Valuation Ideal is Maximal Ideal of Induced Valuation Ring
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
- $(a): \quad \OO$ 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
First it is shown that $\PP$ is an ideal of $\OO$ by applying Test for Ideal.
That is, it is shown that:
- $(1): \quad \PP \ne \O$
- $(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$
- $(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$
By Maximal Left and Right Ideal iff Quotient Ring is Division Ring the statements (a), (b) and (c) above are equivalent.
So then it is shown:
- $(4): \quad \PP$ is a maximal left ideal
$(1): \quad \PP \ne \O$
By Non-Archimedean Norm Axiom $\text N 1$: Positive Definiteness:
- $\norm {0_R} = 0$
Hence:
- $0_R \in \PP \ne \O$
$\Box$
$(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$
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:
- $x + \paren {-y} \in \PP$
$\Box$
$(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$
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 $\OO$.
$(4):\quad \PP$ is a maximal left ideal
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.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction ... (previous) ... (next): $\S 2.4$ Algebra: Proposition $2.4.1$