Let $\nu : K \rightarrow G \cup \{\infty\}$ be a map defined by, where $G$ is a totally ordered group and $g < \infty$ for all $g\in G.$
- $\nu(a) = \infty$ if and only if $a = 0,$
- $\nu(a + b) \geq \min\{\nu(a), \nu(b)\},$
- $\nu(ab) = \nu(a) + \nu(b)$ for all $a, b\in K$
Then $R = \{a\in K: \nu(a) \geq 0\}$ is a valuation domain and $M = \{a\in K: \nu(a) > 0\}$.
I want to show $M$ is the only maximal ideal.
Clearly $0\in M.$ $\nu(-1) = 0, \nu(a) = \nu(-a).$ Let $a\in M$ and $r\in R,$ then $ar\in M$ since $\nu(ar) = \nu(a) + \nu(r) > 0.$ For $x, y\in M,$ then $\nu(x + y)\geq \min \{\nu(x), \nu(y)\} > 0,$ since both greater than $0.$ This shows $x + y\in M.$ So $M$ is an ideal.
Now need to show $M$ is the unique maximal ideal. How do I complete the proof? Thank you.
The ring you're looking at is $$R=\{a\in K:\nu(a)\geq 0\}.$$ Suppose that $I\subseteq R$ is an ideal containing some $a\in I$ with $\nu(a)=0$. Then observe that $$0=\nu(1)=\nu(a^{-1}a)=\nu(a^{-1})+\nu(a)=\nu(a^{-1})$$ and the fact that $\nu(a^{-1})=0$ means that $a^{-1}\in R$ (by the definition of $R$). Since $I$ is an ideal and $a\in I$ and $a^{-1}\in R$, we conclude that $1=a^{-1}a\in I$, and hence $I=R$.
Thus, we've shown that any ideal $I\subseteq R$ containing an element of valuation $0$ must be equal to $R$. Therefore every proper ideal of $R$ contains only elements of valuation $>0$, or in other words, every proper ideal of $R$ is contained in the ideal $M=\{a\in K:\nu(a)>0\}$.