# The Stacks Project

## Tag: 02GL

This tag has label morphisms-lemma-etale-over-field and it points to

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Lemma 25.37.7. Fibres of étale morphisms.
1. Let $X$ be a scheme over a field $k$. The structure morphism $X \to \mathop{\rm Spec}(k)$ is étale if and only if $X$ is a disjoint union of spectra of finite separable field extensions of $k$.
2. If $f : X \to S$ is an étale morphism, then for every $s \in S$ the fibre $X_s$ is a disjoint union of spectra of finite separable field extensions of $\kappa(s)$.

Proof. You can deduce this from Lemma 25.36.11 via Lemma 25.37.5 above. Here is a direct proof.

We will use Algebra, Lemma 9.135.4. Hence it is clear that if $X$ is a disjoint union of spectra of finite separable field extensions of $k$ then $X \to \mathop{\rm Spec}(k)$ is étale. Conversely, suppose that $X \to \mathop{\rm Spec}(k)$ is étale. Then for any affine open $U \subset X$ we see that $U$ is a finite disjoint union of spectra of finite separable field extensions of $k$. Hence all points of $X$ are closed points (see Lemma 25.21.2 for example). Thus $X$ is a discrete space and we win. $\square$

\begin{lemma}
\label{lemma-etale-over-field}
Fibres of \'etale morphisms.
\begin{enumerate}
\item Let $X$ be a scheme over a field $k$.
The structure morphism $X \to \Spec(k)$ is \'etale if
and only if $X$ is a disjoint union of spectra of finite separable
field extensions of $k$.
\item If $f : X \to S$ is an \'etale morphism, then for every $s \in S$ the
fibre $X_s$ is a disjoint union of spectra of finite separable field
extensions of $\kappa(s)$.
\end{enumerate}
\end{lemma}

\begin{proof}
You can deduce this from Lemma \ref{lemma-unramified-over-field}
via Lemma \ref{lemma-etale-smooth-unramified} above.
Here is a direct proof.

\medskip\noindent
We will use Algebra, Lemma \ref{algebra-lemma-etale-over-field}.
Hence it is clear that if $X$ is a disjoint union of spectra of finite
separable field extensions of $k$ then $X \to \Spec(k)$ is \'etale.
Conversely, suppose that $X \to \Spec(k)$ is \'etale. Then for any affine
open $U \subset X$ we see that $U$ is a finite disjoint union of spectra
of finite separable field extensions of $k$. Hence all points of $X$
are closed points (see
Lemma \ref{lemma-algebraic-residue-field-extension-closed-point-fibre}
for example). Thus $X$ is a discrete space and we win.
\end{proof}


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