**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

\[ 0 \to R/\mathfrak p \to R/\mathfrak p \to R/(xR + \mathfrak p) \to 0. \]

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.
$\square$

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