I am reading Eisenbud's Commutative Algebra, p.293, Proof of the Theorem A (p.290) and stuck at some statement.
Theorem A. If $R$ is an affine domain over a field $k$ ( ; i.e., finitely generated $k$-algebra which is a domain ), then $$ \operatorname{dim}R = tr.deg._kR$$ and this number is the length of every maximal chain of primes in $R$.
For now, let's refer to the Noether's Normalization theorem ( His book p.287, Theorem 13.3).
Theorem 13.3 (Noether Normalization) Let $R$ be an affine ring of dimension $d$ over a field $k$. If $I_1 \subset \cdots \subset I_m$ is a chain of ideals of $R$ with dim $I_j=d_j$ and $d_1 >d_2 > \cdots >d_m \ge 0$, then $R$ contains a polynomial ring $S= k[x_1, \dots, x_d]$ in such a way that $R$ is a finitely generated $S$-module and $$ I_j \cap S = (x_{d_j+1},\dots,x_d) $$for $j=1,\dots m$.
Proof of Theorem A . Suppose that $R$ is an affine domain of dimension $d$, and let $S$ be the polynomial subring $k[x_1, \dots, x_d]$ as in Theorem 13.3 ( with no ideals $I_j$ specified ). Clearly, $tr.deg._kS =d$. Since $R$ is finite over $S$, the quotient field of $R$ is finite over that of $S$, so $tr.deg._k R = tr.deg_S R + tr.deg._kS = d$ as well, proving the first statement.
Let $P_0 \subset \cdots \subset P_m \subset R$ be a chain of primes of $R$, so that $m \le d$ ; we must show that if $m <d$, then a new prime can be inserted somewhere in the chain ( If so, then we can show that every maximal chain has length d , so we arrive at our claim ).
First, note that
$$ \operatorname{dim}R = \sup \{ \operatorname{length}((0) \subsetneq P_1 \subsetneq \cdots \subsetneq P_m=M_m) : P_i \in \operatorname{Spec}R ( 0 \le i \le m-1 ), P_m =M_m \in Max(R) ) \}$$
( $R$ is domain so $(0)$ is prime ideal ). So we may assume that $P_0 = (0)$ and $P_m$ is maximal (True?).
Second, choose $S=k[x_0 ,\dots x_d]$ as in Theorem 13.3, taking $I_j = P_j$. If $m <d$, then for some $j$ we shall have
$$ d_{j-1} := \dim P_{j-1} > d_{j-1}-1 > \dim P_j =: d_j. $$
( If such $j$ does not exist, then $d_{j-1} =d_j +1$ for all $j$. This implies $d := \dim R = \dim (0) =: \dim P_0 = \dim P_m + m = 0 +m$ ( $P_m$ is maximal ) so that $d=m$, which contradicts to $m<d$ ).
So the prime $Q:=(x_{d_{j-1}},\dots x_d)$ lies strictly between $Q_{j-1}:= S \cap P_{j-1} = (x_{d_{j-1} +1 } , \dots x_d)$ and $Q_j := S \cap P_j := (x_{d_j+1} ,\dots, x_d)$.
We shall finish the proof by showing that this implies the existence of a prime $P$ strictly between $P_{j-1}$ and $P_j$. ( Next proof is omitted. That argument uses the 'Going down theorem for integral extensions of normal rings ; his book Theorem 13.9. If needed, I will upload proof in detail ).
I am stuck at the bold statement. By brutal force? Can anyone helps?
O.K! The bold statement is true. I think that I solve the problem.
Note that $Q_{j-1} \subset Q \subset Q_j$ so that $\dim Q_{j} \le \dim Q \le \dim Q_{j-1}$. So, $Q_{j-1} \subsetneq Q$ is implied by $\dim Q \lneq \dim Q_{j-1}$ and $Q \subsetneq Q_j$ is implied by $\dim Q_j \lneq \dim Q$. And note that
$$ \dim Q_j := \dim (S/Q_j) = \dim ( \frac{k[x_0, \dots, x_d]}{(x_{d_j+1} , \dots ,x_d )} ) = \dim k[x_0, \dots ,x_{d_j}] = d_j +1$$
$$ \dim Q = \dim ( \frac{k[x_0, \dots ,x_d]}{(x_{d_{j -1} } , \dots, x_d )} ) = \dim k[x_0, \dots, x_{d_{j-1}-1}] = d_{j-1} $$
$$ \dim Q_{j-1} = \cdots = d_{j-1} +1 $$
So, from these, we can show that $\dim Q \lneq \dim Q_{j-1}$ and $\dim Q_j \lneq \dim Q$ ( We have $d_{j-1} -1 > d_j $) and we are done.