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The definition of the structure sheaf of an affine scheme $\operatorname{Spec} R$ is often done by extending a sheaf defined on the standard opens $D(f)$, $f\in R$. From this definition it is not quite clear what the rings $\mathcal{O}(U)$ look like for general $U\subset \operatorname{Spec} R$. As Ravi Vakil notes in the lecture notes, one might hope that $$\mathcal{O}(U)\cong R_S$$ where $$S=\{r\in R\mid \forall \mathfrak{p}\in U: r\not\in \mathfrak{p}\}$$ He also notes that this is not true, and gives an example of two planes intersecting at a point, and then removing the point. More precisely $$X=\operatorname{Spec} \mathbb{C}[w,x,y,z]/(wy,wz,xy,xz), U=X\setminus\{(w,x,y,z)\}$$ But $U$ is clearly the union of two open subsets: $U_1$ the $xw$-plane with the origin removed and $U_2$ the $yz$-plane with the origin removed. So then one can construct sections on $U_1$ and $_2$ separately and glue them to a section on $U$. This makes it clear the $\mathcal{O}(U)$ is not a localisation of $\mathbb{C}[w,x,y,z]/(wy,wz,xy,xz)$.

However, this example is in some sense trivial in that it relies upon the fact that if $U$ is a disjoint union of $U_1$ and $U_2$ then $$\mathcal{O}(U)=\mathcal{O}(U_1)\times \mathcal{O}(U_2)$$ So I wonder if there are examples that do not rely on this fact. So if there is an affine scheme $\operatorname{Spec} R$ with an irreducible open subset $U$ such that $\mathcal{O}(U)$ is not a localisation of $R$.

Edit: another way of looking at the same question is by taking Hartshornes perspective. Sections over $U$ are maps to the stalks that can locally be realised by fractions. So is there an example of an irreducible $U$ and a section $s:U\to \bigcup_{\mathfrak{p}\in U}R_\mathfrak{p}$ so that locally $s=\frac{g}{h}$, but it is not possible to write $s$ like this on all of $U$?

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Here is my example.

Let $$A=k[x,y,z,w]/(xw-yz)$$ and $$U=D(y)\cup D(w).$$

Now to the proof !

Note first that $A$ is an integral domain. Because $k[x,y,z,w]$ is a unique factorization domain and $xw-yz$ is irreducible (clearly not the product of linear factors).

This implies that $U$ is irreducible. Since if there are $p,q\in A$ such that $$pq=0 \ \text{on} \ U$$ then $$ywpq=0\ \text{on} \ Spec \ A$$ and this means $p=0$ or $q=0$.

Now we show that $\mathcal{O}_X(U)$ is not a localization. The fact that $A$ is an integral domain simplifies the definition of localization, we no longer have to multiply by a factor $f^n$, and it implies that maps between localizations are injective. Further if $A_f \subseteq A_g$ then $$\frac{1}{f}=\frac{s}{g}$$ and so $$g=sf,$$ that is $f$ divides $g$. (But $A$ is not a unique factorization domain.)

Now $$\mathcal{O}_X(D(y))=A_y$$ $$\mathcal{O}_X(D(w))=A_w$$ $$\mathcal{O}_X(D(y)\cap D(w))=A_{yw}$$

Thus the restriction $$\mathcal{O}_X(U)=\mathcal{O}_X(D(y)\cup D(w))\rightarrow \mathcal{O}_X(D(y)\cap D(w))=A_{yw}$$ sends $\mathcal{O}_X(U)$ into $A_y\cap A_w$.

Now $$\mathcal{O}_X(U)\neq A$$ since

$$\frac{x}{y}=\frac{z}{w}\in \mathcal{O}_X(U)$$ (this is the core idea).

To finish remark that if $$A_f \subseteq A_y\cap A_w$$ then $A_f=A$. Since by the above remarks we would have

$$y=sf$$ $$w=rf$$ and this implies that $f\in k$.

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    Thanks for your answer. I've been thinking about it a bit today, but still I'm not sure what is really going on here. It seems that the point is to find a rational function $\frac{f}{g}$ on $X$, such that, restricted to $U$, the poles of $g$ cancel with the zero's of $f$, but such that this cancellation of poles does not happen over all of $X$? So it will "seem as if" $\frac{f}{g}$ will have poles on $U$, and hence "shouldn't be" in $\mathcal{O}(U)$, but actually the poles are cancelled by the zero's of $f$, and so all is fine? – 2017-01-30
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    Just to be clear, I get your answer and I see that it works, but I mean that I don't understand what the underlying idea is. – 2017-01-30
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    Also, I get why the restriction of $\frac{x}{y}$ to $D(y)\cap D(w)$ must lie in $A$, but I don't understand which element of $A$ it would actually be? Can this be determined? – 2017-01-30
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    So $\frac{x}{y}$ is NOT in $A$, that is the main point. As for the underlying idea I dont know how to answer, but I can tell you how I came up with the example. You cannot have $U=D(f)$ since that is always a localisation, thus the next thing is to have $U=D(f)\cup D(g)$, in which case you are in $A_f\cap A_g$, so you need that this intersection is bigger than $A$. The rational function $\frac{x}{y}$ has domain of definition exactly equal to my $U$. – 2017-01-30
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    I get that $\frac{x}{y}$ does not lie in $A$, but the restriction of $\frac{x}{y}$ to $D(x)\cap D(w)$ does lie in $A$ correct? Or at least in $A\subset A_{xw}$. So I was wondering which element of $A$ it would correspond to? – 2017-01-30
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    Its not an element of $A$. Elements of $A$ are everywhere defined. – 2017-01-30
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    But didn't you say that the restricion $\mathcal{O}(U)\to \mathcal{O}(D(f)\cap D(g))$ lands in $A_x\cap A_w=A$? So in particular the restriction of $\frac{x}{y}\in\mathcal{O}(U)$ lands in $A$ – 2017-01-30
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    But $A_y\cap A_w$ is not equal to $A$. Because it contains $\frac{x}{y}$. – 2017-01-30
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    Aah of course... I'm still slightly confused, as you can probably tell, but I'll think about this some more. Thanks a lot – 2017-01-30
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    Ok, no problem, let me just remark that $A_y\cap A_w$ is bigger than $A$ but it does not contain any localisation, except $A$. – 2017-01-30
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    Yea I see now, that also agrees with what you actually wrote in your answer. I just did think it through enough. Thanks again, this was very helpful – 2017-01-30