If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $\frac{R}{s}$ has finitely many prime ideals . If   $R$ is a principal ideal domain and  $s$ is a  non-zero prime ideal, then  $\frac{R}{s}$   has finitely  many prime ideals .
How can I prove it?
My attempt:
As $R$ is a PID then $s$ will be maximal ideal so $\frac {R}{s}$ will be field which has precisely two prime ideal . One is {$0$} and another one is $\frac {R}{s}$ .
Can anyone correct me If I have gone wrong anywhere?

The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is $\{0\}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
