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).


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