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Let $G$ and $H$ algebraic linear groups and $\phi : G \to H$ a regular group homomorphism. I wonder if $\overline{\phi(G)}$ (the Zariski closure of $\phi(G)$) is again a subgroup and how this could be proven. First of all, I know Zariski closure $\overline{Z}$ of a set $Z$ is defined as $\overline{Z} = V(I(Z))$. I imagine this definition is also valid for algebraic linear groups, since $G$ is an affine variety. However, I can’t seem to find a strategy for this proof: any ideas?

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  • $\begingroup$ Is $H$ another algebraic group? Is $\phi$ a group homomorphism? Which algebraic formalism are you using (schemes, or something more elementary?) $\endgroup$
    – Aphelli
    Commented Jan 21, 2021 at 16:44
  • $\begingroup$ Thanks, I just edited. I am not using schemes, I am studying "classic" algebraic geometry $\endgroup$
    – cip
    Commented Jan 21, 2021 at 17:01
  • $\begingroup$ The image is automatically closed. Are you not aware of this? $\endgroup$
    – Alex Youcis
    Commented Jan 21, 2021 at 17:38
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    $\begingroup$ @cip I mean the proof in Milne doesn't, at least not explicitly, but if you really just wanna know that you just need to note that since $m:H\times H\to H$ is continuous that since $m(\phi(G),\phi(G))\subseteq \overline{\phi(G)}$ that $m(\overline{\phi(G)},\overline{\phi(G)})\subseteq \overline{\phi(G)}$. The reason is that $\overline{\phi(G)\times \phi(G)}=\overline{\phi(G)}\times \overline{\phi(G)}$ and proofwiki.org/wiki/Continuity_Defined_by_Closure $\endgroup$
    – Alex Youcis
    Commented Jan 21, 2021 at 22:19
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    $\begingroup$ @cip that’s right. $\endgroup$
    – Alex Youcis
    Commented Jan 21, 2021 at 22:47

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