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Let's work over an algebraically closed field $K$. A $1$-dimensional Zariski-closed connected subgroup of ${\mathbf G}_{\mathbf a}^n$ is isomorphic to ${\mathbf G}_{\mathbf a}^1$. If $K$ has characteristic $0$, then a Zariski-closed subgroup $G\subset{\mathbf G}_{\mathbf a}^n$ is simply a $K$-vector space since for any non zero $g\in G$, the line $L$ directed by $g$ intersets $G$ in infinitely many points, implying $L\subset G$.

Are there any references concerning the structure of Zariski-closed subgroups of ${\mathbf G}_{\mathbf a}^n$ when $K$ has characteristic $p$ ?

For instance, if a Zariski-closed connected subgroup $G\subset {\mathbf G}_{\mathbf a}^n$ has dimension $k$, does there exist a group automorphism $\alpha$ of ${\mathbf G}_{\mathbf a}^n$ such that $\alpha(G)\subset {\mathbf G}_{\mathbf a}^k\times\{0\}^{n-k}$?
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  • $\begingroup$ Related: mathoverflow.net/questions/230745 $\endgroup$
    – YCor
    Commented Feb 13, 2016 at 14:16
  • $\begingroup$ The question is a bit weird since an isomorphism of variety does not necessarily map subgroups to subgroups (so a positive answer to the question would not provide a classification of Zariski-closed subgroups of $K^n$) $\endgroup$
    – YCor
    Commented Feb 13, 2016 at 14:17
  • $\begingroup$ What about $\alpha_p:={\rm Spec}K[X]/X^p$? $\endgroup$
    – Damian Rössler
    Commented Feb 13, 2016 at 14:55
  • $\begingroup$ I'd understand the question as a question about Zariski-closed connected subgroups of $K^n$, to avoid reduceness issues. (The phrasing "closed connected subgroup" is indeed misleading.) $\endgroup$
    – YCor
    Commented Feb 13, 2016 at 15:10
  • $\begingroup$ Sorry again: you mean a polynomial group automorphism, and probably should write $K^n$ instead of $G_a^n$ to avoid any ambiguity. $\endgroup$
    – YCor
    Commented Feb 13, 2016 at 15:16

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The answer to your question is yes, assuming your algebraic groups are smooth (equivalently reduced, since your ground field $K$ is algebraically closed).

Indeed, every smooth connected commutative $p$-torsion linear algebraic group over $K = \overline{K}$ of characteristic $p>0$ is a vector group (i.e., direct product of copies of $\mathbf{G}_a$) -- one reference for this classical fact is Lemma B.1.10 in the book Pseudo-reductive groups -- and Proposition B.1.11 of the same book then gives exactly the affirmative answer to the question posed (with equality $\alpha(G) = \mathbf{G}_a^k \times \{0\}^{n-k}$, same as inclusion for smooth connected $G$ by dimension reasons).

The entire Appendix B of that book (at least in its statements of results) is based on Tits' lectures at Yale in the late 1960's, so the result in question may be originally due to Tits but I am not sure.

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    $\begingroup$ Over a perfect field, the statement follows from the description of commutative algebraic groups with zero Verschiebung in terms of modules over k[F] (cf. Demazure Gabriel p.523). The treatment in CGP is more complicated because they are interested in imperfect base fields. $\endgroup$
    – zeno
    Commented Feb 13, 2016 at 18:40
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    $\begingroup$ @zeno: The proof in B.1 seems elementary and self-contained, especially if one runs it over a perfect field. Is that more complicated than the approach via k[F]-modules when one includes the work needed to set up the relationship to k[F]-modules? (Perhaps this is just a matter of taste.) $\endgroup$
    – nfdc23
    Commented Feb 14, 2016 at 6:19

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