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
+\ref{lemma-uniqueness-relative-dualizing},
+\ref{lemma-relative-dualizing-noetherian}, and
+\ref{lemma-base-change-relative-dualizing}.
+This reduces us to the situation discussed in the next paragraph.
+
+\medskip\noindent
+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 \ref{lemma-upper-shriek-is-tensor-functor} and
+\ref{lemma-composition-shriek-algebraic}. Thus $K$ is a relative
+dualizing complex for $R \to B$.
+\end{proof}
+