Valuation Ideal is Maximal Ideal of Induced Valuation Ring

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Theorem

Let $\struct {R, \norm {\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$ and unity $1_R$.


Let $\OO$ be the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:

$\OO = \set{x \in R : \norm x \le 1}$


Let $\PP$ be the valuation ideal induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:

$\PP = \set{x \in R : \norm x < 1}$


Then $\PP$ is an ideal of $\OO$:

$(a):\quad \PP$ is a maximal left ideal
$(b):\quad \PP$ is a maximal right ideal
$(c):\quad$ the quotient ring $\OO / \PP$ is a division ring.


Corollary

$\OO$ is a local ring.


Proof

First it is shown that $\PP$ is an ideal of $\OO$ by applying Test for Ideal.

That is, it is shown that:

$(1): \quad \PP \ne \O$
$(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$
$(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$


(1)

By Non-Archimedean Norm Axiom $\text N 1$: Positive Definiteness:

$\norm {0_R} = 0$

Hence:

$0_R \in \PP \ne \O$

$\Box$


(2)

Let $x, y \in \PP$.

Then:

\(\ds \norm {x + \paren{-y} }\) \(\le\) \(\ds \max \set {\norm x, \norm{-y} }\) Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality
\(\ds \) \(=\) \(\ds \max \set {\norm x, \norm y}\) Norm of Negative
\(\ds \) \(<\) \(\ds 1\) Since $x, y \in \PP$

Hence:

$x + \paren {-y} \in \PP$

$\Box$


(3)

Let $x \in \PP, y \in \OO$.

Then:

\(\ds \norm{x y}\) \(\le\) \(\ds \norm x \norm y\) Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(<\) \(\ds 1\) Since $x \in \PP, y \in \OO$

Hence:

$x y \in \PP$

$\Box$


By Test for Ideal it follows that $\PP$ is an ideal of $\OO$.


By Maximal Left and Right Ideal iff Quotient Ring is Division Ring the statements (a), (b) and (c) above are equivalent.

It is now shown that statement (a) holds.


Let $J$ be a left ideal of $\OO$:

$\PP \subsetneq J \subset \OO$

Let $x \in J \setminus \PP$, then:

$\norm x = 1$

By Norm of Inverse then:

$\norm {x^{-1} } = 1 / \norm x = 1 / 1 = 1$

Hence:

$x^{-1} \in \OO$

Since $J$ is a left ideal then:

$x^{-1} x = 1_R \in J$

Thus:

$\forall y \in \OO: y \cdot 1_R = y \in J$

That is, $J = \OO$

Hence $\PP$ is a maximal left ideal.

The result follows.

$\blacksquare$


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