Let $A$ be a finite dimensional semisimple $K$-algebra. Since $A$ is semisimple, I can write it as a direct sum of its simple components: $$A=A_1\oplus\cdots\oplus A_m$$ where each $A_i$ is of the form Mat$_{n_i}(D^i)$ with $D^i$ a division ring and $n_i\ge1$, by Wedderburn's Theorem.
Question: Why is $D^i$ a $K$-algebra and $K\subset$ Cent$(D^i)?$ What is the easiest way to see this?
Many thanks.
Let's recall how you arrive at the $D^i$.
Each $D^i=End_{A_i}(S_i)$ where $S_i$ is a simple right ideal of $A_i$. Now $K$ is embedded in there too, consiering that $s\mapsto \lambda s$ is an $A_i$ homomorphism from $S_i$ into $S_i$ for each $\lambda\in K$.
If $f$ is any other endomorphism of $S_i$, then $(\lambda f)(s)=\lambda f(s)= f(\lambda s)=(f\lambda)(s)$ since $f$ is a homomorphism. This shows that $K$ is centrally embedded in $End_{A_i}(S_i)$.