66.23 Local irreducibility
A point on an algebraic space has a well defined étale local ring, which corresponds to the strict henselization of the local ring in the case of a scheme. In general we cannot see how many irreducible components of a scheme or an algebraic space pass through the given point from the étale local ring. We can only count the number of geometric branches.
Lemma 66.23.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ be a point. The following are equivalent
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the local ring $\mathcal{O}_{U, u}$ has a unique minimal prime,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ there is a unique irreducible component of $U$ through $u$,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the local ring $\mathcal{O}_{U, u}$ is unibranch,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the local ring $\mathcal{O}_{U, u}$ is geometrically unibranch,
$\mathcal{O}_{X, \overline{x}}$ has a unique minimal prime for any geometric point $\overline{x}$ lying over $x$.
Proof.
The equivalence of (1) and (2) follows from the fact that irreducible components of $U$ passing through $u$ are in $1$-$1$ correspondence with minimal primes of the local ring of $U$ at $u$. Let $a : U \to X$ and $u \in U$ be as in (1). Then $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$ by Lemma 66.22.1. In particular (4) and (5) are equivalent by More on Algebra, Lemma 15.106.5. The equivalence of (2), (3), and (4) follows from More on Morphisms, Lemma 37.36.2.
$\square$
Definition 66.23.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. We say that $X$ is geometrically unibranch at $x$ if the equivalent conditions of Lemma 66.23.1 hold. We say that $X$ is geometrically unibranch if $X$ is geometrically unibranch at every $x \in |X|$.
This is consistent with the definition for schemes (Properties, Definition 28.15.1).
Lemma 66.23.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ be a point. Let $n \in \{ 1, 2, \ldots \} $ be an integer. The following are equivalent
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the number of minimal primes of the local ring $\mathcal{O}_{U, u}$ is $\leq n$ and for at least one choice of $U, a, u$ it is $n$,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the number irreducible components of $U$ passing through $u$ is $\leq n$ and for at least one choice of $U, a, u$ it is $n$,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the number of branches of $U$ at $u$ is $\leq n$ and for at least one choice of $U, a, u$ it is $n$,
for any scheme $U$ and étale morphism $a : U \to X$ and $u \in U$ with $a(u) = x$ the number of geometric branches of $U$ at $u$ is $n$, and
the number of minimal prime ideals of $\mathcal{O}_{X, \overline{x}}$ is $n$.
Proof.
The equivalence of (1) and (2) follows from the fact that irreducible components of $U$ passing through $u$ are in $1$-$1$ correspondence with minimal primes of the local ring of $U$ at $u$. Let $a : U \to X$ and $u \in U$ be as in (1). Then $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$ by Lemma 66.22.1. Recall that the (geometric) number of branches of $U$ at $u$ is the number of minimal prime ideals of the (strict) henselization of $\mathcal{O}_{U, u}$. In particular (4) and (5) are equivalent. The equivalence of (2), (3), and (4) follows from More on Morphisms, Lemma 37.36.2.
$\square$
Definition 66.23.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The number of geometric branches of $X$ at $x$ is either $n \in \mathbf{N}$ if the equivalent conditions of Lemma 66.23.3 hold, or else $\infty $.
Comments (2)
Comment #397 by Matthew Emerton on
Comment #401 by Johan on