Why are the only associative division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Question

Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).

This is an old result proved by Frobenius, but I can't remember how the argument goes. Anyone have a quick proof?

Answer 1

Essentially one first proves that any real division algebra $D$ is a Clifford algebra (i.e. it's generated by elements of some inner product vector space I subject to relations $v^2=\langle v, v\rangle$): first one splits $D$ as $\mathbb R\oplus D_0$ where $D_0$ is the space of elements with $Tr=0$ and then one observes that minimal polynomial of a traceless element has the form $x^2-a=0$ (it's quadratic because it's irreducible and the coefficient of $x$ is zero because it is the trace). Now it remains to find out which Clifford algebras are division algebras which is pretty straightforward (well, and it follows from the classification of Clifford algebras).

This proof is written in Wikipedia.

Answer 2

There is a simple proof of Frobenius's theorem in Lam's book on noncommutative rings, pp. 208--209. He attributes the argument to Palais.

One should consider this theorem to be two theorems: (1) $\mathbb C$ is the only $\mathbb C$-central division algebra and (2) $\mathbb R$ and $\mathbb H$ are the only $\mathbb R$-central division algebras. The reason there are so few choices is that $\mathbb C$ is alg. closed and $\mathbb R$ is nearly so. Division algebras with center equal to a particular field can be created using cyclic Galois extensions* and since $\mathbb Q$ has such extensions of arb. high degree there are $\mathbb Q$-central division alg. of arb. high dimension.

*There are further technical conditions to be satisfied on the cyclic extension in order for the construction of a division algebra to work, e.g., a finite field has a cyclic extension of each degree but there are no central div. alg. of $\dim > 1$ over a finite field. The relevant technical conditions are satisfied when the base field is the rational numbers.

Answer 3

I will write an answer which uses the Brauer group.

As $\mathbb C$ is the algebraic closure of $ \mathbb R$, $Br(\mathbb R)=Br(\mathbb C / \mathbb R)$. Also we know $Br(\mathbb C / \mathbb R) \cong H^2(Gal(\mathbb C / \mathbb R), \mathbb C ^{\times} )\cong \hat H^0(Gal(\mathbb C / \mathbb R),\mathbb C ) $. Here $\hat H $ stands for the Tate cohomology group and the second isomorphism is true because $Gal(\mathbb C / \mathbb R)$ is cyclic (it is $\mathbb Z /2 \mathbb Z $).

So it is clear that the only central division algebras over $\mathbb R $ are $\mathbb{R}$ itself and $ \mathbb H $.

If $D$ is a division algebra over $ \mathbb R$ whose center is not $\mathbb{R}$, then as it is a field algebraic over $\mathbb R$), $ \mathbb C =Z(D) $. So $D $ is a division algebra over $\mathbb C $ which implies $\mathbb C = D $.

Answer 4

There is another proof using the theory of central simple algebras and quaternion algebras.

Denote by $D$ such a division algebra. Note that a division algebra is always simple. Moreover, since the center of a division algebra is a field, $D$ is central simple over either $\mathbb{C}$ or $\mathbb{R}$.

As there is no nontrivial division algebra over an algebraically closed field, $D$ must be $\mathbb{C}$ if $Z(D)=\mathbb{C}$.

Otherwise, $D$ is central simple over $\mathbb{R}$, then $\mathbb{C}$ splits $D$. Hence we obtain the following: $$ \sqrt{\mathbf{dim}_{\mathbb{R}}(D)} = \mathbf{ind}_{\mathbb{R}}(D)|[\mathbb{C}:\mathbb{R}]=2 $$ $D$ is nontrivial, hence the dimension of $D$ is 4. Then $D$ must be $\mathbb{H}$ by Wedderburn's theorem.