open but not affine subscheme?example? [closed]

In the book algebraice geometry by R.Harshorne we always say :an open affine subset U=SpecA of a sheme X.

I was always wondering that whether there existed an open but not affine subset. I want to have a good example.

Thank you all very much!

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closed as too localized by Martin Brandenburg, Andrés Caicedo, Harry Gindi, S. Carnahan♦Nov 27 '10 at 14:44

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A scheme is a locally ringed space covered by affine schemes. This doesn't necessarily mean that all its open subschemes are affine. The example given by Robin below is the most basic example. You should check by way of example to see that this won't work in dimension 1 (if you stick to normal schemes). Also, look at Liu's book which is filled with almost all the basic examples. – Ariyan Javanpeykar Nov 26 '10 at 15:33
Sorry, but I have the impression that the questioner has not thought at all about the question. Besides, this is definitively not a research level question ... -1. – Martin Brandenburg Nov 26 '10 at 21:29
Sorry ,I am lost in the journey of reading the book. So I cannot think clearly,and then write a qestion here. Thank you very much! – user11085 Nov 30 '10 at 15:42

The weirdest example of affineness / non-affineness I know is the following. There is a smooth projective scheme $X$ over $\mathbb{Z}[1/7]$ and a closed subscheme $Z$ (in fact, $Z$ is a smooth irreducible divisor) such that $U = X \setminus Z$ is not affine. However, after tensoring with $\mathbb{Z}/p$ , $$U_p := U \otimes_{\mathbb{Z}} \mathbb{Z}/p$$ is affine for infinitely many $p > 0$ and also non-affine for infinitely many $p > 0$.

For details, see Remark 4.8 in MR2220102, Brenner and Katzman, J. Amer. Math. Soc. 19 (2006), no. 3.

EDIT: As far as I know, it is unknown whether similar examples exist in the equal characteristic setting.

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+1 for this fabulously weird example! – Hailong Dao Nov 26 '10 at 19:43
For an equal characteristic example which exhibits a similar behavior see "Tight closure does not commute with localization" Holger Brenner and Paul Monsky Annals Vol. 171 (2010), No. 1, 571–588 especially Remark 4.2. – Holger Partsch Nov 27 '10 at 11:19
@Holger Partsch, that's awesome. That's an equally weird equal characteristic ($p > 0$) example of a family all of whose closed fibers are affine, but whose generic fiber is not affine! – Karl Schwede Nov 27 '10 at 15:44

The standard example is to let $X$ be the affine plane over a field, and $U=X-\{(0,0)\}$.

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Isn't that the standard example of something else? I would have gone for $U=X=$ projective 1-space :-) – Kevin Buzzard Nov 26 '10 at 15:16
With regards to this example, see Exercise 3.6 in chapter I. In Hartshorne, many important examples and theorems are in the exercises. – Karl Schwede Nov 26 '10 at 17:19

1) Obviously if $X$ is not affine, then $U=X$ is an open subset which is not affine (see Kevin's comment above) but this might feel cheating.

2) A more general example along the lines of Robin's example is the following:

Let $X$ be an arbitrary $S_2$ affine scheme of dimension at least $2$. (If you don't know what $S_2$ is, take normal, or smooth). Let $Z\subset X$ be a non-empty closed subscheme of codimension at least $2$. Then $U=X\setminus Z\subset X$ is a proper open subset which is not affine.

Here is why it's not affine: Assume that $X={\rm Spec} A$ is affine and let $\iota:U\hookrightarrow X$ be the natural embedding. By the $S_2$ property it follows that $\iota_*\mathcal O_U\simeq \mathcal O_X$, and consequently the induced morphism on global sections $\Gamma(U, \mathcal O_U)\to A$ is an isomorphism, but then if $U$ were affine, then this would imply that the embedding $\iota$ is an isomorphism, which is not true by construction.

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To unknown: the point is the nullstellensatz holds for affine sets, i.e. over the complex field say, points of the set correspond to maximal ideals. so if you remove points but don't change the ring of functions, the set is no longer affine. But a rational function on a good (S2) space, has poles in codimension one, so if you remove a smaller set, you cannot acquire any new regular functions. I.e. if f is regular on X-Z where Z is too small to support a pole, then f is regular on X. – roy smith Nov 26 '10 at 22:14