How to reconstruct a valuation from a valuation ring

Question

I have been independently studying valuation rings for a research paper for an undergraduate abstract algebra course. I've been heavily relying on Gathmann's lecture notes, found at http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/chapter-12.pdf.

My question pertains to Construction 12.7 in Gathmann's notes. I understand the set up for the construction - using the First Isomorphism Theorem and why the value group is ordered - however, I'm having a hard time understanding how the canonical homomorphism is a valuation that is isomorphic to the valuation we started with.

For those who don't want to search through the lecture notes, here is an outline of the construction. Let $R$ be the valuation ring for $\nu: K^* \rightarrow G$. Then we can recover $\nu$ up to isomorphism in the following sense:

1. $K$ is the field of fractions of $R$
2. $\ker\nu = U$ where $U$ is the group of units of $R$
3. $\nu(K^*)\cong K^*/U$ by the First Isomorphism Theorem
4. Define the new value map $\nu: K^* \rightarrow K^*/U$ to be the canonical homomorphism
5. The value group $\nu(K^*)$ is ordered since for all $a,b\in K^*$, $\nu(a)\leq\nu(b) \iff ba^{-1}\in R$

I would like for someone to walk me through how the map in $(4)$ functions as a valuation and how it is isomorphic to the map we started with. Thank you in advance.

Answer 1

First, the symbol $\nu$ seems to be used to mean two different things. I will write $\nu:K^*\to G$ for the given valuation and $\nu':K^*\to K^*/U$ for the canonical homomorphism. In order for $\nu'$ to be a valuation, the target $K^*/U$ needs to be an ordered abelian group. We define the order as follows: given $x_1$, $x_2\in K^*/U$, choose coset representatives $r_1$, $r_2\in K^*$, and define $x_1\leq x_2$ if and only if $r_2r_1^{-1}\in R$. It is an exercise to check that the definition is independent of the coset representatives and in fact gives a total order on $K^*/U$.

To say that $\nu'$ is isomorphic to $\nu$ means there is an order-preserving homomorphism of the value groups $\varphi:K^*/U\to G$ such that $\varphi\circ \nu'=\nu$. Since $U=\ker(\nu)$, $\nu$ factors through $K^*/U$, and we can take $\varphi$ to be the induced map.