When we write for an element of, we tacitly assume that is contained in.
We can consider as a finite-dimensional -vector space. Any element of defines an endomorphism of by left-multiplication, we identify with that endomorphism. Therefore, we can speak about the trace of, and its characteristic and minimal polynomials.
For any in define the following real quadratic polynomial:
The claim
The key to the argument is the following Proof of Claim: Let be the dimension of as an -vector space, and pick in with characteristic polynomial. By the fundamental theorem of algebra, we can write We can rewrite in terms of the polynomials : Since, the polynomials are all irreducible over. By the Cayley–Hamilton theorem, and because is a division algebra, it follows that either for some or that for some. The first case implies that is real. In the second case, it follows that is the minimal polynomial of. Because has the same complex roots as the minimal polynomial and because it is real it follows that Since is the characteristic polynomial of the coefficient of in is up to a sign. Therefore, we read from the above equation we have: if and only if, in other words if and only if. So is the subset of all with. In particular, it is a vector subspace. Moreover, has codimension since it is the kernel of a non-zero linear form, and note that is the direct sum of and as vector spaces.
The finish
For in define. Because of the identity, it follows that is real. Furthermore, since, we have: for. Thus is a positive definitesymmetric bilinear form, in other words, an inner product on. Let be a subspace of that generates as an algebra and which is minimal with respect to this property. Let be an orthonormal basis of with respect to. Then orthonormality implies that: If, then is isomorphic to. If, then is generated by and subject to the relation. Hence it is isomorphic to. If, it has been shown above that is generated by subject to the relations These are precisely the relations for. If, then cannot be a division algebra. Assume that. Let. It is easy to see that . If were a division algebra, implies, which in turn means: and so generate. This contradicts the minimality of.
Remarks and related results
The fact that is generated by subject to the above relations means that is the Clifford algebra of. The last step shows that the only real Clifford algebras which are division algebras are and.
As a consequence, the only commutative division algebras are and. Also note that is not a -algebra. If it were, then the center of has to contain, but the center of is. Therefore, the only finite-dimensional division algebra over is itself.
This theorem is closely related to Hurwitz's theorem, which states that the only real normed division algebras are, and the algebra octonions|.
