The Stacks project

Proof. This is so because we can always add a maximal ideal at the end of a chain of prime ideals. $\square$

Proof. Assume $R$ is a Noetherian ring of dimension $0$. By Lemma 10.31.5 the space $\mathop{\mathrm{Spec}}(R)$ is Noetherian. By Topology, Lemma 5.9.2 we see that $\mathop{\mathrm{Spec}}(R)$ has finitely many irreducible components, say $\mathop{\mathrm{Spec}}(R) = Z_1 \cup \ldots \cup Z_ r$. According to Lemma 10.26.1 each $Z_ i = V(\mathfrak p_ i)$ with $\mathfrak p_ i$ a minimal ideal. Since the dimension is $0$ these $\mathfrak p_ i$ are also maximal. Thus $\mathop{\mathrm{Spec}}(R)$ is the discrete topological space with elements $\mathfrak p_ i$. All elements $f$ of the Jacobson radical $\bigcap \mathfrak p_ i$ are nilpotent since otherwise $R_ f$ would not be the zero ring and we would have another prime. By Lemma 10.53.5 $R$ is equal to $\prod R_{\mathfrak p_ i}$. Since $R_{\mathfrak p_ i}$ is also Noetherian and dimension $0$, the previous arguments show that its radical $\mathfrak p_ iR_{\mathfrak p_ i}$ is locally nilpotent. Lemma 10.32.5 gives $\mathfrak p_ i^ nR_{\mathfrak p_ i} = 0$ for some $n \geq 1$. By Lemma 10.52.8 we conclude that $R_{\mathfrak p_ i}$ has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 10.53.6.

If $R$ is an Artinian ring then by Lemma 10.53.6 it is Noetherian. All of its primes are maximal by a combination of Lemmas 10.53.3, 10.53.4 and 10.53.5. $\square$

Proof. This is because $d(R) = 0$ if and only if $R$ has finite length as an $R$-module. See Lemma 10.53.6. $\square$

Proof. This is a combination of Lemmas 10.53.5, 10.53.6, 10.60.5, and 10.60.6. $\square$

Proof. First, assume that $\dim (R) = 1$. Let $\mathfrak p_ i$ be the minimal primes of $R$. Because the dimension is $1$ the only other prime of $R$ is $\mathfrak m$. According to Lemma 10.31.6 there are finitely many. Hence we can find $x \in \mathfrak m$, $x \not\in \mathfrak p_ i$, see Lemma 10.15.2. Thus the only prime containing $x$ is $\mathfrak m$ and hence (3).

If (3) then $\mathfrak m = \sqrt{(x)}$ by Lemma 10.17.2, and hence (4). The converse is clear as well. The equivalence of (4) and (5) follows from directly the definitions.

Assume (5). Let $I = (x)$ be an ideal of definition. Note that $I^ n/I^{n + 1}$ is a quotient of $R/I$ via multiplication by $x^ n$ and hence $\text{length}_ R(I^ n/I^{n + 1})$ is bounded. Thus $d(R) = 0$ or $d(R) = 1$, but $d(R) = 0$ is excluded by the assumption that $0$ is not an ideal of definition.

Assume (2). To get a contradiction, assume there exist primes $\mathfrak p \subset \mathfrak q \subset \mathfrak m$, with both inclusions strict. Pick some ideal of definition $I \subset R$. We will repeatedly use Lemma 10.59.10. First of all it implies, via the exact sequence $0 \to \mathfrak p \to R \to R/\mathfrak p \to 0$, that $d(R/\mathfrak p) \leq 1$. But it clearly cannot be zero. Pick $x\in \mathfrak q$, $x\not\in \mathfrak p$. Consider the short exact sequence

This implies that $\chi _{I, R/\mathfrak p} - \chi _{I, R/\mathfrak p} - \chi _{I, R/(xR + \mathfrak p)} = - \chi _{I, R/(xR + \mathfrak p)}$ has degree $ < 1$. In other words, $d(R/(xR + \mathfrak p)) = 0$, and hence $\dim (R/(xR + \mathfrak p)) = 0$, by Lemma 10.60.6. But $R/(xR + \mathfrak p)$ has the distinct primes $\mathfrak q/(xR + \mathfrak p)$ and $\mathfrak m/(xR + \mathfrak p)$ which gives the desired contradiction. $\square$

Proof. This proof is really just the same as the proof of Lemma 10.60.8. We will prove the proposition by induction on $d$. By Lemmas 10.60.6 and 10.60.8 we may assume that $d > 1$. Denote the minimal number of generators for an ideal of definition of $R$ by $d'(R)$. We will prove the inequalities $\dim (R) \geq d'(R) \geq d(R) \geq \dim (R)$, and hence they are all equal.

