The main ingredients for the following proof are the
Cayley–Hamilton theorem and the
fundamental theorem of algebra.
Introducing some notation • Let be the division algebra in question. • Let be the dimension of . • We identify the real multiples of with . • When we write for an element of , we imply 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: ::Q(z; x) = x^2 - 2\operatorname{Re}(z)x + |z|^2 = (x-z)(x-\overline{z}) \in \mathbf{R}[x]. :Note that if then is
irreducible over .
The claim The key to the argument is the following :
Claim. The set of all elements of such that is a vector subspace of of dimension . Moreover as -vector spaces, which implies that generates as an algebra.
Proof of Claim: Pick in with characteristic polynomial . By the fundamental theorem of algebra, we can write :p(x) = (x-t_1)\cdots(x-t_r) (x-z_1)(x - \overline{z_1}) \cdots (x-z_s)(x - \overline{z_s}), \qquad t_i \in \mathbf{R}, \quad z_j \in \mathbf{C} \setminus \mathbf{R}. We can rewrite in terms of the polynomials : :p(x) = (x-t_1)\cdots(x-t_r) Q(z_1; x) \cdots Q(z_s; x). 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 :p(x) = Q(z_j; x)^k = \left(x^2 - 2\operatorname{Re}(z_j) x + |z_j|^2 \right)^k for some . 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. The
rank–nullity theorem then implies that has dimension since it is the
kernel of \operatorname{tr} : D \to \mathbf{R}. Since and are disjoint (i.e. they satisfy \mathbf R \cap V = \{0\}), and their dimensions sum to , we have that .
The finish For in define . Because of the identity , it follows that is real. Furthermore, since , we have: for . Thus is a
positive-definite symmetric 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: :e_i^2 =-1, \quad e_i e_j = - e_j e_i. The form of then depends on : 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 :e_1^2 = e_2^2 =-1, \quad e_1 e_2 = - e_2 e_1, \quad (e_1 e_2)(e_1 e_2) =-1. These are precisely the relations for . If , then cannot be a division algebra. Assume that . Define and consider . By rearranging the elements of this expression and applying the orthonormality relations among the basis elements we find that . If were a division algebra, implies , which in turn means: and so generate . This contradicts the minimality of . ==Remarks and related results==