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