Lemma 10.122.2 (00PK)—The Stacks project

Lemma 10.122.2. 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 10.18.5. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent

$\overline{\mathfrak q}$ is an isolated point of $F$,
$S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa (\mathfrak p)$,
there exists a $g \in S$, $g \not\in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
$\dim _{\overline{\mathfrak q}}(F) = 0$,
$\overline{\mathfrak q}$ is a closed point of $F$ and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ 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 10.122.1 for the $\kappa (\mathfrak p)$-algebra $S \otimes _ R \kappa (\mathfrak p)$ and the prime $\overline{\mathfrak q}$, hence they are equivalent. $\square$

Comments (3)

Comment #1654 by Matthieu Romagny on September 24, 2015 at 21:11

Suggested slogan: Equivalent conditions for isolated points in fibres

Comment #4629 by Jian Qiu on October 26, 2019 at 13:50

can item 2 be strenghtened to: $S_{\mathfrak{q}}/\mathfrak{p}S_{\mathfrak{q}}$ is module finite over $R_{\mathfrak{p}}$ (i.e. not just over the residue field), by Nakayama?

Comment #4630 by Johan on October 26, 2019 at 16:27

Yes, but I'm not sure that is helpful. Namely, the quotient $S_\mathfrak q/\mathfrak p S_\mathfrak q$ is a module over $R_\mathfrak p$ annihilated by $\mathfrak p$ . So it is really a module over $\kappa(\mathfrak p)$ .

If you meant to ask "strengthened to: $S_\mathfrak q$ finite (as a module) over $R_\mathfrak p$ " then the answer would have been no.

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