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Suppose $X$ is a smooth affine algebraic variety over $\mathbb{C}$ and let $V$ be an algebraic vector field (i.e. an algebraic section of the tangent bundle). If $V$ is locally nilpotent, meaning that for every $f\in\mathbb{C}[X]$ there exists $k\in\mathbb{N}$ such that $V^k(f) = 0$, does it necessarily follow that $V$ is complete? By complete vector field, I mean that the flow $\phi_t(x)$ of $V$ exists for all $x\in X$ and for all $t\in \mathbb{C}$.

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    $\begingroup$ Can't you just define $\phi_t^*$ on $\mathbb{C}[X]$ by $f\mapsto e^{tV}(f)$? This gives a one-parameter family of automorphisms of $X$, and a complete flow for $V$. $\endgroup$
    – abx
    Apr 3, 2016 at 5:51

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Yes, it is complete. The flow is given by the operator $e^{tV}$ acting on the coordinate functions $x^i$. This is defined globally because on Zariski open sets the operator is a unipotent linear transformation, so given by polynomials in $t$.

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