If $A$ is a PID and $p\in A$, I want to prove that the following are equivalent:
a) $p$ is prime
b) $(p)$ is a non-zero prime ideal
c) $p$ is irreducible
d) $(p)$ is a maximal ideal
I assume the key here is proving that $a)\implies b)$ and $c)\implies d)$ and then proving that either $a) \implies c)$ or $b) \implies d)$
Since $A$ is a PID, I use the definition that it is an integral domain where every ideal is principal to use the proof of $p$ prime $\implies$ $p$ irreducible in an integral domain.
(ie. $p=ab$ with $p$ prime $\implies p|a$ or $p|b$ so wlog let $a=pc \implies p=pcb \implies cb=1 \implies b $ is a unit $\implies p$ irreducible.) There is also a proof that in a PID $p$ irreducible $\implies p$ prime, but it is quite long. Is this necessary also? If so, is there a relatively short proof of it?
Using the definitions of prime and prime ideal (resp. irreducible and maximal ideal) I am finding it hard to come up with a proof of their equivalence. It just seems obvious! Is there a neat way of showing it?
$A$- PID, $p\in A$, $(p)$ ideal of $A$ and $a,b\in A$.
maximal ideal: no ideals $d$ of $A$ exist such that $(p) \subset d \subset A$
irreducible: $p=ab \implies a$ or $b$ is a unit
prime ideal: $ab \in (p) \implies a \in (p)$ or $b \in (p)$
prime: $p|ab \implies p|a$ or $p|b$