The Stacks Project


Tag: 06VE

This tag has label descent-lemma-equivalence-quasi-coherent-limits and it points to

The corresponding content:

Lemma 31.7.13. Let $S$ be a scheme. Let $\tau \in \{Zariski, fppf, \acute{e}tale, smooth, syntomic\}$. The functors $$ \textit{QCoh}(\mathcal{O}_S) \longrightarrow \textit{Mod}((\textit{Sch}/S)_\tau, \mathcal{O}) \quad\text{and}\quad \textit{QCoh}(\mathcal{O}_S) \longrightarrow \textit{Mod}(S_\tau, \mathcal{O}) $$ defined by the rule $\mathcal{F} \mapsto \mathcal{F}^a$ seen in Proposition 31.7.11 are
  1. fully faithful,
  2. compatible with direct sums,
  3. compatible with colimits,
  4. right exact,
  5. exact as a functor $\textit{QCoh}(\mathcal{O}_S) \to \textit{Mod}(S_\tau, \mathcal{O})$,
  6. not exact as a functor $\textit{QCoh}(\mathcal{O}_S) \to \textit{Mod}((\textit{Sch}/S)_\tau, \mathcal{O})$ in general,
  7. given two quasi-coherent $\mathcal{O}_S$-modules $\mathcal{F}$, $\mathcal{G}$ we have $(\mathcal{F} \otimes_{\mathcal{O}_S} \mathcal{G})^a = \mathcal{F}^a \otimes_\mathcal{O} \mathcal{G}^a$,
  8. given two quasi-coherent $\mathcal{O}_S$-modules $\mathcal{F}$, $\mathcal{G}$ such that $\mathcal{F}$ is of finite presentation we have $(\mathop{\mathcal{H}\!{\it om}}\nolimits_{\mathcal{O}_S}(\mathcal{F}, \mathcal{G}))^a = \mathop{\mathcal{H}\!{\it om}}\nolimits_\mathcal{O}(\mathcal{F}^a, \mathcal{G}^a)$, and
  9. given a short exact sequence $0 \to \mathcal{F}_1^a \to \mathcal{E} \to \mathcal{F}_2^a \to 0$ of $\mathcal{O}$-modules then $\mathcal{E}$ is quasi-coherent (Warning: This is misleading. See part (6).), i.e., $\mathcal{E}$ is in the essential image of the functor.

Proof. Part (1) we saw in Proposition 31.7.11.

We have seen in Schemes, Section 22.24 that a colimit of quasi-coherent sheaves on a scheme is a quasi-coherent sheaf. Moreover, in Remark 31.7.6 we saw that $\mathcal{F} \mapsto \mathcal{F}^a$ is the pullback functor for a morphism of ringed sites, hence commutes with all colimits, see Modules on Sites, Lemma 17.14.3. Thus (3) and its special case (3) hold.

This also shows that the functor is right exact (i.e., commutes with finite colimits), hence (4).

The functor $\textit{QCoh}(\mathcal{O}_S) \to \textit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^a$ is left exact because an étale morphism is flat, see Morphisms, Lemma 25.37.12. This proves (5).

To see (6), suppose that $S = \mathop{\rm Spec}(\mathbf{Z})$. Then $2 : \mathcal{O}_S \to \mathcal{O}_S$ is injective but the associated map of $\mathcal{O}$-modules on $(\textit{Sch}/S)_\tau$ isn't injective because $2 : \mathbf{F}_2 \to \mathbf{F}_2$ isn't injective and $\mathop{\rm Spec}(\mathbf{F}_2)$ is an object of $(\textit{Sch}/S)_\tau$.

We omit the proofs of (7) and (8).

