33.40 The number of branches

We have defined the number of branches of a scheme at a point in Properties, Section 28.15.

Lemma 33.40.1. Let $X$ be a scheme. Assume every quasi-compact open of $X$ has finitely many irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Let $x \in X$.

1. The number of branches of $X$ at $x$ is the number of inverse images of $x$ in $X^\nu$.

2. The number of geometric branches of $X$ at $x$ is $\sum _{\nu (x^\nu ) = x} [\kappa (x^\nu ) : \kappa (x)]_ s$.

Proof. First note that the assumption on $X$ exactly means that the normalization is defined, see Morphisms, Definition 29.54.1. Then the stalk $A' = (\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in the total ring of fractions of $A_{red}$, see Morphisms, Lemma 29.54.4. Since $\nu$ is an integral morphism, we see that the points of $X^\nu$ lying over $x$ correspond to the primes of $A'$ lying over the maximal ideal $\mathfrak m$ of $A$. As $A \to A'$ is integral, this is the same thing as the maximal ideals of $A'$ (Algebra, Lemmas 10.36.20 and 10.36.22). Thus the lemma now follows from its algebraic counterpart: More on Algebra, Lemma 15.106.7. $\square$

Lemma 33.40.2. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme. Let $K/k$ be an extension of fields. Let $y \in X_ K$ be a point with image $x$ in $X$. Then the number of geometric branches of $X$ at $x$ is the number of geometric branches of $X_ K$ at $y$.

Proof. Write $Y = X_ K$ and let $X^\nu$, resp. $Y^\nu$ be the normalization of $X$, resp. $Y$. Consider the commutative diagram

$\xymatrix{ Y^\nu \ar[r] \ar[d] & X^\nu _ K \ar[r] \ar[d]_{\nu _ K} & X^\nu \ar[d]_\nu \\ Y \ar@{=}[r] & Y \ar[r] & X }$

By Lemma 33.27.5 we see that the left top horizontal arrow is a universal homeomorphism. Hence it induces purely inseparable residue field extensions, see Morphisms, Lemmas 29.45.5 and 29.10.2. Thus the number of geometric branches of $Y$ at $y$ is $\sum _{\nu _ K(y') = y} [\kappa (y') : \kappa (y)]_ s$ by Lemma 33.40.1. Similarly $\sum _{\nu (x') = x} [\kappa (x') : \kappa (x)]_ s$ is the number of geometric branches of $X$ at $x$. Using Schemes, Lemma 26.17.5 our statement follows from the following algebra fact: given a field extension $l/\kappa$ and an algebraic field extension $m/\kappa$, then

$\sum \nolimits _{m \otimes _\kappa l \to m'} [m' : l']_ s = [m : \kappa ]_ s$

where the sum is over the quotient fields of $m \otimes _\kappa l$. One can prove this in an elementary way, or one can use Lemma 33.7.6 applied to

$\mathop{\mathrm{Spec}}(m \otimes _\kappa l) \times _{\mathop{\mathrm{Spec}}(l)} \mathop{\mathrm{Spec}}(\overline{l}) = \mathop{\mathrm{Spec}}(m) \otimes _{\mathop{\mathrm{Spec}}(\kappa )} \mathop{\mathrm{Spec}}(\overline{l}) \longrightarrow \mathop{\mathrm{Spec}}(m) \times _{\mathop{\mathrm{Spec}}(\kappa )} \mathop{\mathrm{Spec}}(\overline{\kappa })$

because one can interpret $[m : \kappa ]_ s$ as the number of connected components of the right hand side and the sum $\sum _{m \otimes _\kappa l \to m'} [m' : l']_ s$ as the number of connected components of the left hand side. $\square$

Lemma 33.40.3. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme. Let $K/k$ be an extension of fields. Let $y \in X_ K$ be a point with image $x$ in $X$. Then $X$ is geometrically unibranch at $x$ if and only if $X_ K$ is geometrically unibranch at $y$.

Proof. Immediate from Lemma 33.40.2 and More on Algebra, Lemma 15.106.7. $\square$

Definition 33.40.4. Let $A$ and $A_ i$, $1 \leq i \leq n$ be local rings. We say $A$ is a wedge of $A_1, \ldots , A_ n$ if there exist isomorphisms

$\kappa _{A_1} \to \kappa _{A_2} \to \ldots \to \kappa _{A_ n}$

and $A$ is isomorphic to the ring consisting of $n$-tuples $(a_1, \ldots , a_ n) \in A_1 \times \ldots \times A_ n$ which map to the same element of $\kappa _{A_ n}$.

If we are given a base ring $\Lambda$ and $A$ and $A_ i$ are $\Lambda$-algebras, then we require $\kappa _{A_ i} \to \kappa _{A_{i + 1}}$ to be a $\Lambda$-algebra isomorphisms and $A$ to be isomorphic as a $\Lambda$-algebra to the $\Lambda$-algebra consisting of $n$-tuples $(a_1, \ldots , a_ n) \in A_1 \times \ldots \times A_ n$ which map to the same element of $\kappa _{A_ n}$. In particular, if $\Lambda = k$ is a field and the maps $k \to \kappa _{A_ i}$ are isomorphisms, then there is a unique choice for the isomorphisms $\kappa _{A_ i} \to \kappa _{A_{i + 1}}$ and we often speak of the wedge of $A_1, \ldots , A_ n$.

Lemma 33.40.5. Let $(A, \mathfrak m)$ be a strictly henselian $1$-dimensional reduced Nagata local ring. Then

$\delta \text{-invariant of }A \geq \text{number of geometric branches of }A - 1$

If equality holds, then $A$ is a wedge of $n \geq 1$ strictly henselian discrete valuation rings.

Proof. The number of geometric branches is equal to the number of branches of $A$ (immediate from More on Algebra, Definition 15.106.6). Let $A \to A'$ be as in Lemma 33.39.2. Observe that the number of branches of $A$ is the number of maximal ideals of $A'$, see More on Algebra, Lemma 15.106.7. There is a surjection

$A'/A \longrightarrow \left(\prod \nolimits _{\mathfrak m'} \kappa (\mathfrak m')\right)/ \kappa (\mathfrak m)$

Since $\dim _{\kappa (\mathfrak m)} \prod \kappa (\mathfrak m')$ is $\geq$ the number of branches, the inequality is obvious.

If equality holds, then $\kappa (\mathfrak m') = \kappa (\mathfrak m)$ for all $\mathfrak m' \subset A'$ and the displayed arrow above is an isomorphism. Since $A$ is henselian and $A \to A'$ is finite, we see that $A'$ is a product of local henselian rings, see Algebra, Lemma 10.153.4. The factors are the local rings $A'_{\mathfrak m'}$ and as $A'$ is normal, these factors are discrete valuation rings (Algebra, Lemma 10.119.7). Since the displayed arrow is an isomorphism we see that $A$ is indeed the wedge of these local rings. $\square$

Lemma 33.40.6. Let $(A, \mathfrak m)$ be a $1$-dimensional reduced Nagata local ring. Then

$\delta \text{-invariant of }A \geq \text{number of geometric branches of }A - 1$

Proof. We may replace $A$ by the strict henselization of $A$ without changing the $\delta$-invariant (Lemma 33.39.6) and without changing the number of geometric branches of $A$ (this is immediate from the definition, see More on Algebra, Definition 15.106.6). Thus we may assume $A$ is strictly henselian and we may apply Lemma 33.40.5. $\square$

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