# Proposition 4.2.24 in Liu Qing‘s AG book

In the book Algebraic Geometry and Arithmetic Curves, Qing Liu states the following proposition:

Proposition 4.2.24. Let $$X$$ be an algebraic variety over an algebraically closed field $$k$$ (i.e. $$X\to \mathrm{Spec}(k)$$ is of finite type). Then $$\mathrm{Reg}(X)$$ (regular locus) is an open subset.

The first step of his proof: since $$U=\{x\in X:O_{X,x}\text{ is a domain}\}\subset X$$ is open, and contains $$\mathrm{Reg}(X)$$, we may reduce to $$X$$ integral.

My question: why can we reduce to this case? $$U$$ is reduced but is it integral?

• $U$ isn’t integral, but we know that it has exactly one irreducible component going at each point. So its connected components are irreducible: as a consequence, its irreducible components are pairwise disjoint, finitely many and closed: so they are open, and we can consider each of them individually. – Mindlack Feb 27 '20 at 9:57
• @Mindlack looks like an answer to me! Would you care to record it below? – KReiser Feb 27 '20 at 10:17
• @Mindlack Dear friend, why a connected component of $U$ is irreducible? See here: stacks.math.columbia.edu/tag/0568, $U=X=\mathrm{Spec}(A)$ is connected, reduced, but not irreducible. – Doug Liu Feb 27 '20 at 10:34

## 1 Answer

I rephrase my comment for clarity, answering the OP’s comment as well.

$$U$$ isn’t integral, but we can find it has an “almost integral structure”.

Indeed, for each $$x \in U$$, $$O_{X,x}$$ is a domain. With a slight back-and-forth with commutative algebra, it implies that there is only one irreducible component of $$U$$ going through any given point.

So the (finitely many) irreducible components of $$U$$ are pairwise disjoint and closed. As a consequence, they are open. Therefore, $$U$$ is a disjoint reunion of integral open subschemes.

• Thanks for your answer. 1. Let $A$ be a ring whose localization at any prime ideal is a domain. Then given a prime $p$, there's only one minimal prime $q\subset p$, as $A_p$ has a unique minimal prime. 2. Let $X$ be a scheme whose stalks are integral, then its irreducible components are disjoint: if $Z_1,Z_2$ two irr components intersecting at $x$, we take an affine open $U$ around $x$, then $Z_i\cap U$ are irr components of $U$, hence equal. And $Z_i\cap U\subset Z_i$ is dense. – Doug Liu Feb 27 '20 at 18:50