# The Stacks Project

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### 9.59. Dimension

Definition 9.59.1. The Krull dimension of the ring $R$ is the Krull dimension of the topological space $\mathop{\rm Spec}(R)$, see Topology, Definition 5.7.1. In other words it is the supremum of the integers $n\geq 0$ such that there exists a chain of prime ideals of length $n$: $$\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n, \quad \mathfrak p_i \not= \mathfrak p_{i + 1}.$$

Definition 9.59.2. The height of a prime ideal $\mathfrak p$ of a ring $R$ is the dimension of the local ring $R_{\mathfrak p}$.

Lemma 9.59.3. The Krull dimension of $R$ is the supremum of the heights of its (maximal) primes.

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

Lemma 9.59.4. A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.

Proof. By Lemma 9.29.5 the space $\mathop{\rm Spec}(R)$ is Noetherian. By Topology, Lemma 5.6.2 we see that $\mathop{\rm Spec}(R)$ has finitely many irreducible components, say $\mathop{\rm Spec}(R) = Z_1 \cup \ldots Z_r$. According to Lemma 9.24.1, 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 $\mathop{\rm Spec}(R)$ is the discrete topological space with elements $\mathfrak p_i$. All elements $f$ of the radical $I = \cap \mathfrak p_i$ are nilpotent since otherwise $R_f$ would not be the zero ring and we would have another prime. Since $I$ is finitely generated we conclude that $I$ is nilpotent, Lemma 9.49.3. By Lemma 9.51.7 $R$ is the product of its local rings. By Lemma 9.50.8 each of these has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 9.51.8.

If $R$ is Artinian then by Lemma 9.51.8 it is Noetherian. All of its primes are maximal by a combination of Lemmas 9.51.3, 9.51.4 and 9.51.7. $\square$

In the following we will use the invariant $d(-)$ defined in Definition 9.58.7. Here is a warm up lemma.

Lemma 9.59.5. Let $R$ be a Noetherian local ring. Then $\dim(R) = 0 \Leftrightarrow d(R) = 0$.

Proof. This is because $d(R) = 0$ if and only if $R$ has finite length as an $R$-module. See Lemma 9.51.8. $\square$

Proposition 9.59.6. Let $R$ be a ring. The following are equivalent:

1. $R$ is Artinian,
2. $R$ is Noetherian and $\dim(R) = 0$,
3. $R$ has finite length as a module over itself,
4. $R$ is a finite product of Artinian local rings,
5. $R$ is Noetherian and $\mathop{\rm Spec}(R)$ is a finite discrete topological space,
6. $R$ is a finite product of Noetherian local rings of dimension $0$,
7. $R$ is a finite product of Noetherian local rings $R_i$ with $d(R_i) = 0$,
8. $R$ is a finite product of Noetherian local rings $R_i$ whose maximal ideals are nilpotent,
9. $R$ is Noetherian, has finitely many maximal ideals and its radical ideal is nilpotent, and
10. $R$ is Noetherian and there are no strict inclusions among its primes.

Proof. This is a combination of Lemmas 9.51.7, 9.51.8, 9.59.4, and 9.59.5. $\square$

Lemma 9.59.7. Let $R$ be a local Noetherian ring. The following are equivalent:

1. $\dim(R) = 1$,
2. $d(R) = 1$,
3. there exists an $x \in \mathfrak m$, $x$ not nilpotent such that $V(x) = \{\mathfrak m\}$,
4. there exists an $x \in \mathfrak m$, $x$ not nilpotent such that $\mathfrak m = \sqrt{(x)}$, and
5. there exists an ideal of definition generated by $1$ element, and no ideal of definition is generated by $0$ elements.

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 9.29.6 there are finitely many. Hence we can find $x \in \mathfrak m$, $x \not \in \mathfrak p_i$, see Lemma 9.14.3. Thus the only prime containing $x$ is $\mathfrak m$ and hence (3).

If (3) then $\mathfrak m = \sqrt{(x)}$ by Lemma 9.16.2, 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 9.58.9. 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 9.59.5. 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. $\square$

Proposition 9.59.8. Let $R$ be a local Noetherian ring. 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 9.59.7. We will prove the proposition by induction on $d$. By Lemmas 9.59.5 and 9.59.7 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 that 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 9.29.6 there are finitely many. Hence we can find $x \in \mathfrak m$, $x \not \in \mathfrak p_i$, see Lemma 9.14.3. 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 9.58.9. 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. $\square$

Let $(R, \mathfrak m)$ be a Noetherian local ring. From the above it is clear that $\mathfrak m$ cannot be generated by fewer than $\dim(R)$ variables. By Nakayama's Lemma 9.18.1 the minimal number of generators of $\mathfrak m$ equals $\dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2$. Hence we have the following fundamental inequality $$\dim(R) \leq \dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2.$$ It turns out that the rings where equality holds have a lot of good properties. They are called regular local rings.

