Is the property of being a torsor for a smooth affine group scheme detectable on the small étale site?

Question

Let $S$ be a scheme, $G$ a smooth $S$-affine $S$-group scheme, and $X$ an $S$-scheme with a ($S$-linear) $G$-action $\alpha : G\times X\rightarrow X$.

Let $h_X$, $h_G$ be the representable sheaves on the small étale site $S_\text{ét}$ associated to $X,G$ respectively. Suppose $h_X$ is a torsor under $h_G$ (i.e., $h_G(T)$ acts simply transitively on $h_X(T)$ for every étale $T\rightarrow S$). Must $X$ be a $G$-torsor? (i.e., $\alpha\times\mathrm{id} : G\times X\rightarrow X\times X$ is an isomorphism).

If necessary, I'm happy to assume $G$ reductive and that $X$ is $S$-smooth and $S$-affine.

If this is true, a pleasing consequence is that it suffices to check that $h_G(\operatorname{Spec} R)$ is simply transitive on $h_X(\operatorname{Spec} R)$ for strict local rings $R$ of $S$.

Answer 1

Negative result: If the scheme $S$ lives in positive characteristic, then the answer is negative. For an example, take $S = \operatorname{Spec}(k)$ where $k$ is an algebraically closed field, and take $G = \mathbb{G}_m$ acting on itself $X = \mathbb{G}_m$ via the equation $t \cdot x = t^p x$.

Positive result: If $S$ is a $\mathbb{Q}$-scheme (i.e. it lives in characteristic $0$), then I believe that it is true (this is under all of the assumptions: $X$ is smooth affine over $S$, and $G$ is a connected reductive group scheme over $S$). The argument is a bit intricate (maybe can be simplified). I hope that there are no glaring mistakes.

Assume the "torsorness" at the level of the small étale site of $S$. First, we show the following.

Claim: For all geometric points $\overline{s} \to S$, the fiber $X_{\overline{s}}$ is $G_{\overline{s}}$-equivariantly isomorphic to $G_{\overline{s}}/H$ for some algebraic subgroup $H \subset G_{\overline{s}}$.

We may suppose that $\overline{s}$ is the algebraic closure of the residue field $\kappa(s)$ of its image $s \in S$. Pick a point $x \in X_{\overline{s}}(\overline{s})$. Set $H = \operatorname{Stab}_{G_{\overline{s}}}(x)$. There is a morphism $f:G_{\overline{s}} / H \to X_{\overline{s}}$. A small argument (valid in any characteristic) shows that in order to prove that $f$ is an isomorphism, it suffices to show that $G_{\overline{s}}(\overline{s})$ acts transitively on $X_{\overline{s}}(\overline{s})$ (here we use the smoothness and separatedness of $X_{\overline{s}}$). Hence, we are reduced to showing that $\alpha \times \mathrm{id}: G_{\overline{s}} \times X_{\overline{s}} \to X_{\overline{s}} \times X_{\overline{s}}$ is surjective at the level of $\overline{s}$-points. Choose an $\overline{s}$-point $y$ of $X_{\overline{s}} \times X_{\overline{s}}$. It is define over a finite separable extension of the ground field $\kappa(s)$. By smoothness of $X\times X \to S$, we may find an étale $S$-scheme $T \to S$ equipped with a lift $\overline{s} \to T$ and a point $\widetilde{y}: T \to X \times X$ such that $\overline{s} \to T \to X \times X$ is the point $y$ we started with. Now, by assumption the morphism on $T$-points $G(T) \times X(T) \to X(T) \times X(T)$ is surjective, and so there is a $T$-point $ \widetilde{z}: T \to G \times X$ that maps to $\widetilde{y}$. The composition $\overline{s} \to T \xrightarrow{\widetilde{z}} G \times X$ maps to the original $y$, and we conclude the desired surjectivity. This concludes the proof of the claim.

Now we use a hammer. Consider the quotient stack $[X/G] \to S$. By affinness of $X$ and (linear) reductivity of $G$, this has a good moduli space $[X/G] \to M \to S$ in the sense of Alper. The morphism $M \to S$ is flat affine and of finite type, because the stack was flat and of finite type over $S$ and $M$ is the relative spectrum of the sheaf of invariant rings. The formation of $M$ commutes with arbitrary base-change on $S$, and it follows from the claim that for all geometric points $\overline{s} \to S$ we have that the base-change $M_{\overline{s}} \to \overline{s}$ is an isomorphism. Hence, $M \to S$ is an isomorphism.

