**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 $x_1, \ldots , x_ n$. 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

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

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 $ < 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.
$\square$

## Comments (2)

Comment #3571 by Herman Rohrbach on

Comment #3695 by Johan on