Let $0 \to \mathcal{F}_1^a \to \mathcal{E} \to \mathcal{F}_2^a \to 0$ be a short exact sequence of $\mathcal{O}$-modules with $\mathcal{F}_1$ and $\mathcal{F}_2$ quasi-coherent on $S$. Consider the restriction $$ 0 \to \mathcal{F}_1 \to \mathcal{E}|_{S_{Zar}} \to \mathcal{F}_2 $$ to $S_{Zar}$. By Proposition 31.7.10 we see that on any affine $U \subset S$ we have $H^1(U, \mathcal{F}_1^a) = H^1(U, \mathcal{F}_1) = 0$. Hence the sequence above is also exact on the right. By Schemes, Section 22.24 we conclude that $\mathcal{F} = \mathcal{E}|_{S_{Zar}}$ is quasi-coherent. Thus we obtain a commutative diagram $$ \xymatrix{ & \mathcal{F}_1^a \ar[r] \ar[d] & \mathcal{F}^a \ar[r] \ar[d] & \mathcal{F}_2^a \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1^a \ar[r] & \mathcal{E} \ar[r] & \mathcal{F}_2^a \ar[r] & 0 } $$ To finish the proof it suffices to show that the top row is also right exact. To do this, denote once more $U = \mathop{\rm Spec}(A) \subset S$ an affine open of $S$. We have seen above that $0 \to \mathcal{F}_1(U) \to \mathcal{E}(U) \to \mathcal{F}_2(U) \to 0$ is exact. For any affine scheme $V/U$, $V = \mathop{\rm Spec}(B)$ the map $\mathcal{F}_1^a(V) \to \mathcal{E}(V)$ is injective. We have $\mathcal{F}_1^a(V) = \mathcal{F}_1(U) \otimes_A B$ by definition. The injection $\mathcal{F}_1^a(V) \to \mathcal{E}(V)$ factors as $$ \mathcal{F}_1(U) \otimes_A B \to \mathcal{E}(U) \otimes_A B \to \mathcal{E}(U) $$ Considering $A$-algebras $B$ of the form $B = A \oplus M$ we see that $\mathcal{F}_1(U) \to \mathcal{E}(U)$ is universally injective (see Algebra, Definition 9.79.1). Since $\mathcal{E}(U) = \mathcal{F}(U)$ we conclude that $\mathcal{F}_1 \to \mathcal{F}$ remains injective after any base change, or equivalently that $\mathcal{F}_1^a \to \mathcal{F}^a$ is injective. $\square$

\begin{lemma}
\label{lemma-equivalence-quasi-coherent-limits}
Let $S$ be a scheme.
Let $\tau \in \{Zariski, \linebreak[0] fppf, \linebreak[0]
\acute{e}tale, \linebreak[0] smooth, \linebreak[0] syntomic\}$.
The functors
$$
\textit{QCoh}(\mathcal{O}_S)
\longrightarrow
\textit{Mod}((\Sch/S)_\tau, \mathcal{O})
\quad\text{and}\quad
\textit{QCoh}(\mathcal{O}_S)
\longrightarrow
\textit{Mod}(S_\tau, \mathcal{O})
$$
defined by the rule $\mathcal{F} \mapsto \mathcal{F}^a$ seen in
Proposition \ref{proposition-equivalence-quasi-coherent}
are
\begin{enumerate}
\item fully faithful,
\item compatible with direct sums,
\item compatible with colimits,
\item right exact,
\item exact as a functor
$\textit{QCoh}(\mathcal{O}_S) \to \textit{Mod}(S_\tau, \mathcal{O})$,
\item {\bf not} exact as a functor
$\textit{QCoh}(\mathcal{O}_S) \to
\textit{Mod}((\Sch/S)_\tau, \mathcal{O})$
in general,
\item given two quasi-coherent $\mathcal{O}_S$-modules
$\mathcal{F}$, $\mathcal{G}$ we have
$(\mathcal{F} \otimes_{\mathcal{O}_S} \mathcal{G})^a =
\mathcal{F}^a \otimes_\mathcal{O} \mathcal{G}^a$,
\item given two quasi-coherent $\mathcal{O}_S$-modules
$\mathcal{F}$, $\mathcal{G}$ such that $\mathcal{F}$
is of finite presentation we have
$(\SheafHom_{\mathcal{O}_S}(\mathcal{F}, \mathcal{G}))^a =
\SheafHom_\mathcal{O}(\mathcal{F}^a, \mathcal{G}^a)$, and
\item given a short exact sequence
$0 \to \mathcal{F}_1^a \to \mathcal{E} \to \mathcal{F}_2^a \to 0$
of $\mathcal{O}$-modules then $\mathcal{E}$ is
quasi-coherent\footnote{Warning: This is misleading. See part (6).}, i.e.,
$\mathcal{E}$ is in the essential image of the functor.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) we saw in
Proposition \ref{proposition-equivalence-quasi-coherent}.

