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Let $K$ be a field, $G_a := (K, +)$ be the additive group of $K$, and $X$ an affine variety.

I found the following claim: if $X$ admits a non-trivial $G_a$-action and $\dim(X) \ge 2$, then the group $\operatorname{Aut}(X)$ is infinite dimensional.

Reference: The automorphism group of a rigid affine variety, by I. Arzhantsev and S. Gaifullin.

Question: How do I see/prove this claim in a simple way?

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  • $\begingroup$ I think this claim is false as stated. Take $K = \mathbb{F}_p$ to be the finite field on $p$ elements, and $X = \mathbb{A}^2_K$ to be the affine plane. Then $K$ acts nontrivially on $X$ by addition in each coordinate, but $\operatorname{Aut}(X)$ can't be infinite-dimensional because there are only finitely many functions $X \to X$. $\endgroup$
    – Martin Skilleter
    Commented Oct 2, 2021 at 4:12
  • $\begingroup$ Looking at your reference, they start the introduction by saying that $K$ is an algebraically closed field of characteristic 0, so the above can't occur. $\endgroup$
    – Martin Skilleter
    Commented Oct 2, 2021 at 4:15
  • $\begingroup$ @MartinSkilleter your counterexample is not correct, because an automorphism of the affine plane over $\mathbf{F}_p$ is not determined by its action on $\mathbf{F}_p^2$, and indeed automorphisms $(x,y)\mapsto (x,y+P(x))$, for $P$ ranging over polynomials in $\mathbf{F}_p[t]$, indicate that it's indeed infinite-dimensional. (Anyway the OP's definition of $\mathbf{G}_\mathrm{a}$ is not really correct either in such a case.) $\endgroup$
    – YCor
    Commented Oct 2, 2021 at 11:10
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    $\begingroup$ Fix such a 1-parameter group of automorphisms $A$ acting as $(t,x)\mapsto u(t,x)$. If the algebra $R^A$ of $A$-invariant functions is infinite-dimensional (as vector space over the ground field), then for each $f\in R^A$ each $x\mapsto u(f(x),x)$ is an automorphism and should depend injectively on $f$. I don't know if $R^A$ is always infinite-dimensional for a $\mathbf{G}_\mathrm{a}$-action on an affine variety of dimension $\ge 2$. $\endgroup$
    – YCor
    Commented Oct 2, 2021 at 11:20
  • $\begingroup$ @YCor you're completely right. I made the classic mistake of thinking polynomials over finite fields are determined by their values at points. $\endgroup$
    – Martin Skilleter
    Commented Oct 2, 2021 at 11:25

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