The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.121.1. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $\mathfrak q$ be a prime of $S$. The following are equivalent:

  1. $\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$,

  2. $S_{\mathfrak q}$ is finite over $k$,

  3. there exists a $g \in S$, $g \not\in \mathfrak q$ such that $D(g) = \{ \mathfrak q \} $,

  4. $\dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S) = 0$,

  5. $\mathfrak q$ is a closed point of $\mathop{\mathrm{Spec}}(S)$ and $\dim (S_{\mathfrak q}) = 0$, and

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

In this case $S = S_{\mathfrak q} \times S'$ for some finite type $k$-algebra $S'$. Also, the element $g$ as in (3) has the property $S_{\mathfrak q} = S_ g$.

Proof. Suppose $\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$, i.e., $\{ \mathfrak q\} $ is open in $\mathop{\mathrm{Spec}}(S)$. Because $\mathop{\mathrm{Spec}}(S)$ is a Jacobson space (see Lemmas 10.34.2 and 10.34.4) we see that $\mathfrak q$ is a closed point. Hence $\{ \mathfrak q\} $ is open and closed in $\mathop{\mathrm{Spec}}(S)$. By Lemmas 10.20.3 and 10.23.3 we may write $S = S_1 \times S_2$ with $\mathfrak q$ corresponding to the only point $\mathop{\mathrm{Spec}}(S_1)$. Hence $S_1 = S_{\mathfrak q}$ is a zero dimensional ring of finite type over $k$. Hence it is finite over $k$ for example by Lemma 10.114.4. We have proved (1) implies (2).

Suppose $S_{\mathfrak q}$ is finite over $k$. Then $S_{\mathfrak q}$ is Artinian local, see Lemma 10.52.2. So $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) = \{ \mathfrak qS_{\mathfrak q}\} $ by Lemma 10.52.6. Consider the exact sequence $0 \to K \to S \to S_{\mathfrak q} \to Q \to 0$. It is clear that $K_{\mathfrak q} = Q_{\mathfrak q} = 0$. Also, $K$ is a finite $S$-module as $S$ is Noetherian and $Q$ is a finite $S$-module since $S_{\mathfrak q}$ is finite over $k$. Hence there exists $g \in S$, $g \not\in \mathfrak q$ such that $K_ g = Q_ g = 0$. Thus $S_{\mathfrak q} = S_ g$ and $D(g) = \{ \mathfrak q \} $. We have proved that (2) implies (3).

Suppose $D(g) = \{ \mathfrak q \} $. Since $D(g)$ is open by construction of the topology on $\mathop{\mathrm{Spec}}(S)$ we see that $\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$. We have proved that (3) implies (1). In other words (1), (2) and (3) are equivalent.

Assume $\dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S) = 0$. This means that there is some open neighbourhood of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(S)$ which has dimension zero. Then there is an open neighbourhood of the form $D(g)$ which has dimension zero. Since $S_ g$ is Noetherian we conclude that $S_ g$ is Artinian and $D(g) = \mathop{\mathrm{Spec}}(S_ g)$ is a finite discrete set, see Proposition 10.59.6. Thus $\mathfrak q$ is an isolated point of $D(g)$ and, by the equivalence of (1) and (2) above applied to $\mathfrak qS_ g \subset S_ g$, we see that $S_{\mathfrak q} = (S_ g)_{\mathfrak qS_ g}$ is finite over $k$. Hence (4) implies (2). It is clear that (1) implies (4). Thus (1) – (4) are all equivalent.

Lemma 10.113.6 gives the implication (5) $\Rightarrow $ (4). The implication (4) $\Rightarrow $ (6) follows from Lemma 10.115.3. The implication (6) $\Rightarrow $ (5) follows from Lemma 10.34.9. At this point we know (1) – (6) are equivalent.

The two statements at the end of the lemma we saw during the course of the proof of the equivalence of (1), (2) and (3) above. $\square$


Comments (2)

Comment #2821 by Dario Weißmann on

Typo in the proof of (2) implies (3): " is a finite -modules" should read " is a finite -module"


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00PJ. Beware of the difference between the letter 'O' and the digit '0'.