Let $A$ be a ring and $P \subsetneq R \subsetneq Q$ be prime ideals of $A[x]$. The following is not true: $P \cap A= Q \cap A =R \cap A$. 
Let $A$ be a commutative ring with $1$ and $P \subsetneq R \subsetneq Q$  be prime ideads of $A[x]$. The following is not true: $P \cap A= Q \cap A =R \cap A$.

My effort: Suppose the conclusion is true then we get 
$A/(P \cap A) [x] = A/(Q \cap A) [x] = A/(R \cap A) [x]$.
Since $P \subset A[x]$ is a prime ideal; $P \cap A \subset A$ is also a prime ideal. So, $A/(P \cap A) [x]$ is an integral domain. Similarly, for other terms. I am not able to proceed further.
Thanks in advance!
 A: Hint:
Let $K=\operatorname{Frac}(A/\mathfrak p)$ be the residue field. The prime ideals $\mathfrak{p}, \:\mathfrak{q}, \:\mathfrak{r}$ correspond bijectively to prime ideals of $K[X]$, and this bjection is order-preserving. However, $K(x)$ is a P.I.D., hence it has Krull dimension $1$.
A: Elaborating Bernad's helpful hint,

Proof)Suppose $P∩A=Q∩A=R∩A$.(they are prime in A since the preimage of prime ideals are prime under ring homomorphisms.) Let $S$ be the set of non zero element of the integral domain $A/A\cap P$. Then $S$ is a multiplicatively closed subset of $(A/P\cap A)[x]$. Localizing $(A/P\cap A)[x]$ with $S$, clearly $S^{-1}((A/P\cap A)[x])=(S^{-1}(A/P\cap A))[x]$. Now let $\overline{P},\overline{Q},\overline{R}$ be the images of $P,Q,R$, respectively in $(A/P\cap A)[x]$ by the natural homomorphism from $A[x]$ to $(A/P\cap A)[x]$. Note $\overline{P}\subsetneq\overline{Q}\subsetneq\overline{R}$ since $P\subsetneq Q\subsetneq R$ all contain $(P∩A)[x]$ and $(A/P\cap A)[x]\cong A[x]/(P\cap A)[x]$(Dummit,Foote Abstract Algebra, page 296, Proposition2 from which we also know $(P∩A)[x]$ is prime.) so $\overline{P}\subsetneq\overline{Q}\subsetneq\overline{R}$ in $A[x]/(P\cap A)[x]$. Also note $\overline{P},\overline{Q},\overline{R}$ are primes in $(A/P\cap A)[x]\cong A[x]/(P\cap A)[x]$ since $P,Q,R$ all contain $(P∩A)[x]$ so for example $\overline{P}\cong P/(P∩A)[x]$ is prime in $(A/P\cap A)[x]\cong A[x]/(P\cap A)[x]$. Finally note $\overline{P}=\{\sum_{i=0}^{n}(a_i+P\cap A)x^i\mid\sum_{i=0}^{n}a_ix^i\in P\}$. Then $(A/P\cap A)\cap \overline{P}$ is the set of constant terms of elements of $\overline{P}$ which is $\{a_0 +P∩A\mid\sum_{i=0}^{n}a_ix^i\in P\}$. But $\{a_0\mid\sum_{i=0}^{n}a_ix^i\in P\}=P\cap A$. So $(A/P\cap A)\cap \overline{P}=(P\cap A)/(P\cap A)=0$. Similarly, $(A/P\cap A)\cap \overline{Q}=(Q\cap A)/(P\cap A)=0$ and $(A/P\cap A)\cap \overline{R}=(R\cap A)/(P\cap A)=0$.(We supposed $P∩A=Q∩A=R∩A$.) This means $\overline{P},\overline{Q},\overline{R}$ all don't meet $S$. Now the homomorphic images of $\overline{P}\subsetneq\overline{Q}\subsetneq\overline{R}$ in $(S^{-1}(A/P\cap A))[x]$ still preserve the strict inclusion as they are primes.(Atiyah, Macdonald Introduction To Commutative algebra, Proposition 3.11) But $(S^{-1}(A/P\cap A))$ is a field by the above construction. So $(S^{-1}(A/P\cap A))[x]$ is a PID and every non-zero prime ideal in a PID is maximal, so since at least the images of $\overline{Q}\subsetneq\overline{R}$ are non-zero, and we get a maximal ideal strictly contained in another maximal ideal, a contradiction.

