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The answer to the question in the title is "yes". This is proved for example here, or here (page 5).

However, I seem to be able to construct a counterexample. Can you help me find the flaw in my argument?

Let $E$ be an elliptic curve, say over $\mathbb C$. Then $E-\{0\}$ is the spectrum of a Dedekind domain $R$, and has class group isomorphic to $E(\mathbb C)$. (Because $\mathrm{Cl}(E)=E(\mathbb C)\times \mathbb Z$, and $\mathrm{Cl}(E-\{0\})$ is the quotient of $\mathrm{Cl}(E)$ by the cyclic subgroup generated by $[0]$.) Now the class group of $\mathrm{Spec}(R)$ is just the ideal class group of $R$. Take a non-torsion point of $E$, considered as a prime ideal $\mathfrak{p}$ of $R$. I claim that $S:=\mathrm{Spec} (R)-\{\mathfrak{p}\}$ is not affine.

Let $T=\Gamma (S,\mathcal O_S)=\{ {{a}\over{b}}\in Q(R) |b\notin \mathfrak{q}\in \mathrm{Spec}(R),\mathfrak{q}\neq \mathfrak{p}\}$. I will show that $T=R$. It suffices to show that every element $x\in \mathfrak {p}$ is also contained in some other prime ideal than $\mathfrak p$. Were it not so, then the principal ideal $(x)$ would be equal to $\mathfrak{p}^n$ for some $n$, by the unique prime decomposition of ideals in Dedekind domains. But the image of $\mathfrak{p}$ in the ideal class group is a non-torsion element, so no power of $\mathfrak{p}$ can be a principal ideal. Therefore $T=R$, and $S\neq\mathrm{Spec}(R)=\mathrm{Spec}(T)$, and hence $S$ is not affine.

($S$ and $\mathrm{Spec}(R)$ really are nonisomorphic, because for instance they have different etale fundamental groups)

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Your description of $T$ is at best very confusing, because an element of $Q(R)$ has several ways to be written as a fraction, and what follows depends heavily on which representative you take. A clearer way to describe it should be the set of functions on the elliptic curve whose poles are restricted to $0$ and $p$.

In your description you are restricting the denominators too much. Maybe you can find a fraction where the denominator has a zero somewhere else, but where the numerator also coincidentally has a zero there.

Indeed if you find two points $A,B$ such that the divisor $A+B$ is linearly equivalent to the divisor $0 + p$, then there is a function whose divisor is $A+B-0-p$, so this is an element of $T$

If your curve is in Weierstrass form and $0$ is the point at infinity, you can for example look at a half of $p$ : If $2q = 0 + p$, then the tangent line at $q$ crosses the elliptic curve at $-p$, so when you divide the equation of the tangent line going through $q,q,-p$ by the equation of the line going through $0,p$ and $-p$ (this is the vertical line $x=x_p$), you obtain an element $f = (ax+by+c)/(x-x_p)$ of $T$ for some $a,b,c \in \Bbb C$.

You cannot find a description of $f$ as a fraction whose denominator only vanish at $p$, but for every other point $q$ you can find a representation with a denominator who doesn't vanish at $q$, it will just have to vanish somewhere else (and be cancelled with the corresponding numerator)

So you can write $T = \{ f \in Q(R) \mid \forall \mathfrak q \neq \mathfrak p, \exists a,b \in R, f = a/b, b \notin \mathfrak q\} = \bigcap_{\mathfrak q \neq \mathfrak p} R_ \mathfrak q$ (where $R_\mathfrak q = (R \setminus \mathfrak q)^{-1}R$) which is pretty close to your original statement, but where it's made clear that $f$ can change its representative to show it is a good function defined everywhere.

mercio
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  • That makes sense, thanks! You are saying that while S is affine, its global section ring is not obtained by a localization of R. Would you then agree that the following statement from Neukirch's "Algebraic Number Theory" (p70) is false? Let A be a domain, X a set of prime ideals. Neukirch says "If X is finite or omits only finitely many prime ideals of A, then only the prime ideals from X survive in A(X)". Here A(X) is the localisation of A at the complement of the union of the ideals in X. – user00000 Mar 25 '16 at 12:24
  • In the German version of the book, he actually claims "If A is a Dedekind domain, only the prime ideals from X survive in A(X)".. – user00000 Mar 25 '16 at 12:26
  • Correction to the first comment: It's page 66, not 70 (it's 70 in the German version). I can't seem to edit the comment. – user00000 Mar 25 '16 at 12:41
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    Yes, I think $T$ isn't a localisation of $R$. if $T$ was $S^{-1}R$ for some $S$, then an element of $S$ would be invertible in $T$, but $T$ doesn't have any more units than $R$ has, since they would be functions whose divisors' supports are subsets of ${0,p}$, from which you get only constant functions due to $(p-0)$ not having finite order. – mercio Mar 25 '16 at 12:50
  • I think there is some confusion about localization here. If $A$ is a Dedekind domain and $0\neq I$ an ideal, the functions on the open set removing $V(I)$ is $\cup_{n\in\mathbb{N}} I^{-n}$. – Mohan Mar 25 '16 at 16:22