Usual argument in algebra?

Question

Let R be a ring with 1, and P a prime ideal of $R[x]$, then $$\phi: \dfrac{R}{P \cap R}[x] \to \dfrac{R[x]}{P}$$ given by $(r_n+P\cap R)x^n+...+(r_0 + P \cap R)x^0 \to (r_nx^n+...+r_0x^0)+P$ is an epimorphism and $$Ker (\phi) = \{(r_n+P\cap R)x^n+...+(r_0 + P \cap R)x^0 \in \dfrac{R}{P \cap R}[x] \left / \right (r_nx^n+...+r_0x^0) \in P \}$$

An easy verification shows that $Ker(\phi) \cap\dfrac{R}{P \cap R} = 0$ and $Ker(\phi)$ is prime in $\dfrac{R}{P \cap R}[x]$.

I am studying an proof and he claims that from this we need only consider the case where $P \cap R = 0$. I am trying to understand why we can claim that.

I know that by isomorphism theorem we have that $\dfrac{\dfrac{R}{P \cap R}[x]}{Ker (\phi)} \cong \dfrac{R[x]}{P}$.

Does $Ker(\phi) \cap\dfrac{R}{P \cap R} = 0$ implies $P \cap R = 0$?

Answer 1

Consider the following statements:

1. For all Jacobson rings $R$ and all prime ideals $P$ of $R[x]$ such that $R\cap P=0$, the ring $R[x]/P$ is semiprimitive.
2. For all Jacobson rings $R$ and all prime ideals $P$ of $R[x]$, the ring $R[x]/P$ is semiprimitive.

To answer your question, we have to prove $1\Rightarrow 2$.

Let $R$ be a Jacobson ring and let $P$ be a prime ideal of $R[x]$. As you pointed out in your question, we have a ring isomorphism $$R'[x]/P'\cong R[x]/P$$ where $R'=R/(P\cap R)$ is a Jacobson ring and $P'=\operatorname{Ker}\varphi$ is a prime ideal of $R'$ satisfying $P'\cap R'=0$. By 1., $R'[x]/P'$ is semiprimitive, hence $R[x]/P$ is semiprimitive as well. Thus 2. is proved.