Proof of prime and irreducible equivalences in PIDs 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 $p|b$
 A: $(a)\Rightarrow(b)$:
If p is a prime, then (p) is not empty,since p is in it.
suppose $ab \in (p)$, then p|ab ,since p is prime, so we have p|a or p|b,which means $a \in (p)$ or $b \in (p)$.
(p) is a prime ideal.
$(b)\Rightarrow(c)$:
suppose (p) is a non-zero prime ideal, but p is not irreducible.
$\exists a,b$ non-unit such that p=ab.
$p \in (a)$ and $p \in (b)$.
$(p) \subseteq (a)$ and $(p) \subseteq (b)$
We notice $ab \in (p)$,so $a \in (p)$ or $b \in (p)$.
so $(a) \subseteq (p)$ or $ (b) \subseteq (p)$
$(a)=(p)$ or $(b)=(p)$
If $(a)=(p)$,then there is a unit u,such that p=au.
since $p=ab$ , so $ab=au$ $\Rightarrow a(b-u)=0$.
A is a domain and $a \neq 0$ , so $b-u=0 \Rightarrow b=u$
Contradiction!!!! Since we assume b is not a unit.
similar proof for case $(b)=(p)$.
p is irreducible.
$(c) \Rightarrow (d)$:
suppose P is irreducible and (p) is not maximal.
$\exists$ an proper ideal (t) such that $(p) \subset (t)$.
So $p \in (t) \Rightarrow \exists$ a non-unit $s$ in A such that $p=ts$, otherwise $(p)=(t)$
$t$ is not a unit,otherwise $(t)=A$,and s is not a unit.
so p is not irreducible.Contradiction!!!!
(p) is a maximal ideal.
$(d) \Rightarrow (a)$:
suppose (p) is a maximal ideal ,but p is not a prime.
Since p is not a prime, we can factor it in a non-trivial way.
That is to say, we can find a prime $p'$ such that $p'|p$ and $p=p'N$, where $N$ is not a unit.
That is to say we find a ideal $(p') \supset (p)$.
Contradiction to the maximality of (p).
p is a prime.
