Can someone give an example of a commutative ring where there are elements $a$ and $b$ such that $a\mid b$ and $b\mid a$ but $a$ and $b$ are not associates (i.e. there is no unit $u$ such that $a = ub$)?
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1Just by chasing definitions around, it seems $a,b$ are associates in a ring $R$ iff $R$ is an integral domain (see: https://en.wikipedia.org/wiki/Divisibility_(ring_theory)). So if you were to start considering rings, excluding those would be helpful – Sidharth Ghoshal Jan 10 '16 at 05:01
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I couldn't find a definition of associates (other what's in @frogeyedpeas' link and there it's the same as $a\mid b$ and $b \mid a$). Can you include a definition in your question? – Rudy the Reindeer Jan 10 '16 at 06:49
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@frogeyedpeas Well, "if" but not really "only if" so much... – rschwieb Jan 10 '16 at 13:03
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For a more thorough answer and other questions related to what you are asking, I recommend the paper "Factorization in Commutative Rings with Zero-Divisors" by Anderson and Valdez-Leon if you are interested in more associate properties in rings with zero-divisors. There is a lot of information about various notions of irreducibles that come from the different notions of associate relations. A more interesting example than $0$ is given in Example 2.3 of the paper (linked below)
https://projecteuclid.org/euclid.rmjm/1181072068
If you cannot access it, the ring is $F[X,Y,Z]/(X-XYZ)$ so that $x=xyz$ implies $(x)=(xy)$ yet there is no unit $\lambda$ such that $\lambda x= xy$.
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