Concerning Hartshorne's proof of the Vanishing Theorem of Grothendieck (Hartshorne III 2.7)

Question

At certain point of the proof of the theorem in Hartshorne's book, he mentioned the following in which I don't understand:

1. 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.

2. 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.

Answer 1

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 $\mathscr{R}\mid_V=d\mathbb{Z}$

Answer 2

In Hartshorne's notation, $B=\bigcup_{U\subset X}\mathcal{F}(U)$. Let $A$ denote the set of all finite subsets of $B$.

1. 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.
2. 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 $d$.