Subschemes in group action

Question

Let $k$ be a field, of any characteristic. Let $G$ be a smooth group scheme over $k$ and let $X$ be a smooth scheme of finite type over $k$. Let $Y\subseteq X$ be a smooth subscheme of $X$, and let $H\subseteq G$ be the group subscheme stabilizing $Y$ (i.e. $gY(R)=Y(R)$ for every $g\in H(R)$ and every $k$-algebra $R$). Assume that the map $Y\times^HG\to X$ is geometrically injective and surjective. Does this imply that this is an isomorphism, even in characteristic $p$ (in characteristic zero this is true much more generally, thanks to ZMT)? In all the concrete examples I have, this is true, but my examples are quite mild. One can assume that $X$ is connected, but $Y$ might not be.

By $Y\times^HG$ I mean $(Y\times G)/H$.

Edit: forgot the assumption $H$ smooth.

Answer 1

I am writing up my comment as an answer. Let $p$ be a prime integer. Let $k$ be field of characteristic $p$. Denote by $\mathbb{A}^2_k$ the $k$-scheme $\text{Spec}\ k[s,t]$ with its usual structure of group $k$-scheme, $$m:\mathbb{A}^2_k \times_{\text{Spec}\ k}\mathbb{A}^2_k \to \mathbb{A}^2_k, \ \ m^*(x) = x\otimes 1 + 1\otimes x, \ \ m^*(y) = y\otimes 1 + 1\otimes y,$$ and where the identity of this group action is the origin, i.e., the point corresponding to the maximal ideal $\mathfrak{m}=\langle x,y\rangle$.

Denote by $\mathbb{P}^1_k$ the $k$-scheme $\text{Proj}\ k[u,v]$ where $u$ and $v$ each have degree $1$. The group $k$-scheme $\mathbb{A}^2_k$ induces a group $\mathbb{P}^1$-scheme by pullback, $$\text{pr}_1:\mathbb{P}^1_k \times_{\text{Spec}\ k} \mathbb{A}^2_k \to \mathbb{P}^1_k, $$ $$(\text{pr}_1,m\circ \text{pr}_{2,3}):\mathbb{P}^1_k\times_{\text{Spec}\ k}\mathbb{A}^2_k \times_{\text{Spec}\ k}\mathbb{A}^2_k \to \mathbb{P}^1_k \times_{\text{Spec}\ k} \mathbb{A}^2_k.$$ Denote by $\Gamma$ the closed subscheme of $\mathbb{P}^1_k\times_{\text{Spec}\ k}\mathbb{A}^2_k$ whose ideal sheaf is generated by $\text{pr}_2^*(\mathfrak{m}^p)$ and the equation $uy-vx$.

The closed subscheme $\Gamma$ is a subgroup $\mathbb{P}^1$-scheme of $\mathbb{P}^1_k\times_{\text{Spec}\ k}\mathbb{A}^2_k,$ that is finite and flat over $\mathbb{P}^1_k$, and there exists a unique finite, flat morphism of group $\mathbb{P}^1$-schemes, $$q:\mathbb{P}^1_k\times_{\text{Spec}\ k}\mathbb{A}^2_k \to X,$$ such that the following induced morphism is an isomorphism, $$\Gamma\times_{\mathbb{P}^1_k} \mathbb{A}^2_k \to \mathbb{A}^2_k\times_X \mathbb{A}^2_k.$$ This claim is easiest to prove Zariski locally on $\mathbb{P}^1_k$. On the dense open subset $D_+(u)\subset \mathbb{P}^1_k$, there is an isomorphism of group $D_+(u)$-schemes, $$\phi_u:D_+(u)\times_{\text{Spec}\ k} \mathbb{A}^2_k \to D_+(u)\times_{\text{Spec}\ k} \mathbb{A}^2_k, \ \ \phi_u^*x = x, \ \phi_u^*y = y+(v/u)x.$$ The inverse image of $\Gamma$ is $D_+(u)\times (\alpha_p\times \{0\}),$ i.e., the copy of $\alpha_p$ in the first factor of $\mathbb{A}^2_k$. A quotient by this inverse image group $D_+(u)$-scheme is the finite flat morphism of group $D_+(u)$-schemes, $$q_u:D_+(u)\times \mathbb{A}^2_k \to D_+(u)\times \mathbb{A}^2_k, \ \ q_u^*(x) = x^p, \ \ q_u^*(y) = y.$$ There is a similar description on $D_+(v)$, and the uniqueness of quotients allows to glue these quotients on $D_+(u)\cap D_+(v)$.

The quotient group morphism $q$ induces an action of the group $\mathbb{P}^1$-scheme $\mathbb{P}^1_k\times_{\text{Spec}\ k} \mathbb{A}^2_k$ on $X$. This is equivalent to an action of $\mathbb{A}^2_k$ on the $k$-scheme $X$ such that the projection morphism, $$\pi:X\to \mathbb{P}^1_k,$$ is $\mathbb{A}^2_k$-invariant. Define $Y$ to be the image of the section of $\pi$ coming from the group identity. The inverse image of $Y$ under $q$ equals $\Gamma$. Thus, the stabilizer subgroup $k$-scheme $H$ of $Y$ equals the subgroup $k$-scheme that maps $\Gamma$ to itself.

In particular, for the fiber $\Gamma_0$ over $\text{Zero}(v)\subset \mathbb{P}^1_k$, this is a subgroup $k$-scheme of $\mathbb{A}^2_k$ whose regular action on $\mathbb{A}^2_k$ maps $\alpha_p\times\{0\}$ to itself, i.e., a subgroup $k$-scheme of $\alpha_p\times\{0\}$. On the other hand, for the fiber $\Gamma_\infty$ over $\text{Zero}(u)\subset \mathbb{P}^1_k$, this is a subgroup $k$-scheme of $\{0\}\times \alpha_p.$ The intersection of these two subgroup $k$-schemes of $\mathbb{A}^2_k$ is the trivial subgroup scheme $\text{Spec}\ k$. Thus, $H$ is the trivial subgroup $k$-scheme.

The induced morphism $(Y\times G)/\Delta(H) \to X$ is simply the morphism $q$. This morphism is a universal homeomorphism. However, the morphism $q$ is not an isomorphism: it is finite and flat of degree $p>1$.