I’m trying to see if there is a proof of the following theorem using Galois theory, or number theory of some kind. Specifically, I’m looking at (iii) implies (i) below.

**Theorem:** Let $ K$ be a field and let $ \mathcal{O}$ be a subring. Let $ l$ be an algebraically closed field, and let $ \phi : \mathcal{O} \rightarrow l$ be a map into $ l$ . Then, for any $ a \in K^\times$ , there is either an extension of $ \phi$ from $ \mathcal{O}[a]$ into $ l$ or an extension of $ \mathcal{O}[a^{-1}]$ into $ l$ .

My intuition for the theorem is as follows. For any subring $ \mathcal{O}$ of $ K$ , we have a partially ordered abelian group $ K^\times / \mathcal{O}^\times$ , where the partial order on $ K^\times$ is given by declaring $ x \leq y$ when there is $ a \in \mathcal{O}$ such that $ x a = y$ , and the partial order on $ K^\times / \mathcal{O}^\times$ is given by declaring $ xU \leq yU$ when there exists $ a \in A$ such that $ xa \mathcal{O}^\times = y \mathcal{O}^\times$ .

This theorem is sort of saying that we can add at least one of the relations $ 0 \leq a$ and $ 0 \geq a$ in $ K^\times / \mathcal{O}^\times$ (mod out by the convex subgroup they generate), collapsing it so that it now corresponds to the ring $ \mathcal{O}[a]$ or $ \mathcal{O}[a^{-1}]$ . A corollary is that we can get $ K^\times / \mathcal{O}^\times$ to be totally ordered this way.

What is going on in this proof? Maybe it could be clarified with some higher number theory or Galois theory.