Ex. 11.7 from Atiyah-Macdonald is basically to prove $\dim A[x]=\dim A+1$ for $A$ noetherian.
From exercise 11.6, we get $\dim A[x]\geq\dim A+1$, so we are left to prove "$\leq$".
I've followed the hint and proved that $\text{ht}(p[x])=\text{ht}(p)$ for every $p\subset A$ prime, but I'm still not sure why this implies $\dim A[x]\leq\dim A+1$.
Here is what I've thought so far:
if $P_1\subsetneq...\subsetneq P_k$ is a maximal chain in $A[x]$, then $P_k$ is maximal and $k=\dim A[x]$. Suppose that $m:=P_k\cap A$ is a maximal ideal of $A$. That way, $A/m$ is a field and $A[x]/m[x]\simeq A/m[x]$ is a P.I.D. Since $P_k$ is maximal, $\overline{P_k}\in A[x]/m[x]$ is maximal, so $\text{ht}(\overline{P_k})=1$, which means $P_k$ is minimal prime of $m[x]$. That way, $k=\text{ht}(P_k)=\text{ht}(m[x])+1=\text{ht}(m)+1\leq\dim A+1$.
The problem with what I did is that it depends on $P_k\cap A$ being maximal. Is this necessarily the case? I feel like this is crucial, otherwise the hint wouldn't work. How do I prove that?
If it isn't true, how can I conclude that $\dim A[x]\leq\dim A+1$?
Consider the quotient map $q : A[x] \rightarrow A[x]/(x) \cong A$. Clearly $q(P_k) = P_k \cap A$, and if $q(P_k)$ not maximal then by Zorn's lemma $q(P_k) \subset Q$ where $Q$ is a maximal ideal of $A$. However, this implies $ P_k \subset q^{-1}(Q)$ where $q^{-1}(Q)$ is a prime ideal, a contradiction.