The orbit set of k-points need not be the k-points of the fppf quotient sheaf

The algebra $A=k[x]\mathrm{/}(x^2-a)$ is a field $K=k[x]\mathrm{/}(x^2-a)$: the polynomial $x^2-a$ has no root in $k$ because $a$ is a nonsquare, hence is irreducible of degree two, and it is separable because its derivative $2x$ is nonzero as $\mathrm{char}k\mathrm{\ne }2$ and $a\mathrm{\ne }0$. Thus $A$ is free of rank two over $k$ and finitely presented, and ${\mathrm{\Omega }}_{A\mathrm{/}k}=Adx\mathrm{/}(2xdx)=0$ because $x$ is a unit with $x^2=a\in k^{\times }$ and $2$ is a unit; by [F3], $X\to \mathrm{Spec}k$ is finite étale of degree two. A $k$-algebra map $A\to k$ would send $x$ to an element $q\in k$ with $q^2=a$, so $X(k)=\mathrm{\varnothing }$ and the orbit set $X(k)\mathrm{/}G(k)$ is empty.

Put $G=\mathrm{Spec}k[t]\mathrm{/}(t^2-1)$ with comultiplication $t\mapsto t\otimes t$, counit $t\mapsto 1$ and antipode $t\mapsto t$, the last well defined because $t^2=1$; for every $k$-algebra $R$ the set $G(R)=\{r\in R:r^2=1\}$ is a group under multiplication, naturally in $R$, so by [F1] these maps make $G$ a finite-type $k$-group scheme with these groups of points. The formula $g\cdot z=gz$ defines a natural action of $G(R)$ on $X(R)=\{z\in R:z^2=a\}$: it is associative, unital, and stays in $X(R)$ because $(gz{)}^2=g^2{z}^2=a$, so by [F2] there is a morphism $\alpha :G{\times }_{k}X\to X$ with $\alpha (g,z)=gz$.

The morphism $\phi :G{\times }_{k}X\to X{\times }_{k}X$, $(g,z)\mapsto (gz,z)$, is an isomorphism. On coordinate rings it is the $k$-algebra map $k[x,y]\mathrm{/}(x^2-a,y^2-a)\to k[t,x]\mathrm{/}(t^2-1,x^2-a)$ with $x\mapsto tx$ and $y\mapsto x$; in the source ring $x$ and $y$ are units with $x^2=y^2=a\in k^{\times }$, and the assignment $t\mapsto x\mathrm{/}y$, $x\mapsto y$ defines an inverse: it respects the relations since $(x\mathrm{/}y{)}^2=x^2\mathrm{/}{y}^2=1$ and $y^2=a$, and the two composites fix each generator.

The relation $R=G{\times }_{k}X\rightrightarrows X$ has $s(g,z)=z$ and $t(g,z)=gz$, so $j=(t,s)$ is the isomorphism $\phi$ of step 2.1, in particular an equivalence relation. The ring $B=A[t]\mathrm{/}(t^2-1)$ is free over $A$ via $s$ with basis $1,t$, so $s$ is finite locally free. The involution $(g,z)\mapsto (g,gz)$ carries $s$ to $t$, so $t$ is finite locally free as well. The invariant ring is $C=k$: for $f=\alpha +\beta x$ one computes $s^{\mathrm{♯}}(f)=\alpha +\beta x$ and $t^{\mathrm{♯}}(f)=\alpha +\beta tx$ in $B\cong K\times K$ (evaluation at $t=1,-1$). Equality is equivalent to $\beta x(t-1)=0$; since $x$ is a unit and $t-1$ has components $(0,-2)$, this forces $\beta =0$. By [F4] the quotient is represented by $\mathrm{Spec}C=\mathrm{Spec}k$, so $(X\mathrm{/}G)(k)=\{\ast \}$.

Consider the identity section ${\mathrm{i}\mathrm{d}}_X\in X(X)$ and its class $[{\mathrm{i}\mathrm{d}}_X]$ in the naive quotient presheaf $P(X)=X(X)\mathrm{/}G(X)$ of [F5]. Its two pullbacks along the projections $X{\times }_{k}X\rightrightarrows X$ are the first and second projections, which differ by the element $u=y\mathrm{/}x\in G(X{\times }_{k}X)$: indeed $u^2=y^2\mathrm{/}{x}^2=a\mathrm{/}a=1$, so $u$ is a ${\mu }_2$-valued point, and $u\cdot {\mathrm{p}\mathrm{r}}_1={\mathrm{p}\mathrm{r}}_2$ because $u\cdot x=y$. Hence the two pullbacks of $[{\mathrm{i}\mathrm{d}}_X]$ in $P(X{\times }_{k}X)$ agree, so the class of the identity section is a compatible family of sections of the naive presheaf over the fppf covering $X\to \mathrm{Spec}k$.

That compatible family does not descend: $P(k)=X(k)\mathrm{/}G(k)$ is empty by step 1.1, so there is no class in $P(k)$ whose pullback along $X\to \mathrm{Spec}k$ could be the class of the identity section. Therefore the naive quotient presheaf is not an fppf sheaf and is not the quotient sheaf $X\mathrm{/}G$; since $X\mathrm{/}G$ is represented by $\mathrm{Spec}k$ with $(X\mathrm{/}G)(k)=\{\ast \}$ by step 3.1, while $X(k)\mathrm{/}G(k)=\mathrm{\varnothing }$, the orbit set does not compute the $k$-points of the quotient sheaf. Sheafification is strictly necessary, and the quotient sheaf here is representable: the counterexample separates the orbit set from the quotient sheaf, not representability from the quotient sheaf.