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
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
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
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
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
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
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
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
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)