Lemma 26.7.8. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. Suppose that

\[ 0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0 \]

is a short exact sequence of sheaves of $\mathcal{O}_ X$-modules. If two out of three are quasi-coherent then so is the third.

**Proof.**
This is clear in case both $\mathcal{F}_1$ and $\mathcal{F}_2$ are quasi-coherent because the functor $M \mapsto \widetilde M$ is exact, see Lemma 26.5.4. Similarly in case both $\mathcal{F}_2$ and $\mathcal{F}_3$ are quasi-coherent. Now, suppose that $\mathcal{F}_1 = \widetilde M_1$ and $\mathcal{F}_3 = \widetilde M_3$ are quasi-coherent. Set $M_2 = \Gamma (X, \mathcal{F}_2)$. We claim it suffices to show that the sequence

\[ 0 \to M_1 \to M_2 \to M_3 \to 0 \]

is exact. Namely, if this is the case, then (by using the mapping property of Lemma 26.7.1) we get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & \widetilde M_1 \ar[r] \ar[d] & \widetilde M_2 \ar[r] \ar[d] & \widetilde M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1 \ar[r] & \mathcal{F}_2 \ar[r] & \mathcal{F}_3 \ar[r] & 0 } \]

and we win by the snake lemma.

The “correct” argument here would be to show first that $H^1(X, \mathcal{F}) = 0$ for any quasi-coherent sheaf $\mathcal{F}$. This is actually not all that hard, but it is perhaps better to postpone this till later. Instead we use a small trick.

Pick $m \in M_3 = \Gamma (X, \mathcal{F}_3)$. Consider the following set

\[ I = \{ f \in R \mid \text{the element }fm\text{ comes from }M_2\} . \]

Clearly this is an ideal. It suffices to show $1 \in I$. Hence it suffices to show that for any prime $\mathfrak p$ there exists an $f \in I$, $f \not\in \mathfrak p$. Let $x \in X$ be the point corresponding to $\mathfrak p$. Because surjectivity can be checked on stalks there exists an open neighbourhood $U$ of $x$ such that $m|_ U$ comes from a local section $s \in \mathcal{F}_2(U)$. In fact we may assume that $U = D(f)$ is a standard open, i.e., $f \in R$, $f \not\in \mathfrak p$. We will show that for some $N \gg 0$ we have $f^ N \in I$, which will finish the proof.

Take any point $z \in V(f)$, say corresponding to the prime $\mathfrak q \subset R$. We can also find a $g \in R$, $g \not\in \mathfrak q$ such that $m|_{D(g)}$ lifts to some $s' \in \mathcal{F}_2(D(g))$. Consider the difference $s|_{D(fg)} - s'|_{D(fg)}$. This is an element $m'$ of $\mathcal{F}_1(D(fg)) = (M_1)_{fg}$. For some integer $n = n(z)$ the element $f^ n m'$ comes from some $m'_1 \in (M_1)_ g$. We see that $f^ n s$ extends to a section $\sigma $ of $\mathcal{F}_2$ on $D(f) \cup D(g)$ because it agrees with the restriction of $f^ n s' + m'_1$ on $D(f) \cap D(g) = D(fg)$. Moreover, $\sigma $ maps to the restriction of $f^ n m$ to $D(f) \cup D(g)$.

Since $V(f)$ is quasi-compact, there exists a finite list of elements $g_1, \ldots , g_ m \in R$ such that $V(f) \subset \bigcup D(g_ j)$, an integer $n > 0$ and sections $\sigma _ j \in \mathcal{F}_2(D(f) \cup D(g_ j))$ such that $\sigma _ j|_{D(f)} = f^ n s$ and $\sigma _ j$ maps to the section $f^ nm|_{D(f) \cup D(g_ j)}$ of $\mathcal{F}_3$. Consider the differences

\[ \sigma _ j|_{D(f) \cup D(g_ jg_ k)} - \sigma _ k|_{D(f) \cup D(g_ jg_ k)}. \]

These correspond to sections of $\mathcal{F}_1$ over $D(f) \cup D(g_ jg_ k)$ which are zero on $D(f)$. In particular their images in $\mathcal{F}_1(D(g_ jg_ k)) = (M_1)_{g_ jg_ k}$ are zero in $(M_1)_{g_ jg_ kf}$. Thus some high power of $f$ kills each and every one of these. In other words, the elements $f^ N \sigma _ j$, for some $N \gg 0$ satisfy the glueing condition of the sheaf property and give rise to a section $\sigma $ of $\mathcal{F}_2$ over $\bigcup (D(f) \cup D(g_ j)) = X$ as desired.
$\square$

## Comments (2)

Comment #5685 by Laurent Moret-Bailly on

Comment #5761 by Johan on

There are also: