# In arbitrary commutative rings, what is the accepted definition of “associates”?

In an integral domain, the following are equivalent:

1. $r \mid s$ and $s \mid r$
2. $r=us$ for some unit $u$

However in arbitrary commutative rings this is no longer the case; in particular, (2) implies (1) but not the converse.

Question. Is there an accepted meaning for the phrase "associates" in an arbitrary commutative ring? And if the accepted meaning is (2), then what do we call (1)?

• It seems that wikipedia (en.wikipedia.org/wiki/Unit_(ring_theory)) takes it to mean (2). Don't know what (1) would be generally called though. – Hayden Jul 10 '14 at 13:01
• @Hayden, thank you. However, its worth mentioning that on finer points like this, wikipedia's conventions should probably not be taken too seriously. I'm mainly interested to hear from people who have studied non-integral domain commutative rings. – goblin Jul 10 '14 at 13:04
• I agree, which is why I added it only as a comment, in the hopes that it would at least help. Good luck. – Hayden Jul 10 '14 at 13:06

In rings with zero-divisors, factorization theory is much more complicated than in domains, e.g. $\rm\:x = (3+2x)(2-3x)\in \Bbb Z_6[x].\:$ Basic notions such as associate and irreducible bifurcate into at least a few inequivalent notions, e.g. see the papers below, where three different notions of associateness are compared:

• $\ a\sim b\$ are  associates  if $\, a\mid b\,$ and $\,b\mid a$
• $\ a\approx b\$ are  strong associates  if $\, a = ub\,$ for some unit $\,u.$
• $\ a \cong b\$ are  very strong associates  if $\,a\sim b\,$ and $\,a\ne 0,\ a = rb\,\Rightarrow\, r\,$ unit

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles.
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

Factorization in Commutative Rings with Zero-divisors.
D.D. Anderson, Silvia Valdes-Leon.
Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480

• Thanks; so it seems that condition 1 is called "associates" and condition 2 is called "strong associates." – goblin Jul 10 '14 at 13:38
• @user18921 The terminology is very far from standard in this area. – Bill Dubuque Jul 10 '14 at 13:40