Dimension of a simple algebra over its center Let $A$ be a finite dimensional simple algebra over a field $k$. Denote by $K$ the center of $A$. Why the dimension of $A$ over $K$ is a square?
 A: This follows from the Artin-Wedderburn classification of central simple algebras, together with a sneaky little argument involving tensor products.
First, the big theorem: let $A$ be a finite dimensional simple (associative, unital) $K$-algebra with precise center $K$.  Then $A$ is isomorphic to $M_n(D)$, where $D$ is a 
$K$-central division algebra.  Here, for any associative algebra $B$, $M_n(B)$ is the $n \times n$ matrix algebra with entries in $B$.  Clearly $\operatorname{dim}_K M_n(B) = n^2 \operatorname{dim}_K B$, so it is enough to show that the dimension of a $K$-central division algebra $D$ is a square.
For this, tensor from $K$ to the algebraic closure $\overline{K}$.  On the one hand, 
$\operatorname{dim}_{\overline{K}} D \otimes \overline{K} = \operatorname{dim}_K D$.  But on the other hand, an algebraically closed field admits no nontrivial finite-dimensional division algebras -- easy exercise -- so $D \otimes \overline{K} \cong M_N(\overline{K})$ and thus $\operatorname{dim}_{\overline{K}} M_N(\overline{K}) = N^2$.
To prove the "easy exercise", let $D$ be a finite-dimensional division algebra over an algebraically closed field $K$, let $x \in D$ and consider the minimal polynomial of 
multiplication by $x$, viewed as a $K$-linear endomorphism of $D$...
