# The Unapologetic Mathematician

## Divisibility

There is an interesting preorder we can put on the nonzero elements of any commutative ring with unit. If $r$ and $s$ are nonzero elements of a ring $R$, we say that $r$ divides $s$ — and write $r|s$ — if there is an $x\in R$ so that $rx=s$. The identity $1$ trivially divides every other nonzero element of $R$.

We can easily check that this defines a preorder. Any element divides itself, since $r1=r$. Further, if $r|s$ and $s|t$ then there exist $x$ and $y$ so that $rx=s$ and $sy=t$, so $r(xy)=t$ and we have $r|t$.

On the other hand, this preorder is almost never a partial order. In fact since $r(-1)=-r$ and $-r(-1)=r$ we see that $r|-r$ and $-r|r$, and most of the time $r\neq-r$. In general, when both $r|s$ and $r|s$ we say that $r$ and $s$ are associates. Any unit $u$ comes with an inverse $u^{-1}$, so we have $u|1$ and $1|u$. If $r=su$ for some unit $u$, then $r$ and $s$ are associates because $s=ru^{-1}$.

We can pull a partial order out of this preorder with a little trick that works for any preorder. Given a preorder $(P,\preceq)$ we write $a\sim b$ if both $a\preceq b$ and $b\preceq a$. Then we can check that $\sim$ defines an equivalence relation on $P$, so we can form the set $P/\sim$ of its equivalence classes. Then $\preceq$ descends to an honest partial order on $P/\sim$.

One place that divisibility shows up a lot is in the ring of integers. Clearly $n$ and $-n$ are associate. If $m$ and $n$ are positive integers with $m|n$, then there is another positive integer $x$ so that $mx=n$. If $x=1$ then $m=n$. Otherwise $m\lneq n$. Thus the only way two positive integers can be associate is if they are the same. The preorder of divisibility on $\mathbb{Z}^\times$ induces a partial order of divisibility on $\mathbb{N}^+$.

April 28, 2007 Posted by | Ring theory | 6 Comments