**Theorem**. (Azumaya – Nakayama, 1947) Let be a central simple -algebra and let be an arbitrary -algebra. If is a two-sided ideal of then

*Proof*. It is obvious for and so we may assume that Let

Consider the natural -algebra homomorphism

*Claim 1*. To see this, let Let be a -basis of which contains 1. Let be such that So for some integer and Then

Thus for some we have and for all But then and hence

*Claim 2*. This is easy to see: since is clearly onto, is a two-sided ideal of Thus, by the theorem we proved in part (3), we must have either which is not true by cliam 1, or

*Claim 3*. This will be a result of claim 2 if we prove But it is obvious that and so we only need to show that Let Let be a -basis for Then for some positive integer and some Then

and thus, since are linearly independent, for all Hence for all and so

*Claim 4*. The reason is that if then because and is an ideal of

**Corollary.** 1) If is a central simple -algebra and is a simple -algebra, then is a simple -algebra.

2) If both and are central simple -algebras, then is a central simple -algebra.

*Proof*. Part 1) is a trivial result of the above theorem. Part 2) follows from part 1) and Lemma 2.

**Example**. Let be the division algebra of quaternions over Then

*Proof*. Let The first part of the above corollary implies that is simple. Moreover, by Lemma 2, and so is a central simple -algebra. Hence, by Remark 2, for some integer Since is a field extension of we have and thus Hence

Hi

I am working on algebras which are not necessarily unital. I would like to now that if there exists a similar results when at least one of algebras is not unital? In my special case $A$ is a central-simple algebra without $1$ and $B$ is a field extension of $k$. I will appreciate any clue.

Thnk you, I found the answer. See [MacCrimmon, A Taste of Jordan Algebras, Thm II.1.7.1]

Thanks.