I have recently studied prime ideals,maximal ideals and divisibility in an integral domain.Suppose $a$ is an irreducible element in an integral domain $D$,can we claim that $(a)$ is a maximal element of $D$.But then $(a)$ is a prime ideal,then $a$ is prime as $a\neq 0$.But the it proves that every irreducible is prime but this is not true.In particular consider $\mathbb Z[\sqrt{-5}]$ and the element $2$,it is irreducible but not prime,So,$(2)$ is not a prime ideal ,so it cannot be maximal as well.But $2$ is irreducible.
Actually I was trying to draw an analogy with the theorem:
In an integral domain $D$,an element $a$ is prime iff $(a)$ is prime and $a\neq 0$.
I was trying to replace prime by irreducible and prime ideal by maximal ideal but you see in the above statement where this does not hold.
So my question is,what is a necessary sufficient condition that $(a)$ is maximal iff $a$ is irreducible.It is clear to me that $(a)$ is maximal and $a\neq 0$ would imply $a$ is irreducible.But when does the converse hold?
The very definition of "irreducible" can be rephrased as "maximal among principal ideals."
So for a given irreducible element $(a)$, we have that $(a)$ is a maximal ideal if and only if every maximal ideal containing $a$ is principal.