We will assume that is a field.

**Definition**. The **opposite** of the ring is the ring where for all

**Remark** **1**. If is a finite dimensional -algebra, then clearly is also a -algebra and

**Theorem**. If is a finite dimensional central simple -algebra and then

*Proof*. Define the map by for all and Note that is well-defined because is obviously a -linear map. It is also clear that is -bilinear. Thus induces a -module homomorphism

which is defined by for all and

*Claim *. is an algebra homomorphism.

*Proof* *of the claim*. For any and we have

We also have

Thus and the claim is proved.

Clearly and so is a proper two-sided ideal of Thus because, by the corollary in this post, is simple. Both and have the same dimension over Thus is surjective and so an isomorphism.

**Example**. Consider the real quaternion algebra It is easy to see that the map defined by for all is an -algebra isomorphism. Thus and so, by the above theorem,

**Remark 2**. The example is just a very special case of this nice result that if is a finite dimensional central simple -algebra of degree then

To prove this result, we need to know a little bit of cohomology of algebras and we will do that later.