Lemma 33.41.6. Let $X$ be a scheme. Let $T \subset X$ be finite set of points. Assume

$X$ has finitely many irreducible components $Z_1, \ldots , Z_ t$, and

$Z_ i \cap T$ is contained in an affine open of the reduced induced subscheme corresponding to $Z_ i$.

Then there exists an affine open subscheme of $X$ containing $T$.

**Proof.**
Using Limits, Proposition 32.11.2 there is an immediate reduction to the case where $X$ is reduced. Details omitted. In the rest of the proof we endow every closed subset of $X$ with the induced reduced closed subscheme structure.

We argue by induction that we can find an affine open $U \subset Z_1 \cup \ldots \cup Z_ r$ containing $T \cap (Z_1 \cup \ldots \cup Z_ r)$. For $r = 1$ this holds by assumption. Say $r > 1$ and let $U \subset Z_1 \cup \ldots \cup Z_{r - 1}$ be an affine open containing $T \cap (Z_1 \cup \ldots \cup Z_{r - 1})$. Let $V \subset X_ r$ be an affine open containing $T \cap Z_ r$ (exists by assumption). Then $U \cap V$ contains $T \cap ( Z_1 \cup \ldots \cup Z_{r - 1} ) \cap Z_ r$. Hence

\[ \Delta = (U \cap Z_ r) \setminus (U \cap V) \]

does not contain any element of $T$. Note that $\Delta $ is a closed subset of $U$. By prime avoidance (Algebra, Lemma 10.14.2), we can find a standard open $U'$ of $U$ containing $T \cap U$ and avoiding $\Delta $, i.e., $U' \cap Z_ r \subset U \cap V$. After replacing $U$ by $U'$ we may assume that $U \cap V$ is closed in $U$.

Using that by the same arguments as above also the set $\Delta ' = (U \cap (Z_1 \cup \ldots \cup Z_{r - 1})) \setminus (U \cap V)$ does not contain any element of $T$ we find a $h \in \mathcal{O}(V)$ such that $D(h) \subset V$ contains $T \cap V$ and such that $U \cap D(h) \subset U \cap V$. Using that $U \cap V$ is closed in $U$ we can use Lemma 33.41.5 to find an element $g \in \mathcal{O}(U)$ whose restriction to $U \cap V$ equals the restriction of $h$ to $U \cap V$ and such that $T \cap U \subset D(g)$. Then we can replace $U$ by $D(g)$ and $V$ by $D(h)$ to reach the situation where $U \cap V$ is closed in both $U$ and $V$. In this case the scheme $U \cup V$ is affine by Limits, Lemma 32.11.3. This proves the induction step and thereby the lemma.
$\square$

## Comments (2)

Comment #3210 by Kollar on

Comment #3212 by Johan on