Normed Division Algebra is Unitary Division Algebra
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Theorem
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra over a field $F$.
Let the unit of $A$ be $1_A$, and the zero of $A$ be $0_A$.
Then $A$ is a unitary division algebra.
Also:
- $\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
Proof
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra as defined by hypothesis.
The fact that $A$ is a unitary algebra is a consequence of the definition of normed divison algebra.
From the definition of a norm, we have that:
- $\forall a \in A: \norm a = 0 \iff a = 0_A$
So, let $a, b \in A \setminus \set {0_A}$.
We have:
| \(\ds \norm a \norm b\) | \(\ne\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \norm {a \oplus b}\) | \(\ne\) | \(\ds 0\) | Definition of Normed Division Algebra: $\norm {a \oplus b} = \norm a \norm b$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus b\) | \(\ne\) | \(\ds 0_A\) | Definition of Norm on Vector Space |
Thus for any arbitrary $a, b \ne 0_A$ we have shown that $a \oplus b \ne 0_A$.
Thus $A$ is a division algebra.
$\Box$
Next:
| \(\ds \) | \(\) | \(\ds \norm a \norm {1_A}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {a \oplus 1_A}\) | Definition of Normed Division Algebra: $\norm {a \oplus b} = \norm a \norm b$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm a\) | Definition of Norm on Vector Space |
demonstrating that $\norm {1_A} = 1$.
$\blacksquare$