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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