Now we use a double-hammer. Since the stack $[X/G] \to S$ is smooth, the etale local structure of good moduli space morphisms (see Section 10.2 in Alper, Hall, and Rydh - The étale local structure of algebraic stacks) tells us that, after replacing $S$ with an étale cover, there is a linearly reductive group $H$ over $S$ and an $H$-equivariant vector bundle $E \to S$ such that $[X/G] \cong [V/H]$. By our fiberwise claim, we see that $V$ must be the zero vector bundle. Hence, we upgrade our claim to conclude that, étale-locally on $S$, we have a $G$-equivariant isomorphism $X \cong G/H$ for some linearly reductive closed subgroup scheme $H \subset G$, which is smooth because the characteristic is 0.

Now, if $H$ was not trivial, then we would have some nonidentity section of $H$ over some $S$-scheme $T$ that would stabilize the tautological $S$-section of $X \cong G/H$. Since $H$ is smooth, we may even take $T$ to be étale. This would contradict the assumption of simple transitivity of the action on étale $S$-schemes. We conclude that $H$ is trivial, and hence $X$ is a torsor.

Answer 2

Upon further thought, I think that may avoid the (very nontrivial) local structure theorems in my answer and even prove a more general result in characteristic $0$. I post this potential argument as a separate answer, since merging everything in one answer would be a bit bulky.

Here is the claim.

Statement. Let $S$ be a Noetherian scheme over $\mathbb{Q}$. Let $G \to S$ be a finite type smooth group scheme, and let $X \to S$ be a smooth scheme. Suppose that $X$ is equipped with an action of $G$ such that the induced morphism $\alpha \times id: G \times_S X \to X \times_S X$ is an isomorphism of sheaves on the small etale site of $S$. Then, $X$ is a $G$-torsor.

To prove this, first we use the same claim as before. The proof is the same as in the other answer. I state here as two lemmas for completeness:

Lemma 1. Let $s \in S$ be a point with residue field $\kappa(s)$, and let $\overline{s}= \mathrm{Spec}(\overline{\kappa(s)}) \to S$ be the geometric point corresponding to a choice of algebraic closure of $\kappa(s)$. Then, the morphism $G_{\overline{s}} \times X_{\overline{s}} \to X_{\overline{s}} \times X_{\overline{s}}$ is surjective at the level of $\overline{s}$-points.

Proof of Lemma 1. Choose an $\overline{s}$-point $y$ of $X_{\overline{s}} \times X_{\overline{s}}$. It is defined over a finite separable extension of the ground field $\kappa(s)$. By smoothness of $X\times X \to S$, we may find an étale $S$-scheme $T \to S$ equipped with a lift $\overline{s} \to T$ and a point $\widetilde{y}: T \to X \times X$ such that $\overline{s} \to T \to X \times X$ is the point $y$ we started with. Now, by assumption the morphism on $T$-points $G(T) \times X(T) \to X(T) \times X(T)$ is surjective, and so there is a $T$-point $ \widetilde{z}: T \to G \times X$ that maps to $\widetilde{y}$. The composition $\overline{s} \to T \xrightarrow{\widetilde{z}} G \times X$ maps to the original $y$, and we conclude the desired surjectivity. QED

Lemma 2. For all geometric points $\overline{s} \to S$, the fiber $X_{\overline{s}}$ is $G_{\overline{s}}$-equivariantly isomorphic to $G_{\overline{s}}/H$ for some closed subgroup $H \subset G_{\overline{s}}$.

