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Let $(M,*,1)$ be a commutative monoid. Define the binary relation $R$ on $M$, such that $xRy$ iff $(\exists z)(x*z=y)$. It is easy to show that, since $M$ is a commutative monoid, the relation $R$ is both transitive and reflexive. I read in a paper the theorem that if additionally $M$ is cancellative and pure, (where pure means $1$ is the only invertible element), then $R$ is also antisymmetric. I am wondering if this hypothesis can be weakened. That is, if $M$ is merely cancellative, is $R$ antisymmetric? And also, if $M$ is merely pure, is $R$ antisymmetric? All of this is given the background assumption that $M$ is a commutative monoid.

user107952
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After some tinkering around, I found the answer to my question. Pure is needed, because otherwise $x | 1$ and $1 | x$. As for cancellative, here is a counter-model I found where pure alone is not sufficient: Let $M = \{0,1,2\}$, let the identity of $M$ be $0$, and let $+$ be defined as $0+x=x+0=x$, $1+1=2+2=1$, and $1+2=2+1=2$. Then $(M,+,0)$ is a commutative pure monoid, where the divisor relation $R$ is not anti-symmetric, as both $1R2$ and $2R1$

user107952
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Today, I replied to another related question which weakens the assumptions in another way: A weak cancellation property for monoids

Essentially, the assumption of cancellativity can be weakened. If the commutative monoid $M$ with identity $1$ has the property that for all $a,b\in M,$ $ab=a$ implies $b=1$ (perhaps called weak cancellativity), then $R$ is antisymmetric.

If a monoid is pure, it need not be antisymmetric. Just look at $\{-1,1\}$ with multiplication.

Can you please post a reference to the article you mentioned? I would be very interested in it.

Vince Matsko
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