Lemma 47.27.4. Let $R \to A$ be a flat ring map of finite presentation. Then

1. there exists a relative dualizing complex $K$ in $D(A)$, and

2. for any ring map $R \to R'$ setting $A' = A \otimes _ R R'$ and $K' = K \otimes _ A^\mathbf {L} A'$, then $K'$ is a relative dualizing complex for $R' \to A'$.

Moreover, if

$\xi : A \longrightarrow K \otimes _ A^\mathbf {L} (A \otimes _ R A)$

is a generator for the cyclic module $\mathop{\mathrm{Hom}}\nolimits _{D(A \otimes _ R A)}(A, K \otimes _ A^\mathbf {L} (A \otimes _ R A))$ then in (2) the derived base change of $\xi$ by $A \otimes _ R A \to A' \otimes _{R'} A'$ is a generator for the cyclic module $\mathop{\mathrm{Hom}}\nolimits _{D(A' \otimes _{R'} A')}(A', K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A'))$

Proof. We first reduce to the Noetherian case. 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 $A = A_0 \otimes _{R_0} R$. By Lemma 47.27.3 there exists a relative dualizing complex $K_0 \in D(A_0)$. Thus if we show (2) for $K_0$, then we find that $K_0 \otimes _{A_0}^\mathbf {L} A$ is a dualizing complex for $R \to A$ and that it also satisfies (2) by transitivity of derived base change. The uniqueness of relative dualizing complexes (Lemma 47.27.2) then shows that this holds for any relative dualizing complex.

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 have to show that $K'$ is a relative dualizing complex for $R' \to A'$. By More on Algebra, Lemma 15.83.5 we see that $K'$ is $R'$-perfect in all cases. By Lemmas 47.25.1 and 47.27.3 if $R'$ is Noetherian, then $K'$ is a relative dualizing complex for $R' \to A'$ (in either sense). Transitivity of derived tensor product shows that $K \otimes _ A^\mathbf {L} (A \otimes _ R A) \otimes _{A \otimes _ R A}^\mathbf {L} (A' \otimes _{R'} A') = K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A')$. Flatness of $R \to A$ guarantees that $A \otimes _{A \otimes _ R A}^\mathbf {L} (A' \otimes _{R'} A') = A'$; namely $A \otimes _ R A$ and $R'$ are tor independent over $R$ so we can apply More on Algebra, Lemma 15.61.2. Finally, $A$ is pseudo-coherent as an $A \otimes _ R A$-module by More on Algebra, Lemma 15.82.8. Thus we have checked all the assumptions of More on Algebra, Lemma 15.83.6. We find there exists a bounded below complex $E^\bullet$ of $R$-flat finitely presented $A \otimes _ R A$-modules such that $E^\bullet \otimes _ R R'$ represents $R\mathop{\mathrm{Hom}}\nolimits _{A' \otimes _{R'} A'}(A', K' \otimes _{A'}^\mathbf {L} (A' \otimes _{R'} A'))$ and these identifications are compatible with derived base change. Let $n \in \mathbf{Z}$, $n \not= 0$. Define $Q^ n$ by the sequence

$E^{n - 1} \to E^ n \to Q^ n \to 0$

Since $\kappa (\mathfrak p)$ is a Noetherian ring, we know that $H^ n(E^\bullet \otimes _ R \kappa (\mathfrak p)) = 0$, see remarks above. Chasing diagrams this means that

$Q^ n \otimes _ R \kappa (\mathfrak p) \to E^{n + 1} \otimes _ R \kappa (\mathfrak p)$

is injective. Hence for a prime $\mathfrak q$ of $A \otimes _ R A$ lying over $\mathfrak p$ we have $Q^ n_\mathfrak q$ is $R_\mathfrak p$-flat and $Q^ n_\mathfrak p \to E^{n + 1}_\mathfrak q$ is $R_\mathfrak p$-universally injective, see Algebra, Lemma 10.99.1. Since this holds for all primes, we conclude that $Q^ n$ is $R$-flat and $Q^ n \to E^{n + 1}$ is $R$-universally injective. In particular $H^ n(E^\bullet \otimes _ R R') = 0$ for any ring map $R \to R'$. Let $Z^0 = \mathop{\mathrm{Ker}}(E^0 \to E^1)$. Since there is an exact sequence $0 \to Z^0 \to E^0 \to E^1 \to Q^1 \to 0$ we see that $Z^0$ is $R$-flat and that $Z^0 \otimes _ R R' = \mathop{\mathrm{Ker}}(E^0 \otimes _ R R' \to E^1 \otimes _ R R')$ for all $R \to R'$. Then the short exact sequence $0 \to Q^{-1} \to Z^0 \to H^0(E^\bullet ) \to 0$ shows that

$H^0(E^\bullet \otimes _ R R') = H^0(E^\bullet ) \otimes _ R R' = A \otimes _ R R' = A'$

as desired. This equality furthermore gives the final assertion of the lemma. $\square$

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