Is all Euclidean domain local?

Question

We say a ring is local iff its all nonunits form an (maximal) ideal.

But the integer set Z with the norm f(n)=|n| is a Euclidean domain and its all nonunits dont form an ideal; since 3, -2 are nonunit elements of Z but 3+(-2)=1 is a unit, thus nonunit elements doesnt form an additive subgroup, and thus an ideal.

Am i right? And then where the local domain is located? Like, we know that UFD > PID > ED > field. Is a local domain located between them? Or is it an independent concept?

Answer 1

Hint $\ $ See the graphs below, from Hutchins' book Examples of Commutative Rings. You may find it an illuminating exercise to prove that those are the only arrows from/to local domains (many interesting (counter)examples are given in the book, which is based upon Hutchins' $1978$ Chicago thesis under Kaplansky).

Answer 2

You are correct: a Euclidean domain need not be local. The condition of a ring being a local domain is independent of being a UFD, a PID, or a Euclidean domain. However, every field is a local domain (try proving this yourself, it's not hard). Also, every local PID is a Euclidean domain, so there is no distinction between PIDs and Euclidean domains among local rings (proof sketch: a local PID has a single prime $p$, and every nonzero element is associate to a power of $p$, so define your Euclidean function to send an element $x$ to the $n$ such that $x$ is associate to $p^n$).