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Let $R$ be a commutative ring. Let $U\subseteq X:=\operatorname{Spec}(R)$ be an arbitrary open subset. Let $S$ be the multiplicative set of elements of $R$ which don't vanish anywhere on $U$. Explicitly, $U$ is open so by definition there exists some ideal $I\subseteq R$ such that $U$ is the prime ideals of $R$ which don't contain $I$, and then $S$ is the complement of the union of these prime ideals.

If $A:=\mathcal{O}_X(U)$ is the value of the structure sheaf at $U$, then restriction of functions gives us a map $R\to A$ and the image of every $s\in S$ is invertible in $A$ (because locally invertible) so we get a map $R[1/S]\to A$.

What is an example where this map is not an isomorphism of rings? I have no reason to believe that it is an isomorphism in general, but (a) if $U=D(f)$ (that is, $I=(f)$) then it is an isomorphism, and (b) if $X$ is affine 2-space over a field $k$ and $U$ is the complement of the origin -- the canonical "non-affine open in an affine" -- then $S$ is the non-zero constants and $R[1/S]=k[x,y]=A$. I need to expand my mental store of examples/counterexamples to answer this one, I think. Can someone help?

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  • $\begingroup$ How about nilpotents $F[x,y,z]/(xz,yz,z^2)$ . . . Normalization in codimension 2 $\{(f,g)\in F[x,y]\times F[a,b]\;|\;f(0,0)=g(0,0)\}=F[x,y,a,b]/(xa,ya,xb,yb)$ $\endgroup$ – Ben Wieland May 13 at 18:52
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An even easier example than the two current answers is the following. (See also this MO answer.)

Example. Let $E$ be an elliptic curve over an algebraically closed field $k$. Let $O \in E(k)$ be the origin, and $P \in E(k)$ a non-torsion point. Let $X = E \setminus \{O\}$ and $U = X \setminus \{P\}$.

If $f \in S$ (i.e. $f \in R = \Gamma(X,\mathcal O_X)$ becomes invertible on $U$), then $\operatorname{div}(f) \subseteq E$ has to be supported on $\{O,P\}$. Because $P$ is non-torsion, the only way this can happen is if $\operatorname{div}(f) = 0$, i.e. $f \in k^\times$. We conclude that $S = R^\times = k^\times$, so $R[1/S] \cong R$. But $\Gamma(U,\mathcal O_U)$ is not isomorphic to $R$. $\square$

This is the canonical "affine open that is not standard affine open".

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  • $\begingroup$ "An even easier example . . ." That is a matter of opinion. Certainly it is faster to say: "the complement of the origin and a non-torsion point on an elliptic curve". However, it takes work to unravel this. Having taught undergrad algebraic geometry classes several times, I prefer: "the complement of two skew lines in the union of two intersecting planes". I can write the equation in one line and prove the presentation of the ring of regular functions in the next line. $\endgroup$ – Jason Starr May 25 at 11:01
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Here is the simplest example that I know. Let $R$ be $k[x,y,z]/\langle xy \rangle$. Let $I$ be $\langle \overline{x}(1+\overline{z}), \overline{y}(1-\overline{z}),\overline{z}^2-1 \rangle$. Then $S$ equals $k^\times$. Yet $A$ is the following countably generated $R$-algebra, $$A=R[u_n,v_n : n\in \mathbb{Z}_{\geq 0}]/J, $$ $$ J := \langle \overline{y}-u_0, (\overline{z}-1)u_{n+1}-u_n, \overline{x}-v_0, (\overline{z}+1)v_{n+1}-v_n, u_nv_n : n\geq 0 \rangle.$$

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The map is not always an isomorphism (assuming there is no stupid mistake in my judgment). If you take an arbitrary commutative unital Noetherian ring and localize it at an arbitrary multiplicative set, the resulting ring is Noetherian (see here). If the map in question is always an isomorphism, then for a Noetherian $R$ and any open set $U\subset X=\mathrm{Spec}\,R$, the ring $\mathcal{O}_X(U)$ would be Noetherian. This need not be true. See Remark 3.7.b in this paper. You also may find this paper useful.

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    $\begingroup$ That is...a whole lot more complicated than I was hoping for :-/ I will take a look. $\endgroup$ – Kevin Buzzard May 9 at 13:57
  • $\begingroup$ I'm finding it really hard to extract an explicit example from these papers, although it seems to me that your answer is fine. $\endgroup$ – Kevin Buzzard May 11 at 18:00

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