The Stacks project

10.61 Applications of dimension theory

We can use the results on dimension to prove certain rings have infinite spectra and to produce more Jacobson rings.

Lemma 10.61.1. Let $R$ be a Noetherian local domain of dimension $\geq 2$. A nonempty open subset $U \subset \mathop{\mathrm{Spec}}(R)$ is infinite.

Proof. To get a contradiction, assume that $U \subset \mathop{\mathrm{Spec}}(R)$ is finite. In this case $(0) \in U$ and $\{ (0)\} $ is an open subset of $U$ (because the complement of $\{ (0)\} $ is the union of the closures of the other points). Thus we may assume $U = \{ (0)\} $. Let $\mathfrak m \subset R$ be the maximal ideal. We can find an $x \in \mathfrak m$, $x \not= 0$ such that $V(x) \cup U = \mathop{\mathrm{Spec}}(R)$. In other words we see that $D(x) = \{ (0)\} $. In particular we see that $\dim (R/xR) = \dim (R) - 1 \geq 1$, see Lemma 10.60.13. Let $\overline{y}_2, \ldots , \overline{y}_{\dim (R)} \in R/xR$ generate an ideal of definition of $R/xR$, see Proposition 10.60.9. Choose lifts $y_2, \ldots , y_{\dim (R)} \in R$, so that $x, y_2, \ldots , y_{\dim (R)}$ generate an ideal of definition in $R$. This implies that $\dim (R/(y_2)) = \dim (R) - 1$ and $\dim (R/(y_2, x)) = \dim (R) - 2$, see Lemma 10.60.14. Hence there exists a prime $\mathfrak p$ containing $y_2$ but not $x$. This contradicts the fact that $D(x) = \{ (0)\} $. $\square$

The rings $k[[t]]$ where $k$ is a field, or the ring of $p$-adic numbers are Noetherian rings of dimension $1$ with finitely many primes. This is the maximum dimension for which this can happen.

Lemma 10.61.2. A Noetherian ring with finitely many primes has dimension $\leq 1$.

Proof. Let $R$ be a Noetherian ring with finitely many primes. If $R$ is a local domain, then the lemma follows from Lemma 10.61.1. If $R$ is a domain, then $R_\mathfrak m$ has dimension $\leq 1$ for all maximal ideals $\mathfrak m$ by the local case. Hence $\dim (R) \leq 1$ by Lemma 10.60.4. If $R$ is general, then $\dim (R/\mathfrak q) \leq 1$ for every minimal prime $\mathfrak q$ of $R$. Since every prime contains a minimal prime (Lemma 10.17.2), this implies $\dim (R) \leq 1$. $\square$

Lemma 10.61.3. Let $S$ be a nonzero finite type algebra over a field $k$. Then $\dim (S) = 0$ if and only if $S$ has finitely many primes.

Proof. Recall that $\mathop{\mathrm{Spec}}(S)$ is sober, Noetherian, and Jacobson, see Lemmas 10.26.2, 10.31.5, 10.35.2, and 10.35.4. If it has dimension $0$, then every point defines an irreducible component and there are only a finite number of irreducible components (Topology, Lemma 5.9.2). Conversely, if $\mathop{\mathrm{Spec}}(S)$ is finite, then it is discrete by Topology, Lemma 5.18.6 and hence the dimension is $0$. $\square$

Lemma 10.61.4. Noetherian Jacobson rings.

  1. Any Noetherian domain $R$ of dimension $1$ with infinitely many primes is Jacobson.

  2. Any Noetherian ring such that every prime $\mathfrak p$ is either maximal or contained in infinitely many prime ideals is Jacobson.

Proof. Part (1) is a reformulation of Lemma 10.35.6.

Let $R$ be a Noetherian ring such that every non-maximal prime $\mathfrak p$ is contained in infinitely many prime ideals. Assume $\mathop{\mathrm{Spec}}(R)$ is not Jacobson to get a contradiction. By Lemmas 10.26.1 and 10.31.5 we see that $\mathop{\mathrm{Spec}}(R)$ is a sober, Noetherian topological space. By Topology, Lemma 5.18.3 we see that there exists a non-maximal ideal $\mathfrak p \subset R$ such that $\{ \mathfrak p\} $ is a locally closed subset of $\mathop{\mathrm{Spec}}(R)$. In other words, $\mathfrak p$ is not maximal and $\{ \mathfrak p\} $ is an open subset of $V(\mathfrak p)$. Consider a prime $\mathfrak q \subset R$ with $\mathfrak p \subset \mathfrak q$. Recall that the topology on the spectrum of $(R/\mathfrak p)_{\mathfrak q} = R_{\mathfrak q}/\mathfrak pR_{\mathfrak q}$ is induced from that of $\mathop{\mathrm{Spec}}(R)$, see Lemmas 10.17.5 and 10.17.7. Hence we see that $\{ (0)\} $ is a locally closed subset of $\mathop{\mathrm{Spec}}((R/\mathfrak p)_{\mathfrak q})$. By Lemma 10.61.1 we conclude that $\dim ((R/\mathfrak p)_{\mathfrak q}) = 1$. Since this holds for every $\mathfrak q \supset \mathfrak p$ we conclude that $\dim (R/\mathfrak p) = 1$. At this point we use the assumption that $\mathfrak p$ is contained in infinitely many primes to see that $\mathop{\mathrm{Spec}}(R/\mathfrak p)$ is infinite. Hence by part (1) of the lemma we see that $V(\mathfrak p) \cong \mathop{\mathrm{Spec}}(R/\mathfrak p)$ is the closure of its closed points. This is the desired contradiction since it means that $\{ \mathfrak p\} \subset V(\mathfrak p)$ cannot be open. $\square$

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