I was woolgathering about the notion of a scheme, and it occurred to me that I know of no non-affine scheme $S$ that is the union of $Spec(O_K)$'s of some number field $K$ (I allow $K$ to vary - so that $S$ might be $Spec(O_K)\cup Spec(O_L)$ for example).

It would an interesting notion if one could patch rings of integers together to form some non-affine $1$-dimensional normal scheme $S$. The fact that I've never seen an example makes me think it's impossible.


Is there a connected non-affine scheme $S$ such that it is the union of open subschemes of it that are $Spec$'s of rings of integers of number fields?

More pointedly, if $Spec(O_K)$ (the ring of integers of some number field $K$) is an open subscheme of a normal scheme $S$ then is it equal to it?

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    $\begingroup$ Are you requiring the scheme to be normal in the question? That seems to be what you wanted in the discussion leading up to it. Without some hypothesis, it is trivially false: take a disjoint union of $Spec(O_K)$ with something else. $\endgroup$ Sep 25, 2011 at 17:58
  • $\begingroup$ Yes, I want it to be connected, and equal to the union of Spec's of rings of integers (this implies normal, of course). I'll edit the question. $\endgroup$ Sep 25, 2011 at 18:01
  • $\begingroup$ Is it possible to glue $Spec (O_K)$ and $Spec(O_{K'})$ along some open set? For example consider the affine line with a double point. $\endgroup$
    – user16974
    Sep 25, 2011 at 18:49
  • $\begingroup$ I'm not sure what you mean, but I want to require that $S$ be separable. $\endgroup$ Sep 25, 2011 at 19:03
  • $\begingroup$ You mean "separated", not "separable" I suppose. $\endgroup$
    – Joël
    Sep 25, 2011 at 20:07

3 Answers 3


If $i: \mathrm{Spec}(O_K)\to S$ is an open immersion into a connected separate scheme $S$, then $i$ is an isomorphism. Indeed, the canonical morphism $\pi : \mathrm{Spec}(O_K)\to \mathrm{Spec}(\mathbb Z)$ is finite (hence proper) and can be decomposed into $i$ followed by the canonical morphism $S\to \mathrm{Spec}(\mathbb Z)$. As the latter is separated, this implies that $i$ is also proper, hence closed. The connectedness of $S$ implies that $i$ is onto.

  • $\begingroup$ Brilliant ! $\endgroup$ Sep 25, 2011 at 23:11
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    $\begingroup$ Also, one can show that any integral noetherian separated scheme whose field of rational functions is a number field is the spectrum of a localization of the corresponding number ring. $\endgroup$
    – Angelo
    Sep 26, 2011 at 7:27
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    $\begingroup$ @Angelo: Perhaps "a localization of an order in the ring of integers", unless you add "normal" to the assumptions. But in fact I suspect that you can remove the noetherian condition. $\endgroup$ Sep 28, 2011 at 8:36
  • $\begingroup$ Dear Laurent, of course you are right. $\endgroup$
    – Angelo
    Sep 29, 2011 at 21:20

I say no.

Let $\xi_K$ be the generic point in $Spec(O_K)$, and $\xi_L$ the generic point in $Spec(O_L)$. Since $Spec(O_K)$ and $Spec(O_L)$ have nonempty intersection, their intersection must be an open set in each, and must contain both generic points. The local ring at $\xi_K$ is $K$, and the local ring at $\xi_L$ is $L$.But no point in $O_K$ has local ring $L$, and no point in $O_L$ has local ring $K$. This is a contradiction.

  • $\begingroup$ Yes, I guess it would have to be the same $K$ for all open affines... Silly that I didn't think of that. $\endgroup$ Sep 25, 2011 at 19:30
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    $\begingroup$ How does this solve the question? Aren't we left with the possibility that several copies of the same number field can be glued? $\endgroup$ Sep 25, 2011 at 19:58
  • $\begingroup$ Spot on, Jack: see my answer. $\endgroup$ Sep 25, 2011 at 21:04

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit I wasn't too happy with this non-separated example when I posted it, but Qing now has proved that it is impossible to find a separated one.

  • $\begingroup$ Yes, but I believe the PO wanted his scheme to be separated. $\endgroup$
    – Joël
    Sep 25, 2011 at 21:20
  • $\begingroup$ @Joël: yes, the OP required separatedness, but only after he had posted his question, in a comment that I hadn't read because I had already started writing my answer . Anyway Qing has now proved that no separated example exists. $\endgroup$ Sep 26, 2011 at 5:06

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