define the map by for all this map is K-bilinear and so it induces a K-module homomorphism defined by

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

and therefore

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.

Thank you!!!

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