# Existence of quotient variety for group implies existence of quotient for normal subgroups

Let $G=G_1\times G_2$ be a product of two linear algebraic groups over an algebraically closed field. Assume that $G$ acts on a variety $X$ such that the quotient $X\rightarrow X/G$ exists in the category of varieties and is a $G$-torsor locally trivial in the etale topology.

Is it true, that $X \rightarrow X/G_1$ exists in the category of varieties?

Is it an etale locally trivial $G_1$-torsor?

Is $X/G_1\rightarrow X/G$ an etale locally trivial $G_2$-torsor?

If not does it help to assume the $G_i$ to be reductive?

What if instead $G_1$ is only a normal subgoup of $G$ and $G_2$ the quotient group?

• Since many (closely related) questions are being asked here, it would be helpful to number them. – Jim Humphreys Sep 25 '13 at 14:59
• I changed the title into a more distinguishable one. – Jan Weidner Sep 26 '13 at 8:07

## 1 Answer

In what follows, we only need $G_1$ to be normal in $G$, as in the last question. Put $Y=X/G$, and $Z=X/G_1$: $Y$ is a variety by assumption, and both $X$ and $Z$ make sense as sheaves on the étale site of $Y$. On this site, $X\to Y$ is locally isomorphic to $Y\times G$ with the obvious action of $G$; it follows that $Z$ is locally isomorphic to $Y\times G_2$, i.e. $Z$ is a $G_2$-torsor over $Y$. But $G_2$ is affine by assumption. So, by flat descent of (quasi-)affine schemes (SGA 1, VIII, 7.9), $Z$ is a scheme, affine over $Y$. Moreover, $X\to Z$ is locally over $Z$ (even over $Y$) isomorphic to $Z\times G_1\to Z$, so it is a $G_1$-torsor.

Everything works just as well with "étale" replaced by "fppf".

Note that the sheaf-theoretic arguments here are completely formal. The key assumption for the representability of $X/G_1$ by a scheme is the fact that $G_2$ is affine. For more general algebraic groups, the "right" viewpoint is to forget about varieties and work with algebraic spaces: $Z$ is always an algebraic space and $X\to Z$ (resp. $Z\to Y$) is a $G_{1,Z}$-torsor (resp. a $G_{2,Y}$-torsor).