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I have a variety $V$ given by polynomial equations. These equations admit a lot of symmetry. This means there are a lot of automorphisms on $V$. I want to get rid of this symmetry. So I somewhat want to form a quotient variety.

However, googling for quotient variety does not really give me what I am looking for. I found articles about geometric invariant theory and read that quotient varieties need not exist in general and I am somewhat wondering if it is even possible in my case.

To give a hint how my problem looks consider the projective curve(hyperbola) $C:XY+Z$. There is an isomorphism $\phi:(x,y,z) \mapsto (y,x,z)$. Is it possible to sort of mod out this isomorphism. I am thinking of this as having a variety $V$ such that there is a surjective map $C \to V$ where two points in $C$ have the same image if they they are mapped to each other by $\phi$.

Is this possible? how can I compute equations for $V$. What is the general theory? Furhtermore if this is not possible, can I find such a $V$ having this property but maybe not being a variety but some other object(say maybe it is possble to form quotients like these if we allow general schemes)?

EDIT: Ok, so I looked up example 11 and indeed it is what I want. But it only covers about half a page. The defintion is via the associated function fields. What about the computational aspect. Can someone point me to a reference where it is explained how to actually compute properties of the quotient variety. For example, dimension, defining equations etc..

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    Take a look at Shafarevich's book "Basic Algebraic Geometry, Varieties in Projective Space", page 30, example 11. He defines the quotient variety for an affine variety.2011-09-21
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    I guess for affine varieties you should look in the direction of subrings.2011-11-21
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    If $V = Spec(A)$ and $G$ is a finite group acting on $A$, then the quotient scheme exists and is isomorphic to $Spec(A^G)$, where $A^G$ is the ring of invariants. Now $A/A^G$ is an integral extension, which allows some conclusions to be drawn (e.g. $dim(V) = dim(V^G)$).2011-11-21
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    I do not really know; but I have a feeling that you will find the answer in the book "Ideals, Varieties and Algorithms"..2011-09-22

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