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.  $\Box$