# 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?

## 1 Answer

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).