In a commutative ring with identity, if $p$ is irreducible, is ($p$) a maximal ideal?

Question

In a Euclidean Domain, $D$, if we mod out by an irreducible, $p$, we get the field $D/(p)$. I can see that this follows since we are going to be able to write $1$ as a linear combination of $p$ and some $q \in D$. Same reason $\mathbb Z/(5)$ is a field.

Now suppose we are not in a Euclidean Domain. Suppose we just have a commutative ring with identity, $R$.

Question $1$: Is the ideal generated by an irreducible element $p$ always maximal?

If so, then if we mod out by $p$, since $(p)$ is maximal, then $R/(p)$ is a field.

Assuming question one is true:

Question $2$: What's the reasoning behind $R/(p)$ being a field? Will we be able to write $1$ as a linear combination of $p$ and some $q$ in a non-Euclidean Domain?

Answer 1

The result that $p$ irreducible in $R$ implies $(p)$ maximal holds in the more general case of PIDs. To see this, suppose $p$ is an irreducible, and that $(p)\subset I$ for some ideal $I$ in $R$. Since $R$ is a PID, $I=(k)$ for some $k$. In particular, $p\in(k)$, so $kr=p$ for some $r\in R$. Since $p$ is irreducible, either $k$ or $r$ is a unit, leading to the cases $(k)=R$ and $(k)=(p)$ respectively.

This is not true in more general rings. For example, $X$ is irreducible in $\Bbb{Z}[X]$, but $(X)$ it is not maximal as $(X)\subsetneq (2,X)\subsetneq \Bbb{Z}[X]$. (This works because $\Bbb{Z}[X]$ is not a PID, although it is a UFD.)

Answer 2

Consider the ring $\mathbb Z[\sqrt 5i]$ Here $2$ is an irreducible element but $\langle 2\rangle =\{2(a+b\sqrt 5i\:a,b\in \mathbb Z\} $ is not a maximal ideal as it is properly contained in the ideal $J=\{a+b\sqrt 5i:b\in \mathbb Z ,a\in \mathbb 2Z\}$

Answer 3

Generally speaking, we only talk about irreducible or reducible elements in integral domains. But even for integral domains, the property you posted in Question 1 is not always satisfied. I know the Bézout domains (that is, the integral domains of which all finitely generated ideals are principal, or equivalently, the GCD domains of which all coprime pairs of elements generate $1$) satisfy the such property.

However, I think it's hard to give a comprehensive description to the class of integral domains such that all irreducible elements generate maximal ideals, given the fact that all the integral domains that admit no irreducible elements lie in this class.