Question on a proof of dimension property In this note proposition 1.3 claimed that :

If $k$ is a field, $R$ is a finitely generated $k$-algebra which is a domain then $\dim R$ is finite and any saturated chain of prime ideals has length equal to $\dim R$.

I understand mostly the proof, however, I do not understand the following sentences:

Let $P_0\subset P_1\subset...\subset P_m$ be a saturated chain of prime ideals in $R$. 

So, what is the $m$ here ? Or it is just a typo mistake, it should be $n$ ? 
After proving $\dim B=n-1$, the author wrote :

By induction, we are done.

So, what did he want to prove ? If he want to prove  any saturated chain of prime ideals has length equal to $\dim R$ by induction on $\dim R$, what is the induction hypothesis ? I do not understand how can we get any saturated chain of prime ideals in $R$ has length equal to $\dim R$ from any  any saturated chain of prime ideals in $R$ has length equal to $\dim R-1$. 
Please help me point it out. Thanks.
 A: I haven't looked at the text in question, but here is a sketch of how such arguments go:
We want to prove that for all f.g. $k$-algebras, some quantity $q(R)$ attached to $R$ is equal to $\dim R$.  (In this case, the quantity in question is the length of a saturated chain of prime ideals.)
One proves it for all $R$ at once, by induction on dimension of $R$, i.e. the inductive hypothesis is that for every f.g. $k$-algebra of dim'n $< n$, the statement is true, and one goes on to prove the statement for every f.g. $k$-algebra of dimension $n$.
Thus we give ourselves an arbitrary f.g. $R$ algebra of dimension $n$, and try to prove that $q(R) = n$.
One typical approach is to find a quotient $B$ of $R$ of dimension $n -1$, and prove that the quantity in question for $B$, i.e. $q(B)$, is $1$ less than the quantity in question for $R$, i.e. $q(R)$.  Assuming we've done this: then by the inductive hypothesis one has 
$q(B) = \dim B = n - 1$ (since $B$ is a f.g. $k$-algebra of dimension $n -1$, which is $< n$), and so $q(R) = q(B) + 1$ (as I already wrote, I'm assuming that we've proved this in some way; this is where the heart of the proof lies) $ = \dim B + 1 = (n-1) + 1 = n,$ and we're done.
Based on what is written by the OP, I think the above proof schema is the one being used.

My impression is that the OP's confusion comes from imagining the the ring $R$ is fixed (and that the induction somehow involves the length of chains of prime ideals in this fixed ring $R$ --- which it can't, because all saturated chains of primes ideals will be of the same length, and so there is no variable $n$ to induct on in this setting), whereas it is not: the statement we are proving by induction involves all $R$ at once.
