Frobenius's Theorem/Lemma 3

Lemma

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

Then:

$A = \R \oplus U$

Proof

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.

$\blacksquare$