First, assume that $\dim (R) = d$. Let $\mathfrak p_ i$ be the minimal primes of $R$. According to Lemma 10.31.6 there are finitely many. Hence we can find $x \in \mathfrak m$, $x \not\in \mathfrak p_ i$, see Lemma 10.15.2. Note that every maximal chain of primes starts with some $\mathfrak p_ i$, hence the dimension of $R/xR$ is at most $d-1$. By induction there are $x_2, \ldots , x_ d$ which generate an ideal of definition in $R/xR$. Hence $R$ has an ideal of definition generated by (at most) $d$ elements.

Assume $d'(R) = d$. Let $I = (x_1, \ldots , x_ d)$ be an ideal of definition. Note that $I^ n/I^{n + 1}$ is a quotient of a direct sum of $\binom {d + n - 1}{d - 1}$ copies $R/I$ via multiplication by all degree $n$ monomials in $x_1, \ldots , x_ d$. Hence $\text{length}_ R(I^ n/I^{n + 1})$ is bounded by a polynomial of degree $d-1$. Thus $d(R) \leq d$.

Assume $d(R) = d$. Consider a chain of primes $\mathfrak p \subset \mathfrak q \subset \mathfrak q_2 \subset \ldots \subset \mathfrak q_ e = \mathfrak m$, with all inclusions strict, and $e \geq 2$. Pick some ideal of definition $I \subset R$. We will repeatedly use Lemma 10.59.10. First of all it implies, via the exact sequence $0 \to \mathfrak p \to R \to R/\mathfrak p \to 0$, that $d(R/\mathfrak p) \leq d$. But it clearly cannot be zero. Pick $x\in \mathfrak q$, $x\not\in \mathfrak p$. Consider the short exact sequence

This implies that $\chi _{I, R/\mathfrak p} - \chi _{I, R/\mathfrak p} - \chi _{I, R/(xR + \mathfrak p)} = - \chi _{I, R/(xR + \mathfrak p)}$ has degree $ < d$. In other words, $d(R/(xR + \mathfrak p)) \leq d - 1$, and hence $\dim (R/(xR + \mathfrak p)) \leq d - 1$, by induction. Now $R/(xR + \mathfrak p)$ has the chain of prime ideals $\mathfrak q/(xR + \mathfrak p) \subset \mathfrak q_2/(xR + \mathfrak p) \subset \ldots \subset \mathfrak q_ e/(xR + \mathfrak p)$ which gives $e - 1 \leq d - 1$. Since we started with an arbitrary chain of primes this proves that $\dim (R) \leq d(R)$.

Reading back the reader will see we proved the circular inequalities as desired. $\square$

Proof. Proof of (1). If $\mathfrak p$ is minimal over $x$, then the only prime ideal of $R_\mathfrak p$ containing $x$ is the maximal ideal $\mathfrak p R_\mathfrak p$. This is true because the primes of $R_\mathfrak p$ correspond $1$-to-$1$ with the primes of $R$ contained in $\mathfrak p$, see Lemma 10.17.5. Hence Lemma 10.60.8 shows $\dim (R_\mathfrak p) = 1$ if $x$ is not nilpotent in $R_\mathfrak p$. Of course, if $x$ is nilpotent in $R_\mathfrak p$ the argument gives that $\mathfrak pR_\mathfrak p$ is the only prime ideal and we see that the height is $0$.

Proof of (2). By part (1) we see that $\mathfrak q/\mathfrak p$ is a prime of height $1$ or $0$ in $R/\mathfrak p$. This immediately implies there cannot be a prime strictly between $\mathfrak p$ and $\mathfrak q$. $\square$

Proof. Proof of (1). If $\mathfrak p$ is minimal over $f_1, \ldots , f_ r$, then the only prime ideal of $R_\mathfrak p$ containing $f_1, \ldots , f_ r$ is the maximal ideal $\mathfrak p R_\mathfrak p$. This is true because the primes of $R_\mathfrak p$ correspond $1$-to-$1$ with the primes of $R$ contained in $\mathfrak p$, see Lemma 10.17.5. Hence Proposition 10.60.9 shows $\dim (R_\mathfrak p) \leq r$.

Proof of (2). By part (1) we see that $\mathfrak q/\mathfrak p$ is a prime of height $\leq r$. This immediately implies the statement about chains of primes between $\mathfrak p$ and $\mathfrak q$. $\square$

Proof. If $x_1, \ldots , x_{\dim R/xR} \in R$ map to elements of $R/xR$ which generate an ideal of definition for $R/xR$, then $x, x_1, \ldots , x_{\dim R/xR}$ generate an ideal of definition for $R$. Hence the inequality by Proposition 10.60.9. On the other hand, if $x$ is not contained in any minimal prime of $R$, then the chains of primes in $R/xR$ all give rise to chains in $R$ which are at least one step away from being maximal. $\square$

Proof. Follows either from the proof of Proposition 10.60.9, or by using induction on $d$ and Lemma 10.60.13. $\square$