I am reading Goldschmidt's Algebraic functions and projective curves. From the book:
Let $K$ be a field. An integral domain $\mathcal{O}\subset K$ is a valuation ring if for all $x\in K$ either $x$ or $x^{-1}$ is in $\mathcal{O}$. The valuation afforded by $\mathcal{O}$ is the natural map $\nu: K^\times\rightarrow K^\times/\mathcal{O}^\times$, where $\cdot^\times$ is the group of units. Accordingly we define a total order on $V:=K^\times/\mathcal{O}^\times$ by $\underline{a\mathcal{O}^\times\le b\mathcal{O}^\times}$ if $\underline{a^{-1}b\in\mathcal{O}}$.
Now my question: I wish to verify that $\nu(a+b) \ge \min\{\nu(a),\nu(b)\}$, but I am a bit lost in which operations take place in which structure.
I think the addition in $\nu(a+b)$ means that we write $K^\times$ additively. However the first underlined part seems to indicate that we write $K^\times$ multiplicative. Furthermore in what structure does the multiplication in the second underlined part take place?
Can somebody help to clarify the constructions after which I hope to be able to prove that $\nu(a+b) \ge \min\{\nu(a),\nu(b)\}$. Thanks in advance!
Edit: The book states that it writes $V$ additively.
Just to have this answered.
You want to show that $v(a+b)\geqslant\min\{v(a),v(b)\}$. WLOG you can assume $v(a)\leqslant v(b)$. By definition, this means that $a^{-1}b\in\mathcal{O}$. To show that $v(a+b)\geqslant v(a)$, it suffices to show that $a^{-1}(a+b)\in\mathcal{O}$. But, $a^{-1}(a+b)=1+a^{-1}b$. But, $1,a^{-1}b\in\mathcal{O}$ so $1+a^{-1}b\in\mathcal{O}$.