The Unapologetic Mathematician

Mathematics for the interested outsider


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