# Uniform Quotient vs Universal Quotient

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.
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. –  BCnrd Sep 13 '10 at 19:05
@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. –  jlk Sep 13 '10 at 21:16
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. –  Angelo Sep 15 '10 at 18:26