## Opposite of central simple algebras

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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We will assume that $k$ is a field.

Definition. The opposite $A^{op}$ of the ring $(A,+, \cdot)$ is the ring $(A, +, *),$ where $a * b = b \cdot a$ for all $a, b \in A.$

Remark 1. If $A$ is a finite dimensional $k$-algebra, then clearly $A^{op}$ is also a $k$-algebra and $\dim_k A^{op} = \dim_k A.$

Theorem. If $A$ is a finite dimensional central simple $k$-algebra and $\dim_k A = m,$ then $A \otimes_k A^{op} \cong M_m(k).$

Proof. Define the map $f : A \times A^{op} \longrightarrow End_k(A)$ by $f(a,a')(b)=aba',$ for all $a,b \in A$ and $a' \in A^{op}.$ Note that $f$ is well-defined because $f(a,a')$ is obviously a $k$-linear map. It is also clear that $f$ is $k$-bilinear. Thus $f$ induces a $k$-module homomorphism

$g : A \otimes_k A^{op} \longrightarrow End_k(A)$

which is defined by $g(a \otimes_k a')(b)=aba',$ for all $a, b \in A$ and $a' \in A^{op}.$

Claim . $g$ is an algebra homomorphism.

Proof of the claim. For any $a_1, a_2 \in A, \ a_1', a_2' \in A^{op}$ and $b \in A$ we have

$g((a_1 \otimes_k a_1')(a_2 \otimes_k a_2'))(b)=g(a_1a_2 \otimes_k a_2'a_1')(b)=a_1a_2ba_2'a_1'.$

We also have

$g(a_1 \otimes_k a_1')g(a_2 \otimes_k a_2')(b)=g(a_1 \otimes_k a_1')(a_2ba_2')=a_1a_2ba_2'a_1'.$

Thus $g((a_1 \otimes_k a_1')(a_2 \otimes_k a_2')) = g(a_1 \otimes_k a_1')g(a_2 \otimes_k a_2')$ and the claim is proved.

Clearly $g \neq 0$ and so $\ker g$ is a proper two-sided ideal of $A \otimes_k A^{op}.$ Thus $\ker g = (0)$ because, by the corollary in this post, $A \otimes_k A^{op}$ is simple. Both $A \otimes_k A^{op}$ and $End_k(A) \cong M_m(k)$ have the same dimension $m^2$ over $k.$ Thus $g$ is surjective and so an isomorphism. $\Box$

Example. Consider the real quaternion algebra $\mathbb{H}.$ It is easy to see that the map $\varphi : \mathbb{H} \longrightarrow \mathbb{H}^{op}$ defined by $\varphi(a+bi+cj+dk)=a-bi-cj-dk,$ for all $a,b,c,d \in \mathbb{R},$ is an $\mathbb{R}$-algebra isomorphism. Thus $\mathbb{H} \cong \mathbb{H}^{op}$ and so, by the above theorem, $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \cong M_4(\mathbb{R}).$

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

$\underbrace{A \otimes_k A \otimes_k \ldots \otimes_k A}_{\text{n times}} \cong M_{n^n}(k).$

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