Tensor product of simple algebras (2)

Posted: January 22, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this post $k$ is a field. Recall that the center of a simple ring is a field (see Remark 1 in here).

Definition. A ring $A$ is called a central simple $k$-algebra if $A$ is simple and $Z(A)=k.$ If the dimension of $A,$ as a vector space over $k,$ is finite, then $A$ is called a finite dimensional central simple $k$-algebra.

Remark 1. Let $A$ be a finite dimensional central simple $k$-algebra and let $I_1 \supset I_2 \supset \cdots$ be a descending chain of left ideals of $A.$ Each $I_j$ is clearly a $k$-vector subspace of $A$ and therefore $\dim_k I_j < \infty.$ Thus, since $\dim_k I_1 > \dim_k I_2 > \cdots ,$ the chain must stop at some point. This shows that every finite dimensional central simple $k$-algebra $A$ is (left) Artinian. Hence $A$ is simple and Artinian and thus, by the Wedderburn-Artin theorem, $A = M_n(D),$ for some positive integer $n$ and some division ring $D.$ Note that

$k = Z(A)=Z(M_n(D)) \cong Z(D).$

Let $Z=Z(D).$ Then, since $\dim_k A < \infty$ and

$\dim_k A = \dim_Z M_n(D)=(\dim_Z D)(\dim_D M_n(D))=n^2 \dim_Z D,$

we have $\dim_Z D < \infty.$

Note. Since $Z \cong k,$ we may assume that $k$ is the center of $D$ and write $\dim_k D$ instead of $\dim_Z D.$

Remark 2. If $k$ is algebraically closed and $A$ is a finite dimensional central simple $k$-algebra, then $A \cong M_n(k),$  for some $n.$ To see this, we first apply Remark 1 to get a division ring $D$ with $Z=Z(D) \cong k$ and $\dim_Z D < \infty$ such that $A = M_n(D).$ So $Z$ is algebraically closed. Now if $d \in D,$ then $Z[d]$ is an algebraic field extension of $Z$ because $\dim_Z D < \infty.$ Thus $Z[d]=Z,$ because $Z$ is algebraically closed. So $D =Z \cong k$ and hence $A \cong M_n(k).$

Lemma. Let $A$ and $B$ be two $k$-algebras and let $C:=A \otimes_k B.$ Then

1) $Z(C)=k$ if and only if $Z(A)=Z(B)=k.$

2) If $A$ is central simple, then the centralizer of $A$ in $C$ is $B.$

Proof. 1) If $Z(A)=Z(B)=k,$ then by Lemma 2

$Z(C)=Z(A) \otimes_k Z(B)=k \otimes_k k = k.$

Conversely, if $Z(C)=k,$ then

$1 = \dim_k Z(C) =\dim_k Z(A) \otimes_k Z(B) = (\dim_k Z(A))(\dim_k Z(B)),$

and hence $\dim_k Z(A)=\dim_k Z(B)=1.$ Therefore $Z(A)=Z(B)=k.$

2) Just to avoid any possible confusion, I mention that we have identified $A$ and $B$ with $A \otimes_ k 1$ and $1 \otimes_k B$ here. It is obvious that every element of $1 \otimes_k B$ commutes with every element of $A \otimes_k 1.$ So the centralizer of $A \otimes_k 1$ in $C$ contains $1 \otimes_k B.$ Conversely, let $c \in C$ be in the centralizer of $A \otimes_k 1$ in $C.$ Let $\{b_i\}$ be a $k$-basis for $B.$ Then there exist $a_i \in A$ such that $c = \sum a_i \otimes_k b_i.$ Since $c$ is in the centralizer of $A \otimes_k 1,$ we must have $(a \otimes_k 1)c = c(a \otimes_k 1),$ for all $a \in A.$
Thus $\sum (aa_i-a_ia) \otimes_k b_i=0$ and so by Lemma 1, $aa_i=a_ia,$ for all $i.$ Hence $a_i \in Z(A)=k$ and so $c = \sum a_i \otimes_k b_i = \sum 1 \otimes_k a_ib_i = 1 \otimes_k \sum a_ib_i \in 1 \otimes_k B. \ \Box$