The Stacks project

Lemma 61.29.3. Let $X$ be a weakly contractible affine scheme. Let $\Lambda $ be a Noetherian ring and let $I \subset \Lambda $ be an ideal. Let $K$ be an object of $D_{cons}(X, \Lambda )$ such that the cohomology sheaves of $K \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I}$ are locally constant. Then there exists a finite disjoint open covering $X = \coprod U_ i$ and for each $i$ a finite collection of finite projective $\Lambda ^\wedge $-modules $M_ a, \ldots , M_ b$ such that $K|_{U_ i}$ is represented by a complex

\[ (\underline{M^ a})^\wedge \to \ldots \to (\underline{M^ b})^\wedge \]

in $D(U_{i, {pro\text{-}\acute{e}tale}}, \Lambda )$ for some maps of sheaves of $\Lambda $-modules $(\underline{M^ i})^\wedge \to (\underline{M^{i + 1}})^\wedge $.

Proof. We freely use the results of Lemma 61.29.2. Choose $a, b$ as in that lemma. We will prove the lemma by induction on $b - a$. Let $\mathcal{F} = H^ b(K)$. Note that $\mathcal{F}$ is a derived complete sheaf of $\Lambda $-modules by Proposition 61.21.1. Moreover $\mathcal{F}/I\mathcal{F}$ is a locally constant sheaf of $\Lambda /I$-modules of finite type. Apply Lemma 61.28.7 to get a surjection $\rho : (\underline{\Lambda }^\wedge )^{\oplus t} \to \mathcal{F}$.

If $a = b$, then $K = \mathcal{F}[-b]$. In this case we see that

\[ \mathcal{F} \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I} = \mathcal{F}/I\mathcal{F} \]

As $X$ is weakly contractible and $\mathcal{F}/I\mathcal{F}$ locally constant, we can find a finite disjoint union decomposition $X = \coprod U_ i$ by affine opens $U_ i$ and $\Lambda /I$-modules $\overline{M}_ i$ such that $\mathcal{F}/I\mathcal{F}$ restricts to $\underline{\overline{M}_ i}$ on $U_ i$. After refining the covering we may assume the map

\[ \rho |_{U_ i} \bmod I : \underline{\Lambda /I}^{\oplus t} \longrightarrow \underline{\overline{M}_ i} \]

is equal to $\underline{\alpha _ i}$ for some surjective module map $\alpha _ i : \Lambda /I^{\oplus t} \to \overline{M}_ i$, see Modules on Sites, Lemma 18.43.3. Note that each $\overline{M}_ i$ is a finite $\Lambda /I$-module. Since $\mathcal{F}/I\mathcal{F}$ has tor amplitude in $[0, 0]$ we conclude that $\overline{M}_ i$ is a flat $\Lambda /I$-module. Hence $\overline{M}_ i$ is finite projective (Algebra, Lemma 10.78.2). Hence we can find a projector $\overline{p}_ i : (\Lambda /I)^{\oplus t} \to (\Lambda /I)^{\oplus t}$ whose image maps isomorphically to $\overline{M}_ i$ under the map $\alpha _ i$. We can lift $\overline{p}_ i$ to a projector $p_ i : (\Lambda ^\wedge )^{\oplus t} \to (\Lambda ^\wedge )^{\oplus t}$1. Then $M_ i = \mathop{\mathrm{Im}}(p_ i)$ is a finite $I$-adically complete $\Lambda ^\wedge $-module with $M_ i/IM_ i = \overline{M}_ i$. Over $U_ i$ consider the maps

\[ \underline{M_ i}^\wedge \to (\underline{\Lambda }^\wedge )^{\oplus t} \to \mathcal{F}|_{U_ i} \]

By construction the composition induces an isomorphism modulo $I$. The source and target are derived complete, hence so are the cokernel $\mathcal{Q}$ and the kernel $\mathcal{K}$. We have $\mathcal{Q}/I\mathcal{Q} = 0$ by construction hence $\mathcal{Q}$ is zero by Lemma 61.28.6. Then

\[ 0 \to \mathcal{K}/I\mathcal{K} \to \underline{\overline{M}_ i} \to \mathcal{F}/I\mathcal{F} \to 0 \]

is exact by the vanishing of $\text{Tor}_1$ see at the start of this paragraph; also use that $\underline{\Lambda }^\wedge /I\overline{\Lambda }^\wedge $ by Modules on Sites, Lemma 18.42.4 to see that $\underline{M_ i}^\wedge /I\underline{M_ i}^\wedge = \underline{\overline{M}_ i}$. Hence $\mathcal{K}/I\mathcal{K} = 0$ by construction and we conclude that $\mathcal{K} = 0$ as before. This proves the result in case $a = b$.

If $b > a$, then we lift the map $\rho $ to a map

\[ \tilde\rho : (\underline{\Lambda }^\wedge )^{\oplus t}[-b] \longrightarrow K \]

in $D(X_{pro\text{-}\acute{e}tale}, \Lambda )$. This is possible as we can think of $K$ as a complex of $\underline{\Lambda }^\wedge $-modules by discussion in the introduction to Section 61.20 and because $X_{pro\text{-}\acute{e}tale}$ is weakly contractible hence there is no obstruction to lifting the elements $\rho (e_ s) \in H^0(X, \mathcal{F})$ to elements of $H^ b(X, K)$. Fitting $\tilde\rho $ into a distinguished triangle

\[ (\underline{\Lambda }^\wedge )^{\oplus t}[-b] \to K \to L \to (\underline{\Lambda }^\wedge )^{\oplus t}[-b + 1] \]

we see that $L$ is an object of $D_{cons}(X, \Lambda )$ such that $L \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I}$ has tor amplitude contained in $[a, b - 1]$ (details omitted). By induction we can describe $L$ locally as stated in the lemma, say $L$ is isomorphic to

\[ (\underline{M^ a})^\wedge \to \ldots \to (\underline{M^{b - 1}})^\wedge \]

The map $L \to (\underline{\Lambda }^\wedge )^{\oplus t}[-b + 1]$ corresponds to a map $(\underline{M^{b - 1}})^\wedge \to (\underline{\Lambda }^\wedge )^{\oplus t}$ which allows us to extend the complex by one. The corresponding complex is isomorphic to $K$ in the derived category by the properties of triangulated categories. This finishes the proof. $\square$

[1] Proof: by Algebra, Lemma 10.32.7 we can lift $\overline{p}_ i$ to a compatible system of projectors $p_{i, n} : (\Lambda /I^ n)^{\oplus t} \to (\Lambda /I^ n)^{\oplus t}$ and then we set $p_ i = \mathop{\mathrm{lim}}\nolimits p_{i, n}$ which works because $\Lambda ^\wedge = \mathop{\mathrm{lim}}\nolimits \Lambda /I^ n$.

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 09C3. Beware of the difference between the letter 'O' and the digit '0'.