proving that $\dim A[X] = \dim A + 1$ (Matsumura) Let $A$ be a Noetherian ring and $X$ an indeterminate over $A$.
I am having trouble understanding Matsumura's proof (Commutative Ring Theory, Theorem 15.4) that $\dim A[X] = \dim A + 1$. Below, i provide all the necessary details.
As an auxiliary result, Theorem 15.1(ii) says that if $f:A \rightarrow B$ is a faithfully flat morphism of Noetherian rings, $P$ a prime ideal of $B$ and $p$ a prime ideal of $A$ such that $P \cap A=p$, then $ht(P) = ht(p) + \dim B_P/pB_P$. Let's call this theorem $(*)$. So far so good.
Now, Matsumura says that if $p$ is a prime ideal of $A$ then the fibre over $p$ given by $A[x] \otimes_A \kappa(p) \cong \kappa(p)[X]$ is a one-dimensional ring, since it is principal, which i can also see.
I can further see that $A[X]$ is free over $A$ and so faithfully flat, hence the condition of $(*)$ is satisfied.
What i don't see is that the statement $\dim A[X] = \dim A + 1$ follows immediately from $(*)$, as Matsumura claims. In particular, if $P$ is a prime ideal of $A[X]$ lying over the prime ideal $p$ of $A$, then from $(*)$ we have $ht(P) = ht(p) + \dim A[X]_P/pA[X]_P$. Now i have two concerns: i) is $A[X]_P/pA[X]_P$ the same as $\kappa(p)[X]$? ii) even if we could show that $ht(P)=ht(p)+1$ why does this prove that $\dim A[X] = \dim A +1$, since the map $\operatorname{Spec} B \rightarrow \operatorname{Spec} A$ need not be surjective?
 A: Obviously $\dim A[X]\ge\dim A+1$. 
For the converse take $P\subset A[X]$ a prime ideal, $\mathfrak p=P\cap A$ and use   $$\operatorname{ht}(P)=\operatorname{ht}(\mathfrak p) + \dim A[X]_P/\mathfrak pA[X]_P.$$ Note that $A[X]_P/\mathfrak pA[X]_P\simeq (A[X]/\mathfrak p[X])_{P/\mathfrak p[X]}$ whose dimension is $\operatorname{ht}(P/\mathfrak p[X])$. 
(If you prefer, at this point can go further with the isomorphism and obtain $(A[X]/\mathfrak p[X])_{P/\mathfrak p[X]}\simeq (A/\mathfrak p)[X]_Q$, where $Q$ is a prime ideal in $(A/\mathfrak p)[X]$ lying over (or equal to) $(0)$. Moreover, $(A/\mathfrak p)[X]_Q$ is isomorphic to a localization of $\kappa(\mathfrak p)[X]$, so its dimension is $\le 1$.) 
But $\operatorname{ht}(P/\mathfrak p[X])\le 1$, so $\operatorname{ht}(P)\le\operatorname{ht}(\mathfrak p) + 1$. If $\dim A[X]=n$, then there exists a chain $P_0\subset P_1\subset\cdots\subset P_n$ of prime ideals in $A[X]$. In particular $\operatorname{ht}P_n=n$ and therefore $\operatorname{ht}\mathfrak p_n\ge n-1$, so $\dim A\ge n-1$, qed.
