Let $K$ be a field, $G$ a totally ordered group and $v:K \rightarrow G\cup\{\infty\}$ be a valuation on $K$.
I am trying to show that $m_v=\{k \in K: v(k)>0\}$ is a (the?) maximal ideal in $R_v=\{k \in K: v(k)\geq0\}$ the valuation ring of $v$.
I thought of trying to show by using the definition of an ideal being maximal in $R_v$ but I am not sure how to complete the proof that way. My second method has been to try to show that $R_v/m_v$ is a field. As $R_v$ is a commutative ring with identity, $R_v/m_v$ is also a commutative ring with identity and $(1+m_v)\neq (0+m_v)$ for if so then it would mean $1 \in m_v$ so $m_v=R_v$ but $v(1)=0$ so $1 \in R_v$ but $1 \notin m_v$. So $m_v$ is non-trivial. I am trying to find the inverse of some non-zero $(a+m_v) \in R_v/m_v$. I have that the statement $$(a+m_v)(x+m_v)=(1+m_v)$$ is true if and only if $ax-1 \in m_v$. I call $ax-1=m$ and I see that $x=a^{-1}(m+1)$. I am now trying to show that $x \in R_v$. If $a^{-1} \in R_v$ it looks like $x \in R_v$. The case I am working on now is if $a^{-1}\notin R_v$. In this case $v(a^{-1})<0$ and I am having some trouble now showing that $v(x)\geq 0$ for $x \in R_v$. I see that $v(m+1)\geq 0$ and that $v(a^{-1}(m+1))\geq\min(v(a^{-1}m,v(a^{-1}))=v(a^{-1})$ as $v(a^{-1}m)=v(a^{-1})+v(m)>v(a^{-1})$. Any hints from here or another method I can try?
**After looking at this a bit too I feel that I am goofing up on some fundamental things involving logic with my use of $m$. Any help?
Hint: If $A$ is a ring and $I\neq A$ is an ideal such that every $x\in A\setminus I$ is a unit of $A$, then $A$ is a local ring and $I$ is its maximal ideal. (This is Proposition 1.6i in Atiyah-Macdonald.)
(This is a complementary hint to Alex Youcis's answer.)