Proof of Lemma 2. We may suppose that $\overline{s}$ is the algebraic closure of the residue field $\kappa(s)$ of its image $s \in S$. Pick a point $x \in X_{\overline{s}}(\overline{s})$. Set $H = \operatorname{Stab}_{G_{\overline{s}}}(x)$. There is an induced morphism $f:G_{\overline{s}} / H \to X_{\overline{s}}$. By transitivity of the action of $G_{\overline{s}}$ on $\overline{s}$-points and the definition of $H$, we see that the preimage of every closed point of $X_{\overline{s}}$ is a reduced point. Note that $X_{\overline{s}}$ is smooth by assumption, and that $G_{\overline{s}}$ is smooth by descent. Since $G_{\overline{s}}$ is of finite type, the morphism of $f$ is of finite type, and hence by Chevalleys theorem the image is constructible. Since the image of $f$ contains all closed points, at least one of the (smooth integral) components of $G_{\overline{s}}/H$ dominates a (smooth integral) component of $X_{\overline{s}}$ (the image contains an open subset). By translation, every (smooth integral) component of $G_{\overline{s}}/H$ dominates a component of $X_{\overline{s}}$. Now we are in the context to apply (locally) the miracle flatness theorem: we have a dominant morphism f smooth integral schemes with all fibers points. So we have that $G_{\overline{s}}/H \to X_{\overline{s}}$ is flat, hence etale by the assumption on the fibers at closed points. The openness of $f$ and the surjectivity on closed points imply a posteriori that $X_{\overline{s}}$ is also finite type and the morphism $f$ is surjective. Since it induces an injection on closed points, one can see that it $f$ indeed a (surjective) open immersion, hence an isomorphism. QED

Lemma 3. The quotient stack $[X/G]$ is a gerbe over $S$.

Proof of Lemma 3. We first show that $[X/G]$ is a gerbe over an $S$-algebraic space $Y$. By https://stacks.math.columbia.edu/tag/06QJ, to show this it suffices to prove that the relative inertia of $[X/G] \to S$ is flat. This would follow from the morphism $G \times_S X \to X \times_S X$ being flat. By the fiberwise criterion for flatness https://stacks.math.columbia.edu/tag/039E, it suffices to check that for all $s \in S$, we have that the induced morphism on $s$-fibers $G_s \times X_s \to X_s \times X_s$ is flat. We may even check this after passing to geometric points $\overline{s} \to S$, and by Lemma 1 we are reduced to checking flatness of the corresponding morphism when $X_{\overline{s}} = G_{\overline{s}}/H$, which is true. Now consider the factorization $[X/G] \to Y \to S$, where the first morphism is a gerbe. Since $[X/G] \to Y$ is flat and $[X/G] \to S$ is smooth, it follows that $Y \to S$ is smooth. Furthermore, the formation of $Y$ commutes with base change on $S$, and over every geometric point $\overline{s} \to S$ we have that the fiber $Y_{\overline{s}}$ is $\overline{s}$ (because the fiber $[X_{\overline{s}}/G_{\overline{s}}]$ is $BH$, which is a gerbe over $\overline{s}$). Hence we see that $Y \to S$ is a universally injective etale morphism, and hence it is an open immersion. Since $Y \to S$ is also surjective, we conclude that $Y=S$. QED

Proof of the Statement. At this point we know that $[X/G] \to S$ is a gerbe, and by the smoothness of $X$ it admits sections etale locally on $S$. After passing to an etale cover of $S$, we may fix a section of $X \to S$ and we obtain an isomorphism $[X/G] \cong BH$ for some flat subgroup scheme $H \subset G$ (the stabilizer of the section). At this point, to conclude that $X$ is a $G$-torsor we just need to show that $H \to S$ is the trivial group scheme. Since we are in characteristic $0$, the fibers of $H \to S$ are automatically regular, and hence $H \to S$ is smooth. Suppose for the sake of contradicton that $H \to S$ was nontrivial. Then there would be an etale $S$-scheme $T \to S$ and a $T$ point $h \in H(T)$ that is not the identity. Note that $h$ would fix the corresponding distinguished section $x \in X(T)$ induced by the trivial section $S \to BH$ under the isomorphism $[X/G] \cong BH$. Viewing $h \in G(T)$, this would contradict the freeness of the action of $G(T)$ on $X(T)$.QED

Edited: Added necessary hypothesis that $G$ is of finite type, needed in the proof of Lemma 2. Otherwise the statement is false (take $S$ the spectrum of an algebraically closed field $k$, take $X=\mathbb{A}^1$, and let $G$ be disjoint copies of the $k$-points of $\mathbb{A}^1$ acting by translation). I added more details to the proof of Lemma 2 to be extra careful; non quasi-compact things are subtle.