Quotient of a PID by a prime ideal is a PID too.

Question

I am trying to prove that if $R$ is a PID and $P$ is a prime ideal of $R$ then $R/P$ is also a PID. This is question 8.2.3 from Dummit & Foote.

My thoughts and what I know : $P$ is also maximal. $R/P$ is hence a field. Also is an integral domain due $P$ being prime. $P$ is generated by by some element of $R$, say $a$, written as $P=(a)$. My conjecture is that an ideal $I$ of $R/P$ is generated by the coset $a+P$, ie. $I=(a+P)$.

I am having some trouble understanding that the set $I$ is equal to a set generated by a coset. I'm not sure how to show the inclusions to show that those two sets are the same. Any help is appreciated. Thanks for your time.

Answer 1

As you noted all prime ideals in a PID are maximal and hence the quotient R/P is a field. We know that all fields are Euclidean domains and hence PIDs.

Answer 2

As an alternative to the other given answer, you can split this into two parts.

First show that in any quotient of a PID all ideals are principal (just take the image of a generator under the canonical homomorphism).

Next, the quotient by a prime ideal is always an integral domain.

Putting these together, we see that if $R$ is a ring where all ideals are principal and $P$ is a prime ideal of $R$ then $R/P$ is a PID (so we could even relax the condition on the original ring a bit).