\medskip\noindent
We have seen in
Schemes, Section \ref{schemes-section-quasi-coherent}
that a colimit of quasi-coherent sheaves on a scheme is a quasi-coherent
sheaf. Moreover, in
Remark \ref{remark-change-topologies-ringed-sites}
we saw that $\mathcal{F} \mapsto \mathcal{F}^a$ is the pullback functor
for a morphism of ringed sites, hence commutes with all colimits, see
Modules on Sites, Lemma
\ref{sites-modules-lemma-exactness-pushforward-pullback}.
Thus (3) and its special case (3) hold.

\medskip\noindent
This also shows that the functor is right exact (i.e., commutes with
finite colimits), hence (4).

\medskip\noindent
The functor $\textit{QCoh}(\mathcal{O}_S) \to
\textit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$,
$\mathcal{F} \mapsto \mathcal{F}^a$
is left exact because an \'etale morphism is flat, see
Morphisms, Lemma \ref{morphisms-lemma-etale-flat}.
This proves (5).

\medskip\noindent
To see (6), suppose that $S = \Spec(\mathbf{Z})$.
Then $2 : \mathcal{O}_S \to \mathcal{O}_S$ is injective but the associated
map of $\mathcal{O}$-modules on $(\Sch/S)_\tau$ isn't
injective because $2 : \mathbf{F}_2 \to \mathbf{F}_2$ isn't injective
and $\Spec(\mathbf{F}_2)$ is an object of $(\Sch/S)_\tau$.

\medskip\noindent
We omit the proofs of (7) and (8).

\medskip\noindent
Let $0 \to \mathcal{F}_1^a \to \mathcal{E} \to \mathcal{F}_2^a \to 0$
be a short exact sequence of $\mathcal{O}$-modules with $\mathcal{F}_1$
and $\mathcal{F}_2$ quasi-coherent on $S$. Consider the restriction
$$
0 \to \mathcal{F}_1 \to \mathcal{E}|_{S_{Zar}} \to \mathcal{F}_2
$$
to $S_{Zar}$. By
Proposition \ref{proposition-same-cohomology-quasi-coherent}
we see that on any affine $U \subset S$ we have
$H^1(U, \mathcal{F}_1^a) = H^1(U, \mathcal{F}_1) = 0$.
Hence the sequence above is also exact on the right. By
Schemes, Section \ref{schemes-section-quasi-coherent}
we conclude that $\mathcal{F} = \mathcal{E}|_{S_{Zar}}$ is
quasi-coherent. Thus we obtain a commutative diagram
$$
\xymatrix{
& \mathcal{F}_1^a \ar[r] \ar[d] &
\mathcal{F}^a \ar[r] \ar[d] &
\mathcal{F}_2^a \ar[r] \ar[d] & 0 \\
0 \ar[r] &
\mathcal{F}_1^a \ar[r] &
\mathcal{E} \ar[r] &
\mathcal{F}_2^a \ar[r] & 0
}
$$
To finish the proof it suffices to show that the top row is also
right exact. To do this, denote once more $U = \Spec(A) \subset S$
an affine open of $S$. We have seen above that
$0 \to \mathcal{F}_1(U) \to \mathcal{E}(U) \to \mathcal{F}_2(U) \to 0$
is exact. For any affine scheme $V/U$,
$V = \Spec(B)$ the map $\mathcal{F}_1^a(V) \to \mathcal{E}(V)$
is injective. We have $\mathcal{F}_1^a(V) = \mathcal{F}_1(U) \otimes_A B$
by definition. The injection
$\mathcal{F}_1^a(V) \to \mathcal{E}(V)$ factors as
$$
\mathcal{F}_1(U) \otimes_A B \to
\mathcal{E}(U) \otimes_A B \to \mathcal{E}(U)
$$
Considering $A$-algebras $B$ of the form $B = A \oplus M$
we see that $\mathcal{F}_1(U) \to \mathcal{E}(U)$ is
universally injective (see
Algebra, Definition \ref{algebra-definition-universally-injective}).
Since $\mathcal{E}(U) = \mathcal{F}(U)$ we conclude that
$\mathcal{F}_1 \to \mathcal{F}$ remains injective after any base change,
or equivalently that $\mathcal{F}_1^a \to \mathcal{F}^a$ is injective.
\end{proof}
    

To cite this tag (see How to reference tags), use:

\cite[\href{http://stacks.math.columbia.edu/tag/06VE}{Tag 06VE}]{stacks-project}

Comments (0)

There are no comments yet for this tag.

Add a comment on tag 06VE

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 lower-right corner).




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 box. So in case this is tag 0321 you just have to write 0321. This captcha seems more appropriate than the usual illegible gibberish, right?