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I was reading this old paper which asserts that if $E$ is an euclidean domain with unique division we have $E\approx \mathbb{F}$ or $E\approx \mathbb{F}[X]$ where $\mathbb{F}$ is a field.

The author first proves $g: E\rightarrow \mathbb{Z}_{\geq 0}$ is such that $g(a+b)\leq \text{max}\{g(a),g(b)\}$ if division is to be unique where $g$ is the "norm function" in the euclidean domain.

He then goes on to build a certain $g'$ with $g'(1_E)=0$. He takes $g'(x)=g(x)-g(1_E)$ and I understand that is okay, because $g( 1_E)\leq g(1_Ea)=g(a)$ so $g': E \rightarrow \mathbb{Z}_{\geq 0}$.

He then goes on to build a certain $\mathbb{F}$ but I am completely lost there. He says "it is enough to show that a nonzero sum of units is a unit", but I do not know what a unit is. I am sorry if this is too easy, but I only know the definition of a field being a ring endowed with multiplicative inverse for every non zero element.

Dan Asimov
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Kadmos
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    $u$ is called a unit when there is $v$ such that $uv=vu=1$. Elements with inverses. – NDB Jul 08 '23 at 00:48
  • Ow! So he considers the subset of $E$ with invertible elements and that is clearly a field (if they are addition closed)! Thanks. But I still do not follow why $g'(u_1+u_2)=0$ implies $u_1+u_2$ is invertible (if each $u_i$ is invertible). – Kadmos Jul 08 '23 at 00:49
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    If $g(u)=0$, apply the Euclidean condition as in "dividing $1$ by $u$". The condition tells that there is $q$ and $r$ with $1=qu+r$ and either $r=0$ or $g(r)<g(u)=0$. Since the latter cannot happen, then $r=0$. So, $1=qu$ and $u$ has inverse $q$. – NDB Jul 08 '23 at 00:55

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$g'(1) = 0$

Let $u$ be a unit. Here we have $uq = 1$ for some $q$. $$g'(u) \leq g'(uq) = g'(1) = 0 \implies g'(u) = 0$$

Now based on comments by NDB:

If $g'(u) = 0$, then $$1 = uq + r \implies g'(r) < g'(u) = 0 \implies r = 0$$ and hence $u$ is a unit.

Hence $u$ is a unit $\iff$ $g'(u) = 0$.

let $u_1,u_2,..$ be units. Now $$g'(\sum_i u_i) \leq \max(g'(u_i)) = 0 \implies g'(\sum_i u_i) = 0 \implies \sum_i u_i \ is \ a \ unit$$.

Balaji sb
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