## 10.59 Dimension

Definition 10.59.1. The Krull dimension of the ring $R$ is the Krull dimension of the topological space $\mathop{\mathrm{Spec}}(R)$, see Topology, Definition 5.10.1. 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 10.59.2. The height of a prime ideal $\mathfrak p$ of a ring $R$ is the dimension of the local ring $R_{\mathfrak p}$.

Lemma 10.59.3. 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 10.59.4. A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.

Proof. By Lemma 10.30.5 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 $\mathop{\mathrm{Spec}}(R) = Z_1 \cup \ldots Z_ r$. According to Lemma 10.25.1, 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 $I = \cap \mathfrak p_ i$ 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 10.31.5. By Lemma 10.52.5 $R$ is the product of its local rings. By Lemma 10.51.8 each of these has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 10.52.6.

If $R$ is Artinian then by Lemma 10.52.6 it is Noetherian. All of its primes are maximal by a combination of Lemmas 10.52.3, 10.52.4 and 10.52.5. $\square$

In the following we will use the invariant $d(-)$ defined in Definition 10.58.8. Here is a warm up lemma.

Lemma 10.59.5. 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 10.52.6. $\square$

Proposition 10.59.6. 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 $\mathop{\mathrm{Spec}}(R)$ 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 Jacobson radical ideal is nilpotent, and

10. $R$ is Noetherian and there are no strict inclusions among its primes.

Lemma 10.59.7. 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 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. Thus the only prime containing $x$ is $\mathfrak m$ and hence (3).

If (3) then $\mathfrak m = \sqrt{(x)}$ by Lemma 10.16.2, 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 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 $d(R/\mathfrak p) \leq 1$. 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 10.59.5. 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 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 $\mathfrak p \subset \mathfrak q \subset \mathfrak q_2 \subset \ldots \subset \mathfrak q_ e = \mathfrak m$, with all inclusions strict, and $e \geq 2$. Pick some ideal of definition $I \subset R$. We will repeatedly use Lemma 10.58.10. 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. $\square$

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 10.19.1 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 10.59.9. 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 following lemmas are clear from the proofs of the lemmas and proposition above, but we spell them out so we have convenient references.

Lemma 10.59.10. Let $R$ be a Noetherian ring. Let $x \in R$.

1. If $\mathfrak p$ is minimal over $(x)$ then the height of $\mathfrak p$ is $0$ or $1$.

2. If $\mathfrak p, \mathfrak q \in \mathop{\mathrm{Spec}}(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, x)$, then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$.

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 10.16.5. Hence Lemma 10.59.7 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 $\mathfrak q$. $\square$

Lemma 10.59.11. Let $R$ be a Noetherian ring. Let $f_1, \ldots , f_ r \in R$.

1. If $\mathfrak p$ is minimal over $(f_1, \ldots , f_ r)$ then the height of $\mathfrak p$ is $\leq r$.

2. If $\mathfrak p, \mathfrak q \in \mathop{\mathrm{Spec}}(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, f_1, \ldots , f_ r)$, then every chain of primes between $\mathfrak p$ and $\mathfrak q$ has length at most $r$.

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 10.16.5. Hence Proposition 10.59.8 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 $\mathfrak q$. $\square$

Lemma 10.59.12. 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 10.59.8. 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 10.59.13. 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. Follows either from the proof of Proposition 10.59.8, or by using induction on $d$ and Lemma 10.59.12. $\square$

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).