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In what follows let all variables be in some Euclidean Domain $R$.

  1. Suppose $n = kq + r$ s.t. $\deg(r) < \deg(k)$ or $r = 0$.
  2. Now suppose that $k \mid n$ so that $n = kq_1$ for some $q_1 \in R$.

Question 1: Since we have that $n = kq + r = kq_1 + 0$, can we somehow conclude that $r = 0$ and $q_1 = q$?

Question 2: If yes, do we even need $R$ to be a Euclidean Domain, or can it just be a UFD?

Community
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user115803
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2 Answers2

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It is known that if $D$ is a euclidean domain with euclidean algorithm having unique quotient and remainder, then $D$ is field or $D = F[x]$ for a field $F.$ For proofs see

M. A. Jodeit, Uniqueness in the division algorithm, Amer. Math. Monthly 74 (1967), 835-836.

T. S. Rhai, A characterization of polynomial domains over a field, Amer. Math. Monthly 69 (1962), 984-986.

Bill Dubuque
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You have $$kq_1=n=kq+r$$ Rewrite as $$k(q_1-q)=r$$ But you also have some condition relating $r$ and $k$ involving degree. I'm a bit unsure what you mean by "deg" since in a general Euclidean domain it's a Euclidean function not degree; but whatever you've got implies that $q_1-q$ must be zero to avoid contradiction.

In answer to question 2, absent a Euclidean function you have no contradiction, since $k$ and $r$ might be associates, and $q-q_1$ a unit.

vadim123
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