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 $R/P$.

 A: 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 $\,2\not\in I.$
A: There are many cases that I'm tempted to call degenerate: $R/P$ could be a field, for example. For concreteness, consider $R = \mathbf Z$, $P = 2\mathbf Z$, and $a = 3$.
Maybe it's helpful to think geometrically: if $k$ is a field of characteristic $\neq 2$ we can describe a parabola in the $k$-plane via the ring $A = k[x, y]/(y - x^2)$. The element $y - 1$ is certainly prime in $k[x, y]$, but
\[
A/(y - 1)A \approx k[x]/(1 - x^2) \approx k^2
\]
is not a domain, so the image of $y - 1$ in $A$ is not prime. The picture is that the line $y = 1$ and the parabola $y = x^2$ intersect in two points.
A: I think thats wrong. If $ a \in P $ holds, then $ a + P = 0 + P \in R/P$ and therefore $a+P$ not prime.
