Frobenius's Theorem/Lemma 4

Lemma
Let $\struct {A, \oplus}$ be a quadratic real algebra.

If $A$ is also a division algebra, then every non-zero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.

Proof
This follows directly from the definition of division algebra.

Also, we have that $u^2 \in \R$.

$u^2 = 0$.

Since $u$ is non-zero and since $A$ is a division algebra, there exists an inverse $u^{-1}$ such that $u u^{-1} = 1$.

But then $u = u 1 = u u u^{-1} = 0 \cdot u^{-1} = 0$, which cannot be since $u$ was assumed to be non-zero.

Now suppose $u^2 > 0$.

Then there exists an $\alpha \in \R$ such that $\alpha^2 = u^2$.

But then $\paren {u - \alpha} \paren {u + \alpha} = u^2 - \alpha^2$ would be $0$.

Since $A$ is assumed to be a division algebra, this would imply that either $u = \alpha$ or $u = -\alpha$.

This is impossible, since $u \in U$.

So:
 * $u^2 < 0$

and so:
 * $u^2 = -\alpha^2$

with $0 \ne \alpha \in \R$.

Thus, $v = \alpha^{-1} u$ is a desired element.