The prime spectrum of a Dedekind Domain Let $A$ be a Dedekind Domain, let $X = \operatorname{Spec}(A)$. Are all open sets in $X$ basic open sets? Thinking about the Zariski topology (in the classical sense) of a non-singular affine curve, if I had to guess, I would say "yes" but I can't think of a proof or counterexample. So far I have only been able to prove the result is true for PID's. 
Any help is much appreciated.
Cheers
 A: Claim: Every open subset of $X$ is a basic open set if and only if $\text{Cl}(A)$, the ideal class group of $A$, is torsion. (In particular, this is true whenever $A$ is the ring of integers of a number field, for then $\text{Cl}(A)$ is finite.)
The question boils down to knowing whether, for any finite set of primes $\mathfrak{p}_1, \dots, \mathfrak{p}_n$, there is an $f\in A$ such that $\mathfrak{p}_1, \dots, \mathfrak{p}_n$ are exactly the primes of $A$ containing $f$.
Therefore, suppose $\text{Cl}(A)$ is torsion. Consider the ideal $I=\mathfrak{p}_1 \cdots \mathfrak{p}_n$. This has finite order in the class group, hence there is an integer $m$ such that $(g)=\mathfrak{p}_1^m \cdots \mathfrak{p}_n^m$ is principal. Then $f=g$ does the trick.
Conversely, if there is an element of infinite order in $\text{Cl}(A)$, then there must be such an element which is represented by a prime $\mathfrak p$ of $A$ (since the primes of $A$ generate the class group, by unique factorization of ideals). The complement of $\mathfrak p$ in $\text{Spec}(A)$ could not be a principal open set; otherwise, there would be $f$ such that $(f)=\mathfrak p^m$ for some $m$, contradicting that $\mathfrak p$ has infinite order in $\text{Cl}(A)$.
Note: Georges' pretty counterexample exhibits this idea: the group structure on an elliptic curve is none other than its (degree $0$) Picard group. Hence to find a Dedekind domain whose class group contains an element of infinite order, it sufficies to find an elliptic curve with a point of infinite order...
A: No, it is not true that for a Dedekind ring $A$ all open subsets of $X = \operatorname{Spec}(A)$ are principal. Here is a counter-example:
a) Let $\bar X$  be an elliptic curve  over $\mathbb C $ and  $P\in \bar X$ a non-torsion point  (this means that for all $n\geq 1 $ we have $n\cdot P\neq O$ in the abelian group $\bar X$, where $O$ is the zero element of $\bar X$ ).
Let $X=\bar X\setminus \lbrace O \rbrace$.
The ring $A=\Gamma(X,\mathcal O) $ is Dedekind-because $X$ is affine and smooth of dimension one-and we have $X=\operatorname{Spec}(A)$.
b)  I claim however  that $X_P= X\setminus \lbrace P \rbrace$ is a non-principal open subset of $X$.
Indeed if we had $X_P=D(f)$ for some $f\in A$, the divisor of $f$ seen as a rational function on $\bar X$ would be of the form $div(f)=nP-nO\in \operatorname {Div}(\bar X)$.
But then, by Abel-Jacobi's theorem, we would have in the group $\bar X$ the relation $n\cdot P-n\cdot O=n\cdot P=O$, contradicting the  choice of $P$.
Edit
While I was typing my answer, Bruno gave a perfect criterion for every open subset of $X = \operatorname{Spec}(A)$ to be principal: that $Cl(A)=Pic(X)$ be torsion.
So, for which Dedekind rings $A$ is that true?
My counter-example of course is one whereit is not true.
Actually for any elliptic curve $\bar X$, only denumerably many points of  the group $\bar X ( \mathbb C)$ are torsion and this implies that $Pic(X) \quad (X=\bar X\setminus \lbrace O \rbrace)$ has continuum many non-torsion elements.
But this is small beer.
Now comes the real surprise: Claborn has proved that given any abelian group $G$ whatsoever , there exists a Dedekind ring $A$ with $Cl(A)=G$.
So, take any abelian group $G$ with a non-torsion element and Claborn's theorem provides you with a Dedekind ring $A$ whose spectrum    $\operatorname{Spec}(A)$ has a non-principal open subset!
