There is a binary relation $\mid^*$ defined on any commutative ring as follows: $a \mid^* b$ iff $ak=b$ for some $k \in R$ that is not a zero divisor. This is always transitive, and it is reflexive except on the trivial ring. We also have: $$(*) \qquad x \mid^* 0 \rightarrow x=0.$$

Now in an integral domain, the relation $\mid^*$ almost completely agrees with the more usual $\mid$, with the exception that whereas $x \mid 0$ is always true, nonetheless by $(*)$ the statement $x \mid^* 0$ is never true for $x \neq 0$.

However in other commutative rings it can behave differently. For a simple example, observe that whereas $1 \mid x$ is always true, nonetheless $1 \mid^* x$ is equivalent to $x$ not being a zero-divisor. For another example of how they can diverge, observe that in $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, we have $(1,1,0) \mid (1,0,0)$. But it can't be the case that $(1,1,0) \mid^* (1,0,0)$. In general, I think that $\mid^*$ behaves much more like an equivalence relation than $\mid$ does. Its almost as if $a \mid^* b$ is 25% of the way to saying that there exists a unit $u$ such that $au=b$.

Soft Question. Does the relation $\mid^*$ have any interesting applications for understanding the structure of commutative rings that aren't integral domains? If so, what are some examples of not-entirely trivial applications of $\mid^*$?

  • $\begingroup$ @ThomasAndrews, I don't quite know what you mean, but I've changed it to $\mathbb{Z} \times \mathbb{Z}$ to better illustrate. $\endgroup$ – goblin Mar 24 '15 at 14:28
  • $\begingroup$ Basically, in any ring, it is true that if $a\mid^* 0$ then $a=0$. You first illustrate this in $\mathbb Z$, then you illustrate it in $\mathbb Z\times\mathbb Z$. That isn't "different." $\endgroup$ – Thomas Andrews Mar 24 '15 at 14:29
  • $\begingroup$ @ThomasAndrews, good point. I fixed it. $\endgroup$ – goblin Mar 24 '15 at 14:39
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    $\begingroup$ @ThomasAndrews, and stuffed it up. But now I think I fixed it :) $\endgroup$ – goblin Mar 24 '15 at 14:39
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    $\begingroup$ This gives a disconnected poset, which is interesting. For example, $\mathbb Z$ has two components, while $\mathbb Z^n$ has $2^n$ components. $\mathbb Z/n\mathbb Z$ has $\tau(n)$ components. In all cases, there is a natural order amongst the components when you consider them under the standard $\mid$ relation. $\endgroup$ – Thomas Andrews Mar 24 '15 at 14:51

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