# Prime elements in a noncommutative ring

Is there a reasonable definition of prime element in a noncommutative ring? The definition from wikipedia makes the assumption of commutativity and I'd like to know how necessary this condition is. Given a ring $R$, define:

We say $p$ is an right prime in $R$ iff for all $x,y,k\in R$, $kp=xy\to\exists m\in R,mp\in\{x,y\}$. We say $p$ is a left prime in $R$ if it is a right prime in $R^{op}$, and $p$ is a two-sided prime if it is both left and right prime.

Note that the condition of being right prime is the same as $p\mid xy\to p\mid x\lor p\mid y$, where $|$ is the right divisibility relation (I spelled it out for clarity). Obviously all three conditions are equivalent to being a prime in $R$ if $R$ is commutative. Also, $p$ is right prime iff the right principal ideal $(p)_r$ generated by $p$ is a prime ideal. (Not certain if $p$ is two-sided prime iff the two-sided principal ideal is a prime ideal.)

Is there a reason such a definition is not popular? Does it inherit any of the properties of primes in commutative rings? I'm not really sure what nontrivial theorems you can say about primes though in an arbitrary commutative ring - most of the interesting math seems to come from stronger conditions that use primes in their definitions, i.e. "every element is a product of primes" (factorization domain) or "every prime ideal contains a prime" (UFD), which always assume that $R$ is commutative anyway.

• What would you like to do with such a definition? In the presence of noncommutativity there is essentially no hope of recovering any reasonable analogue of prime factorization. – Qiaochu Yuan Feb 7 '15 at 22:35
• @QiaochuYuan I guess part of the problem is that I don't know; my interest is in building a library of results in a formal system, which usually means I want to derive anything useful that is derivable. Certainly the FD condition makes sense in noncommutative rings, and the Kaplansky characterization of UFDs that I quoted also makes sense; and I'm not convinced that there is no noncommutative analogue of the "traditional" definition of UFDs. – Mario Carneiro Feb 7 '15 at 22:43
• @QiaochuYuan Given that there are various papers on non-commutative analogues of UFDs, I consider your assertion as somewhat surprising. – quid Feb 7 '15 at 22:44
• @quid No kidding! At least P.M. Cohn's papers deserve mention on this topic. It's to be expected though: many people raised in the glow of commutativity frequently expect nothing good of noncommutativity :) – rschwieb Feb 8 '15 at 0:15