Suppose a ring R is a PID. What does this say about possible orders on the set of R?

Question

More explicitly, is there always some partial or total order we can apply to the set of R?

Answer 1

Every finite set can be totally (even well) ordered. If you assume the Axiom of Choice then you can drop "finite" assumption. Also note that every set can be equipped with trivial "$x\leq y$ iff $x=y$" ordering (thanks to @AsafKaragila comment). But I assume you want the ordering to preserve the ring structure.

For the purpose of this answer I will use strict $<$ ordering instead of $\leq$ (these are equivalent). We shall consider the following condition:

$$\forall_{a,x,y\in R}\ x<y \Rightarrow a+x<a+y$$

This is in many cases impossible. For example consider $R=\mathbb{Z}_n$ (or any finite ring). Assume there is an ordering of $R$ that is compatible with the ring structure as defined above. If $0<x$ for some $x\in R$ then by induction $0<x<2x<3x<\ldots$. But this sequence has to repeat itself eventually (we have finite number of elements in $R$). Thus we obtain that $kx<kx$ for some $k\in\mathbb{N}$. This is a contradiction. In particular $0$ cannot be compared with any other element. But $0$ is not special, we can generalize it: if $x<y$ for some $x,y$ then $x-x<y-x$ (again by the property) and thus $0<y-x$ which is a contradiction by the previous statement.

In particular the only (strict) partial ordering on $R$ that is compatible with addition is when no two elements can be (strictly) compared. This is boring.

Note that for $n=p$ prime the ring $\mathbb{Z}_p$ is a field thus PID.