valuation domain is local

Theorem.

Every valuation domain is a local ring.

Proof.  Let $R$ be a valuation domain and $K$ its field of fractions.  We shall show that the set of all non-units of $R$ is the only maximal ideal of $R$.

Let $a$ and $b$ first be such elements of $R$ that $a-b$ is a unit of $R$; we may suppose that  $ab\ne 0$  since otherwise one of $a$ and $b$ is instantly stated to be a unit.  Because $R$ is a valuation domain in $K$, therefore e.g.  $\frac{a}{b}\in R$.  Because now  $\frac{a-b}{b}=1-\frac{a}{b}$  and  ${(a-b)}^{-1}$  belong to $R$, so does also the product  $\frac{a-b}{b}\cdot {(a-b)}^{-1}=\frac{1}{b}$,  i.e. $b$ is a unit of $R$.  We can conclude that the difference $a-b$ must be a non-unit whenever $a$ and $b$ are non-units.

Let $a$ and $b$ then be such elements of $R$ that $ab$ is its unit, i.e.  $a^{-1}{b}^{-1}\in R$.  Now we see that

$$a^{-1}=b\cdot a^{-1}{b}^{-1}\in R,b^{-1}=a\cdot a^{-1}{b}^{-1}\in R,$$

and consequently $a$ and $b$ both are units.  So we conclude that the product $ab$ must be a non-unit whenever $a$ is an element of $R$ and $b$ is a non-unit.

Thus the non-units form an ideal $𝔪$.  Suppose now that there is another ideal $𝔫$ of $R$ such that  $𝔪\subset 𝔫\subseteq R$.  Since $𝔪$ contains all non-units, we can take a unit $\epsilon$ in $𝔫$.  Thus also the product ${\epsilon }^{-1}\epsilon$, i.e. 1, belongs to $𝔫$, or  $R\subseteq 𝔫$.  So we see that $𝔪$ is a maximal ideal.  On the other hand, any maximal ideal of $R$ contains no units and hence is contained in $𝔪$; therefore $𝔪$ is the only maximal ideal.