https://stacks.math.columbia.edu/tag/00KD

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 s…