Lemma 37.52.2. In Lemma 37.52.1 assume in addition that $f$ is locally of finite type. Then for $s \in S$ the fibre $\pi ^{-1}(\{ s\} ) = \{ s_1, \ldots , s_ n\}$ is finite and the field extensions $\kappa (s_ i)/\kappa (s)$ are finite.

Proof. Recall that there are no specializations among the points of $\pi ^{-1}(\{ s\} )$, see Algebra, Lemma 10.36.20. As $f'$ is surjective, we find that $|X_ s| \to \pi ^{-1}(\{ s\} )$ is surjective. Observe that $X_ s$ is a quasi-separated scheme of finite type over a field (quasi-compactness was shown in the proof of the referenced lemma). Thus $X_ s$ is Noetherian (Morphisms, Lemma 29.15.6). A topological argument (omitted) now shows that $\pi ^{-1}(\{ s\} )$ is finite. For each $i$ we can pick a finite type point $x_ i \in X_ s$ mapping to $s_ i$ (Morphisms, Lemma 29.16.7). We conclude that $\kappa (s_ i)/\kappa (s)$ is finite: $x_ i$ can be represented by a morphism $\mathop{\mathrm{Spec}}(k_ i) \to X_ s$ of finite type (by our definition of finite type points) and hence $\mathop{\mathrm{Spec}}(k_ i) \to s = \mathop{\mathrm{Spec}}(\kappa (s))$ is of finite type (as a composition of finite type morphisms), hence $k_ i/\kappa (s)$ is finite (Morphisms, Lemma 29.16.1). $\square$

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).