Let $G/k$ be a finite group scheme over a field $k$ and $X$ be $k$-scheme of finite type. An action of $G$ on $X$ is a $k$-morphism $\mu : G \times_k X \rightarrow X$ satisfying the usual conditions. In SGA3-V-4 and 5, it states that the quotient $X/G$ exists if $\mu$ is a finite flat morphism with other conditions. But in somewhere, I saw that the quotient $X/G$ exists if $G$ is a finite group scheme over $k$, but it doesn't require that the action morphism $\mu$ to be finite or flat. My question is that if we need $\mu$ to be finite flat to ensure the existence of the quotient $X/G$.

Now I will show that under the assumption that $X$ is geometrically reduced, an action of a finite group scheme over $k$ on $X$ is always finite flat. So the above question reduces to if we really need the geometrically reduced assumption. This assumption enables us to get a scheme morphism from a "morphism" on its closed points. Also, even under this assumption, is there another way to show that the finiteness of $\mu$ without working on the base change to the algebraically closure of $k$ first.

Notice that a finite $k$-scheme $G$ is a finite disjoint union of $ U_g := \mathrm{Spec}(A_g)$ with each $A_g$ being a finite $k$-algebra ( $\mathrm{dim}(A_g) = 0$ ) such that $U_g$ is a one-point set and each $U_g$ is both open and closed in $G$. For each point $g \in G$, we denote the one point set $g$ by $U_g$, or simply by $g$, and $(U_g)_\mathrm{red}$ by $\overline{g}$.

We also denote the point in $G_{\mathrm{red}}$ which corresponds to the point $g$ in $G$ by $\overline{g}$.

Let first assume that $k$ is algebraically closed. In this case, each point $g$ of $G$ is a $k$-rational point and that $\overline{g}$ is an open affine subset of $G_{\mathrm{red}}$, which is $k$-isomorphic to $\mathrm{Spec}(k)$. For each $\overline{g}$, we consider the natural morphism $\overline{g} \times_k X = (U_g)_\mathrm{red} \times_k X \rightarrow X$. From the conditions of a group scheme acting on a scheme, we know that each point $\overline{g}$ gives an isomorphism from $X(k)$ to $X(k)$. The assumption on geometrically reduced tells us that it in fact gives an $k$-isomorphism of $X$. Hence the composition $\mu_{\overline{g}} : {\overline{g}} \times_k X \rightarrow g \times_k X \rightarrow X$ is a $k$-isomorphism. Using the fact that the natural morphism $ i : Y_{\mathrm{red}} \rightarrow Y$ has a property that $i(U)$ is an open affine subset of $Y$ for any open affine subset $U$ of $Y_{\mathrm{red}}$ ( for any noetherian scheme $Y$ ), one sees that $\mu_{\overline{g}}$ is an affine morphism. For any open affine subset $\mathrm{Spec}(A)$ of $X$, let $\mathrm{Spec}(B)$ be its inverse image under $\mu_g : g \times_k X \rightarrow X$. We have that the composition of $A \leftarrow B \leftarrow A$ is $\mathrm{id}_A$. Also notice that $A \leftarrow B$ is the quotient of $B$ by a nilpotent ideal. This implies that for each $b \in b$, there exists $a \in A$ such that $b-a$ is nilpotent. Hence $B \leftarrow A$ is an integral homomorphism. It's easy to see that $\mu_g$ is of finite type. Hence $B \leftarrow A$ is in fact finite. This proves that $\mu_g : g \times_k X \rightarrow X$ is finite flat. Since $\mu : G \times_k X \rightarrow X$ is the finite disjoint union of $\mu_g$, we know that $\mu$ is finite flat.

Now for $k$ not necessary being algebraically close, we take the base change to $\overline{k}$ and apply the above result. So the question is that, if the base change $\overline{\mu}$ of $\mu : G \times_k X \rightarrow X$ to $\overline{k}$ is finite flat, then could we conclude that $\mu$ is finite flat. Notice that $\mu$ is of finite type. One can show that $\mu$ is separated and universally closed from the separateness of $\overline{\mu}$ which is at the same time universally closed ( since $\overline{\mu}$ is finite). So one concludes that $\mu$ is proper. It's easy to show that $\mu$ is quasi-finite hence $\mu$ is finite. Finally, it's easy to show that $\mu$ is flat.


1 Answer 1


If $G$ is finite over $k$, then $\mu$ is automatically finite and flat. Indeed the morphism $G\to\mathrm{Spec}\;k$ is finite and flat and hence so is the projection $p:G\times_k X\to X$. But $\mu$ only differs from $p$ by an automorphism of $G\times_k X$ and hence it is also finite and flat. Namely $\mu=p\circ \alpha$ for $\alpha(g,x)=(g,gx)$.

Let me add the remark that more generally for a groupoid $X_1 \overset{d_1}{\underset{d_0}{\rightrightarrows}} X_0$ as in SGA3.V there is an automorphism $\alpha$ of $X_1$ such that $d_1=d_0\circ \alpha$. This removes the a priori strange asymmetry in for example Theorem 4.1 of loc.cit., where only the map $d_1$ is required to be finite locally free.

  • 1
    $\begingroup$ That´s so easy, why I didn´t figure out this $\endgroup$
    – user565739
    Jul 16, 2011 at 21:44

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