
Equivalent conditions for isolated points in fibres

Lemma 10.121.2. Let $R \to S$ be a ring map of finite type. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $F = \mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ be the fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$, see Remark 10.16.8. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent

1. $\overline{\mathfrak q}$ is an isolated point of $F$,

2. $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa (\mathfrak p)$,

3. there exists a $g \in S$, $g \not\in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,

4. $\dim _{\overline{\mathfrak q}}(F) = 0$,

5. $\overline{\mathfrak q}$ is a closed point of $F$ and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and

6. the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.

Proof. Note that $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = (S \otimes _ R \kappa (\mathfrak p))_{\overline{\mathfrak q}}$. Moreover $S \otimes _ R \kappa (\mathfrak p)$ is of finite type over $\kappa (\mathfrak p)$. The conditions correspond exactly to the conditions of Lemma 10.121.1 for the $\kappa (\mathfrak p)$-algebra $S \otimes _ R \kappa (\mathfrak p)$ and the prime $\overline{\mathfrak q}$, hence they are equivalent. $\square$

Comment #1654 by Matthieu Romagny on

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