Valuation Ring is Local

Theorem
Let $R$ be a valuation ring.

Then $R$ is a local ring.

Proof
Let us verify the definition.

Firstly, $1\ne 0$ in $R$, since $R$ is an integral domain.

Therefore $R$ is nontrivial.

Secondly, let $x, y \in R$ be non-units.

We shall show that $x + y$ is non-unit.

If $x = 0$ or $y = 0$, then the claim is trivial.

In the following, we assume that $x \ne 0$ and $y \ne 0$.

In particular, $x^{-1}$ and $y^{-1}$ exist in $K$, where $K$ is the field of quotients of $R$.

As $R$ is a valuation ring, either $x y^{-1} \in R$ or $x^{-1} y \in R$ must be true.

In view of the symmetry of $x$ and $y$, we may assume $x y^{-1} \in R$.

there exists a $u \in R$ such that:
 * $u \paren {x + y} = 1$

Then:
 * $y^{-1} = y^{-1} u \paren {x + y} = u \paren {x y^{-1} + 1} \in R$

This contradicts the fact that $y$ is a non-unit.

Therefore $x + y$ is a non-unit.