Let $k$ be a field, and let $A$ be a $k$-algebra of finite type that is an integral domain. Show that any maximal ascending chain of prime ideals in $A$ has length equal to $\operatorname {dim}A$.
I'll use this result, that is the proposition 3.3.8 from Bosch's Algebraic Geometry and Commutative Algebra:
Let $k$ be a field, and let $A$ be a $k$-algebra of finite type that is an integral domain. Then $\operatorname{coht}\mathfrak p+\operatorname{ht}\mathfrak p=\operatorname {dim}A$ for every prime ideal $\mathfrak p\subset A$, while $\operatorname{ht}\mathfrak m=\operatorname {dim}A$ for every maximal ideal $\mathfrak m\subset A$.
Let $d:=\operatorname {dim}A$, and suppose that $\mathfrak p_1\subset \dots \subset \mathfrak p_r\subseteq A$ is a chain of prime ideals (so clearly $r\le d$). Trivially, for every $i\le r$, holds $\operatorname{ht}\mathfrak p_i\gt\operatorname{ht}\mathfrak p_{i-1}$; now suppose $\operatorname{ht}\mathfrak p_i\gt\operatorname{ht}\mathfrak p_{i-1}+1$ for some $i\le r$.
The $k$-algebra $A/\mathfrak p_{i-1}$ is again an integral domain of finite type, thus $\operatorname {dim}A/\mathfrak p_{i-1}=d-\operatorname{ht}\mathfrak p_{i-1}$; observe also that $\operatorname{coht}\mathfrak p_i/\mathfrak p_{i-1}=\operatorname{coht}\mathfrak p_i$. Consider the following equalities: $$\operatorname {dim}A/\mathfrak p_{i-1}=\operatorname{coht}\mathfrak p_i/\mathfrak p_{i-1}+\operatorname{ht}\mathfrak p_i/\mathfrak p_{i-1}$$ $$d-\operatorname{ht}\mathfrak p_{i-1}=\operatorname{coht}\mathfrak p_i +\operatorname{ht}\mathfrak p_i/\mathfrak p_{i-1}$$ $$\operatorname{ht}\mathfrak p_i-\operatorname{ht}\mathfrak p_{i-1}=\operatorname{ht}\mathfrak p_i/\mathfrak p_{i-1}.$$ The last equality should mean that, if $\operatorname{ht}\mathfrak p_i\gt\operatorname{ht}\mathfrak p_{i-1}+1$, we can insert $\operatorname{ht}\mathfrak p_i-\operatorname{ht}\mathfrak p_{i-1}-1$ prime ideals between $\mathfrak p_{i-1}$ and $\mathfrak p_i$; hence if the initial chain is maximal, $\operatorname{ht}\mathfrak p_i=\operatorname{ht}\mathfrak p_{i-1}+1$ for every $i\le r$. Also, $\mathfrak p_r$ is maximal, so $\operatorname{ht}\mathfrak p_r=d$; since at every step the heights of the prime ideals increase of exactly $1$, we conclude that $r=d$.