How to show that a valuation ring has a unique maximal ideal?

Question

A subring $R$ of a field $K$ is said to be a valuation ring of $K$ if for each $x$ $\in$ $K^{*}$ we have either $x$ $\in$ R or $x^{-1}$ $\in$ $R$. How can I show that the valuation ring has a unique maximal ideal?

Answer 1

1. In an evaluation ring, all the ideals are linearly ordered.

This in turn is equivalent to another property:

2. the principal ideals are linearly ordered.

I'll leave the equivalence of 1 and 2 as an exercise.

Proving 2) holds in our ring is a snap: let $a$ and $b$ be any nonzero elements. Then according to the assumption $ab^{-1}\in R$ or $ba^{-1}\in R$. In the former case, $(a)\subseteq (b)$ and in the second case $(b)\subseteq(a)$.

Now, every ring with identity has a maximal ideal, and the ideals are linearly ordered, hence there is only a single maximal ideal.

Answer 2

It should be clear what set $m$ should be the maximal ideal of $R$ (what is it?), and with this definition, that every element of $R$ outside of $m$ is a unit. This immediately implies that if $m$ is an ideal, then it is the only maximal ideal. It remains to show that $m$ is an ideal.

It is easy to prove that for $r \in R$ and $a \in m$, we have $ra \in m$ as well. What is perhaps difficult is to show that if $a,b \in m$, then $a+b$ is in $m$. Of course, we may assume that both $a$ and $b$ are nonzero. That $R$ is a valuation ring implies that $a/b$ or $b/a$ is in $R$. Now use $$a+b = b(1+a/b) \quad \text{and} \quad a+b = a(1+b/a)$$ depending on the two cases.