Variety that is affine and projective is a finite number of points

Question

I was trying to proof the following without any luck. I would appreciate good hints.

A projective variety that is isomorphic to an affine variety is a finite number of points.

Answer 1

In the context of complex analytic geometry one can prove your statement as follows:

If a projective-algebraic variety is biholomorphically equivalent to a subvariety $X \subset \mathbb C^n$, then $X$ is compact. Each connected component $U$ of $X$ is compact, too. Hence each coordinate function of $\mathbb C^n$ is constant on $U$ by the maximum principle for holomorphic functions. Therefore $U$ is a singleton. Due to compactness of $U$ there are only finitely many connected components, q.e.d.

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In the context of algebraic geometry with an algebraically closed base field $k$ the proof may follow KReiser's hint:

For a projective algebraic variety $X \subset \mathbb P^n$ one has $\mathscr O(X) = k$ (see Hartshorne, Chap. I, Theor. 3.4). On the other hand, for an affine algebraic variety $Y \subset \mathbb A^n$ one has

$$\mathscr O(Y) = k[X_1, ..., X_n]/Id(Y).$$

Hence $\mathscr O(Y) = k$ iff the ideal $Id(Y)$ is maximal. The latter property is equivalent to $Y$ being a singleton (see Hartshorne, Chap. I, Ex. 1.4.4.), q.e.d.