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Are there integral domains which admit ordinal-valued Euclidean functions but not $\mathbb{N}$-valued Euclidean functions?

flag	asked Oct 21 at 1:52 Sridhar Ramesh 2,361●10●20

	Nice question! A possible candidate: the omnific integers or some other ring associated closely with the ordinals. The ordinals themselves support division in the form $\forall\alpha,\beta\exists\gamma,r\ \beta=\alpha\gamma+r$, where $r\lt\alpha$, since one takes as many copies of $\alpha$ that fit into $\beta$, and $r$ is the leftover part. I'm not sure, however, if this feature extends fully to the Omnific integers with the symmetric arithmetic operations of the surreal numbers. – Joel David Hamkins Oct 21 at 2:18
	Or the polynomials over a field, but allowing transfinite ordinal degrees, which add by symmetric ordinal addition. – Joel David Hamkins Oct 21 at 3:01
	As the floor function is defined in the surreals (floor(x)=\{x−1\|x+1\}) is an omnific integer, Euclidian division is indeed possible – Feldmann Denis Oct 21 at 3:30
	I don't think the omnific integers can admit an ordinal-valued (as opposed to omnific integer-valued) Euclidean function; an ordinal-valued Euclidean function should still lead to being a unique factorization domain, in the usual way, but the omnific integers lack unique factorization (e.g., 2, 3, 5, 7, ..., all remain prime, but $\omega$ is divisible by all infinitely many of them). – Sridhar Ramesh Oct 21 at 18:53

Properly “transfinitely” Euclidean domains