I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse norm.
$\mathbb{Z}/n\mathbb{Z}$ with $n$ composite is an example of a principal ideal ring that is not a domain. However, I was unable to think of a euclidean ring (a ring that admits an Euclidean function), but is not a domain.
- Is there a Euclidean ring that is not a domain?
- Is there a finite Euclidean ring? (It seems unplausible because it would be hard to give order.)
Added:
I was referring to the usual euclidean function as the definition does not seem to require the ring to be domain. If $R$ is a commutative unital ring,
a euclidean function is $N:R \backslash \{ 0 \} \to \mathbb{Z}_{> 0}$ such that for all nonzero $a,b \in R$, there exists $q,r \in R$ such that $a = bq +r$ and either $r=0$ or $N(r) <N(b)$.
I think $R = \mathbb{F}_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), \quad 1+x = 0x + (1 + x) = 1(1+x).$$ Also $x = (1+x)x = 1x.$ I think that covers it.