# centre and automorphism groups of finite group schemes

Let $G$ be a group scheme over a scheme $X$ with centre $Z(G)$, automorphism group $\mathrm{Aut}(G)$ and outer automorphism group $\mathrm{Out}(G)$ (viewed as group schemes on $X$).

1. If $G$ is finite flat over $X$, then are $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ also finite flat over $X$?
2. If $G$ is finite étale over $X$, then are $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ also finite étale over $X$?
• If $G$ is $\mu_p$ over $\text{Spec}(\mathbb{Z}_p)$, then the central fiber of $\text{Aut}(G)$ has length divisible by $p$, because it is non-reduced, yet the generic fiber has length $p-1$. – Jason Starr Aug 30 '17 at 11:36
• Here is my second attempt. The $p$-torsion over $\text{Spec}(\mathbb{Z}_p)$ of an elliptic curve with supersingular reduction is a group scheme $M$ whose closed fiber is $\mu_p\times \alpha_p$. Let $G$ be $M\times_{\text{Spec}(\mathbb{Z}_p)}M$. Then the closed fiber is a product $\mu_p^2 \times \alpha_p^2$. The automorphism group scheme of $\alpha_p^2$ is positive dimensional: it contains a copy of $\textbf{GL}_2$. – Jason Starr Aug 30 '17 at 12:02
• Those functors need not be representable for $G$ not finite fppf (e.g.,${\rm{Aut}}_{\mathbf{G}_a/R}$ isn't for $R$ not a $\mathbf{Q}$-algebra). Assume $G$ is finite fppf. The Aut-scheme exists as affine of finite presentation (coordinate ring for $G$ is vector bundle, etc.), so $Z(G)$ exists as $\ker(G\to{\rm{Aut}}_{G/X})$. If $Z(G)$ is flat (can fail!) then $G/Z(G)$ exists and $G/Z(G)\to{\rm{Aut}}_{G/X}$ is finite and monic, so a closed immersion, so ${\rm{Out}}_{G/X}$ exists. That being said, #2 is "yes" by reducing to constant $G$, but ${\rm{Aut}}_{\alpha^n_p/\mathbf{F}_p}={\rm{GL}}_n$. – nfdc23 Aug 30 '17 at 12:07
• Here is an example with non-flat center in case #1. Consider the standard inclusion $j:\alpha_p \hookrightarrow {\rm{GL}}_2$ into the upper-triangular unipotent subgroup $\mathbf{G}_a$. Over $R=\mathbf{F}_p[t]$ we define $i:\alpha_p \rightarrow {\rm{GL}}_{2/R}$ by $x \mapsto j_R(tx)$ and then form the semi-direct product $G :=\alpha_p \ltimes (\alpha_p \times \alpha_p)$ over $R$ using the action through $i$. Then $Z(G)|_{t \ne 0} = \alpha_p \times (\alpha_p \times \{0\})$ but $Z(G)_0 = Z(G)$, so due to such jumping of fiber-rank we see that the finite $Z(G)$ is not flat. – nfdc23 Aug 30 '17 at 12:13
• @JasonStarr: The automorphism scheme of $\mu_n$ is $(\mathbf{Z}/(n))^{\times}$ over any base, by Cartier duality, and the Aut-scheme of $\alpha_p^n$ (over any $\mathbf{F}_p$-scheme) is ${\rm{GL}}_n$ on the nose because the endomorphism scheme is ${\rm{Mat}}_n$ (which in turn is reduced to the case $n=1$ that is seen by bare hands). – nfdc23 Aug 30 '17 at 12:16

Here is an example of a finite flat group scheme where $Z(G)$ is not finite flat: Let $k$ be algebraically closed of odd characteristic $p$ and $X=\mathbf A^1$. Let $G$ be the closed subscheme of $GL_{3,X}$ given by matrices of the form $$A=\pmatrix{a&b&x\\sb&a&y\\0&0&1}$$ subject to the relations $\det A=a^2-sb^2=1$ and $a^p-1=b^p=x^p=y^p=0$. Here $s$ is the parameter on $X$. For $s\ne0$, the fiber of $G$ is the semidirect product $\mu_p\ltimes\alpha_p^2$ with $\mu_p$ acting by non-trivial characters. In particular, the center is trivial. For $s=0$ the group scheme is unipotent and its center is nontrivial.
A sketch of a proof is as follows: $G$, being finite étale, is étale locally on $X$ isomorphic to a constant finite group scheme. Therefore, by a standard descent argument, it suffices to prove the result for constant finite group schemes. However here $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ are all clearly finite étale, as required.