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.


  • $\begingroup$ On a positive side-note, every open subscheme of an affine Dedekind scheme is affine. More generally, every open affine subscheme of a 1-dimensional noetherian integral scheme X whose normalization is a finite morphism is affine. $\endgroup$ – Harry Jul 23 '12 at 20:30
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    $\begingroup$ Dear Harry: in your second sentence you write that an open affine subscheme of... is affine. Surely, you mean something else! $\endgroup$ – Georges Elencwajg Jul 23 '12 at 22:20

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...

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    $\begingroup$ +1 for your great characterization of the rings for which the question has a positive answer. $\endgroup$ – Georges Elencwajg Jul 23 '12 at 21:39
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    $\begingroup$ Dear Bruno, indeed you can be proud of having discovered this result on your own: congratulations! (By the way, at 92 years of age, you must be the senior member of this site: bravo for that too) $\endgroup$ – Georges Elencwajg Jul 23 '12 at 23:44
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    $\begingroup$ Thanks @GeorgesElencwajg! I'm going to have to admit, however, that I'm not really 92 years of age. However, I hope to still be thinking about math then! Regards, $\endgroup$ – Bruno Joyal Jul 24 '12 at 1:20
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    $\begingroup$ This is really interesting stuff Bruno. Thanks for your help! $\endgroup$ – user32259 Jul 24 '12 at 9:41
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    $\begingroup$ Very nice characterizaton indeed. $\endgroup$ – Andrea Mori Jul 27 '12 at 20:11

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 $P$.

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!

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  • $\begingroup$ Cheers, Georges. This looks really interesting. Having read Intro to Commutative Algebra (Atiyah - MacDonald) I have started to learn some algebraic geometry. I feel I am perhaps a couple of days a way from completely understanding this example. I'll get back to you when I do! $\endgroup$ – user32259 Jul 24 '12 at 9:52
  • $\begingroup$ I wish you success in understanding this and much, much more commutative algebra: good luck! $\endgroup$ – Georges Elencwajg Jul 24 '12 at 9:55
  • $\begingroup$ Dear Georges; your reference to the theorem of Claborn brings me to ask the following question, which is probably very difficult: can every finite abelian group be realised as the class group of the ring of integers of a number field? Regards, $\endgroup$ – Bruno Joyal Jul 24 '12 at 22:30
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    $\begingroup$ Dear @Bruno, the question you ask is indeed difficult since it was open in 1975: see here. I found this on the Internet. As the Romans said: doctus cum Googlo (=erudite with Google. They actually wrote "cum libro", with a book). $\endgroup$ – Georges Elencwajg Jul 24 '12 at 22:49
  • $\begingroup$ Dear @GeorgesElencwajg, thank you for the reference! I need to practice using Google to search for mathematical results... Regards, $\endgroup$ – Bruno Joyal Jul 25 '12 at 15:58

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