Locally constant group schemes

Question

Let $k$ be a field of characteristic $p$, then the group scheme $\mu_\ell$, if $p$ does not divide $\ell$, is étale-locally isomorphic to $\mathbb{Z}/\ell\mathbb{Z}$. I have two quick questions which I don't feel very certain. Is $\mu_p$ fppf locally isomorphic to $\mathbb{Z}/p\mathbb{Z} $? Also, is $\alpha_p $ fppf locally constant?

Answer 1

I guess if $\mu_p$ were locally isomorphic to $\mathbb{Z}/p\mathbb{Z}$, there would be a $k$-algebra $A$ such that $\mu_{p,A}\simeq \mathbb{Z}/p\mathbb{Z}_A$. But for a field $K$ which is a $A$-algebra, $\mu_{p,A}(K)=1$ but $\mathbb{Z}/p\mathbb{Z}_A(K)$ is the abstract group $\mathbb{Z}/p\mathbb{Z}$

Answer 2

AlexL's answer is a good one, I just want to point out another thought process.

Let $G$ be a finite group scheme over $X$ any scheme. If $G$ is fppf locally on $X$ constant then, in particular, the morphism $G\to X$ is fppf locally on $X$ etale since, after all, the constant group scheme is etale over its base. But, whether a morphism is etale can be checked fppf locally (e.g. see [1, Tag02VN]). Thus, if $G$ is locally constant then $G\to X$ is etale. The schemes $\mu_p\to\text{Spec}(k)$ and $\alpha_p\to\mathrm{Spec}(k)$ are not etale if $p = \mathrm{char}(k)$ and thus cannot be locally constant.

Remark 1: Of course, you can also replace 'fppf' by 'fpqc' as the above link shows.

Remark 2: The moral of the discussion in the comments to this answer/the answer of Qiaochu's is that if $k$ is a field then something happening fppf/fpqc locally on $k$ implies that this thing already happens over $\overline{k}$ (in fact it's iff for fpqc and iff fppf for most reasonable things that behave well with limits--one gets a similar statement where something which plays nicely with limits happens smooth locally, e.g. in the sense of [1, Tag021Z], over $k$ iff it happens over $k^\mathrm{sep}$).

Thus, if two group schemes $G$ and $G'$ are isomorphic fppf/fpqc locally on $\mathrm{Spec}(k)$ then, in fact, $G_{\overline{k}}$ and $G'_{\overline{k}}$ are isomorphic. The point is then that this clearly doesn't happen if $G=\underline{\mathbb{Z}/p\mathbb{Z}}$ and $G'=\mu_p$ or $G'=\alpha_p$ as can be checked in a variety of ways:

1. They don't have isomorphic $\overline{k}$-points (e.g. $G(\overline{k})$ has size $p$ but $G'(\overline{k})$ is trivial for either choice of $G'$).
2. The schemes $G_{\overline{k}}$ and $G'_{\overline{k}}$ aren't even isomorphic--$G_{\overline{k}}$ is reduced but $G'_{\overline{k}}$ is not reduced (for either choice of $G'$).
3. ...

The above answer I gave works more generally for any base whereas this argument in the comments to this answer/Qiaochu's answer aren't really generalizable in an easy way.

[1] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/

Answer 3

Let me see if I've unwound the definitions properly: it sounds to me like to check whether two things are fppf-locally isomorphic over $\text{Spec } k$ it suffices to check whether or not there exists some finitely presented $k$-algebra $A$ (flatness and surjectivity of $\text{Spec } A \to \text{Spec } k$ being automatic) such that the extensions of scalars to $A$ are isomorphic. Do I have that right?

If so, we can show that $\mu_{p, A} \cong \text{Spec } A[x]/(x^p - 1)$ is never isomorphic to the constant group scheme $\text{Spec } A^{\mathbb{Z}/p\mathbb{Z}}$; in fact already they are never isomorphic as $A$-schemes, forgetting the group structure. I will be very explicit here for those reading along who, like me, are not true algebraic geometers! Since every finitely presented $A$ admits a map to $\bar{k}$, it suffices to show this for $A = \overline{k}$, and then we can just take $\bar{k}$-points and observe that $\mu_{p, \overline{k}}$ has only one $\bar{k}$-point but $\text{Spec } \bar{k}^{\mathbb{Z}/p\mathbb{Z}}$ has, of course, $p$ of them.

Similarly we can show that $\alpha_p$ is not locally constant by checking that $\alpha_{p, A} \cong \text{Spec } A[x]/x^p$ is never isomorphic to any constant group scheme, which is already true forgetting the group structure. Again it suffices to show this for $A = \bar{k}$ and then again taking $\bar{k}$-points shows that there's one of them, so if $\alpha_{p, A}$ were locally constant it would have to be locally isomorphic to a point and it's not. (And of course since $x^p - 1 \cong (x - 1)^p$ we have that $\mu_p$ and $\alpha_p$ have the same underlying scheme.)