Does $x$ irreducible in ring $R$ imply $(x)$ maximal ideal of $R$?

Question

I was studying for my final exam of abstract algebra and, after seeing that $p$ prime element of a ring $R$ is equivalent as saying the ideal $(p)\unlhd R$ is prime, I came up with the assumption that something similar might happend with the irreducible elements and the maximal ideals, but this is not said anywhere in my course book so I may be wrong. So my question is, is this statement true?

Being $R$ a commutative unital ring, and $0\neq x\in\mathbb{R\setminus R^\times}$, then: $$x \text{ irreducible in $R$} \Longleftrightarrow (x) \text{ is maximal ideal of $R$}$$

Maybe this is only verified under certain extra conditions of $R$ (for example, being PID or UFD or some kind of special ring). Is my assumption true in general? If not, could you give me a counter example? It would be nice to have a proof if it's true. Any help will be appreciated, thanks in advance.

Answer 1

$x \in R$ is irreducible if and only if the ideal $(x)$ is maximal among proper principal ideals. The proof is straightforward.

Thus if there exist proper ideals in $R$ which are not principal, then there will exist irreducible elements whose principal ideals are not maximal ideals.

Answer 2

The simplest counterexample is $R=\mathbb Z[x]$. Then $(x) \subset (2,x)$ is not maximal.