**Proof.**
Let $U$ be an affine scheme and let $a : U \to X$ be an étale morphism. We have to show that the fibres of $a$ are universally bounded. By assumption (1) the scheme $U$ has finitely many irreducible components. Let $u_1, \ldots , u_ n \in U$ be the generic points of these irreducible components. Let $\{ x_1, \ldots , x_ m\} \subset |X|$ be the image of $\{ u_1, \ldots , u_ n\} $. Each $x_ j$ is a point of codimension $0$. By assumption (2) we may choose a monomorphism $\mathop{\mathrm{Spec}}(k_ j) \to X$ representing $x_ j$. By Properties of Spaces, Lemma 65.11.1 we have

\[ U \times _ X \mathop{\mathrm{Spec}}(k_ j) = \coprod \nolimits _{a(u_ i) = x_ j} \mathop{\mathrm{Spec}}(\kappa (u_ i)) \]

This is a scheme finite over $\mathop{\mathrm{Spec}}(k_ j)$ of degree $d_ j = \sum _{a(u_ i) = x_ j} [\kappa (u_ i) : k_ j]$. Set $n = \max d_ j$.

Observe that $a$ is separated (Properties of Spaces, Lemma 65.6.4). Consider the stratification

\[ X = X_0 \supset X_1 \supset X_2 \supset \ldots \]

associated to $U \to X$ in Lemma 67.8.2. By our choice of $n$ above we conclude that $X_{n + 1}$ is empty. Namely, if not, then $a^{-1}(X_{n + 1})$ is a nonempty open of $U$ and hence would contain one of the $x_ i$. This would mean that $X_{n + 1}$ contains $x_ j = a(u_ i)$ which is impossible. Hence we see that the fibres of $U \to X$ are universally bounded (in fact by the integer $n$).
$\square$

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