A polynomial ring in n variables over a field has dimension n

Let $D$ be any finite-type $k$-domain and $Q⊊Q^{\mathrm{\prime }}$ primes of $D$. Set $E=D\mathrm{/}Q$, a nonzero finite-type $k$-domain, and $J=Q^{\mathrm{\prime }}\mathrm{/}Q$, a nonzero prime of $E$. By [L1], fix a module-finite normalization $B=k[z_1,\dots ,z_d]\subset E$. Its fraction field extension is algebraic, so [L2] gives ${\mathrm{trdeg}}_k\mathrm{Frac}(E)=d$.

By incomparability in [L1], $q:=J\cap B$ is nonzero: otherwise $(0)⊊J$ would be comparable primes of $E$ with the same contraction $(0)$ in $B$. If $d=0$, then $B=k$ is a field and has no nonzero proper prime, so the assumed strict pair cannot exist. For a strict pair one therefore has $d\ge 1$. A nonzero polynomial $f\in q$ gives a nontrivial algebraic relation among the images of $z_1,\dots ,z_d$ in the domain $B\mathrm{/}q$. Among this finite set of images, take a maximal algebraically independent subset; because the full set is dependent, it has at most $d-1$ members. Maximality makes each omitted image algebraic over the field generated by the subset. Since all images generate $\mathrm{Frac}(B\mathrm{/}q)$, this proves ${\mathrm{trdeg}}_k\mathrm{Frac}(B\mathrm{/}q)\le d-1$.

The quotient $E\mathrm{/}J$ is integral over $B\mathrm{/}q$, hence its fraction field is algebraic over $\mathrm{Frac}(B\mathrm{/}q)$. By [L2] and step 2.1, ${\mathrm{trdeg}}_k\mathrm{Frac}(D\mathrm{/}{Q}^{\mathrm{\prime }})\le d-1<d={\mathrm{trdeg}}_k\mathrm{Frac}(D\mathrm{/}Q)$. Thus transcendence degree drops at least one along every strict prime inclusion in a finite-type domain.

Apply this drop to any strict chain $P_0⊊\cdots ⊊P_r$ in $k[x_1,\dots ,x_n]$. The fraction field of the initial quotient is generated by at most $n$ images of variables, so its transcendence degree is at most $n$; the final transcendence degree is nonnegative. Hence $r\le n$. For $n\ge 1$, the coordinate ideals $(0)⊊(x_1)⊊\cdots ⊊(x_1,\dots ,x_n)$ are prime and form a chain of length $n$. When $n=0$, the sole prime is $(0)$ and both bounds are zero. Therefore $\dim k[x_1,\dots ,x_n]=n$ for every $n\ge 0$.