Timeline for Etale coverings of certain open subschemes in Spec O_K
Current License: CC BY-SA 4.0
19 events
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| Jun 12, 2020 at 16:50 | history | edited | Ariyan Javanpeykar | CC BY-SA 4.0 |
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| Jul 11, 2011 at 12:47 | comment | added | Ariyan Javanpeykar | This argument shows the following. Let $X$ be an integral 1-dimensional affine noetherian scheme such that the normalization of $X$ is a finite morphism. Then any open subset of $X$ is affine. (I think the technical term is "Japanese" scheme.) | |
| Jul 11, 2011 at 12:44 | comment | added | Ariyan Javanpeykar | The answer is yes. In fact, the morphism from Spec O_K to Spec R is finite and surjective. By the above, for any U open in Spec R, the base change $U \times_R O_K$ is affine. Moreover, the morphism $U\times_R O_K \rightarrow U$ is finite and surjective (stable by base change). By Chevalley's theorem (Hartshorne Exercise III.4.1), we have that U is affine. | |
| Jul 11, 2011 at 12:17 | comment | added | Tommaso Centeleghe | I raise here a small question, hoping that you guys see it: let $R$ be an order in the ring of integers of a number field $K$. Is it true that every open of $Spec(R)$ is affine even when $R$ is not maximal? | |
| May 1, 2010 at 20:45 | answer | added | Quetzalcoatl | timeline score: 2 | |
| May 1, 2010 at 17:02 | vote | accept | Ariyan Javanpeykar | ||
| Apr 30, 2010 at 8:37 | comment | added | BCnrd | Bjorn, that's nice. To rigorously justify that the formation of what you call $A[1/S]$ is compatible with localization on $A$, which is implicit in your argument, it seems simplest to work with the quasi-coherent pushforward of structure sheaf of the open $U$. | |
| Apr 29, 2010 at 23:30 | comment | added | Bjorn Poonen | @BCnrd: Any nonempty open U is the complement of a finite set S of closed points. Let A[1/S] be the set of x in Frac(A) whose valuation at each prime outside S is nonnegative. Then Spec A[1/S] --> Spec A is an open immersion with image U since this can be checked on a covering by basic open affines on which the primes in S become principal. | |
| Apr 28, 2010 at 22:46 | answer | added | Georges Elencwajg | timeline score: 2 | |
| Apr 28, 2010 at 21:33 | comment | added | BCnrd | @Giovanni: false alarm. By Serre's cohom. criterion, $U$ is affine. Consider extension $E$ of $O_U$ by nonzero coherent ideal $I$, so $E$ is rank-2 vector bundle on $U$. Extend $E$ to coherent $E'$ on $X = {\rm{Spec}}(A)$, and wlog kill torsion so $E'$ is vector bundle. Doing local calculation at $m$, can "intersect" $I$ with $E'$ to get saturated line bundle $L$ in $E'$ extending $I$ in $E$, so $N = E'/L$ is also line bundle. (I am using Dedekindness of $X$ all over the place.) Then on affine $X$ this exact sequence splits; restrict back to $U$. QED Huh. Is there an elementary proof? | |
| Apr 28, 2010 at 21:00 | comment | added | BCnrd | @Giovanni: for general Dedekind domain $A$, why should every open subset of the spectrum be affine? Why is the complement of the closed point corresponding to a maximal ideal $m$ affine (obvious if $m$ torsion in class group, but more generally...)? A good example: for complement of origin in an elliptic curve, removing a rational non-torsion point yields an open affine (via Riemann-Roch) but not basic open affine (= inverting some nonzero element). This is an embarrassingly elementary (but "useless") question about general Dedekind domains. I checked with two colleagues, as stumped as me. | |
| Apr 28, 2010 at 19:53 | answer | added | Lars | timeline score: 4 | |
| Apr 28, 2010 at 19:36 | comment | added | Kevin Buzzard | @Ariyan: by definition if S-->T is finite and T is affine, then S is affine. | |
| Apr 28, 2010 at 19:36 | answer | added | Cam McLeman | timeline score: 5 | |
| Apr 28, 2010 at 18:38 | comment | added | xuros | In general any open subset of the spectrum of a Dedekind domain is necessarily affine, and finite étale covers of affine schemes (being finite) are necessarily affine. You can work out this kind of question by thinking of a few things : since you're concerned with bases $S = Spec \mathcal{O}_K$, you know that all finite śtale covers $X\to S$ are going to be $Spec$ of some finite flat $O_K$-algebra... Then go back to the definition of unramified (i.e. calculate the stalk map for $Spec(B)\to Spec(A)$, it's just a the localization of $A\to B$ at a prime of $B$). | |
| Apr 28, 2010 at 18:21 | history | edited | Ariyan Javanpeykar | CC BY-SA 2.5 |
Added analogy with function fields; added 10 characters in body
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| Apr 28, 2010 at 18:19 | comment | added | Ariyan Javanpeykar | Why? Why is $V$ even affine? | |
| Apr 28, 2010 at 18:14 | comment | added | Kevin Buzzard | V is just O_K[1/2], right? | |
| Apr 28, 2010 at 17:53 | history | asked | Ariyan Javanpeykar | CC BY-SA 2.5 |