Norm of Unit of Normed Division Algebra

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.