If $a$ in $R$ is prime, then $(a+P)$ is prime in $R/P$. Let $R$ be a UFD and $P$ a prime ideal. Here we are defining a UFD with primes and not irreducibles. 

Is the following true and what is the justification?


  If $a$ in $R$ is prime, then $(a+P)$ is prime in $R/P$.
The bijection between ideals of $\,R/I\,$ and ideals of $R$ containing $I$ restricts to prime ideals, hence $\,(a+P)\,$ is prime in $\,R/P\,$ iff $\,I = (a)+P\,$ is prime in $\,R.\,$ But generally this is not true, e.g. $\,I=1\,$ for $\,a\nmid P\,$ primes in $\,\Bbb Z,\,$ or $\,I = (4\!-\!x)+(x) = (4,x)\,$ is not prime in $\,\Bbb  Z[x],\,$ by $\,2^2\in I\,$ but $\,2\not\in I.$
I think thats wrong. If $ a \in P $ holds, then $ a + P = 0 + P \in R/P$ and therefore $a+P$ not prime.