Let $G$ be a connected simple algebraic group over an algebraically closed field $C$. What I infer from this definition is that the defining polynomials of $G$ have coefficients in $C$ while $G$ may have points in $K^n$ where $K$ is a field extension of $C$. We denote the $C$-points in $G$ by $G_C$.
Now, we have a result that the Lie algebra of such a group $G$ is simple. ( I am still trying to figure out the proof using Milne's notes as I had to read the basics like what is actually meant by a Lie algebra of such a group.) Any help here will be appreciated.
Moreover, in a paper I am reading, I came across another fact that there exists a simply connected algebraic group $K$ such that $Lie(G) \cong Lie(K)$.
I have not seen the proof of this and welcome some useful suggestions so that I can atleast get an idea of the proof with some basic algebraic geometry.
I had the following doubts :
Is that algebraic group $K$ which we claim to exist via the above result defined over $C$ (I mean its defining equations have coefficients in $C$)?
Further, they say that $G$ is isomorphic to $K/H$ where $H$ is a normal algebraic subgroup of $K$. I am not able to realise this conclusion.
Since $K$ is simply connected, $K_C$ is also simply connected. How to realise this ?