Norm of Unit of Normed Division Algebra
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Theorem
Let $\struct {A_F, \oplus}$ be a normed division algebra.
Let the unit of $\struct {A_F, \oplus}$ be $1_A$.
Then:
- $\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
Proof
By definition:
- $\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$
So:
- $\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$
So $\norm {1_A} \in \R$ is idempotent under real multiplication.
From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill.
But $\norm {1_A}$ can not be $0$ as that would make:
- $\forall a \in A_F: \norm a = \norm {1_A \oplus a} = \norm {1_A} \norm a = 0 \norm a = 0$
which contradicts the definition of the norm.
Hence the result.
$\blacksquare$