Lemma 28.27.3. Let $X$ be a scheme. Suppose that there exist finitely many elements $f_1, \ldots , f_ n \in \Gamma (X, \mathcal{O}_ X)$ such that

1. each $X_{f_ i}$ is an affine open of $X$, and

2. the ideal generated by $f_1, \ldots , f_ n$ in $\Gamma (X, \mathcal{O}_ X)$ is equal to the unit ideal.

Then $X$ is affine.

Proof. Assume we have $f_1, \ldots , f_ n$ as in the lemma. We may write $1 = \sum g_ i f_ i$ for some $g_ j \in \Gamma (X, \mathcal{O}_ X)$ and hence it is clear that $X = \bigcup X_{f_ i}$. (The $f_ i$'s cannot all vanish at a point.) Since each $X_{f_ i}$ is quasi-compact (being affine) it follows that $X$ is quasi-compact. Hence we see that $X$ is quasi-affine by Lemma 28.27.1 above. Consider the open immersion

$j : X \to \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X)),$

see Lemma 28.18.4. The inverse image of the standard open $D(f_ i)$ on the right hand side is equal to $X_{f_ i}$ on the left hand side and the morphism $j$ induces an isomorphism $X_{f_ i} \cong D(f_ i)$, see Lemma 28.18.3. Since the $f_ i$ generate the unit ideal we see that $\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X)) = \bigcup _{i = 1, \ldots , n} D(f_ i)$. Thus $j$ is an isomorphism. $\square$

Comment #4868 by DatPham on

I think there is a simpler argument as follows. Let $j: X\to \mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$ be the morphism induced from the identity map on global sections. Then we can check easily that $j^{-1}(D(f_i))=X_{f_i}$. So by our assumption, the collection $\{D(f_i)\}$ forms an affine open cover of $\mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$ such that $j^{-1}(D(f_i))$ is affine for all $i$. It follows that $j$ is an affine morphism. Hence $X$ is affine since $\mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$ is so.

Comment #5152 by on

@#4868: We cannot yet use affine morphisms as we haven't defined them yet. Also, in the discussion of affine morphisms, you prove this fact directly, so it wouldn't be easier (in some sense).

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