Height of a prime

Let $R$ be a commutative ring and let $\mathfrak p$ be a prime ideal (i.e., $\mathfrak p\in\operatorname{Spec}(R)$, the prime spectrum ).

The height of $\mathfrak p$, denoted $\operatorname{ht}(\mathfrak p)$, is the supremum of integers $n\ge 0$ such that there exists a strictly increasing chain of prime ideals

in $R$.

Equivalently,

where $R_{\mathfrak p}$ is the localization at \mathfrak p and $\dim$ denotes Krull dimension . In particular, the dimension of $R$ is the supremum of the heights of its prime ideals.

Examples

1. The ring of integers.
   In $R=\mathbb{Z}$, one has $\operatorname{ht}((0))=0$, and for any prime number $p$,

   $$(0)\subsetneq (p)$$

   is a maximal chain, so $\operatorname{ht}((p))=1$.

2. A polynomial ring in two variables.
   In $k[x,y]$ (for a field $k$), the prime ideals satisfy

   $$(0)\subsetneq (x)\subsetneq (x,y),$$

   so $\operatorname{ht}((0))=0$, $\operatorname{ht}((x))=1$, and $\operatorname{ht}((x,y))=2$.

3. Dedekind domains.
   If $R$ is a Dedekind domain (for instance, the ring of integers in a number field), then every nonzero prime ideal has height $1$. In a DVR , the unique nonzero prime ideal also has height $1$, reflecting that these rings are “one-dimensional.”