TITLE: Division algebra over a an algebraically closed field
QUESTION [5 upvotes]: I am reading my notes and I stumbled upon a proof that I dont fully understand, and I was hoping maybe someone could clear the details. 
The main goal was to show that if $k$ is algebraically closed, then the group ring $kG\cong M_{n_1}(k)\times....\times M_{n_r}(k)$ (where the characteristic of the field does not divide $|G|$). 
First we applied Maschkes theorem and obtained $kG\cong M_{n_1}(D_1)\times...\times M_{n_r}(D_r)$, second we wanted to prove that in this case each $D_i$ equals $k$. 
This are my poorly taken notes: 
Let $x\in D$ then $k[x]\in D$ is a finite dimensional $k$-algebra which is a domain (because $D$ is division): Either $k[x]/k$ is integral or $\forall y\in k[x], y\neq 0$ the map induced by left multiplication by $y$ is an injection, hence a surjection, hence an isomorphism. So $k[x]/k$ is finite with $k$ algebraically closed then $k=k[x]$ so $x\in k$. 
So on the above proof I see if $k[x]/k$ was integral then its finite and then we are done, but I dont understand the "or" part. 
Thanks

REPLY [9 votes]: I don't really understand what you wrote either. Here is the statement and then the proof: we claim that if $D$ is a finite-dimensional division algebra over an algebraically closed field $k$, then in fact $D \cong k$. As proof, if $x \in D$, then consider the inverse closed subring of $D$ generated by $k$ and $x$. This is a finite, hence algebraic, extension of $k$, hence must be equal to $k$.