Frobenius's Theorem/Lemma 1

Lemma
Let $\left({A, \oplus}\right)$ be a quadratic real algebra.

Then:
 * $(1): \quad U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$ is a linear subspace of $A$


 * $(2): \quad \forall u, v \in U: u v + v u \in \R$


 * $(3): \quad A = \R \oplus U$


 * $(4): \quad$ If $A$ is also a division algebra, then every nonzero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.

Proof of First Assertion
$U$ is closed under scalar multiplication.

We have to show that $u, v \in U$ implies $u + v \in U$.

Let $\left\{{u, v, 1}\right\}$ be a linearly dependent set.

Then there exist constants $a, b \in \R$ such that $u = av + b$.

Then $u^2 = a^2 v^2 + 2abv + b^2$.

Since $u^2$, $a^2 v^2$, and $b^2$ are all real, it follows that $2abv \in \R$, i.e. $v \in \R$.

Since $0$ is the only real element of $U$, it follows that $v = 0$.

Reversing $u$ and $v$ in the preceding argument shows that also $u = 0$, so that $u + v = 0 \in U$.

Now let $\left\{{u, v, 1}\right\}$ be a linearly independent set.

We have:
 * $\left({u + v}\right)^2 + \left({u - v}\right)^2 = 2u^2 + 2v^2 \in \R$

On the other hand, as $A$ is quadratic there exist $\lambda, \mu \in \R$ such that:
 * $\left({u + v}\right)^2 - \lambda \left({u + v}\right) \in \R$
 * $\left({u - v}\right)^2 - \mu \left({u - v}\right) \in R$

Hence:
 * $\lambda \left({u + v}\right) + \mu \left({u - v}\right) \in \R$

However, we have that $\left\{{u, v, 1}\right\}$ is linearly independent.

Therefore $\lambda + \mu = \lambda - \mu = 0$, and so $\lambda = \mu = 0$.

This proves that $u \pm v \in U$.

Thus $U$ is indeed a subspace of $A$.

Proof of Second Assertion
Accordingly, from the result of $(1)$:
 * $\forall u, v \in U: u v + v u = \left({u + v}\right)^2 - u^2 - v^2 \in \R$.

Proof of Third Assertion
Let $a \in A \setminus \R$.

Then:
 * $\exists \nu \in \R: a^2 - \nu a \in \R$

Therefore, if we set
 * $u = a - \dfrac \nu 2 \in U$

then $u^2 = a^2 - \nu a + \nu^2/4 \in \R$, so
 * $a = \dfrac \nu 2 + u \in \R \oplus U$

which proves the assertion.

Proof of Fourth Assertion
This follows directly from the definition of division algebra.

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

Suppose $u^2 = 0$. Since $u$ is nonzero and since $A$ is a division algebra, there exists an inverse $u^{-1}$ such that $uu^{-1} = 1$. But then $u = u1 = uuu^{-1} = 0\cdot u^{-1} = 0$, which cannot be since $u$ was assumed to be nonzero.

Now uppose $u^2 > 0$. Then there exists an $\alpha \in \R$ such that $\alpha^2 = u^2$. But then $\left({u - \alpha}\right) \left({u + \alpha}\right) = u^2 - \alpha^2$ would be $0$. Since $A$ is assumed to be a division algebra, this would imply that either $u = \alpha$ or $u = -\alpha$, which 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.