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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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up vote 32 down vote accepted

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 $f=\frac{-v}{x}=\frac{y+ux}{y^2}$ 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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2  
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 Kahle Jan 16 '12 at 19:32
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How do you show $D(X,Y)$ is not principal? – Jonathan Wise Dec 29 '14 at 6:12
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@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. – Isac Hedén May 15 '15 at 19:36

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.

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I clarified my question... Sorry for being a bit vague. But your answer doesn't work, obviously. – Wanderer Nov 29 '09 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 '09 at 20:20
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But it's not affine, and that's really the point of the question! – Wanderer Nov 29 '09 at 20:22
    
Ahh, now I see what you mean. – Charles Siegel Nov 29 '09 at 20:24

Jeez this thread is old but I found it useful.

Here is another example:

Consider the spectrum $X$ of a Dedekind domain. Remove a divisor with nontrivial class. Then this open set is non-principal in $X$.

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2  
No, you need more: your divisor class should not be torsion, exactly as in Peter McNamara's example. This is why it is very natural to look for algebraic curves. – abx Jan 22 at 15:10

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