The Stacks project

Lemma 29.20.2. 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

  1. $x$ is a closed point of its fibre, and

  2. 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 26.18.5 to prove the first statement we may replace $X$ by $X_ s$ and $S$ by $\mathop{\mathrm{Spec}}(\kappa (s))$. Thus we may assume that $S = \mathop{\mathrm{Spec}}(k)$ is the spectrum of a field. In this case, let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be any affine open containing $x$. The point $x$ corresponds to a prime ideal $\mathfrak q \subset A$ such that $\kappa (\mathfrak q)/k$ is an algebraic field extension. By Algebra, Lemma 10.35.9 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. $\square$

Comments (2)

Comment #2265 by yangan on

In the second line, the inclusion is reversed, should be k(s)\subset k(x).

Comment #2294 by on

Actually, I have been trying to use consistently the notation for field extensions. So I fixed it that way. Thanks!

Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01TE. Beware of the difference between the letter 'O' and the digit '0'.