The Stacks project

Proof. Let $V_ i$ be any affine open containing $\eta _ i$ and disjoint from the closed set $Z_1 \cup \ldots \hat Z_ i \ldots \cup Z_ n$. Since $X$ is quasi-separated for each $i$ the union $W_ i = \bigcup _{j, j \not= i} V_ i \cap V_ j$ is a quasi-compact open of $V_ i$ not containing $\eta _ i$. We can find open neighbourhoods $U_ i \subset V_ i$ containing $\eta _ i$ and disjoint from $W_ i$ by Algebra, Lemma 10.26.4. Finally, $U$ is affine since it is the spectrum of the ring $R_1 \times \ldots \times R_ n$ where $R_ i = \mathcal{O}_ X(U_ i)$, see Schemes, Lemma 26.6.8. $\square$

Proof. Say $X = \bigcup _{i = 1, \ldots , n} U_ i$ is a union of $n$ affine open subschemes. We will prove the lemma by induction on $n$. It is trivial for $n = 1$. Let $V' \subset \bigcup _{i = 1, \ldots , n - 1} U_ i$ be a separated dense open subscheme, which exists by induction hypothesis. Consider

It is clear that $V$ is separated and a dense open subscheme of $X$. $\square$

Proof. Suppose that $x \in Z_1 \cap \ldots \cap Z_ r$ and $x \not\in Z_{r + 1}, \ldots , Z_ n$. Then we may choose an affine open $W \subset X$ such that $x \in W$ and $W \cap Z_ i = \emptyset $ for $i = r + 1, \ldots , n$. Note that clearly $\eta _ i \in W$ for $i = 1, \ldots , r$. By Lemma 28.29.1 we may choose affine opens $U_ i \subset X$ which are pairwise disjoint such that $\eta _ i \in U_ i$ for $i = r + 1, \ldots , n$. Since $X$ is quasi-separated the opens $W \cap U_ i$ are quasi-compact and do not contain $\eta _ i$ for $i = r + 1, \ldots , n$. Hence by Algebra, Lemma 10.26.4 we may shrink $U_ i$ such that $W \cap U_ i = \emptyset $ for $i = r + 1, \ldots , n$. Then the union $U = W \cup \bigcup _{i = r + 1, \ldots , n} U_ i$ is disjoint and hence (by Schemes, Lemma 26.6.8) a suitable affine open. $\square$

Proof. By Properties, Definition 28.18.1 a quasi-affine scheme is a quasi-compact open subscheme of an affine scheme. Any affine scheme $\mathop{\mathrm{Spec}}(R)$ is isomorphic to $\text{Proj}(R[X])$ where $R[X]$ is graded by setting $\deg (X) = 1$. By Proposition 28.26.13 if $X$ has an ample invertible sheaf then $X$ is isomorphic to an open subscheme of $\text{Proj}(S)$ for some graded ring $S$. Hence, it suffices to prove the lemma in case (4). (We urge the reader to prove case (2) directly for themselves.)

Thus assume $X \subset \text{Proj}(S)$ is a locally closed subscheme where $S$ is some graded ring. Let $T = \overline{X} \setminus X$. Recall that the standard opens $D_{+}(f)$ form a basis of the topology on $\text{Proj}(S)$. Since $E$ is finite we may choose finitely many homogeneous elements $f_ i \in S_{+}$ such that

Suppose that $E = \{ \mathfrak p_1, \ldots , \mathfrak p_ m\} $ as a subset of $\text{Proj}(S)$. Consider the ideal $I = (f_1, \ldots , f_ n) \subset S$. Since $I \not\subset \mathfrak p_ j$ for all $j = 1, \ldots , m$ we see from Algebra, Lemma 10.57.6 that there exists a homogeneous element $f \in I$, $f \not\in \mathfrak p_ j$ for all $j = 1, \ldots , m$. Then $E \subset D_{+}(f) \subset D_{+}(f_1) \cup \ldots \cup D_{+}(f_ n)$. Since $D_{+}(f)$ does not meet $T$ we see that $X \cap D_{+}(f)$ is a closed subscheme of the affine scheme $D_{+}(f)$, hence is an affine open of $X$ as desired. $\square$

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. $\square$

Proof. The proof of this lemma has a lot in common with the proof of Algebra, Lemma 10.15.2. Say $E = \{ x_1, \ldots , x_ n\} $. If $E = W = \emptyset $, then $s = 0$ works. If $W \not= \emptyset $, then we may assume $E \not= \emptyset $ by adding a point if necessary. Thus we may assume $n \geq 1$. We will prove the lemma by induction on $n$.

Base case: $n = 1$. After replacing $W$ by an affine open neighbourhood of $x_1$ in $W$, we may assume $W$ is affine. Combining Lemmas 28.27.1 and Proposition 28.26.13 we see that every quasi-coherent $\mathcal{O}_ X$-module is globally generated. Hence there exists a global section $s$ of $\mathcal{L}$ which does not vanish at $x_1$. On the other hand, let $Z \subset X$ be the reduced induced closed subscheme on $X \setminus W$. Applying global generation to the quasi-coherent ideal sheaf $\mathcal{I}$ of $Z$ we find a global section $f$ of $\mathcal{I}$ which does not vanish at $x_1$. Then $s' = fs$ is a global section of $\mathcal{L}$ which does not vanish at $x_1$ such that $X_{s'} \subset W$. Then $X_{s'}$ is affine by Lemma 28.26.4.

Induction step for $n > 1$. If there is a specialization $x_ i \leadsto x_ j$ for $i \not= j$, then it suffices to prove the lemma for $\{ x_1, \ldots , x_ n\} \setminus \{ x_ i\} $ and we are done by induction. Thus we may assume there are no specializations among the $x_ i$. By either Lemma 28.29.5 or Lemma 28.29.6 we may assume $W$ is affine. By induction we can find a global section $s$ of $\mathcal{L}$ such that $X_ s \subset W$ is affine and contains $x_1, \ldots , x_{n - 1}$. If $x_ n \in X_ s$ then we are done. Assume $s$ is zero at $x_ n$. By the case $n = 1$ we can find a global section $s'$ of $\mathcal{L}$ with $\{ x_ n\} \subset X_{s'} \subset W \setminus \overline{\{ x_1, \ldots , x_{n - 1}\} }$. Here we use that $x_ n$ is not a specialization of $x_1, \ldots , x_{n - 1}$. Then $s + s'$ is a global section of $\mathcal{L}$ which is nonvanishing at $x_1, \ldots , x_ n$ with $X_{s + s'} \subset W$ and we conclude as before. $\square$

Proof. After translation into algebra, this follows from Algebra, Lemma 10.31.9. $\square$