In proof of Hartshorne book Theorem III.2.7 I can't understand the following statment:
Let $F$ be a subsheaf of $\mathbb{Z}$ on $X$, where $\mathbb{Z}$ is the constant sheaf. Suppose that $F \neq 0$ and let $d$ be the least positive integer which occurs in any of the group $F_x$ for each $x \in X$. Then there is a nonempty open subset $U$ such that $F_U \cong d\cdot \mathbb{Z}|_U $ as a subsheaf of $\mathbb{Z}|_U$ and hence $F_U \cong \mathbb{Z}_U$.
Notation: $F_U=i_!(F|_U)$ where $i : U \rightarrow X$ inclusion $i_!(F|_U)$ be the sheaf associated to the presheaf $V \mapsto F(V)$ if $V \subseteq U$ $V \mapsto 0$ otherwise.
Help me...
There is some nonempty open subset $U$ with $d \in F(U)$. Hence, we have $d \cdot \mathbb{Z}|_U \subseteq F_U$. That this is an isomorphism can be checked stalkwise. Outside of $U$ this is clear. At $p \in U$ we have $d \cdot \mathbb{Z} \subseteq F_p \subseteq \mathbb{Z}$. Choose $e>0$ with $F_p = e \mathbb{Z}$ for some $e>0$. Then $e | d$. Minimality of $d$ gives us $e=d$, qed.