## 28.15 Local irreducibility

Recall that in More on Algebra, Section 15.106 we introduced the notion of a (geometrically) unibranch local ring.

Definition 28.15.1. Let $X$ be a scheme. Let $x \in X$. We say $X$ is unibranch at $x$ if the local ring $\mathcal{O}_{X, x}$ is unibranch. We say $X$ is geometrically unibranch at $x$ if the local ring $\mathcal{O}_{X, x}$ is geometrically unibranch. We say $X$ is unibranch if $X$ is unibranch at all of its points. We say $X$ is geometrically unibranch if $X$ is geometrically unibranch at all of its points.

To be sure, it can happen that a local ring $A$ is geometrically unibranch (in the sense of More on Algebra, Definition 15.106.1) but the scheme $\mathop{\mathrm{Spec}}(A)$ is not geometrically unibranch in the sense of Definition 28.15.1. For example this happens if $A$ is the local ring at the vertex of the cone over an irreducible plane curve which has ordinary double point singularity (a node).

Proof. This follows from the definitions. Namely, a scheme is normal if the local rings are normal domains. It is immediate from the More on Algebra, Definition 15.106.1 that a local normal domain is geometrically unibranch. $\square$

Lemma 28.15.3. Let $X$ be a Noetherian scheme. The following are equivalent

1. $X$ is geometrically unibranch (Definition 28.15.1),

2. for every point $x \in X$ which is not the generic point of an irreducible component of $X$, the punctured spectrum of the strict henselization $\mathcal{O}_{X, x}^{sh}$ is connected.

Proof. More on Algebra, Lemma 15.106.5 shows that (1) implies that the punctured spectra in (2) are irreducible and in particular connected.

Assume (2). Let $x \in X$. We have to show that $\mathcal{O}_{X, x}$ is geometrically unibranch. By induction on $\dim (\mathcal{O}_{X, x})$ we may assume that the result holds for every nontrivial generalization of $x$. We may replace $X$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. In other words, we may assume that $X = \mathop{\mathrm{Spec}}(A)$ with $A$ local and that $A_\mathfrak p$ is geometrically unibranch for each nonmaximal prime $\mathfrak p \subset A$.

Let $A^{sh}$ be the strict henselization of $A$. If $\mathfrak q \subset A^{sh}$ is a prime lying over $\mathfrak p \subset A$, then $A_\mathfrak p \to A^{sh}_\mathfrak q$ is a filtered colimit of étale algebras. Hence the strict henselizations of $A_\mathfrak p$ and $A^{sh}_\mathfrak q$ are isomorphic. Thus by More on Algebra, Lemma 15.106.5 we conclude that $A^{sh}_\mathfrak q$ has a unique minimal prime ideal for every nonmaximal prime $\mathfrak q$ of $A^{sh}$.

Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the minimal primes of $A^{sh}$. We have to show that $r = 1$. By the above we see that $V(\mathfrak q_1) \cap V(\mathfrak q_ j) = \{ \mathfrak m^{sh}\}$ for $j = 2, \ldots , r$. Hence $V(\mathfrak q_1) \setminus \{ \mathfrak m^{sh}\}$ is an open and closed subset of the punctured spectrum of $A^{sh}$ which is a contradiction with the assumption that this punctured spectrum is connected unless $r = 1$. $\square$

Definition 28.15.4. Let $X$ be a scheme. Let $x \in X$. The number of branches of $X$ at $x$ is the number of branches of the local ring $\mathcal{O}_{X, x}$ as defined in More on Algebra, Definition 15.106.6. The number of geometric branches of $X$ at $x$ is the number of geometric branches of the local ring $\mathcal{O}_{X, x}$ as defined in More on Algebra, Definition 15.106.6.

Often we want to compare this with the branches of the complete local ring, but the comparison is not straightforward in general; some information on this topic can be found in More on Algebra, Section 15.108.

Lemma 28.15.5. Let $X$ be a scheme and $x \in X$. Let $X_ i$, $i \in I$ be the irreducible components of $X$ passing through $x$. Then the number of (geometric) branches of $X$ at $x$ is the sum over $i \in I$ of the number of (geometric) branches of $X_ i$ at $x$.

Proof. We view the $X_ i$ as integral closed subschemes of $X$, see Schemes, Definition 26.12.5 and Lemma 28.3.4. Observe that the number of (geometric) branches of $X_ i$ at $x$ is at least $1$ for all $i$ (essentially by definition). Recall that the $X_ i$ correspond $1$-to-$1$ with the minimal prime ideals $\mathfrak p_ i \subset \mathcal{O}_{X, x}$, see Algebra, Lemma 10.26.3. Thus, if $I$ is infinite, then $\mathcal{O}_{X, x}$ has infinitely many minimal primes, whence both $\mathcal{O}_{X, x}^ h$ and $\mathcal{O}_{X, x}^{sh}$ have infinitely many minimal primes (combine Algebra, Lemmas 10.30.5 and 10.30.7 and the injectivity of the maps $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^ h \to \mathcal{O}_{X, x}^{sh}$). In this case the number of (geometric) branches of $X$ at $x$ is defined to be $\infty$ which is also true for the sum. Thus we may assume $I$ is finite. Let $A'$ be the integral closure of $\mathcal{O}_{X, x}$ in the total ring of fractions $Q$ of $(\mathcal{O}_{X, x})_{red}$. Let $A'_ i$ be the integral closure of $\mathcal{O}_{X, x}/\mathfrak p_ i$ in the total ring of fractions $Q_ i$ of $\mathcal{O}_{X, x}/\mathfrak p_ i$. By Algebra, Lemma 10.25.4 we have $Q = \prod _{i \in I} Q_ i$. Thus $A' = \prod A'_ i$. Then the equality of the lemma follows from More on Algebra, Lemma 15.106.7 which expresses the number of (geometric) branches in terms of the maximal ideals of $A'$. $\square$

Lemma 28.15.6. Let $X$ be a scheme. Let $x \in X$.

1. The number of branches of $X$ at $x$ is $1$ if and only if $X$ is unibranch at $x$.

2. The number of geometric branches of $X$ at $x$ is $1$ if and only if $X$ is geometrically unibranch at $x$.

Proof. This lemma follows immediately from the definitions and the corresponding result for rings, see More on Algebra, Lemma 15.106.7. $\square$

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