Lemma 76.36.2. In Lemma 76.36.1 assume in addition that $f$ is locally of finite type and $Y$ affine. Then for $y \in Y$ the fibre $\pi ^{-1}(\{ y\} ) = \{ y_1, \ldots , y_ n\}$ is finite and the field extensions $\kappa (y_ i)/\kappa (y)$ are finite.

Proof. Recall that there are no specializations among the points of $\pi ^{-1}(\{ y\} )$, see Algebra, Lemma 10.36.20. As $f'$ is surjective, we find that $|X_ y| \to \pi ^{-1}(\{ y\} )$ is surjective. Observe that $X_ y$ is a quasi-separated algebraic space of finite type over a field (quasi-compactness was shown in the proof of the referenced lemma). Thus $|X_ y|$ is a Noetherian topological space (Morphisms of Spaces, Lemma 67.28.6). A topological argument (omitted) now shows that $\pi ^{-1}(\{ y\} )$ is finite. For each $i$ we can pick a finite type point $x_ i \in |X_ y|$ mapping to $y_ i$ (Morphisms of Spaces, Lemma 67.25.6). We conclude that $\kappa (y_ i)/\kappa (y)$ is finite: $x_ i$ can be represented by a morphism $\mathop{\mathrm{Spec}}(k_ i) \to X_ y$ of finite type (by our definition of finite type points) and hence $\mathop{\mathrm{Spec}}(k_ i) \to y = \mathop{\mathrm{Spec}}(\kappa (y))$ is of finite type (as a composition of finite type morphisms), hence $k_ i/\kappa (y)$ is finite (Morphisms, Lemma 29.16.1). $\square$

## Comments (1)

Comment #8715 by Torsten on

Is the condition that $Y$ is affine really needed?

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