As in part (1), is a field, is a finite dimensional central simple -algebra and is a simple -subalgebra of We will also be using notations and the result in the lemma in part (1), i.e. is the unique simple -module, and Finally, as usual, we will denote the center of any algebra by We now prove a nice relationship between dimensions.

**Lemma**.

*Proof*. We have and for some integer Thus as -modules, and hence

We also have

and

Eliminating in these three identities will give us the result.

**Theorem**. If is a central simple -algebra, then is a central simple -algebra too and we have

*Proof*. By the lemma in part (1), is simple and thus, since is central simple, is a simple algebra. Thus the map defined by is a -algebra isomorphism. Hence and so

by the above lemma. Therefore Clearly if is in the center of then commutes with every element of both and Hence commutes with every element of and thus So i.e. is a central simple -algebra.

**The Double Centralizer Theorem**.

*Proof*. Obviously So, to prove that we only need to show that Applying the above lemma to gives us

Plugging which is true by the above lemma, into finishes the proof.

**Corollary**. If is a subfield of then

*Proof*. We first need to notice a couple of things. First, since is commutative, Since is commutative, Thus, by the above theorem, and so is a central simple -algebra. Now, by the lemma

If $A$ is finite dimensional central simple k-algebra, and $B$ is finite dimensional simple k-subalgebra of $A$ and $C$ is the centralizer of $B$ in $A$, then $Z(B)=Z(C)=B\cap C$. .

That’s right. Of course, and always holds for any subalgebra of any algebra . But for the inclusion , which follows from the double centralizer theorem, we will need the conditions that you mentioned.