Definition 9.59.9. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$.

1. A system of parameters of $R$ is a sequence of elements $x_1, \ldots, x_d \in \mathfrak m$ which generates an ideal of definition of $R$,
2. if there exist $x_1, \ldots, x_d \in \mathfrak m$ such that $\mathfrak m = (x_1, \ldots, x_d)$ then we call $R$ a regular local ring and $x_1, \ldots, x_d$ a regular system of parameters.

The folllowing two lemmas are clear from the proofs of the lemmas and proposition above, but we spell them out so we have convenient references.

Lemma 9.59.10. Let $R$ be a Noetherian ring.

1. Let $x\in R$, $\mathfrak p, \mathfrak q\in \mathop{\rm Spec}(R)$. Suppose that $\mathfrak p \subset (\mathfrak p, x) \subset \mathfrak q$ and $\mathfrak q$ minimal over $(\mathfrak p, x)$. Then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$.
2. If $x\in R$ and $x \in \mathfrak p$ is minimal over $(x)$ then the height of $\mathfrak p$ is $0$ or $1$.

Proof. Consider the situation of the first assertion. The primes containing $\mathfrak p$ and contained in $\mathfrak q$ correspond to primes of $R_{\mathfrak q}/\mathfrak pR_{\mathfrak q}$, and the primes containing $x$ correspond to the ones containing the image of $x$. Thus we may assume $R$ is a Noetherian local domain, $\mathfrak p = (0)$ and $\mathfrak q$ maximal. Now since $\sqrt{(x)}$ is the intersection of the prime ideals containing it, and since $\mathfrak q$ is the only prime containing $x$ by minimality, we see that $\sqrt{(x)} = \mathfrak q$. Hence Lemma 9.59.7 applies. The second assertion follows from the first. $\square$

Lemma 9.59.11. Suppose that $R$ is a Noetherian local ring and $x\in \mathfrak m$ an element of its maximal ideal. Then $\dim R \leq \dim R/xR + 1$. If $x$ is not contained in any of the minimal primes of $R$ then equality holds. (For example if $x$ is a nonzerodivisor.)

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 9.59.8. 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. $\square$

Lemma 9.59.12. Let $(R, \mathfrak m)$ be a Noetherian local ring. Suppose $x_1, \ldots, x_d \in \mathfrak m$ generate an ideal of definition and $d = \dim(R)$. Then $\dim(R/(x_1, \ldots, x_i)) = d - i$ for all $i = 1, \ldots, d$.

Proof. Clear from the proof of Proposition 9.59.8, or use induction on $d$ and Lemma 9.59.11 above. $\square$

\section{Dimension}
\label{section-dimension}

\begin{definition}
\label{definition-Krull}
The {\it Krull dimension} of the ring $R$ is the
Krull dimension of the topological space $\Spec(R)$, see
Topology, Definition \ref{topology-definition-Krull}.
In other words it is the supremum of the integers $n\geq 0$
such that there exists a chain of prime ideals of length $n$:
$$\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n, \quad \mathfrak p_i \not= \mathfrak p_{i + 1}.$$
\end{definition}

\begin{definition}
\label{definition-height}
The {\it height} of a prime ideal $\mathfrak p$ of
a ring $R$ is the dimension of the local ring $R_{\mathfrak p}$.
\end{definition}

\begin{lemma}
\label{lemma-dimension-height}
The Krull dimension of $R$ is the supremum of the
heights of its (maximal) primes.
\end{lemma}

\begin{proof}
This is so because we can always add a maximal ideal at the end of a chain
of prime ideals.
\end{proof}

\begin{lemma}
\label{lemma-Noetherian-dimension-0}
A Noetherian ring of dimension $0$ is Artinian.
Conversely, any Artinian ring is Noetherian of dimension zero.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-Noetherian-topology} the space $\Spec(R)$
is Noetherian. By Topology, Lemma \ref{topology-lemma-Noetherian} we see
that $\Spec(R)$ has finitely many irreducible
components, say $\Spec(R) = Z_1 \cup \ldots Z_r$.
According to Lemma \ref{lemma-irreducible}, 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 $\Spec(R)$
is the discrete topological space with elements $\mathfrak p_i$.
All elements $f$ of the radical $I = \cap \mathfrak p_i$
are nilpotent since otherwise $R_f$ would not be the zero ring
and we would have another prime. Since $I$ is finitely generated
we conclude that $I$ is nilpotent, Lemma \ref{lemma-Noetherian-power}.
By Lemma \ref{lemma-product-local} $R$ is the product of its
local rings. By Lemma \ref{lemma-length-finite} each of these
has finite length over $R$. Hence we conclude that $R$
is Artinian by Lemma \ref{lemma-artinian-finite-length}.

