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Feb 19 at 3:46 comment added Caleb Stanford For reference, the theorem is that: "$a$ very strongly irreducible $\implies$ $a$ is $m$-reducible $\implies$ $a$ is strongly irreducible $\implies$ $a$ is weakly irreducible. Moreover, none of these implications can be reversed." I've posted a related answer on mathSE.
Feb 19 at 3:45 comment added Caleb Stanford m-irreducible), which holds only for $a \ne 0$. With the definition in this answer, things work out better, very strong irreducible implies m-irreducible for all $a \in R$. Second, there seems to be an error in Theorem 2.13: they say that $p$ is strongly irreducible implies that $a$ is both weakly irreducible and prime. But this is false for integral domains. That is, it is well-known that there are examples of irreducible (hence very strongly irreducible) elements that are not prime. Other than this, I think the theorem is correct.
Feb 19 at 3:43 comment added Caleb Stanford This is an excellent answer! I just took a look at Anderson and Valdes-Leon's paper and there are a couple of "gotchas" I wanted to point out in case it is helpful for future readers. First, AV define $a \cong b$ slightly differently than in this answer: they add the fact that $0 \cong 0$. But this seems like a mistake in definitions; it means their "$a$ is very strongly irreducible" only agrees with the usual def. of irreducible when $a \ne 0$. And this mistake shows up later in Theorem 2.13 (very strong irreducible implies (contd)
Aug 16, 2012 at 11:06 history edited CPM CC BY-SA 3.0
reference correction
Aug 16, 2012 at 9:49 history edited CPM CC BY-SA 3.0
Added to original answer by adding more information regarding a comment by the original poster
Aug 16, 2012 at 9:29 comment added CPM This is a very good paper as well! Bouvier and Allard also have written a lot of papers on this subject; however, I have only been able to find French versions of their papers, which might pose some problems. D.D. Anderson and Frazier's paper linked by Guntram has a pretty extensive bibliography. This is unfortunately why it gets a bit complicated. Many authors use different terminology for similar content. I have run out of characters, so I will edit my answer above to add a few comments there about this. (I am new, not sure what protocol is here, sorry if this is bad form to do)
Aug 16, 2012 at 7:57 vote accept Tom De Medts
Aug 16, 2012 at 7:57 comment added Tom De Medts Thanks for your answer and for the references! While Googling for the papers, I bumped into another related paper "Unique Factorization Rings with Zero Divisors" by S. Galovich (Mathematics Magazine 51, No. 5, 1978) that also seems interesting.
Aug 15, 2012 at 22:01 history edited CPM CC BY-SA 3.0
fixed a typo
Aug 15, 2012 at 21:44 history answered CPM CC BY-SA 3.0