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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 = Spec A, I mean anything of the form D(f) = {P in Spec A : f is not in P}, where f is an element of A.

edited Nov 29 2009 at 20:26

asked Nov 29 2009 at 19:53

Wanderer
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Peter McNamara · Accepted Answer · 2009-11-29 21:21:33Z UTC

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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.

answered Nov 29 2009 at 21:21

Peter McNamara
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	Beautiful - do you think this is the "simplest" example we can get? (As far as "simple" is well-defined in this context...) – Wanderer Nov 30 2009 at 0:12
	Its the simplest compact example, because the curve of genus 0 has no open affine subschemes which aren't principal. – Greg Muller Nov 30 2009 at 1:00

Hailong Dao · Answer 2 · 2009-12-20 08:07:14Z UTC

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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 $f=-V/X=(Y+UX)/Y^2$ and $g=-U/Y=(X+VY)/X^2$ are defined on $U$. But $Yf+Xg=1$, so $U$ is affine!

Cheers,

edited Dec 20 2009 at 8:07

answered Nov 30 2009 at 1:57

Hailong Dao
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finally an easy example :-) – Martin Brandenburg Dec 29 2009 at 1:23

Thanks, Martin! – Hailong Dao Dec 29 2009 at 1:43

You have used $U$ twice... but well. – Thomas Kahle Jan 16 2012 at 19:32

Why does $Yf+Xg=1$ imply that $U$ is affine? – benblumsmith Nov 27 at 1:19

@benblumsmith see Hartshorne excercise II.2.17 – solbap Jan 31 at 9:09

Martin Brandenburg · Answer 3 · 2011-06-04 14:37:00Z UTC

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

Charles Siegel · Answer 4 · 2009-11-29 20:06:33Z UTC

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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.

answered Nov 29 2009 at 20:06

Charles Siegel
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I clarified my question... Sorry for being a bit vague. But your answer doesn't work, obviously. – Wanderer Nov 29 2009 at 20:18

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. – Charles Siegel Nov 29 2009 at 20:20

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But it's not affine, and that's really the point of the question! – Wanderer Nov 29 2009 at 20:22

Ahh, now I see what you mean. – Charles Siegel Nov 29 2009 at 20:24

Open affine subscheme of affine scheme which is not principal

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