Definition 10.60.1. Let $R$ be a ring. A chain of prime ide…

Definition 10.60.1. Let $R$ be a ring. A chain of prime ideals is a sequence $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ of prime ideals of $R$ such that $\mathfrak p_ i \not= \mathfrak p_{i + 1}$ for $i = 0, \ldots , n - 1$. The length of this chain of prime ideals is $n$.
10.60 Dimension
Please compare with Topology, Section 5.10.
Recall that we have an inclusion reversing bijection between prime ideals of a ring $R$ and irreducible closed subsets of $\mathop{\mathrm{Spec}}(R)$, see Lemma 10.26.1.
Definition 10.60.2. 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 $R$ has a chain of prime ideals of length $n$.
Definition 10.60.2. 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 $R$ has a chain of prime ideals
of length $n$.
Definition 10.60.3. The height of a prime ideal $\mathfrak p$ of a ring $R$ is the dimension of the local ring $R_{\mathfrak p}$.
Definition 10.60.3. The height of a prime ideal $\mathfrak p$ of a ring $R$ is the dimension of the local ring $R_{\mathfrak p}$.
Lemma 10.60.4. The Krull dimension of $R$ is the supremum of the heights of its (maximal) primes.
Lemma 10.60.4. 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 10.60.5. A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.
Lemma 10.60.5. A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.
Proof. Assume $R$ is a Noetherian ring of dimension $0$. By Lemma 10.31.5 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 $\mathop{\mathrm{Spec}}(R) = Z_1 \cup \ldots \cup Z_ r$. According to Lemma 10.26.1 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 $\bigcap \mathfrak p_ i$ are nilpotent since otherwise $R_ f$ would not be the zero ring and we would have another prime. By Lemma 10.53.5 $R$ is equal to $\prod R_{\mathfrak p_ i}$. Since $R_{\mathfrak p_ i}$ is also Noetherian and dimension $0$, the previous arguments show that its radical $\mathfrak p_ iR_{\mathfrak p_ i}$ is locally nilpotent. Lemma 10.32.5 gives $\mathfrak p_ i^ nR_{\mathfrak p_ i} = 0$ for some $n \geq 1$. By Lemma 10.52.8 we conclude that $R_{\mathfrak p_ i}$ has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 10.53.6.
If $R$ is an Artinian ring then by Lemma 10.53.6 it is Noetherian. All of its primes are maximal by a combination of Lemmas 10.53.3, 10.53.4 and 10.53.5. $\square$
In the following we will use the invariant $d(-)$ defined in Definition 10.59.8. Here is a warm up lemma.
Lemma 10.60.6. Let $R$ be a Noetherian local ring. Then $\dim (R) = 0 \Leftrightarrow d(R) = 0$.
Lemma 10.60.6. 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 10.53.6. $\square$
Proposition 10.60.7. Let $R$ be a ring. The following are equivalent:
$R$ is Artinian,
$R$ is Noetherian and $\dim (R) = 0$,
$R$ has finite length as a module over itself,
$R$ is a finite product of Artinian local rings,
$R$ is Noetherian and $\mathop{\mathrm{Spec}}(R)$ is a finite discrete topological space,
$R$ is a finite product of Noetherian local rings of dimension $0$,
$R$ is a finite product of Noetherian local rings $R_ i$ with $d(R_ i) = 0$,
$R$ is a finite product of Noetherian local rings $R_ i$ whose maximal ideals are nilpotent,
$R$ is Noetherian, has finitely many maximal ideals and its Jacobson radical ideal is nilpotent, and
$R$ is Noetherian and there are no strict inclusions among its primes.
Proposition 10.60.7. Let $R$ be a ring. The following are equivalent:
$R$ is Artinian,
$R$ is Noetherian and $\dim (R) = 0$,
$R$ has finite length as a module over itself,
$R$ is a finite product of Artinian local rings,
$R$ is Noetherian and $\mathop{\mathrm{Spec}}(R)$ is a finite discrete topological space,
$R$ is a finite product of Noetherian local rings of dimension $0$,
$R$ is a finite product of Noetherian local rings $R_ i$ with $d(R_ i) = 0$,
$R$ is a finite product of Noetherian local rings $R_ i$ whose maximal ideals are nilpotent,
$R$ is Noetherian, has finitely many maximal ideals a…