Lemma 47.22.2. Let $A \to B$ be a flat map of Noetherian rings. Let $I \subset A$ be an ideal such that $A/I = B/IB$ and such that $IB$ is contained in the Jacobson radical of $B$. Let $\omega _ A^\bullet$ be a dualizing complex. Then $\omega _ A^\bullet \otimes _ A B$ is a dualizing complex for $B$.

Proof. It is clear that $\omega _ A^\bullet \otimes _ A B$ is in $D^ b_{\textit{Coh}}(B)$. By More on Algebra, Lemma 15.99.2 we see that

$R\mathop{\mathrm{Hom}}\nolimits _ B(K \otimes _ A B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ) \otimes _ A B$

for any $K \in D^ b_{\textit{Coh}}(A)$. For any ideal $IB \subset J \subset B$ there is a unique ideal $I \subset J' \subset A$ such that $A/J' \otimes _ A B = B/J$. Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension by More on Algebra, Lemma 15.69.6. Finally, we also have

$R\mathop{\mathrm{Hom}}\nolimits _ B(\omega _ A^\bullet \otimes _ A B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \otimes _ A B = A \otimes _ A B = B$

as desired. $\square$

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