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Matsumura gives us the following definition of an additive valuation.

A map $\nu: K \rightarrow H \cup \{\infty\}$ from a field $K$ to $H \cup \{\infty\}$ is called an additive valuation or just a valuation of $K$ if it satisfies the conditions

(1) $\nu(xy) = \nu(x) + \nu(y)$

(2) $\nu(x+y) \geq \text{min}\{(\nu(x), \nu(y))\}$

(3) $\nu(x) = \infty \iff x=0$

At the end of the page (p.75), Matsumura asks us to prove the following:

Let $\nu: K \rightarrow G \cup \{\infty\}$ and $\nu' : K \rightarrow G' \cup \{\infty\}$ be two additive valuations where $K$ is a field and $G, G'$ are ordered groups. Suppose that both $\nu$ and $\nu'$ have the same valuation ring $R$. Then I want to prove that there is an order-isomorphism $\phi: H \rightarrow H'$ with $\nu' = \phi\nu$, where $H$ and $H'$ are the images of $\nu$ and $\nu'$, respectivaly.

This was my initial thought:

Let $x \in H$, then $\exists \space a \in K$ with $\nu(a) = x$. So define $\phi: H \rightarrow H'$ by $x \mapsto \nu'(a)$.

Suppose that $\nu(a)=x$ and $\nu(b)=y$. Then $\phi(x) + \phi(y)= \nu'(a) + \nu'(b) = \nu'(ab) = \phi(x + y)$ since $x + y =\nu(a) + \nu(b) = \nu(ab)$. So $\phi$ is obviously a group homomorphism.

Also if $\nu(a) \geq \nu(b)$, then $\nu(ab^{-1}) = \nu(x) - \nu(y) \geq 0 \implies ab^{-1} \in R$. Since $\nu$ and $\nu'$ have the same valuation ring, $\nu'(ab^{-1}) \geq 0$ as well.

But...I later realized that $\nu$ is not necessarily injective. So it may be that two elements in $K$ are sent to the same element in $G$...so then $\phi$ would not be well-defined. However, I cannot see any other way to define $\phi$, so I would appreciate if anybody helped me out.

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Suppose there are elements $a,b\in K$ such that $\nu(a)=\nu(b) = x$, but $\nu'(a)<\nu'(b)$. Then $\nu(ab^{-1}) = x-x = 0$, so $ab^{-1}\in R$. However, $\nu'(ab^{-1}) = \nu'(a)-\nu'(b)<0$, which means that $ab^{-1}\notin R$. This is a contradiction.

This implies that your map is well defined and injective.

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