Sheaf of regular functions I am trying to show that if $X$ is an (irreducible) affine variety, then $O_X$, which is the set of rings $O_X(U)$ of regular functions on open subsets of $X$, with the obvious restriction maps, is really a sheaf. I am able to verify that 


*

*$\rho_{U,U}$ is the identity

*for any inclusion $U\subset V\subset W$, we have $\rho_{V,U}\circ\rho_{W,V}=\rho_{W,U}$.


Also, I can show the glueing property. However, I can't show that $\mathcal{F}(\emptyset)=0$. How would the definition work, seeing that
$$\mathcal{F}(\emptyset)=O_X(\emptyset)=\bigcap_{P\in\emptyset}O_{X,P}.$$
So how does one show that $\mathcal{F}(\emptyset)=0$?
 A: By your definition of regular functions ("A regular function on $U$ is a rational function that is well-defined at all points of $U$"), they indeed do not form a sheaf, as $O_X(\emptyset)$ is the entire field of rational functions on $X$, rather than $0$.  An easy (if inelegant) way to fix this is to say that your definition is only the definition of a regular function on a nonempty open subset of $X$, and by definition the ring of regular functions on the empty set is $0$.
There are various more elegant definitions that achieve this same result.  For instance, a commonly used definition is that a regular function on $U$ is a function $\varphi:U\to k$ such that there exists a rational function $f\in K(X)$ which is regular at each point of $U$ and such that $\varphi(P)=f(P)$ for each $P\in U$.  By this definition, the formula $O_X(U)=\bigcap_{P\in U} O_{X,P}$ for $U$ nonempty becomes a somewhat nontrivial theorem, and it is false if $U$ is the empty set.  For by this definition, a regular function on the empty set is first and foremost a function $\varphi:\emptyset\to k$, and there is only one such function (even though there are many different $f\in K(X)$ such that $\varphi(P)=f(P)$ for all $P\in \emptyset$!).  So $O_X(\emptyset)$ is a ring with one element, i.e. the zero ring.
