Lemma 29.20.3. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. Assume $f$ is locally of finite type. Then $x$ is a closed point of its fibre if and only if $\kappa (s) \subset \kappa (x)$ is a finite field extension.

Proof. If the extension is finite, then $x$ is a closed point of the fibre by Lemma 29.20.2 above. For the converse, assume that $x$ is a closed point of its fibre. Choose affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ such that $f(U) \subset V$. By Lemma 29.15.2 the ring map $R \to A$ is of finite type. Let $\mathfrak q \subset A$, resp. $\mathfrak p \subset R$ be the prime ideal corresponding to $x$, resp. $s$. Consider the fibre ring $\overline{A} = A \otimes _ R \kappa (\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. The assumption that $x$ is a closed point of its fibre implies that $\overline{\mathfrak q}$ is a maximal ideal of $\overline{A}$. Since $\overline{A}$ is an algebra of finite type over the field $\kappa (\mathfrak p)$ we see by the Hilbert Nullstellensatz, see Algebra, Theorem 10.34.1, that $\kappa (\overline{\mathfrak q})$ is a finite extension of $\kappa (\mathfrak p)$. Since $\kappa (s) = \kappa (\mathfrak p)$ and $\kappa (x) = \kappa (\mathfrak q) = \kappa (\overline{\mathfrak q})$ we win. $\square$

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