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.
$\blacksquare$