## Tag `01TE`

Chapter 28: Morphisms of Schemes > Section 28.19: Quasi-finite morphisms

Lemma 28.19.2. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $\kappa(x)/\kappa(s)$ is an algebraic field extension, then

- $x$ is a closed point of its fibre, and
- if in addition $s$ is a closed point of $S$, then $x$ is a closed point of $X$.

Proof.The second statement follows from the first by elementary topology. According to Schemes, Lemma 25.18.5 to prove the first statement we may replace $X$ by $X_s$ and $S$ by $\mathop{\rm Spec}(\kappa(s))$. Thus we may assume that $S = \mathop{\rm Spec}(k)$ is the spectrum of a field. In this case, let $\mathop{\rm Spec}(A) = U \subset X$ be any affine open containing $x$. The point $x$ corresponds to a prime ideal $\mathfrak q \subset A$ such that $k \subset \kappa(\mathfrak q)$ is an algebraic field extension. By Algebra, Lemma 10.34.9 we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a closed point. Since the affine opens form a basis of the topology of $X$ we conclude that $\{x\}$ is closed. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 3153–3164 (see updates for more information).

```
\begin{lemma}
\label{lemma-algebraic-residue-field-extension-closed-point-fibre}
Let $f : X \to S$ be a morphism of schemes.
Let $x \in X$ be a point. Set $s = f(x)$.
If $\kappa(x)/\kappa(s)$
is an algebraic field extension, then
\begin{enumerate}
\item $x$ is a closed point of its fibre, and
\item if in addition $s$ is a closed point of $S$, then
$x$ is a closed point of $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
The second statement follows from the first by elementary topology.
According to Schemes, Lemma \ref{schemes-lemma-fibre-topological}
to prove the first statement
we may replace $X$ by $X_s$ and $S$ by $\Spec(\kappa(s))$.
Thus we may assume that $S = \Spec(k)$ is the spectrum of a field.
In this case, let $\Spec(A) = U \subset X$ be any affine open
containing $x$. The point $x$ corresponds to a prime ideal
$\mathfrak q \subset A$ such that $k \subset \kappa(\mathfrak q)$
is an algebraic field extension. By
Algebra, Lemma \ref{algebra-lemma-finite-residue-extension-closed}
we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a
closed point. Since the affine opens form
a basis of the topology of $X$ we conclude that $\{x\}$ is closed.
\end{proof}
```

## Comments (2)

## Add a comment on tag `01TE`

Your email address will not be published. Required fields are marked.

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.