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. $\square$

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