So, I'm trying to show that the ring of regular functions on an affine algebraic set forms a sheaf. I have no problem showing that they form a presheaf, that much is clear, but the two additional things you have to prove I find particularly annoying. To quote Hartshorne, p. 61:
(4) if $U$ is an open set, if $\{ V_i \}$ is an open covering of $U$, and if we have elements $s_i \in \mathcal{F}(V_i)$ for each $i$ with the property that for each $i,j$, $s_i |_{V_i \cap V_j} = s_j |_{V_i \cap V_j}$, then there is an element $s \in \mathcal{F}(U)$ such that $s|_{V_i} = s_i$.
Fair enough I figure, only to then ask the question, what if the overlap between the different $V_i$ are always the empty set? After all, is there anything compelling $U$ to not be disjoint?
Consider then the following schematic:
$U \subset X$ has an open covering of $\{ V_1 , V_2, V_3 \}$.">
Since $V_i \cap V_j = \varnothing$ for all $i,j$, it trivially follows that $\mathcal{F}(V_i \cap V_j) = \{ 0 \}$ and so $s_i |_{V_i \cap V_j} = s_j |_{V_i \cap V_j} = 0$ for all $s_i \in V_i, s_j \in V_j$. But that then would imply that the restriction morphisms $\rho_{UV_i}$ must all necessarily be surjective. And that just strikes me as odd and deeply suspicious.
What, and where, am I doing something that must be terribly wrong?
Look forward to your responses!
There is nothing suspicious about the restriction maps being surjective. In fact, a sheaf with this property has a special name: flasque sheaf.
As for the open cover having trivial intersections, the sheaf condition becomes vacuously true as you point out. Note that if the underlying topological space is irreducible, then open subsets are always dense, and therefore intersections of nontrivial open subsets aren't empty.