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.
$g'(1) = 0$
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 $$g'(\sum_i u_i) \leq \max(g'(u_i)) = 0 \implies g'(\sum_i u_i) = 0 \implies \sum_i u_i \ is \ a \ unit$$.