Lemma 28.29.6. Let $X$ be a scheme. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. Let

\[ E \subset W \subset X \]

with $E$ finite and $W$ open in $X$. Then there exists an $n > 0$ and a section $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is affine and $E \subset X_ s \subset W$.

**Proof.**
The reader can modify the proof of Lemma 28.29.5 to prove this lemma; we will instead deduce the lemma from it. By Lemma 28.29.5 we can choose an affine open $U \subset W$ such that $E \subset U$. Consider the graded ring $S = \Gamma _*(X, \mathcal{L}) = \bigoplus _{n \geq 0} \Gamma (X, \mathcal{L}^{\otimes n})$. For each $x \in E$ let $\mathfrak p_ x \subset S$ be the graded ideal of sections vanishing at $x$. It is clear that $\mathfrak p_ x$ is a prime ideal and since some power of $\mathcal{L}$ is globally generated, it is clear that $S_{+} \not\subset \mathfrak p_ x$. Let $I \subset S$ be the graded ideal of sections vanishing on all points of $X \setminus U$. Since the sets $X_ s$ form a basis for the topology we see that $I \not\subset \mathfrak p_ x$ for all $x \in E$. By (graded) prime avoidance (Algebra, Lemma 10.57.6) we can find $s \in I$ homogeneous with $s \not\in \mathfrak p_ x$ for all $x \in E$. Then $E \subset X_ s \subset U$ and $X_ s$ is affine by Lemma 28.26.4.
$\square$

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