\medskip\noindent
If $R$ is Artinian then by Lemma \ref{lemma-artinian-finite-length}
it is Noetherian. All of its primes are maximal by a combination
of Lemmas \ref{lemma-artinian-finite-nr-max},
\end{proof}

\noindent
In the following we will use the invariant $d(-)$ defined
in Definition \ref{definition-d}. Here is a warm up lemma.

\begin{lemma}
\label{lemma-dimension-0-d-0}
Let $R$ be a Noetherian local ring.
Then $\dim(R) = 0 \Leftrightarrow d(R) = 0$.
\end{lemma}

\begin{proof}
This is because $d(R) = 0$ if and only if $R$ has finite
length as an $R$-module. See Lemma \ref{lemma-artinian-finite-length}.
\end{proof}

\begin{proposition}
\label{proposition-dimension-zero-ring}
Let $R$ be a ring. The following are equivalent:
\begin{enumerate}
\item $R$ is Artinian,
\item $R$ is Noetherian and $\dim(R) = 0$,
\item $R$ has finite length as a module over itself,
\item $R$ is a finite product of Artinian local rings,
\item $R$ is Noetherian and $\Spec(R)$ is a
finite discrete topological space,
\item $R$ is a finite product of Noetherian local rings
of dimension $0$,
\item $R$ is a finite product of Noetherian local rings
$R_i$ with $d(R_i) = 0$,
\item $R$ is a finite product of Noetherian local rings
$R_i$ whose maximal ideals are nilpotent,
\item $R$ is Noetherian, has finitely many maximal
ideals and its radical ideal is nilpotent, and
\item $R$ is Noetherian and there are no strict inclusions
among its primes.
\end{enumerate}
\end{proposition}

\begin{proof}
This is a combination of Lemmas
\ref{lemma-product-local},
\ref{lemma-artinian-finite-length},
\ref{lemma-Noetherian-dimension-0}, and
\ref{lemma-dimension-0-d-0}.
\end{proof}

\begin{lemma}
\label{lemma-height-1}
Let $R$ be a local Noetherian ring.
The following are equivalent:
\begin{enumerate}
\item
\label{item-dim-1}
$\dim(R) = 1$,
\item
\label{item-d-1}
$d(R) = 1$,
\item
\label{item-Vx}
there exists an $x \in \mathfrak m$, $x$ not nilpotent
such that $V(x) = \{\mathfrak m\}$,
\item
\label{item-x}
there exists an $x \in \mathfrak m$, $x$ not nilpotent
such that $\mathfrak m = \sqrt{(x)}$, and
\item
\label{item-ideal-1}
there exists an ideal of definition generated by $1$ element,
and no ideal of definition is generated by $0$ elements.
\end{enumerate}
\end{lemma}

\begin{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 \ref{lemma-Noetherian-irreducible-components}
there are finitely many. Hence we can find $x \in \mathfrak m$,
$x \not \in \mathfrak p_i$, see Lemma \ref{lemma-silly}.
Thus the only prime containing $x$ is $\mathfrak m$ and
hence (\ref{item-Vx}).

\medskip\noindent
If (\ref{item-Vx}) then $\mathfrak m = \sqrt{(x)}$ by
Lemma \ref{lemma-Zariski-topology}, and hence (\ref{item-x}).
The converse is clear as well.
The equivalence of (\ref{item-x}) and (\ref{item-ideal-1}) follows
from directly the definitions.

\medskip\noindent
Assume (\ref{item-ideal-1}).
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.

\medskip\noindent
Assume (\ref{item-d-1}). 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 \ref{lemma-hilbert-ses-chi}. 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 \ref{lemma-dimension-0-d-0}. 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.
\end{proof}

\begin{proposition}
\label{proposition-dimension}
Let $R$ be a local Noetherian ring.
The following are equivalent:
\begin{enumerate}
\item
\label{item-dim-d}
$\dim(R) = d$,
\item
\label{item-d-d}
$d(R) = d$,
\item
\label{item-ideal-d}
there exists an ideal of definition generated by $d$ elements,
and no ideal of definition is generated by fewer than $d$ elements.
\end{enumerate}
\end{proposition}

\begin{proof}
This proof is really just the same as the proof of Lemma
\ref{lemma-height-1}. We will prove the proposition by induction
on $d$. By Lemmas \ref{lemma-dimension-0-d-0} and \ref{lemma-height-1}
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 that the inequalities
$\dim(R) \geq d'(R) \geq d(R) \geq \dim(R)$,
and hence they are all equal.

