Welcome to MathOverflow

Question

0

Let G, X and Y are algebraic schemes over k.(k:field) Assume that G is affine, and that the action is proper. Then f:X -> Y is affine. This is the Proposition0.7 in 'GIT(mumford & Fogarty)' I don't understand the part of the proof of this Prop.

g: G×Y -> X is a proper morphism. P_2: G×Y -> Y is the second projection and affine morphism. and f·g=p_2. The author says "by Chevalley's Theorem(EGA 2, Theorem 6.7.1), f is affine". I can't draw it. Please, help me...

(*Chevalley's Thm: X: affine scheme, Y: noetherian pre-scheme, f: x -> Y is a finite surjective morphism Then Y is also affine.)

flag	asked Nov 11 2011 at 10:12 google 23●1

	Which action of $G$ on what? – Martin Brandenburg Nov 11 2011 at 10:30
	This is not well-posed in my opinion. Do you mean that $G$ is to be an affine group scheme and that furthermore $G$ acts on $X$ and that $Y$ is the quotient of $X$ modulo $G$? Maybe you should first edit the question in order to make it precise. – Sebastian Petersen Nov 25 2011 at 11:36

a-fortiori · Answer 1 · 2011-11-11 10:29:52Z UTC

1

$g$ is also affine (EGA II, 1.6.1 (v)), hence finite (EGA III, 4.4.2).

answered Nov 11 2011 at 10:29

a-fortiori
2,786●5●11

affine morphism

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.