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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #4674 by Bogdan on
Comment #4803 by Johan on
There are also: