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 $k \subset \kappa (\mathfrak q)$ 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. $\square$

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 $K/L$ for field extensions. So I fixed it that way. Thanks!

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).