The Stacks project

Lemma 10.122.1. 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 $\kappa (\mathfrak q)/k$ 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 10.35.2 and 10.35.4) we see that $\mathfrak q$ is a closed point. Hence $\{ \mathfrak q\} $ is open and closed in $\mathop{\mathrm{Spec}}(S)$. By Lemmas 10.21.3 and 10.24.3 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 10.115.4. We have proved (1) implies (2).

Suppose $S_{\mathfrak q}$ is finite over $k$. Then $S_{\mathfrak q}$ is Artinian local, see Lemma 10.53.2. So $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) = \{ \mathfrak qS_{\mathfrak q}\} $ by Lemma 10.53.6. 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 10.60.7. 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 10.114.6 gives the implication (5) $\Rightarrow $ (4). The implication (4) $\Rightarrow $ (6) follows from Lemma 10.116.3. The implication (6) $\Rightarrow $ (5) follows from Lemma 10.35.9. 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"

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