Tensor product of simple algebras (2)

Posted: January 22, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

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


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