The Stacks project

Lemma 33.42.6. Let $X$ be a scheme. Let $T \subset X$ be finite set of points. Assume

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

  2. $Z_ i \cap T$ 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 10.15.2), 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 33.42.5 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

Johan, you may be interested in the following which is Cor 48 in AUTHOR = {Koll{\'a}r, J{\'a}nos}, TITLE = {Quotients by finite equivalence relations}, BOOKTITLE = {Current developments in algebraic geometry}, SERIES = {Math. Sci. Res. Inst. Publ.}, VOLUME = {59}, PAGES = {227--256},

Let be an finite and surjective morphism of excellent, separated algebraic spaces. Assume that every finite subset of is contained in an open affine subspace. Then the same holds for .

Comment #3212 by on

Yes, I know about your very interesting paper! It would be very appropriate to add the results from this paper to the Stacks project. However, the particular type of statement you mention comes much later in the theory and should go in some chapter on algebraic spaces, maybe in the chapter on pushouts of algebraic spaces.

The current lemma here is only used to prove that a finite set of codimension 1 points on a separated Noetherian scheme can always be put in an affine open, see Proposition 33.42.7.

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