Lemma 47.15.10. Let $A$ be a Noetherian ring. If $\omega _ A^\bullet $ is a dualizing complex, then $\omega _ A^\bullet \otimes _ A A[x]$ is a dualizing complex for $A[x]$.
Proof. Set $B = A[x]$ and $\omega _ B^\bullet = \omega _ A^\bullet \otimes _ A B$. It follows from Lemma 47.3.10 and More on Algebra, Lemma 15.69.5 that $\omega _ B^\bullet $ has finite injective dimension. Since $H^ i(\omega _ B^\bullet ) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map
\[ B \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ B(\omega _ B^\bullet , \omega _ B^\bullet ) \]
is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.99.2. $\square$
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