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In Mumfords book on abelian varieties there is a theorem (on page 111) whose hypothesis is "Let G be a finite group scheme acting on a scheme X such that the orbit of any point is contained in an affine open subset of X".

Is there a well known example for when this condition is not fulfullied and in general, where should I look for the existance of quotients of schemes via actions of finite groups?

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Mumford's book on geometric invariant theory has an example, the group in question being cyclic of order 2. The same book of Mumford is also the answer to your other question. –  Kevin Buzzard Aug 20 '10 at 14:25
    
Cool, thanks. I actually had a quick flick through GIT but it didn't seem to give a quick easy criteria for when quotients by finite groups exist. Is this because a such a criteria doesn't exist or just because I haven't given the book the time it deserves? I definitely will read GIT thoroughly one day but at the moment I just want to move on with the problem at hand... –  anon Aug 20 '10 at 14:31
    
To handle qts by free actions of finite flat gp schemes (which I don't believe are handled by Mumford's book) over any base scheme, under the same "orbit in an affine" hypothesis see SGA3, Expose V. To weaken hypothesis on orbits but retaining the freeness condition (appropriately defined), you need to work with algebraic spaces; in fact, the freeness condition can be dropped as well with more input (Keel-Mori theorem on coarse moduli spaces, applied to quotient stacks). The Mumford example to which Kevin alludes is one for which the quotient does exist as an algebraic space. –  BCnrd Aug 20 '10 at 14:37
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Dear anon: for purpose of quotient questions like you are asking, it isn't necessary to read GIT thoroughly at all (not that there's anything wrong with that...). There is an awe-inspiring theorem of M. Artin that, under mild finiteness hypotheses, the category of algebraic spaces (defined relative to the etale topology) is stable under the formation of quotients by fppf equivalence relations. This is a reason why alg. spaces are a natural setting for such problems (but always nice when the qt is a scheme...). –  BCnrd Aug 20 '10 at 14:42
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2 Answers 2

The standard example is as follows: Take a $3$-fold $X$; for example let $X=\mathbb{P}^3$. Let $\sigma$ be an automorphism of $X$ of order $n$; for example, $(x_0:x_1:x_2:x_3) \mapsto (x_1:x_2:x_3:x_0)$. Let $C_1$, $C_2$, ..., $C_n$ be an $n$-gon of genus $0$ curves, with $C_i$ meeting $C_{i-1}$ and $C_{i+1}$ transversely and disjoint from the other $C_j$'s, with $\sigma(C_i)=C_{i+1}$. For example, $C_1 = \{(*:*:0:0) \}$, $C_2 = \{(0:*:*:0) \}$, $C_3 = \{(0:0:*:*) \}$ and $C_4 = \{(*:0:0:*) \}$. Let $p_i=C_{i-1} \cap C_{i}$.

Using this input data, we make our example. Take $X \setminus \{ p_1, p_2, \ldots, p_n \}$ and blow up the $C_i$. Call the result $X'$. Also, take a neighborhood $U_i$ of $p_i$, small enough to not contain any other $p_j$. Blow up $C_i \cap U_i$, then blow up the proper transform of $C_{i-1} \cap U_i$. Call the result $U'_i$. Glue together $X'$ and the $U'_i$'s to make a space $Y$. Clearly, $\sigma$ lifts to an action on $Y$.

There is a map $f: Y \to X$. The preimage $f^{-1}(p_i)$ consists of two genus zero curves, $A_i$ and $B_i$, meeting at a node $q_i$. The $q_i$ form an orbit for $\sigma$. We claim that there is no affine open $W$ containing the $q_i$.

Suppose otherwise. The complement of $W$ must be a hypersurface, call it $K$. Since $K$ does not contain $q_i$, it must meet $A_i$ and $B_i$ in finitely many points. Since $W$ is affine, $W$ cannot contain the whole of $A_i$ or the whole of $B_i$, so $K$ meets $A_i$ and $B_i$. This means that the intersection numbers $K \cdot A_i$ and $K \cdot B_i$ are all positive. We will show that there is no hypersurface $K$ with this property.

Proof: Let $x$ be a point on $C_i$, not equal to $p_i$ or $p_{i+1}$. Then $f^{-1}(x)$ is a curve in $Y$. As we slide $x$ towards $p_i$, that curve splits into $A_i \cup B_i$. As we slide $x$ towards $p_{i+1}$, that curves becomes $A_{i+1}$. (Assuming that you chose the same convention I did for which one to call $A$ and which to call $B$.) Thus, $[A_i] + [B_i]$ is homologous to $A_{i+1}$. So $\sum [A_i] = \sum [A_i] + [B_i]$, $\sum [B_i] = 0$, and $\sum B_i \cdot K = 0$. Thus, it is impossible that all the $B_i \cdot K$ are positive. QED

I think this example is discussed in one of the appendices to Hartshorne.

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It is, I believe as an example of a complete variety which is not projective. –  Charles Siegel Aug 20 '10 at 14:36
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Right. I should probably mention that any example is necessarily not quasiprojective (at least over an infinite field, and I think for finite fields too). In projective space, it is easy to build a hypersurface that avoids finitely many points by varying the coefficients of the defining equation, and the complement of any (nonempty) hypersurface in projective space is affine. –  David Speyer Aug 20 '10 at 15:16
    
Dear David, On line 2 of para. 2, should $C_i\cup U_i$ read $C_i\cap U_i$? –  Emerton Aug 20 '10 at 15:26
    
Thanks for the correction. –  David Speyer Aug 20 '10 at 15:55
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Oh, I guess we can just take the affine line with a double origin....:S

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