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 (x)/\kappa (s) 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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