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](http://books.google.com/books?id=dFlv3zn_2-gC&printsec=frontcover&dq=geometric+invariant+theory&hl=en&ei=SpaNTInmIY2MnQfwmdDUCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q&f=false)  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.



Uniform Quotient vs Universal Quotient
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.


