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I am confused by the last line in the proof that $\mathbb{P}^1$ is not affine, as presented in Ravi Vakil's algebraic geometry notes.

First, he computes the ring of global sections. It turns out that $\Gamma(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})=k$. This is all fine. What confuses me is the next line. He says: "If $\mathbb{P}^1$ were affine, then it would be $\operatorname{Spec}\Gamma(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})=\operatorname{Spec}k$, i.e., one point. But it isn't -- it has lots of point."

I don't know what to make of that sentence. In the second equality, isn't he just taking $\operatorname{Spec}$ of both sides of the first equality? If so, what does this have anything to do with assuming $\mathbb{P}^1$ is affine? Further, why do we know $\operatorname{Spec}\Gamma(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})$ should have "lots of points"? In short, what am I missing here?

Edit: The question linked does not answer my question. I am asking about a specific line in this proof given by Vakil. The linked questions only asks why $\mathbb{P}^1$ is not affine in general. Further, none of the given answers there address my concern. The chosen answer uses dimension theory, which is clearly not what Vakil had in mind since dimension theory hasn't been discussed yet

Arctic Char
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ponchan
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    Given any affine scheme $X =\operatorname{Spec}(A)$, we have $\Gamma (X, O_X) = A$. Thus you can recover $X$ from the global section. – Arctic Char Sep 07 '20 at 19:02
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    @ArcticChar That looks like an answer to me! Would you care to record it below? – KReiser Sep 07 '20 at 19:06
  • @ArcticChar So, if $X$ were affine, given the fact what $\Gamma(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})=k$ we would have $\mathbb{P}^1=\operatorname{Spec}k$ which is false. – ponchan Sep 07 '20 at 19:22
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    Yes, exactly. That's why he says "it has lots of points" it refers to $\mathbb P^1$ here. – Arctic Char Sep 07 '20 at 19:24

1 Answers1

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This has come up a lot of times on math.SE. $\text{Spec } \Gamma(X, \mathcal{O}_X)$ is the affinization of $X$; it's the universal affine scheme to which $X$ maps, so $X$ is affine iff this universal map $X \to \text{Spec } \Gamma(X, \mathcal{O}_X)$ is an isomorphism. This follows from general properties of adjunctions; see, for example, this answer.

(Compare: the abelianization $G/[G, G]$ of a group is the universal abelian group to which $G$ maps, so a group is abelian iff the universal map $G \to G/[G, G]$ is an isomorphism.)

Ravi computes that the affinization of $\mathbb{P}^1$ is a point, so $\mathbb{P}^1$ is affine iff the canonical map to a point is an isomorphism iff it's a point. But $\mathbb{P}^1$ has $|L|+1$ points over any field extension $L$ of $k$ (I assume $k$ is a field for simplicity but similar things can be said in general).

Qiaochu Yuan
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