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Reference: http://www.math.u-psud.fr/~colliot/mumbai04.pdf

Proposition 4.3. on page 18 in the above reference reads as follows: Assume $k = \overline{k}$. If $V$ is a finite dimensional vector space over $k$ and $G \subset GL(V)$ is an (abstract) abelian group consisting of semisimple elements, then $k(V)^G$ is pure.

I would like to find an abelian group $G \subset GL(V)$ such that $k(V)^G$ is not pure (if it exists it would need to be infinite due to Fischer's theorem, and not a connected solvable group according to Proposition 4.4).

Thanks.

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    $\begingroup$ I think you are unlikely to find what you are looking for. Let $\bar G$ be the Zariski closure of $G$ in $GL(V)$. Then it seems clear that $\bar G$ is abelian and that the invariants of $G$ and $\bar G$ are the same. Now, divide by the connected component $\bar G^0$. By prop.4.4. which you quoted, $k(V)^{\bar G^0}$ is pure over $k$. Finally, you need to divide by the finite abelian group $H=\bar G/\bar G^0$. To conclude that $k(V)^G$ is pure, by Fischer's theorem, it is sufficient to show that $H$ acts linearly. While I can not see it right away, the possibility of an example seems remote. $\endgroup$
    – VA.
    Commented Mar 20, 2010 at 17:44
  • $\begingroup$ This is useful VA, thank you. How do you show that $H$ acts linearly? $\endgroup$
    – Portland
    Commented Mar 21, 2010 at 0:46

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