## Tag `00PK`

Chapter 10: Commutative Algebra > Section 10.121: Quasi-finite maps

**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 $F = \mathop{\rm Spec}(S \otimes_R \kappa(\mathfrak p))$ be the fibre of $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$, see Remark 10.16.8. 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 $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 28877–28901 (see updates for more information).

```
\begin{lemma}
\label{lemma-isolated-point-fibre}
\begin{slogan}
Equivalent conditions for isolated points in fibres
\end{slogan}
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 $F = \Spec(S \otimes_R \kappa(\mathfrak p))$
be the fibre of $\Spec(S) \to \Spec(R)$, see
Remark \ref{remark-fundamental-diagram}.
Denote $\overline{\mathfrak q} \in F$ the point corresponding to
$\mathfrak q$. The following are equivalent
\begin{enumerate}
\item $\overline{\mathfrak q}$ is an isolated point of $F$,
\item $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over
$\kappa(\mathfrak p)$,
\item there exists a $g \in S$, $g \not \in \mathfrak q$ such that
the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
\item $\dim_{\overline{\mathfrak q}}(F) = 0$,
\item $\overline{\mathfrak q}$ is a closed point of $F$ and
$\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
\item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
is finite and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-isolated-point}
for the $\kappa(\mathfrak p)$-algebra $S \otimes_R \kappa(\mathfrak p)$
and the prime $\overline{\mathfrak q}$, hence they are equivalent.
\end{proof}
```

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