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Why is $\mathbb{P}^{1} \times \mathbb{A}^{1}$ not isomorphic to an affine variety?

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up vote 3 down vote accepted

Every closed subset of an affine variety is affine.
But here the closed subset $\mathbb P^1\times \lbrace 0 \rbrace \subset \mathbb{P}^{1} \times \mathbb{A}^{1}$ (why is it closed?) is isomorphic to $\mathbb P^1$ and thus not affine ( why?)

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What's the definition of "affine"? – Jonathan May 3 '12 at 21:50
Dear Jonathan, the notion "affine" is basic in algebraic geometry: check the first pages of any book on the subject. – Georges Elencwajg May 3 '12 at 21:55
Dear user, yes it is a theorem, relying on the fact that $\mathbb P^2$ is normal. (It is the analogue of Hartogs's theorem in complex analysis). However here the situation is far simpler: given two points on an affine variety, there is a regular function taking different values at the two points. This fails for $\mathbb P^1$, on which all regular functions are constant. So $\mathbb P^1$ is not affine! – Georges Elencwajg May 3 '12 at 23:07
Dear user, look at Ravi Vakil's wonderful online notes (, starting page 285, 12.3.9 – Georges Elencwajg May 3 '12 at 23:43
Dear @GeorgesElencwajg I am looking at Proposition 3.1.9 of Qing Liu which says $\text{Proj} (B\otimes_A C) \cong \text{Proj} B\times_{\text{Spec} A } \text{Spec} C$. Does this mean $\Bbb{P}^2 \cong \Bbb{P}^1 \times_k \Bbb{A}^1$? – user38268 Oct 21 '13 at 6:04

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