Timeline for Is the field of invariants $k(V)^G$ purely transcendental over $k$?
Current License: CC BY-SA 3.0
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| Jun 19, 2016 at 13:43 | history | edited | Sean Lawton | CC BY-SA 3.0 |
Minor edits, added tag.
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| Mar 21, 2010 at 0:46 | comment | added | Portland | This is useful VA, thank you. How do you show that $H$ acts linearly? | |
| Mar 20, 2010 at 17:44 | comment | added | VA. | 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. | |
| Mar 19, 2010 at 0:52 | history | asked | Portland | CC BY-SA 2.5 |