Concerning Hartshorne's proof of the Vanishing Theorem of Grothendieck (Hartshorne III 2.7) At certain point of the proof of the theorem in Hartshorne's book, he mentioned the following in which I don't understand:


*

*In Step 3, he said we can reduce our consideration to a sheaf $\mathscr{F}$ of abelian group which is generated by a single section over certain open set $U\subset X$ ($X$ is irreducible topological space). Then in that case $\mathscr{F}$ $\textbf{is a quotient of the sheaf}$ $\textbf{Z}_U$ (where $\textbf{Z}$ is a constant sheaf on $X$).


Question: I couldn't see how the statement is true.


*Assume the things mentioned in (1) is correct, then we will be able to write an exact sequence of $\mathscr{F}$ as
$$0\to\mathscr{R}\to\textbf{Z}_U\to\mathscr{F}\to 0$$
where $\mathscr{R}$ is of course, the kernel of the map from $\textbf{Z}_U$ to $\mathscr{F}$. Then for any $x\in U$, $\mathscr{R}_x$ is a subgroup of $\textbf{Z}$ and we have certain $d\in\textbf{Z}$ which is the least positive integer which occurs in any of the groups $\mathscr{R}_x$, no problem. But the problem comes when he said there exist another open subset $V\subseteq U$ such that $\mathscr{R}|_V\cong d\cdot\textbf{Z}|_V$ as a subsheaf of $\textbf{Z}|_V$.


Question: Explicitly why there is such an existence?
Thank you in advance and please let me know if there is any unclear point in my question.
 A: For the first question : It is a general (and easy) fact that the maps from a constant sheaf $\Gamma$ to a sheaf $\mathscr{F}$ are in bijection with $\operatorname{Hom}(\Gamma(X),\mathscr{F}(X))$. This means that over a connected scheme $X$, to give a map $\mathbb{Z}\to \mathscr{F}$, you have to give a global section of $\mathscr{F}$. If this global section generates $\mathscr{F}$, then the corresponding map from $\mathbb{Z}\to \mathscr{F}$ would be surjective and hence $\mathscr{F}$ would be a quotient of $\mathbb{Z}$.
For the second question : Over an irreducible space $X$ all the open subsets are connected because every two open subsets have non-empty intersection. This means that for every open $V$, $\mathbb{Z}(V)=\mathbb{Z}$ now, because $d$ appears in some stalk it should be in $\mathscr{R}(V)$ for some open $V$ and hence for every $V'\subset V$ we have $d\in \mathscr{R}(V')$, and because you have assumed that $d$ is the smallest integer appearing in the stalks, $d$ should generate $\mathscr{R}(V')$(because $\mathbb{Z}$ is a p.i.d) and hence $\mathscr{R}\mid_V=d\mathbb{Z}$
A: In Hartshorne's notation, $B=\bigcup_{U\subset X}\mathcal{F}(U)$. Let $A$ denote the set of all finite subsets of $B$.

*

*We know that $\mathcal{F}=\varinjlim \mathcal{F}_\alpha$ where $\mathcal{F}_\alpha$ is the sheaf generated by the elements of $\alpha \in A$ and $\alpha$ ranges over $A$. Using the fact that cohomology commutes with direct limits, we have $\varinjlim H^i(X,\mathcal{F}_\alpha)=H^i(X,\mathcal{F})$ and so it will suffice to show that $H^i(X,\mathcal{F}_\alpha)=0$ for all $\alpha$ and for all $i\ge n+1$ where $n=\dim X$. Hartshorne argues by the induction on the cardinality of $\alpha$. Let's suppose we have proven the result for $\#\alpha=1$ and we assume the result for $\# \alpha \le n-1$. If $\alpha'\subset \alpha$ and $\#\alpha-1=\#\alpha'$ then
$$
0\to \mathcal{F}_{\alpha'}\to \mathcal{F}_\alpha \to \mathcal{G}\to 0
$$
is an exact sequence where $\mathcal{G}$ is generated by a single section $s\in \mathcal{F}(U)$ for some $U$. $s$ is the element of $\alpha\setminus \alpha'$. To understand the answer to your question, we should think about what it means to be generated by some sections. Given $s_1,\ldots, s_n$ so that $s_i\in \mathcal{F}(U_i)$ for each $i$, what we mean by $\mathcal{H}$ is generated by $s_1,\ldots, s_n$ is that $\mathcal{H}$ fits into an exact sequence
$$0\to \mathcal{K}\to\bigoplus_{i=1}^n\Bbb{Z}_{U_i}s_i\to \mathcal{H}\to 0.$$
So, since $\mathcal{G}$ is generated by $s_n$, there is an exact sequence
$$
0\to \mathcal{R}\to \Bbb{Z}_{U_n}\xrightarrow{\times s_n} \mathcal{G}\to 0.
$$
The argument then continues as on p.211 of Hartshorne.

*Now, let's specialize our above exact sequence by taking stalks at $x\in U$. This preserves exactness and so $0\to \mathcal{R}_x\to \Bbb{Z}$ is exact, and in particular $\mathcal{R}_x$ can be thought of as a subgroup of $\Bbb{Z}$. We know all of the subgroups of $\Bbb{Z}$. They are of the form $d\Bbb{Z}$ for $d\in \Bbb{Z}_{\ge0}$. So, $\mathcal{R}_x$ is cyclic. In particular, by definition of $\mathcal{R}_x$ there exists a germ $(s,V)$ where $s\in \mathcal{R}(V)$ so that $\mathcal{R}_x\cong d\Bbb{Z}$ by the map $(s,V)\mapsto d$. Actually, this isomorphism is valid not only on the stalk but also on $V$. So, we get $\mathcal{R}_V\cong d\Bbb{Z}_V$ by sending the generator $s$ to $d$.

