Lemma 47.26.1. Let $A \to B$ be a faithfully flat map of Noetherian rings. If $K \in D(A)$ and $K \otimes _ A^\mathbf {L} B$ is a dualizing complex for $B$, then $K$ is a dualizing complex for $A$.
Proof. Since $A \to B$ is flat we have $H^ i(K) \otimes _ A B = H^ i(K \otimes _ A^\mathbf {L} B)$. Since $K \otimes _ A^\mathbf {L} B$ is in $D^ b_{\textit{Coh}}(B)$ we first find that $K$ is in $D^ b(A)$ and then we see that $H^ i(K)$ is a finite $A$-module by Algebra, Lemma 10.83.2. Let $M$ be a finite $A$-module. Then
\[ R\mathop{\mathrm{Hom}}\nolimits _ A(M, K) \otimes _ A B = R\mathop{\mathrm{Hom}}\nolimits _ B(M \otimes _ A B, K \otimes _ A^\mathbf {L} B) \]
by More on Algebra, Lemma 15.99.2. Since $K \otimes _ A^\mathbf {L} B$ has finite injective dimension, say injective-amplitude in $[a, b]$, we see that the right hand side has vanishing cohomology in degrees $> b$. Since $A \to B$ is faithfully flat, we find that $R\mathop{\mathrm{Hom}}\nolimits _ A(M, K)$ has vanishing cohomology in degrees $> b$. Thus $K$ has finite injective dimension by More on Algebra, Lemma 15.69.2. To finish the proof we have to show that the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K)$ is an isomorphism. For this we again use More on Algebra, Lemma 15.99.2 and the fact that $B \to R\mathop{\mathrm{Hom}}\nolimits _ B(K \otimes _ A^\mathbf {L} B, K \otimes _ A^\mathbf {L} B)$ is an isomorphism. $\square$
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