Does anyone know why the quotient of a linear algebraic Q-group by a finite subgroup is again a (or contained in a) linear algebraic Q-group?

This is not even close to homework.

Does anyone know why the quotient of a linear algebraic Q-group by a finite subgroup is again a (or contained in a) linear algebraic Q-group?

This is not even close to homework.

Try this. It’s short.

http://www.springerlink.com/content/n207r57m34172v62/fulltext.pdf

Cohomology of Pro-Q-Groups, I love it when they talk dirty!

I don’t see anything about quotients in that…

Huh!

Maybe Bricker will come by and help out.

Speaking of which, Bricker never even came back and laughed at my K-theory joke in the "own a number thread. It took me two days to set up that punch line.:dubious:

Link?

Bricker?? Bricker’s a lawyer, no?

Is this a whoosh or does Bricker have math talents of which I’m unaware? Are you saying he’s a closet Laurence Tribe?

Thanks!

What can I tell you?

East Coast is whack.

It’s not my area either, but I suspect it’s for a really easy reason.

I think this theorem might do it. I was about to ask “Are finite linear algebraic groups necessarily normal in some larger linear algebraic group?” but then I realized *my* subgroup is. (After all my quotient is a group!) And of course a finite group is closed…

Indeed. My only concern is the **Q-**group part; I’m not sure what that means, but I’m guessing it’s that all the polynomials involved have rational coefficients? The statement of the theorem doesn’t seem to address that, but perhaps it’s obvious from the proof.

I think it is the underlying field, actually. Heck, I’m not really sure.