If $p$ is a prime ideal then $p[X]$ is a prime ideal 
If $Z$ is a ring and $p$ is a prime ideal of $Z$ then $p[X]$ is a prime ideal of $Z[X]$. Is it true or false? 

I believe that it is true and I try to prove it like that:
Take $f(x)\in p[X]$ and suppose that $f(x)=g(x)h(x)$ then $n=\deg f=\deg g+\deg h=m+r$.
Put $g(x)=a_{m}x^{m}+...+a_{1}x+a_{0}$ and $h(x)=b_{r}x^{r}+...+b_{1}x+b_{0}$
+ $a_0b_0\in p$ then either $a_{0}\in p$ or $b_{0}\in p$ (choose $a_{0}\in p$)
+ $a_{1}b_{0}+b_{1}a_{0}\in p$ then either $a_{1}\in p$ or $b_{0}\in p $
-If $a_{1}\in p$ then $a_{2}\in p$ because $a_{2}b_{0}+b_{2}a_{0}+a_{1}b_{1}\in p$
-If $b_{0}\in p$ then either $a_{1}\in p$ or $b_{1}\in p$,go back to the first one case.
But I think that my proof is quite impossible. Could someone help me?


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*One more thing, just a little bit question I want to ask that is about the Krull dimension. If we have a chain $P_{0} \subseteq P_{1}...$ do we have $P_{0}=0$ because $0$ is also a prime ideal and it is the smallest. In particular, in PID $\dim K=1$ mean that $0\subseteq P$ with $P$ being a prime ideal


Since $R[x]/P[x]\simeq (R/P)[x]$, we only have to show that $A$ is a domain $\Rightarrow A[x]$ is a domain. If $f,g\in A[x]$ and both are nonzero, then consider their minimal term with nonzero coefficient.
