TITLE: Is there a name for those elements $x$ of a commutative ring $R$ such that $Rx$ is maximal among all proper ideals?
QUESTION [5 upvotes]: Ever since learning basic ring theory, I've always felt kind of confused about the fact that:

maximal ideals are prime (because every field is an integral domain), but
irreducible elements needn't be prime.

Recently, the reason for this suddenly snapped into focus: the problem, in short, is that irreducibility can only "see" principal ideals, and therefore isn't usually strong enough to imply full-blown primeness, except in a principal ideal ring. Indeed:

Proposition. Let $R$ denote a commutative ring. Then for all $x \in R,$ the following are equivalent.

$x$ is irreducible.
$Rx$ is maximal among all proper principal ideals.

Corollary. Let $R$ denote a principal ideal commutative ring. Then for all $x \in R,$ the following are equivalent.

$x$ is irreducible.
$Rx$ is maximal among all proper ideals.


Motivated by this realization, I was just wondering:

Question. Is there a name for those elements $x$ of a commutative ring $R$ such that $Rx$ is maximal among all proper ideals?

REPLY [6 votes]: These elements are called m-irreducible or i-atomic sometimes.  In rings with zero-divisors, these are actually distinct from irreducible elements.  I recommend the paper by D.D. Anderson and Valdez Leon called Factorization in Commutative Rings with Zero-divisors if you want to read more.  In a domain which is not a field, $(0)$ is not maximal among principal ideals (obviously) despite being irreducible, so the version you state is not quite even true in domains, although it is for non-zero elements. 
The paper I suggest gives you some non-trivial examples of all the various relations between the notions of irreducible as well as examples which show the reverse inclusions are not true.  
Edit: I figured I would add the example just in case people cannot access that paper.  In $\mathbb{Z} \times \mathbb{Z}$, the element $(0,1)$ is prime and therefore irreducible, but not maximal among principal ideals since it is properly contained in $(2,1)$.