Prime ideals in R[x] having the same contraction in R

Question

Let $R$ be a commutative ring, $Q\subsetneq Q'$ primes in $R[x]$. If $Q'\cap R=Q\cap R$, then $Q=(Q\cap R)[x]$.

One inclusion is clear but I couldn't prove the other and I am not sure if the statement is true.

Answer 1

Yes, equality holds.

Let $P=Q\cap R=Q'\cap R$.

The inclusion $P[x]\subseteq Q$ is immediate.

By Theorem 37 of the text

Kaplansky -- Commutative Rings, 2nd Ed (1974)

there cannot exist a chain of three distinct prime ideals of $R[x]$ with the same contraction in $R$.

But $P[x]$ is a prime ideal of $R[x]$, hence, since we have the chain $$P[x]\subseteq Q\subset Q'$$ of three prime ideals of $R[x]$ with the same contraction in $R$ (namely $P$), it follows that the inclusion $P[x]\subseteq Q$ can't be proper.

Therefore $Q=P[x]$.