for all this map now is also an algebra homomorphism because
also is injective because is a (central) simple algebra and so which is an ideal of must be trivial. finally
Note: some facts that are used here:
1) for any dimensional vector space we know this from elementary linear algebra.
2) the tensor product of a central simple algebra and a simple algebra is a simple algebra. also the tensor product of two central simple algebras is central simple.
3) for any finite dimensional vector spaces we have
finally note that cannot be replaced by because then would not be a ring homomorphism.
first note that if is any field extension, then is an -algebra and now if we assume that is the algebraic closure of and put then2. If A is a finite dimensional central simple algebra over a field K, then
the dimension of A over K is a perfect square.
is a finite dimensional simple algebra. thus is semisimple and so by Artin-Wedderburn we have for some finite dimensional division algebra and some positive integer
since is algebraically closed and is algebraic over we must have thus and therefore