I've been reading some commutative algebra, but have been struggling with this idea for a while.
Let $k$ be a field, and let $A=k[x_1,\dots,x_n]$ be a finitely generated integral domain, such that $\operatorname{tr. deg}_k(k(x_1,\dots,x_n))=r$. I want to know why for any maximal chain of (nonempty) irreducible closed sets $P_1\subset P_2\subset\cdots\subset P_m=\operatorname{Spec}(A)$, with $P_i\neq P_j$ when $i\neq j$, then $m=r+1$.
I know that since the $P_i$ are closed and irreducible, then each $P_i=Z(p_i)$, the set of zeroes for some prime ideal $p_i$. So I tried looking at a maximal chain $ Z(p_1)\subset Z(p_2)\subset\cdots\subset Z(p_m)=\operatorname{Spec}(A). $ I also know that $Z(a)\subset Z(b)\iff\text{rad }a\supset\text{rad }b$, so this gives a maximal chain of prime ideals $ p_1\supset p_2\supset\cdots\supset p_m=(0). $
I've not seen a way to relate this back to the transcendence degree to conclude that $m=r+1$. I thought about assuming $m
I've been looking around, but haven't found a very digestible proof.