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?

  • 1
    $\begingroup$ $U$ 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. $\endgroup$ – Mindlack Feb 27 '20 at 9:57
  • $\begingroup$ @Mindlack looks like an answer to me! Would you care to record it below? $\endgroup$ – KReiser Feb 27 '20 at 10:17
  • $\begingroup$ @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. $\endgroup$ – Doug Liu Feb 27 '20 at 10:34

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.

  • $\begingroup$ 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. $\endgroup$ – Doug Liu Feb 27 '20 at 18:50

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