Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:24:17Z http://mathoverflow.net/feeds/question/76349 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76349/is-there-a-connected-non-affine-scheme-s-such-that-it-is-the-union-of-rings-of Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? James D. Taylor 2011-09-25T17:41:21Z 2011-09-26T05:01:44Z <p>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).</p> <p>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.</p> <h3>Question</h3> <p>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?</p> <p>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?</p> http://mathoverflow.net/questions/76349/is-there-a-connected-non-affine-scheme-s-such-that-it-is-the-union-of-rings-of/76356#76356 Answer by Will Sawin for Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? Will Sawin 2011-09-25T19:02:00Z 2011-09-25T19:02:00Z <p>I say no.</p> <p>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.</p> http://mathoverflow.net/questions/76349/is-there-a-connected-non-affine-scheme-s-such-that-it-is-the-union-of-rings-of/76365#76365 Answer by Georges Elencwajg for Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? Georges Elencwajg 2011-09-25T21:03:39Z 2011-09-26T05:01:44Z <p>Consider $X= Spec( \mathcal O_K)$ and an open subset $U \subset X \quad (U\neq \emptyset, X)$.<br> Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.<br> You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.<br> 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. </p> <p><strong>Edit</strong> 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.</p> http://mathoverflow.net/questions/76349/is-there-a-connected-non-affine-scheme-s-such-that-it-is-the-union-of-rings-of/76368#76368 Answer by Qing Liu for Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? Qing Liu 2011-09-25T22:15:40Z 2011-09-25T22:15:40Z <p>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. </p>