Valuation Ideal is Maximal Ideal of Induced Valuation Ring/Corollary 1

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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:

$(a): \quad \OO$ is a local ring.
$(b): \quad \PP$ is the unique maximal left ideal of $\OO$
$(c): \quad \PP$ is the unique maximal right ideal of $\OO$


Proof

By Valuation Ideal is Maximal Ideal of Induced Valuation Ring then:

$\PP$ is a maximal left ideal of $\OO$.


Let $J \subsetneq \OO$ be any maximal left ideal of $\OO$.

Let $x \in \OO \setminus \PP$.


Aiming for a contradiction, suppose $x \in J$.

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:

$J = \OO$

This contradicts the assumption that $J \ne \OO$.


So:

$x \notin J$

Hence:

$\paren {\OO \setminus \PP} \cap J = \O$

That is:

$J \subseteq \PP$


Since $J$ and $\PP$ are both maximal left ideals then:

$J = \PP$

Hence:

$\PP$ is the unique maximal left ideal of $\OO$.

By Definition of Local Ring:

$\OO$ is a local ring


From Maximal Left and Right Ideal iff Quotient Ring is Division Ring:

every maximal right ideal of $\OO$ is a maximal left ideal

Hence:

$\PP$ is the unique maximal right ideal of $\OO$.


The result follows.

$\blacksquare$


Sources

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