Throughout k is a field, K is the algebraic closure of k and A is a finite dimensional central simple k-algebra.

Lemma. A \otimes_k K \cong M_n(K), for some integer n.

Proof. Let S:=A \otimes_k K. By the first part of the corollary in this post we know that S is simple. We also have

Z(S) = Z(A) \otimes_k K = k \otimes_k K \cong K.

It is easy to see that if \{a_i \} is a k-basis for A, then \{a_i \otimes_k 1 \} is an K-basis for S. Thus \dim_K S = \dim_k A. So S is a finite dimensional central simple K-algebra and hence, since K is algebraically closed, S \cong M_n(K), for some n, by Remark 2 in this post. \Box

Theorem. If A is a finite dimensional central simple k-algebra, then \dim_k A is a perfect square.

Proof.  By the lemma, there exists an integer n such that A \otimes_k K \cong M_n(K). Thus

\dim_k A = \dim_K A \otimes_k K = \dim_K M_n(K) = n^2. \ \Box

Definition. The degree of A is defined by \deg A = \sqrt{\dim_k A}.

Remark. Let R be a finite dimensional k-algebra. Then R is reduced if and only if R is a finite direct product of finite dimensional division k-algebras. In this case, \dim_k R=n^2\dim_k Z(R) for some integer n \geq 1.

Proof. obviously R is (left) Artinian because \dim_k R < \infty and so J(R) is nilpotent. Thus J(R)=(0) because R is reduced and so R is semisimple. The result now follows from the Artin-Wedderburn theorem. The converse is trivial. Finally, the fact that, by the above theorem, \dim_{Z(D)}D is a perfect square for any finite dimensional division algebra D, proves the last part of the remark.  \Box


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