Valuation Ring is Local

Theorem
Let $R$ be a valuation ring.

Then $R$ is a local ring.

Proof
Let us verify the definition.

Firstly, $R$ is nontrivial, since $1\in R$.

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 $x\neq 0$ and $y\neq 0$.

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 $xy^{-1}\in R$ or $x^{-1}y\in R$ must be true.

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

here is a $u\in R$ such that
 * $u(x+y)=1$

Let $z:=u(xy^{-1} +1)$.

Then $z\in R$ and
 * $zy=u(xy^{-1} +1)y=u(x+y)=1$

This contradicts to that $y$ is a non-unit.