Frobenius's Theorem

Theorem
An algebraic associative real division algebra $A$ is isomorphic to $\R, \C$ or $\Bbb H$.

Proof
An algebra $A$ is said to be quadratic if it is unital and the elements $1, x, x^2$ are linearly dependent for every $x \in A$.

Thus, for every $x \in A$ there exist $t \left({x}\right), n \left({x}\right) \in \R$ such that $x^2 - t \left({x}\right) x + n \left({x}\right) = 0$.

Obviously, $t \left({x}\right)$ and $n \left({x}\right)$ are uniquely determined if $x \notin \R$.

Setting $t \left({\lambda}\right) = 2 \lambda$ and $n(\lambda) = \lambda^2$ for $\lambda \in \R$, we can then consider $t$ and $n$ as maps from $A$ into $\R$.

(In this way $t$ becomes a linear functional).

We call $t \left({x}\right)$ and $n \left({x}\right)$ the trace and the norm of $x$ respectively.

From $x^2 - \left({x + x^*}\right) x + x^* x = 0$ we see that all algebras $\Bbb A_n$ are quadratic.

Further, every real division algebra $A$ that is algebraic and power-associative (this means that every subalgebra generated by one element is associative) is automatically quadratic.

Indeed, if $x \in A$ then there exists a nonzero polynomial $f(X) \in \R[X]$ such that $f(x) = 0$.

Writing $f(X)$ as the product of linear and quadratic polynomials in $\R[X]$ it follows that $p(x) = 0$ for some $p(X) \in \R[X]$ of degree 1 or 2.

In particular, algebraic alternative (and hence associative) real division algebras are quadratic.

Finally, if $A$ is a real unital algebra, i.e., an algebra over $\R$ with unity $1$, then we shall follow a standard convention and identify $\R$ with $\R1$.

Thus we shall write $\lambda$ for $\lambda 1$, where $\lambda \in \R$.

Lemma 3
We have from above that $A$ is quadratic.

We may assume that $n = \dim A \ge 2$.

By Lemma 1: Assertion 4 we can ﬁx $i \in A$ such that $i^2 = -1$.

Thus, $A \cong \C$ if $n = 2$.

Let $n > 2$.

By Lemma 3:
 * $\exists j \in A: j^2 = -1, i j = -j i$

Set $k = ij$.

It can immediately be checked that:
 * $k^2 = -1$
 * $k i = j = -i k$
 * $j k = i = -k j$
 * $i, j, k$ are linearly independent.

Therefore $A$ contains a subalgebra isomorphic to $\Bbb H$.

Finally, suppose $n > 4$.

By Lemma 3 there would exist $e \in A, e \ne 0$ such that:
 * $(a): \quad e i = -i e$
 * $(b): \quad e j = -j e$
 * $(c): \quad e k = -k e$

However, from the $(a)$ and $(b)$ it follows that $e i j = -i e j = i j e$.

Since $i j = k$, this contradicts $(c)$.

It follows that $n \le 4$, and so $\Bbb H$ is the highest order