\medskip\noindent
First, assume that $\dim(R) = d$.
Let $\mathfrak p_i$ be the minimal primes of $R$.
According to Lemma \ref{lemma-Noetherian-irreducible-components}
there are finitely many. Hence we can find $x \in \mathfrak m$,
$x \not \in \mathfrak p_i$, see Lemma \ref{lemma-silly}.
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.

\medskip\noindent
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 $d(R) \leq d$.

\medskip\noindent
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 \ref{lemma-hilbert-ses-chi}. 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 $\dim(R) \leq d(R)$.

\medskip\noindent
inequalities as desired.
\end{proof}

\noindent
Let $(R, \mathfrak m)$ be a Noetherian local ring.
From the above it is clear that $\mathfrak m$ cannot be
generated by fewer than $\dim(R)$ variables.
By Nakayama's Lemma \ref{lemma-NAK} the minimal number
of generators of $\mathfrak m$ equals $\dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2$. Hence we have the following
fundamental inequality
$$\dim(R) \leq \dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2.$$
It turns out that the rings where equality holds
have a lot of good properties. They are called
regular local rings.

\begin{definition}
\label{definition-regular-local}
Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$.
\begin{enumerate}
\item A {\it system of parameters of $R$} is a sequence of elements
$x_1, \ldots, x_d \in \mathfrak m$ which generates an ideal of
definition of $R$,
\item if there exist $x_1, \ldots, x_d \in \mathfrak m$
such that $\mathfrak m = (x_1, \ldots, x_d)$ then we call
$R$ a {\it regular local ring} and $x_1, \ldots, x_d$ a {\it regular
system of parameters}.
\end{enumerate}
\end{definition}

\noindent
The folllowing two lemmas are clear from the proofs of the
lemmas and proposition above, but we spell them out so we have
convenient references.

\begin{lemma}
\label{lemma-minimal-over-1}
Let $R$ be a Noetherian ring.
\begin{enumerate}
\item Let $x\in R$, $\mathfrak p, \mathfrak q\in \Spec(R)$.
Suppose that $\mathfrak p \subset (\mathfrak p, x) \subset \mathfrak q$ and $\mathfrak q$ minimal over $(\mathfrak p, x)$.
Then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$.
\item If $x\in R$ and $x \in \mathfrak p$ is minimal over $(x)$
then the height of $\mathfrak p$ is $0$ or $1$.
\end{enumerate}
\end{lemma}

\begin{proof}
Consider the situation of the first assertion.
The primes containing $\mathfrak p$ and contained
in $\mathfrak q$ correspond to primes of
$R_{\mathfrak q}/\mathfrak pR_{\mathfrak q}$, and
the primes containing $x$ correspond to the ones containing
the image of $x$. Thus we may assume $R$ is a Noetherian local domain,
$\mathfrak p = (0)$ and $\mathfrak q$ maximal. Now since
$\sqrt{(x)}$ is the intersection of the prime ideals
containing it, and since $\mathfrak q$ is the only prime
containing $x$ by minimality, we see that $\sqrt{(x)} = \mathfrak q$.
Hence Lemma \ref{lemma-height-1} applies.
The second assertion follows from the first.
\end{proof}

\begin{lemma}
\label{lemma-one-equation}
Suppose that $R$ is a Noetherian local ring and $x\in \mathfrak m$ an
element of its maximal ideal. Then $\dim R \leq \dim R/xR + 1$.
If $x$ is not contained in any of the minimal primes of $R$
then equality holds. (For example if $x$ is a nonzerodivisor.)
\end{lemma}

\begin{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 \ref{proposition-dimension}.
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.
\end{proof}

\begin{lemma}
\label{lemma-elements-generate-ideal-definition}
Let $(R, \mathfrak m)$ be a Noetherian local ring.
Suppose $x_1, \ldots, x_d \in \mathfrak m$ generate an
ideal of definition and $d = \dim(R)$. Then
$\dim(R/(x_1, \ldots, x_i)) = d - i$ for all $i = 1, \ldots, d$.
\end{lemma}

\begin{proof}
Clear from the proof of Proposition \ref{proposition-dimension},
or use induction on $d$ and Lemma \ref{lemma-one-equation} above.
\end{proof}


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