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What is a quotient of an affine scheme that is not a universal quotient? Let's recall some terminology.

Suppose that $k$ is an algebraically closed field and $G$ is a reductive group acting on an affine scheme $X$. Theorem 1.1 of Geometric Invariant Theory states that the uniform categorical quotient $X//G$ of $X$ exists.

In other words, $X \to X//G$ is universal with respect to $G$-invariant morphisms out of $X$ and this property persists under base change by a flat morphism $T \to X//G$.

When $\text{char}(k)=0$, the theorem states that $X \to X//G$ is a universal categorical quotient, so that the universal property persists under base change by an arbitrary morphism $T \to X//G$.

What is an example where $X \to X//G$ is not a universal quotient?

I'd be particularly interested in the case where the stabilizers of the action on $X$ are all linearly reductive.

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Here is an example, which is in some sense the simplest one. Suppose that $k$ has characteristic $p > 0$; set $X := \mathop{\rm Spec} k[x,y]$. Let $G$ be a cyclic group of order $p$ acting via $(x,y) \mapsto (x, x+y)$. The ring of invariants is $k[u,v] := k[x, y^p - x^{p-1}y]$. Consider the point $\mathop{\rm Spec} k = \mathop{\rm Spec} k[u,v]/(u,v)$ of $X/G = \mathop{\rm Spec} k[u,v]$; the inverse image $Y$ in $X$ is $\mathop{\rm Spec} k[x,y]/(x, y^p)$; it is immediate to check that the action of $G$ on $Y$ is trivial, so $Y/G = Y \neq \mathop{\rm Spec} k$.

If you want an example with a connected group, embed $G$ into $\mathrm{GL}_n$ and take the induced action.

I don't know any example with linearly reductive stabilizers, and I suspect that they don't exist.

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    $\begingroup$ Angelo's suspicion is right. By thm of Nagata (Ch.IV, Thm. 3.6 in Demazure-Gabriel), in char. $p > 0$ a smooth affine $k$-gp is lin. red. iff its comp. group has order not divisible by $p$ and id. component is torus. Hence, enough to treat tori and finite gps of order prime to $p$. The latter is universal for affine $X$ in char. $p$ via averaging. For tori $S$, can restrict to noetherian base change and then by noetherian induction $S$-invariants in coordinate rings upstairs and downstairs are $S[n]$-invariants for sufficiently divisible $n$ coprime to $p$. So reduced back to the first case. $\endgroup$
    – BCnrd
    Commented Sep 13, 2010 at 19:05
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    $\begingroup$ Brian, this is for the case of actions of linearly reductive stabilizers, but jlk was asking for the case when the stabilizers are linearly reductive, and I think this is much harder $\endgroup$
    – Angelo
    Commented Sep 13, 2010 at 20:28
  • $\begingroup$ @Angelo: Thanks. Your interpretation is correct: I would particularly like a proof/example in the case where the stabilizers are all linearly reductive, but the group $G$ is not. $\endgroup$
    – jlk
    Commented Sep 13, 2010 at 21:16
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    $\begingroup$ Dear Torsten, of course the approach would be the one you mention. There is a formal slice theorem for group actions with linearly reductive stabilizers, but it is very hard to get consequences from it. This is discussed in a paper of Jarod Alper, On the local quotient structure of Artin stacks <arxiv.org/abs/0904.2050>. $\endgroup$
    – Angelo
    Commented Sep 14, 2010 at 6:13
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    $\begingroup$ Consider the action of $\mathrm{PGL}_2$ on the space of unordered 4-tuples of points of $\mathbb P^1$ (a.k.a. $\mathbb P^4$). The generic stabilizer is isomorphic to the product of two cyclic groups of order 2 (the Klein group); but the stabilizer of a point corresponding to a double points and two distinct point of $\mathbb P^1$ is cyclic of order 2. Clearly, this means that you can't have a slice around this point, since the generic stabilizer is not conjugate to a subgroup of the special stabilizer. $\endgroup$
    – Angelo
    Commented Sep 15, 2010 at 18:26

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