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I'm not sure whether this is non-trivial or not, but do there exist simple examples of an affine scheme $X$ having an open affine subscheme $U$ which is not principal in $X$? By a principal open of $X = \mathrm{Spec} \ A$, I mean anything of the form $D(f) = \{\mathfrak p \in \mathrm{Spec} \ A : f \notin \mathfrak p\}$, where $f$ is an element of $A$.

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Let X be an elliptic curve with the identity element O removed. Let U=X-P where P is a point of infinite order. Then U is affine by a Riemann-Roch argument. Now suppose that U=D(f). Then on the entire elliptic curve, the divisor of f must be supported at P and O only. This implies that P is a torsion point, a contradiction.

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  • $\begingroup$ Beautiful - do you think this is the "simplest" example we can get? (As far as "simple" is well-defined in this context...) $\endgroup$ – Wanderer Nov 30 '09 at 0:12
  • $\begingroup$ Its the simplest compact example, because the curve of genus 0 has no open affine subschemes which aren't principal. $\endgroup$ – Greg Muller Nov 30 '09 at 1:00
  • $\begingroup$ @GregMuller is this example compact? Does not he remove the origin from the elliptic curve (making it non-compact, presumably)? $\endgroup$ – user137767 Apr 9 at 14:26
  • $\begingroup$ It's not compact. I think it's the simplest geometric example. $\endgroup$ – Peter McNamara Apr 10 at 0:37
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For a simple, really concrete example you can also look at:

$A=k[x,y,u,v]/(xy+ux^2+vy^2)$, $X =Spec(A)$, $I=(x,y)$, $U = D(I)$.

Then the functions $f=\frac{-v}{x}=\frac{y+ux}{y^2}$ and $g=\frac{-u}{y}=\frac{x+vy}{x^2}$ are defined on $U$. But $yf+xg=1$, so $U$ is affine!

Cheers,

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    $\begingroup$ finally an easy example :-) $\endgroup$ – Martin Brandenburg Dec 29 '09 at 1:23
  • $\begingroup$ You have used $U$ twice... but well. $\endgroup$ – Thomas Kahle Jan 16 '12 at 19:32
  • $\begingroup$ Why does $Yf+Xg=1$ imply that $U$ is affine? $\endgroup$ – benblumsmith Nov 27 '12 at 1:19
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    $\begingroup$ How do you show $D(X,Y)$ is not principal? $\endgroup$ – Jonathan Wise Dec 29 '14 at 6:12
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    $\begingroup$ @Alison Miller: Take f_1=X, f_2=Y with notation from Hartshorne exercise II.2.17. What we need, is that U_X and U_Y are affine, and this obviously holds since U_X is the principal open subset D(X) of Spec(A), and similarly for U_Y. $\endgroup$ – Isac Hedén May 15 '15 at 19:36
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I just want to remark that there is a purely categorical characterization of the ideals $I \subseteq A$ such that the corresponding open subscheme $D(I) = V(I)^c$ of $\text{Spec}(A)$ is affine, namely that the ideal $I$ is codisjunctable. This means that there is a universal homomorphism $A \to B$ satisfying $IB=B$. This notion is studied in

Yves Diers, Codisjunctors and Singular Epimorphisms in the Category of Commutative Rings, Journal of Pure and Applied Algebra, 53, 1988, pp. 39 - 57

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  • $\begingroup$ What is $B$ in this context? $\endgroup$ – Babai Aug 29 '16 at 14:45
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    $\begingroup$ @Babai: Consider the functor $\mathsf{CRing} \to \mathsf{Set}$ mapping $B$ to the set of homomorphisms $A \to B$ such that $IB=B$. Then, $I$ is called codisjunctable iff this functor is representable. $\endgroup$ – Martin Brandenburg Oct 21 '16 at 13:58
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I'm not entirely sure what you mean, but if you mean whose complement isn't principal, take $\mathbb{A}^2\setminus\{0\}$ which is an open subscheme of $\mathbb{A}^2$. Now, if you mean that the open subscheme isn't cut out by a single equation, any open subscheme other than the whole space will do for an irreducible scheme, because the only open set which is cut out by an ideal is the whole scheme.

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  • $\begingroup$ I clarified my question... Sorry for being a bit vague. But your answer doesn't work, obviously. $\endgroup$ – Wanderer Nov 29 '09 at 20:18
  • $\begingroup$ My answer still works. There is no function $f\in k[x,y]$ such that $f=0$ iff $x=y=0$, so $\mathbb{A}^2\setminus\{0\}$ is not determined by the nonvanishing of a single function. $\endgroup$ – Charles Siegel Nov 29 '09 at 20:20
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    $\begingroup$ But it's not affine, and that's really the point of the question! $\endgroup$ – Wanderer Nov 29 '09 at 20:22
  • $\begingroup$ Ahh, now I see what you mean. $\endgroup$ – Charles Siegel Nov 29 '09 at 20:24
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What about the generic point of the DVR $X=\text{Spec }k[x]_{(x)}$? Any nonzero $f\in k[x]_{(x)}$ is a unit so $D(f)$ is the original space $X$. Removing the only closed point the generic point has a natural open sub-scheme structure so it must be an affine scheme.

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  • $\begingroup$ No. This open subscheme is affine indeed, but equal to $D(x)$. $\endgroup$ – abx Dec 10 '18 at 21:06

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