# Transfinite induction and valuation rank

In Engler's valued fields, exercise 3.5.2 goes as follows

Construct valuations on $\Bbb{C}$ of rank $\kappa$ for every cardinal $\kappa \leq 2^{\aleph_0}$.

The idea behind this (for any countably infinite rank) is that there are two kinds of transcendental valuation extensions, the Gauss extension, which does not change the rank of the value group, and the one that increases the rank of the value group by $1$. Also, algebraic extensions do not change the rank of the value group. So since the only extension on $\Bbb{Q}$ is the $p$-adic extension of rank $1$, we can use induction to obtain valuations on $\Bbb{C}$ for any countably infinite rank by taking a transcendence basis for $\Bbb{R}/\Bbb{Q}$.

I'm not very familiar with transfinite induction however. So how does one use transfinite induction to get valuations of uncountably infinite rank?

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What is the rank of a valuation? – Asaf Karagila Jun 15 '12 at 21:52
@AsafKaragila A valuation is defined as a surjective map from a field $K$ to an ordered abelian group (union $\infty$). The rank of the ordered abelian group is the number of its proper convex subgroups. – Eugene Jun 15 '12 at 21:59
Well, you say that there is a way to extend this rank by one at each step. Try a direct/inverse limit at limit stages? (I'm just shooting from the hip here...) – Asaf Karagila Jun 15 '12 at 22:03
@AsafKaragila I know I'm supposed to use some sort of transfinite induction here though. The order continuum case should be no different from the countably infinite case. – Eugene Jun 15 '12 at 22:08
The rank is just the Krull dimension of the valuation ring. – Martin Brandenburg Jun 22 '12 at 9:07