# Open affine subscheme of affine scheme which is not principal

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.

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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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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 '09 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 '09 at 1:00

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,

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finally an easy example :-) –  Martin Brandenburg Dec 29 '09 at 1:23
Thanks, Martin! –  Hailong Dao Dec 29 '09 at 1:43
You have used $U$ twice... but well. –  Thomas Jan 16 '12 at 19:32
Why does $Yf+Xg=1$ imply that $U$ is affine? –  benblumsmith Nov 27 '12 at 1:19
@benblumsmith see Hartshorne excercise II.2.17 –  solbap Jan 31 '13 at 9:09

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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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.
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 '09 at 20:20