The Stacks project

46.9 Higher exts of quasi-coherent sheaves on the big site

It turns out that the module-valued functor $\underline{I}$ associated to a pure injective module $I$ gives rise to an injective object in the category of adequate functors on $\textit{Alg}_ A$. Warning: It is not true that a pure projective module gives rise to a projective object in the category of adequate functors. We do have plenty of projective objects, namely, the linearly adequate functors.

Lemma 46.9.1. Let $A$ be a ring. Let $\mathcal{A}$ be the category of adequate functors on $\textit{Alg}_ A$. The injective objects of $\mathcal{A}$ are exactly the functors $\underline{I}$ where $I$ is a pure injective $A$-module.

Proof. Let $I$ be an injective object of $\mathcal{A}$. Choose an embedding $I \to \underline{M}$ for some $A$-module $M$. As $I$ is injective we see that $\underline{M} = I \oplus F$ for some module-valued functor $F$. Then $M = I(A) \oplus F(A)$ and it follows that $I = \underline{I(A)}$. Thus we see that any injective object is of the form $\underline{I}$ for some $A$-module $I$. It is clear that the module $I$ has to be pure injective since any universally exact sequence $0 \to M \to N \to L \to 0$ gives rise to an exact sequence $0 \to \underline{M} \to \underline{N} \to \underline{L} \to 0$ of $\mathcal{A}$.

Finally, suppose that $I$ is a pure injective $A$-module. Choose an embedding $\underline{I} \to J$ into an injective object of $\mathcal{A}$ (see Lemma 46.4.2). We have seen above that $J = \underline{I'}$ for some $A$-module $I'$ which is pure injective. As $\underline{I} \to \underline{I'}$ is injective the map $I \to I'$ is universally injective. By assumption on $I$ it splits. Hence $\underline{I}$ is a summand of $J = \underline{I'}$ whence an injective object of the category $\mathcal{A}$. $\square$

Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. Let $M$ be an $A$-module. We will use the notation $M^ a$ to denote the quasi-coherent sheaf of $\mathcal{O}$-modules on $(\mathit{Sch}/U)_\tau $ associated to the quasi-coherent sheaf $\widetilde{M}$ on $U$. Now we have all the notation in place to formulate the following lemma.

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 0707. Beware of the difference between the letter 'O' and the digit '0'.