Degree of a central simple algebra

Remarks

Let $\bar{K}$ be an algebraic closure of $K$. The tensor product $A \otimes_K \bar{K}$ is central simple over $\bar{K}$, hence isomorphic to $M_n(\bar{K})$ for some $n$ by Wedderburn's theorem. So $\dim_K A = \dim_{\bar{K}} M_n(\bar{K}) = n^2$ is a square.

Example: $\deg M_n(K) = n$, and the quaternion algebra $\mathbb{H}$ over $\R$ has degree $2$, since $\dim_\R \mathbb{H} = 4$.

Non-example: $\C$ has dimension $2$ over $\R$, which is not a square, so it has no degree as an algebra over $\R$; it is not central over $\R$.

If $A \cong M_m(D)$ with $D$ a division algebra, then $\deg A = m \deg D$. Degree is multiplicative: $\deg(A \otimes_K B) = \deg A \cdot \deg B$.