Valuation Ring is Local

Theorem

Let $R$ be a valuation ring.

Then $R$ is a local ring.

Proof

By definition of local ring:

$R$ is a local ring if and only if:

$R$ is non-trivial

the sum of two arbitrary non-units is a non-unit.

First we recall that as a valuation ring is an integral domain, then a fortiori:

$R$ is non-trivial

$R$ has no (proper) zero divisors

$R$ is a commutative and unitary ring.

Let $0$ and $1$ be the zero and unity of $R$ respectively.

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

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

Without loss of generality, let $x = 0$.

Then:

$x + y = y$

and so $x + y$ is a non-unit by hypothesis.

Similarly for $y = 0$.

Let $x \ne 0$ and $y \ne 0$.

By definition of non-unit, neither $x$ nor $y$ has an inverse in $R$.

Let $K$ be the field of quotients of $R$.

In particular, $x^{-1} \in K$ and $y^{-1} \in K$.

First note that:

\(\ds \paren {x y^{-1} }^{-1}\)

\(=\)

\(\ds y x^{-1}\)

Inverse of Group Product

\(\ds \)

\(=\)

\(\ds x^{-1} y\)

By definition of valuation ring, either:

$x y^{-1} \in R$

or:

\(\ds \paren {x y^{-1} }^{-1}\)

\(=\)

\(\ds y x^{-1}\)

Inverse of Group Product

\(\ds \)

\(=\)

\(\ds x^{-1} y\)

Ring Axiom $\text C$: Commutativity of Ring Product

\(\ds \)

\(\in\)

\(\ds R\)

Without loss of generality, let $x y^{-1} \in R$.

Aiming for a contradiction, suppose there exists a $u \in R$ such that:

$u \paren {x + y} = 1$

Then:

\(\ds y^{-1}\)

\(=\)

\(\ds u \paren {x + y} y^{-1}\)

Definition of Unity of Ring

\(\ds \)

\(=\)

\(\ds u \paren {x y^{-1} + y y^{-1} }\)

Ring Axiom $\text D$: Distributivity of Product over Addition

\(\ds \)

\(=\)

\(\ds u \paren {x y^{-1} + 1}\)

Definition of Inverse Element

\(\ds \)

\(\in\)

\(\ds R\)

Ring Axiom $\text A0$: Closure under Addition and Ring Axiom $\text M0$: Closure under Product

That is, the inverse of $y$ is in $R$.

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

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

$\blacksquare$