I recently learned that the structure sheaf on the spectrum of a ring $\mathrm{Spec}(R)$ is first defined on the distinguished open subsets $D_f$ for $f\in R$ so that $\mathcal{O}(D_f)=R_f$ where $R_f$ is the localization of the ring $R$ at $f$. Then the structure sheaf is extended to arbitrary open sets by taking limits. I was wondering if it was possible to define the structure sheaf more directly, by giving an explicit definition over an arbitrary open set. First I ran into the question
Why is the structure sheaf for the spectrum of a ring defined locally?
where the author tries to do something similar by defining, for an arbitrary open set
$U=\mathrm{Spec}(R)-V(I),$ the ring $\mathcal{O}(U)$ to be the localization of $R$ at $I$. This definition fails because some of the functions $g\in I$ actually vanish in $U$, so that $U\cap V(g)\neq \varnothing.$ Then we end up allowing division by the function $g$ even though it is zero somewhere in $U$.
My idea, for arbitrary open $U$, was to define $S=\{f\in R: U\cap V(f)=\varnothing\}.$ Then $S$ is multiplicatively closed because if $f,g\in S$ then $U\cap V(fg)=U\cap (V(f)\cup V(g))=\varnothing$ so $fg\in S.$ Then we could define $\mathcal{O}(U)=S^{-1}R.$
Intuitively, this allows us to divide by any function that does not vanish over $U,$ and it agrees with the standard definition of the structure sheaf over the distinguished open sets. This is a bit different from the standard definition of the structure sheaf which only requires that a section over an open set $U$ does not divide by any function that vanishes "locally." If it gives a sheaf then it will actually be the same as the standard structure sheaf, since a sheaf will be uniquely determined by its sections over the distinguished open sets. I suspect that there may be some situations where this pre-sheaf fails gluability, but I can't think of any. Does this pre-sheaf fail the sheaf axioms in some cases?
Best Answer
Let $k$ be a field, $R=k[x,y,z,w]/(xy-zw)$, and $U=D_y\cup D_z$. Note that the elements $w/y\in R_y$ and $x/z\in R_z$ are the same in $R_{yz}$ and so should glue to give an element of $\mathcal{O}(U)$. However, it can be shown that this element cannot be represented by a fraction whose denominator does not vanish on $U$. (Any denominator for this element has to be in the ideal $(y,z)$, but there is no single element of $(y,z)$ which vanishes only on $V(y,z)$.) So, your definition will fail the gluing axiom in this case.