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Let $G$ be an irreducible algebraic group with corresponding Lie algebra $\mathfrak{g}$ , which is unipotent. I'm trying to prove that $G$ is unipotent in this case.

What I know is that exponential map $\mathbf{exp}: \mathfrak{g} \rightarrow G$ is a morphism, and if $\mathbf{exp}\;\mathfrak{g}$ is closed in Zarisky topology then due to irreducibility of $G$ it is closed under left shifts in $G$, and since it ($\mathbf{exp}\;\mathfrak{g}$) contains identity element of $G$, it follows that $\mathbf{exp}\;\mathfrak{g}=G$. So that, all that i need is to show that $\mathbf{exp}\;\mathfrak{g}$ is closed. Can someone help?

Many thanks in advance!

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It is very likely the consequence of isomorphism of nilpotent and unipotent operators (as varieties) induced by $\mathbf{exp}$.

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