Fibres of the orbit map and the scheme-theoretic stabilizer as a closed subgroup scheme

Since $X$ is separated over $k$, the point $x$ is a closed immersion by [F1], and $H=G{\times }_{X,{\varrho }_x,x}\mathrm{Spec}k$ is the pullback of this closed immersion along ${\varrho }_x$, hence a closed subscheme of $G$ by [F4].

For every $k$-algebra $R$ the universal property in [F2] identifies the fiber of $(g,z)\mapsto (gz,z)$ over $(y,x)$ with $\{(g,z)\in G(R)\times X(R):gz=y_R,z=x_R\}=\{g\in G(R):g{x}_R=y_R\}=F_y(R)$; these identifications are natural in $R$ and therefore identify the two schemes. This is (c).

If $y=g_{0}x$ with $g_0\in G(k)$, left translation by $g_0$ is an automorphism of $G$ carrying $H$ onto the fiber $F_y$: for every $R$ it identifies $H(R)=\{h:h{x}_R=x_R\}$ with $\{g_{0}h:(g_{0}h)x_R=g_0{x}_R=y_R\}=F_y(R)$, and conversely $g{x}_R=y_R$ implies $(g_0^{-1}g)x_R=x_R$. Similarly, for $g\in G(k)$ and every $R$ one has $G_{gx}(R)=gH(R)g^{-1}$, because $g^{\mathrm{\prime }}(g{x}_R)=g{x}_R$ is equivalent to $(g^{-1}{g}^{\mathrm{\prime }}g)x_R=x_R$; the closed subschemes $G_{gx}$ and $gH{g}^{-1}$ have the same functor of points and hence coincide.

The morphism $\phi :G\times H\to G{\times }_{X}G$, $(g,h)\mapsto (g,gh)$, is well defined because $(gh)x=g(hx)=gx$; the morphism $\psi :G{\times }_{X}G\to G\times H$, $(g,g^{\mathrm{\prime }})\mapsto (g,g^{-1}{g}^{\mathrm{\prime }})$, is well defined because equal images $gx=g^{\mathrm{\prime }}x$ imply $(g^{-1}{g}^{\mathrm{\prime }})x=x$, that is $g^{-1}{g}^{\mathrm{\prime }}\in H$. On $R$-points for every $k$-algebra $R$ the two composites are the identity: $\psi \phi (g,h)=(g,g^{-1}gh)=(g,h)$ and $\phi \psi (g,g^{\mathrm{\prime }})=(g,g(g^{-1}{g}^{\mathrm{\prime }}))=(g,g^{\mathrm{\prime }})$. Hence $\phi$ and $\psi$ are inverse isomorphisms of $k$-schemes. This proves the first assertion of (d).

A fiber can be nonempty without having a $k$-point. Let $k=\mathbb{R}$, let $G={\mathbf{G}}_m=\mathrm{Spec}\mathbb{R}[t,t^{-1}]$ with the comultiplication $t\mapsto t\otimes t$ of [F6], let $X={\mathbf{G}}_m$, and let $G$ act on $X$ by $g\cdot z=g^{2}z$, an action because $(gh{)}^2=g^2{h}^2$ and $1^2=1$. Take $x=1$, $y=-1$. The fiber $F_y=G{\times }_X\mathrm{Spec}\mathbb{R}$ over $y$ has coordinate ring $\mathbb{R}[t,t^{-1}]{\otimes }_{\mathbb{R}[s,s^{-1}]}\mathbb{R}$, where the right factor is the residue field at $s=-1$; this tensor product is $\mathbb{R}[t,t^{-1}]\mathrm{/}(t^2+1)\cong \mathbb{R}[t]\mathrm{/}(t^2+1)$, since $t^2=-1$ makes $t$ invertible. That ring is nonzero because $t^2+1$ is nonconstant, hence not a unit of $\mathbb{R}[t]$ by [F7]. But $F_y(\mathbb{R})=\mathrm{\varnothing }$: an $\mathbb{R}$-point would give $z\in \mathbb{R}$ with $z^2=-1$, whereas $z^2\ge 0$ for every $z$ by the ordered-field fact of [F7] while $-1<0$, since $0-(-1)=1$ is positive. Thus a nonempty fiber need not have a $k$-point.

For every $k$-algebra $R$ the set $H(R)=\{g:g{x}_R=x_R\}$ is a subgroup of $G(R)$: it contains the identity; if $g{x}_R=x_R$ and $g^{\mathrm{\prime }}{x}_R=x_R$ then $(g{g}^{\mathrm{\prime }})x_R=g(g^{\mathrm{\prime }}{x}_R)=g{x}_R=x_R$; and if $g{x}_R=x_R$ then $g^{-1}{x}_R=g^{-1}(g{x}_R)=x_R$. The subsets are natural in $R$, so by the valued point criterion [F3] the closed subscheme $H$ of step 1.1 carries the unique structure of a closed subgroup scheme with this functor of points, which is (a).

For every $k$-algebra $R$ the map $F_y(R)\times H(R)\to F_y(R)\times F_y(R)$, $(g,h)\mapsto (g,gh)$, is a bijection: it is injective since $g=g^{\mathrm{\prime }}$ and $gh=g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ give $h=h^{\mathrm{\prime }}$, and a pair $(g,g^{\mathrm{\prime }})\in F_y(R)\times F_y(R)$ has $g^{-1}{g}^{\mathrm{\prime }}\in H(R)$ because $(g^{-1}{g}^{\mathrm{\prime }})x_R=g^{-1}(g^{\mathrm{\prime }}{x}_R)=g^{-1}(g{x}_R)=x_R$, with $(g,g^{-1}{g}^{\mathrm{\prime }})\mapsto (g,g^{\mathrm{\prime }})$. By Yoneda this exhibits the action morphism $F_y{\times }_{k}H\to F_y{\times }_k{F}_y$ as an isomorphism. Consequently, if $F_y(L)$ is nonempty for some field extension $L\mathrm{/}k$, base change to $L$ identifies $(F_y{)}_L$ with $H_L$.

Suppose $F_y$ is nonempty. Then $F_y$ has a nonempty affine open $\mathrm{Spec}A$ with $A\mathrm{\ne }0$ a finitely generated $k$-algebra; choose a maximal ideal $\mathfrak{m}\subset A$ and put $L=A\mathrm{/}\mathfrak{m}$. By [F5] the field $L$ is a finite extension of $k$, and the resulting $L$-point of $F_y$ is an element of $F_y(L)$. Step 2.2 then identifies $(F_y{)}_L$ with $H_L$; since a finite field extension is a faithfully flat and finitely presented base change, $F_y$ is an fppf right $H$-torsor, trivialized by the fppf cover $\mathrm{Spec}L\to \mathrm{Spec}k$, and it has a $k$-point exactly when $F_y\cong H$. Step 1.5 shows that the latter can fail, since the fiber $F_{-1}$ constructed there is nonempty and the trivialization just produced applies to it.

The remaining clause of (d) follows because a locally closed immersion is a monomorphism by [F4]: if ${\varrho }_x$ factors through $O_x\hookrightarrow X$, then for every test scheme a pair of points of $G$ has equal images in $O_x$ if and only if it has equal images in $X$, so $G{\times }_{O_x}G=G{\times }_{X}G$ and step 1.4 identifies it with $G\times H$. Statement (c) is step 1.2; statement (a) is step 2.1; statement (b) consists of step 1.3, step 2.2 and step 3.1; the first assertion of (d) is step 1.4. This completes the proof.