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Let $X$ be a variety. Show that if $X$ is irreducible, then the constant abelian presheaf $\mathcal{F}$ with $\mathcal{F}(U)=\mathbb{Z}$ for every nonempty open subset $U\subseteq X$ and $\mathcal{F}(\emptyset)=0$ is a sheaf. Any leads? What does the word "constant" means here?

Vinyl_cape_jawa
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    That $X$ is a variety is irrelevant: the problem is purely topological (I say this in order that you don't lose your time trying to apply results from algebraic geometry). – Georges Elencwajg Jan 02 '15 at 21:30
  • Thanks, but now I am even more confused because I don't have any results on algebraic topology to apply here. I mean this is the first time I stumble on the word topology on this course. – Vinyl_cape_jawa Jan 03 '15 at 12:52
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    It is only general topology, not algebraic topology. Krish's answer illustrates my point: he only talks about open sets and not about rings, localization, ...You can just read his answer and forget about my comments... :-) – Georges Elencwajg Jan 03 '15 at 13:49

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Hint: Since $X$ is irreducible, any two non-empty open set always intersects. And for any inclusion $V \subseteq U$ of non-empty open sets, the restriction map $\mathcal{F}(U) \rightarrow \mathcal{F}(V)$ is the identity map.

Sheaf Property: (1) Let $U \subseteq X$ be open and let $\{U_i\}$ be an open cover of $U.$ Suppose $s \in \mathcal{F}(U)$ and $s|_{U_i} = 0, \forall i.$ The restriction maps are identity maps. So we have $s = 0.$

(2) Let $U \subseteq X$ be open and let $\{U_i\}$ be an open cover of $U.$ Suppose for each $i,$ we have $s_i \in \mathcal{F}(U_i)$ such that $s_i|_{U_I \cap U_j} = s_j|_{U_i \cap U_j}, \forall i, j.$ Since $X$ is irreducible, $U_i \cap U_j \neq \emptyset, \forall i, j$ and the restriction maps $\mathcal{F}(U_i) \rightarrow \mathcal{F}(U_i \cap U_j)$ are identity maps, this shows that $s_i = s_j \in \mathbb{Z}, \forall i, j.$ Take $s \in \mathcal{F}(U)$ to be $s_i.$

Krish
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  • Can I ask, why does the fact that $X$ is irreducible imply that the intersection $U_i \cap U_j$ is nonempty? –  Jan 14 '15 at 21:19
  • @Sodan: Let $X$ be an irreducible topological space and let $U, V$ be two non-empty open subsets of $X.$ Suppose $U \cap V = \emptyset.$ Let $F_1 = X -U, F_2 = X-V.$ Then both $F_1, F_2$ are non-empty proper closed subsets of $X$ and $F_1 \cup F_2 = X,$ contradicting the fact that $X$ is irreducible. – Krish Jan 15 '15 at 05:29