Throughout this post is a commutative ring with identity. We will keep the notation for centralizers in this post.

**Remark**. Given a -algebra and a -subalgebra of we can give a structure of a right module by defining for all and The only thing we need to check is the associativity of product of elements of by elements of This is easy to see. We have

and

**Theorem**. Let be a -algebra and let be a subalgebra of Let Consider as a right -module, as explained in the above remark. Then

*Proof*. Define the map by

for all and We are going to prove that is a -algebra isomorphism.

i) is well-defined. So we have to show that for all Let Let Then

The last identity in (1) is true because We also have

Now (1) and (2) proves that is an -endomorphism of and so is well-defined.

ii) is a homomorphism. For, if and then

iii) is injective. For, if for some then

iv) is surjective. To see this, let Let Then for every we have

Also

Thus and so Finally, for every we have

Thus which proves that is surjective.

There is an extra c after ‘Also’ in the second equality.

Thanks.