Opposite of central simple algebras

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

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.


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