# What makes a Principal Ideal Domain a Euclidean Domain?

The definitions of Principal Ideal Domain (PID) and Euclidean Domain (ED) are both from integral domain:

A non-trivial ring $$R$$ with no zero divisors is said to be entire; a commutative entire ring is called an integral domain.

Let $$R$$ be an integral domain. $$R$$ is a principal ideal domain (PID) if every ideal of $$R$$ is principal.

If $$R$$ is an integral domain, a Euclidean function on $$R$$ is a function $$f$$ from $$R\setminus\{0\}$$ to the non-negative integers satisfying the following fundamental division-with-remainder property:

(EF1) If $$a$$ and $$b$$ are in $$R$$ and $$b$$ is nonzero, then there exist $$q$$ and $$r$$ in $$R$$ such that $$a = bq + r$$ and either $$r = 0$$ or $$f (r) < f (b)$$.

A Euclidean domain is an integral domain which can be endowed with at least one Euclidean function.

And it's known that each ED is a PID, but not every PID is an ED.

However, what confused me is seems the requirements for an integral domain to be an PID or an ED are in different direction, and then surprisingly, walking a bit further from PID, one arrives ED.

Is there a sufficient and necessary condition, such that, if a PID fulfills it then it is an ED? i.e. something like:

If $$R$$ is a PID, $$R$$ is a Euclidean domain if it fulfills condition A) ...

• You need to be more precise about "this sufficient and necessary condition". We always can "add" a condition, which more or less is the same as being Euclidean, so that is not what you want. You could also ask "The definitions of odd integer (ODD) and prime numbers $p>2$ (PRIME) are both from integers. Is there a sufficient and necessary condition to ensure for an odd integer that it is a prime?" Since every prime $p>2$ is odd, but conversely not every odd integer is a prime. May 16, 2021 at 19:10
• @DietrichBurde sorry i wasn't clear. If I may try again.. seems $PID := \text{integral domain} + A$, and $ED := \text{integral domain} + B$, and $PID < ED$, so $A<B$. Is there a $C$ such that $ED := PID + C = \text{integral domain} + A + C = \text{integral domain} + B$? seems I can just take $C = B$ but this is a bit trivial... May 16, 2021 at 19:35
• As I said. It also seems that ODD=integer +A and PRIME = integer +B and $A<B$. But in fact, it is hard to add a $C$ as you want, both for ODD, PRIME and for PID, ED. May 16, 2021 at 19:37
• @DietrichBurde indeed. I just found it mysterious. Say John and Tom both started to see the world, John heading east and Tom heading west, of course they reached different places. If someone suddenly said, John could just march on a little bit, then he'll meet Tom, how surprising it is! Of course this could be explained by the fact that the earth is round. I wonder if there's something on top of $PID$ could lead to $ED$, so to explain the minor gap in between -- minor as, seems most $PID$ are actually $ED$. May 16, 2021 at 19:44