The Stacks project

Lemma 46.9.2. Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. Let $M$, $N$ be $A$-modules. For all $i$ we have a canonical isomorphism

\[ \mathop{\mathrm{Ext}}\nolimits ^ i_{\textit{Mod}(\mathcal{O})}(M^ a, N^ a) = \text{Pext}^ i_ A(M, N) \]

functorial in $M$ and $N$.

Proof. Let us construct a canonical arrow from right to left. Namely, if $N \to I^\bullet $ is a pure injective resolution, then $M^ a \to (I^\bullet )^ a$ is an exact complex of (adequate) $\mathcal{O}$-modules. Hence any element of $\text{Pext}^ i_ A(M, N)$ gives rise to a map $N^ a \to M^ a[i]$ in $D(\mathcal{O})$, i.e., an element of the group on the left.

To prove this map is an isomorphism, note that we may replace $\mathop{\mathrm{Ext}}\nolimits ^ i_{\textit{Mod}(\mathcal{O})}(M^ a, N^ a)$ by $\mathop{\mathrm{Ext}}\nolimits ^ i_{\textit{Adeq}(\mathcal{O})}(M^ a, N^ a)$, see Lemma 46.7.6. Let $\mathcal{A}$ be the category of adequate functors on $\textit{Alg}_ A$. We have seen that $\mathcal{A}$ is equivalent to $\textit{Adeq}(\mathcal{O})$, see Lemma 46.5.3; see also the proof of Lemma 46.7.3. Hence now it suffices to prove that

\[ \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(\underline{M}, \underline{N}) = \text{Pext}^ i_ A(M, N) \]

However, this is clear from Lemma 46.9.1 as a pure injective resolution $N \to I^\bullet $ exactly corresponds to an injective resolution of $\underline{N}$ in $\mathcal{A}$. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0709. Beware of the difference between the letter 'O' and the digit '0'.