I want to solve the following question:
Suppose that the division ring $\Delta$ is a $K$-algebra with $(\Delta:K)$ finite. Prove that $\Delta=K$ if $K$ is algebraically closed. Deduce that if $K$ is algebraically closed and $R$ is a finite-dimensional $K$-algebra, then $R$ is semisimple if and only of $R$ is a product of matrix rings with coefficients from $K$.
Suppose $K$ is algebraically closed, $\Delta$ a finite dimensional $K$-algebra, and let $x\in\Delta$. Let us prove that $x\in K$.
Since $\Delta$ is finite dimensional, we know that the family $$(1,x,x^2,\dots,x^n,\dots)$$ is not free, so this gives a polynomial relation $P(x)=0$ for some polynomial $P\in K[X]$ of degree $d$.
Since $\Delta$ is a division ring, we know that $P$ has at most $d$ roots in $\Delta$, but since $K$ is algebraically closed, $P$ already has $d$ roots in $K$, so all the $\Delta$ roots of $P$ are in $K$, and in particular, $x\in K$.
For the second part, it is enough to use Wedderburn theorem which says that every simple finite dimensional algebra is a matrix ring over a division algebra. Since $K$ is the only (finite dimensional) division algebra, the only simple algebras are matrix rings, hence the result on semi-simple algebras.