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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]

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    $\begingroup$ For you, is an algebraic group always reduced? $\endgroup$
    – Jason Starr
    Commented Feb 10, 2016 at 12:14
  • $\begingroup$ @JasonStarr By algebraic group, I only mean the intersection of zero sets of finitely many polynomials. Does that answer your question ? $\endgroup$
    – Drike
    Commented Feb 10, 2016 at 12:42
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    $\begingroup$ No, that does not answer the question. $\endgroup$
    – Jason Starr
    Commented Feb 10, 2016 at 12:46
  • $\begingroup$ @JasonStarr: I would guess it does answer the question: since the OP is viewing linear algebraic groups in the "classical sense", they are always smooth (i.e. have reduced coordinate algebra). Or? $\endgroup$
    – Tom De Medts
    Commented Feb 10, 2016 at 12:48
  • $\begingroup$ @Drike: I don't expect an "elementary proof" for the case that you consider. Notice, for instance, that the proof of the fact that every $1$-dimensional smooth connected linear algebraic group is isomorphic to either $\mathbf{G}_a$ or $\mathbf{G}_m$ as given in Springer's book "Linear algebraic groups" uses essentially the question that you ask as part of its proof (see Corollary 3.4.8). $\endgroup$
    – Tom De Medts
    Commented Feb 10, 2016 at 12:56

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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 $2$.

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    $\begingroup$ The point is that characteristic zero is the easy case since then the Zariski-closed subgroups of $K^n$ are the linear subspaces by your argument. Characteristic $p$ is more rich: for instance in $K^2$ you have the graph of the Frobenius map as a closed 1-dimensional Zariski-closed subgroup (and indeed defining a reduced algebraic subgroup isomorphic to the additive group). $\endgroup$
    – YCor
    Commented Feb 12, 2016 at 22:55

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