Why is the set of elements in a valuation ring $R$ with positive valuation the unique maximal ideal of $R$?

Question

Definition. Let $K$ be a field and let $G$ be a totally ordered abelian group. A valuation of $K$ with values in $G$ is a map $v: K-\{0\} \rightarrow G$ such that for all $x, y \in K, x, y \neq 0,$ we have $$ \begin{array}{l} \text { (1) } v(x y)=v(x)+v(y) \\ \text { (2) } v(x+y) \geqslant \min (v(x), v(y)) . \end{array} $$

If $v$ is a valuation,show that :

1. $R=\{x \in K \mid v(x) \geqslant 0\} \cup\{0\}$ is a subring of $K$.

2. $m=\{x \in K \mid v(x)>0\} \cup \{0\}$ is unique maximal ideal of $R$.

I can solve 1), but how can we show $m$ is an ideal of $R$ and also the unique maximal ideal of $R$?

Answer 1

That $m$ is an ideal follows readily from the definition of the valuation since:

• if $x$, $y$ are in $m$ then $v(x+y)\geq\min(v(x),v(y))>0$ so that $x+y\in m$;
• if $x\in m$ and $r\in R$ then $v(xr)=v(x)+v(r)>0$ so that $xr\in m$.

To show that $m$ is maximal, observe that if $0\neq r\in R$ but $r\notin m$ then $v(r)=0$. Thus $v(r^{-1})=-v(r)=0$ and $r^{-1}\in R$, i.e. $r$ is invertible.

Then you can use the following well known fact: If $R$ is a ring and $m\subset R$ an ideal such that all elements in $R$ not in $m$ are invertible, then $m$ is a maximal ideal.