Non-archimedean valuation

Question

In the definition of non-archimedean valuation the value group is the group of real numbers $\mathbb R$. Can we replace $\mathbb R$ by any totally ordered group different from $\mathbb R$ ? Thanks

Answer 1

I'm not sure which definition you're talking about, but in general, a valuation can take values in an arbitrary totally ordered abelian group. That this group is abelian is a consequence of the fact that the multiplication in the ring is abelian. I suppose you could consider non-abelian valuation groups for non-commutative rings, but I've never heard of anything like that.

In fact, any totally ordered abelian group $A$ is a valuation group: if you consider the set $k((X^A))=\{f\colon A\to k\mid \operatorname{supp}(f)\textrm{ is well-ordered}\}$ where $k$ is an arbitrary field and $\operatorname{supp}(f)$ is the support, that is, the set of arguments with nonzero values, it has a natural field structure and it has a natural valuation $v(f)=\min(\operatorname{supp}(f))$, and by the definition $A$ is the valuation group. The well-orderedness is used to assure that multiplication is well-defined (so that we only have to sum over finitely many terms).

This definition is a natural extension of the concept of formal Laurent series $k((X))=k((X^{\bf Z}))$ with their natural integer-valued valuation, and the proof that this is a field and $v$ is a valuation is much the same in this case and in general.