I am searching for a possible analogue of a result in algebraic groups in a non-commutative setting, so I am looking for different proofs of the following :



> Let $K$ be an algebraically closed field. A connected (affine) algebraic subgroup of $(K,+)^n$ having dimension $1$ is isomorphic to $(K,+)$.


**Any ideas for elementary proofs ?**


[Edit: let's stick here to characteristic 0; the characteristic $p$ case is [in a separate question](https://mathoverflow.net/questions/231048/)]



One-dimension Algebraic groups
Maybe I missed something, but in the case where the characteristic is zero, it seems to be simple as follow: if $G$ is a closed subgroup of $(K,+)^n$ which is not trivial, then it contains an element $g$, and thus contains all powers of $g$, corresponding to $m\cdot g$, $m\in \mathbb{Z}$ (view $(K,+)^n$ as a vector space). The Zariski closure of this set is then $\{\lambda \cdot g\mid \lambda \in K\}$, which is a closed algebraic subgroup of dimension $1$, isomorphic to $(K,+)$. If $G$ is not equal to this group, it contains another element $h$, linearly independent with $g$ so contains $\{\lambda \cdot g +\mu h\mid \lambda,\mu \in K\}$, which has dimension $2$.


