The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Proposition 10.59.8. Let $R$ be a local Noetherian ring. Let $d \geq 0$ be an integer. The following are equivalent:

  1. $\dim (R) = d$,

  2. $d(R) = d$,

  3. there exists an ideal of definition generated by $d$ elements, and no ideal of definition is generated by fewer than $d$ elements.

Proof. This proof is really just the same as the proof of Lemma 10.59.7. We will prove the proposition by induction on $d$. By Lemmas 10.59.5 and 10.59.7 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 10.30.6 there are finitely many. Hence we can find $x \in \mathfrak m$, $x \not\in \mathfrak p_ i$, see Lemma 10.14.2. 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 $\mathfrak p \subset \mathfrak q \subset \mathfrak q_2 \subset \ldots \subset \mathfrak p_ e = \mathfrak m$, with all inclusions strict, and $e \geq 2$. Pick some ideal of definition $I \subset R$. We will repeatedly use Lemma 10.58.10. 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 (1)

Comment #3571 by Herman Rohrbach on

There is a typo in the opening line of the fourth paragraph of the proof. It reads "Consider a chain of primes ", but the final term of the chain should be .


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00KQ. Beware of the difference between the letter 'O' and the digit '0'.