# Quotient varieties

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..

• 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. Sep 21, 2011 at 15:38
• I do not really know; but I have a feeling that you will find the answer in the book "Ideals, Varieties and Algorithms".. Sep 22, 2011 at 8:55
• I guess for affine varieties you should look in the direction of subrings. Nov 21, 2011 at 10:40
• 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. $dim(V) = dim(V^G)$). Nov 21, 2011 at 11:33

If $V$ is a quasi-projective variety and $G$ is a finite group acting on $V$, then the quotient $V/G$ exists. The idea is the following: because $V$ is quasi-projective, any finite subset of $V$ lies in an affine open, hence every $G$-orbit lies in an affine open. Now a small argument shows that $V$ may be covered by $G$-invariant affine opens. For each $G$-invariant affine open $U$, we will construct the quotient $U/G$, and then glue them to form $V/G$.

If $U =$ Spec $A$, then the $G$-action on $U$ gives a $G$-action on $A$, and by definition $U/G =$ Spec $A^G$, where $A^G$ is the subring of $G$-invariant elements in $A$.

So to compute $V/G$, you have to (a) find a cover by $G$-invariant $U$; this shouldn't be too hard if your variety $V$, your group $G$, and your $G$-action on $V$ are explicit; (b) compute the various $G$-invariants $A^G$ --- this is probably the hardest part, although I imagine the right software can handle it in cases that aren't too complicated; (c) glue the various Spec $A^G$s together --- this is easy in principle, although it means that you end up with $V/G$ described in a somewhat abstract way; while $V/G$ will again be quasi-projective, you don't see this directly from this gluing procedure.

Incidentally, the dimension of $V/G$ will be the same as that of $V$ (because $G$ is finite). For things like singularities, the description of $V$ by gluing is perhaps not so bad, as you can check what is happening in one affine open at a time.

• i know this is an old question and answer, but i still would like to ask you the following: Thanks for the nice and clear answer. Could you tell me what a reference for this could be, i.e. where did you learn these things? Thanks! Apr 1, 2013 at 15:35
• @Joachim: Dear Joachim, I think this particular construction I learned in grad school when I was the TA for Brian Conrad's algebraic geometry course at Harvard. (It was on one of the problem sets I had to grade.) My vague memory is that I was already somewhat familiar with it from discussions in papers where it is used, although I can't remember now which one(s); I used to skim through many papers on arithmetic and algebraic geometry back then (both research papers and foundational material by Grothendieck and others) in order to learn things. Regards, Apr 2, 2013 at 1:38
• thanks for clearing that up. I was trying to link this to example 2.3.5 in Hartshorne, where he defines glueing of schemes. As an example i took $G = \mathbb{Z}/2$ acting of $\mathbb{A}_{\mathbb{C}}^1$ by $\phi: x \mapsto -x$. I wanted to see that taking the quotient by the isomorphism $\phi$ in the sense of the mentioned example and your description above are essentially the same, but it is starting to dawn on me that they are not. (Of the course the quotient is $\mathbb{A}^1$ in both cases anyway, but i just used it to picture the more general case). ... Apr 2, 2013 at 11:13
• An obvious reason is that you run into trouble when $|G|>2$, a more subtle one is that by following Hartshorne one does not really get an isomorphism of global sections of the quotients by your ring of invariants. Anyway, to state it in question form, would you be able to state your construction in a more "topological way", using a quotient topological space and a quotient structure sheaf? ... Apr 2, 2013 at 11:17
• @Joachim: Dear Joachim, Let's restrict to the case of Spec $A$ with a $G$-action (since the general case is obtained by gluing these cases, so hopefully you can see why this is sufficient to answer your question). Let's also suppose that $A$ is f.t. over a field $k$. Then you have to show that maximal ideals in $A^G$ are in bijection with $G$-orbits of maximal ideals in $A$. In the case that $A$ is a Dedekind domain, this is part of the Galois theory of Dedekind domains (the theory of decomposition groups, etc.), and the general case is similar. Maybe you could try to prove this. Regards, Apr 2, 2013 at 11:23