## Tag `08AL`

Chapter 59: Decent Algebraic Spaces > Section 59.13: Decent spaces

Lemma 59.13.5. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram $$ \xymatrix{ \mathop{\mathrm{Spec}}(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S } $$ Assume that the image point $s \in S$ of $\mathop{\mathrm{Spec}}(k) \to S$ is a closed point and that $\kappa(s) \subset k$ is finite. Then $\mathop{\mathrm{Spec}}(k) \to X$ is finite morphism. If $\kappa(s) = k$ then $\mathop{\mathrm{Spec}}(k) \to X$ is closed immersion.

Proof.By Lemma 59.13.4 the image point $x \in |X|$ is closed. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{x\}$ (Properties of Spaces, Lemma 57.12.4). Note that $Z$ is a decent algebraic space by Lemma 59.6.5. By Lemma 59.13.2 we see that $Z = \mathop{\mathrm{Spec}}(k')$ for some field $k'$. Of course $k \supset k' \supset \kappa(s)$. Then $\mathop{\mathrm{Spec}}(k) \to Z$ is a finite morphism of schemes and $Z \to X$ is a finite morphism as it is a closed immersion. Hence $\mathop{\mathrm{Spec}}(k) \to X$ is finite (Morphisms of Spaces, Lemma 58.45.4). If $k = \kappa(s)$, then $\mathop{\mathrm{Spec}}(k) = Z$ and $\mathop{\mathrm{Spec}}(k) \to X$ is a closed immersion. $\square$

The code snippet corresponding to this tag is a part of the file `decent-spaces.tex` and is located in lines 2843–2857 (see updates for more information).

```
\begin{lemma}
\label{lemma-finite-residue-field-extension-finite}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Consider a commutative diagram
$$
\xymatrix{
\Spec(k) \ar[rr] \ar[rd] & & X \ar[ld] \\
& S
}
$$
Assume that the image point $s \in S$ of $\Spec(k) \to S$ is
a closed point and that $\kappa(s) \subset k$ is finite.
Then $\Spec(k) \to X$ is finite morphism. If $\kappa(s) = k$
then $\Spec(k) \to X$ is closed immersion.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-algebraic-residue-field-extension-closed-point}
the image point $x \in |X|$ is closed. Let $Z \subset X$ be the
reduced closed subspace with $|Z| = \{x\}$ (Properties of Spaces,
Lemma \ref{spaces-properties-lemma-reduced-closed-subspace}).
Note that $Z$ is a decent algebraic space by
Lemma \ref{lemma-representable-named-properties}.
By Lemma \ref{lemma-when-field} we see that $Z = \Spec(k')$
for some field $k'$. Of course $k \supset k' \supset \kappa(s)$.
Then $\Spec(k) \to Z$ is a finite morphism of schemes
and $Z \to X$ is a finite morphism as it is a closed immersion.
Hence $\Spec(k) \to X$ is finite (Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-composition-integral}).
If $k = \kappa(s)$, then $\Spec(k) = Z$ and $\Spec(k) \to X$
is a closed immersion.
\end{proof}
```

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