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

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

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.
$\square$

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