Lemma 47.3.10. Let $A$ be a Noetherian ring. Let $E$ be an injective $A$-module. Then $E \otimes _ A A[x]$ has injective-amplitude $[0, 1]$ as an object of $D(A[x])$. In particular, $E \otimes _ A A[x]$ has finite injective dimension as an $A[x]$-module.

Proof. Let us write $E[x] = E \otimes _ A A[x]$. Consider the short exact sequence of $A[x]$-modules

$0 \to E[x] \to \mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) \to \mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) \to 0$

where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the second map sends $\varphi$ to $f \mapsto \varphi (xf) - x\varphi (f)$. The second map is surjective because $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) = \prod _{n \geq 0} E[x]$ as an abelian group and the map sends $(e_ n)$ to $(e_{n + 1} - xe_ n)$ which is surjective. As an $A$-module we have $E[x] \cong \bigoplus _{n \geq 0} E$ which is injective by Lemma 47.3.7. Hence the $A[x]$-module $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x])$ is injective by Lemma 47.3.4 and the proof is complete. $\square$

Comment #4674 by Bogdan on

A typo, the sentence "Hence the $A[x]$-module $\mathrm{Hom}_A(A[x],I[x])$ is injective" should read "Hence the $A[x]$-module $\mathrm{Hom}_A(A[x],E[x])$ is injective".

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