Lemma 10.60.1. Let $R$ be a Noetherian local domain of dimension $\geq 2$. A nonempty open subset $U \subset \mathop{\mathrm{Spec}}(R)$ is infinite.

## 10.60 Applications of dimension theory

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

**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 10.59.12. Let $\overline{y}_2, \ldots , \overline{y}_{\dim (R)} \in R/xR$ generate an ideal of definition of $R/xR$, see Proposition 10.59.8. 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 10.59.13. 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 10.60.2. 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 10.60.1. 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 10.59.3. 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 10.16.2), this implies $\dim (R) \leq 1$.
$\square$

Lemma 10.60.3. 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 10.25.2, 10.30.5, 10.34.2, and 10.34.4. 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 10.60.4. Noetherian Jacobson rings.

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

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 10.34.6.

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 10.25.1 and 10.30.5 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 10.16.5 and 10.16.7. Hence we see that $\{ (0)\} $ is a locally closed subset of $\mathop{\mathrm{Spec}}((R/\mathfrak p)_{\mathfrak q})$. By Lemma 10.60.1 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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