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Let $k$ be a field of characteristic $p> 0$.

Let $\mu_{p^n}$ denote the group scheme $\mu_{p^n}(R) =\{ x \in R: x^{p^n}=1 \}$. Then there are natural maps from $\mu_{p^n}$ to $\mu_{p^{n-1}}$. I'm interested in understanding what is the projective limit of this system of finite group schemes $\underset{n}{\varprojlim} \ \mu_{p^n}$? Is there any computation where this profinite group scheme appears? Or is this profinite group scheme well-known?

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  • $\begingroup$ If you really mean to ask about $\{x^{p^n}=1\}$ in characteristic $p>0$, I would not call it $\mu_p$ because it is not smooth and does not consist of roots of unity in a meaningful sense, and besides, I would translate by $1$ and call it $\{y^{p^n}=0\}$ for $y=x-1$. $\endgroup$
    – Gro-Tsen
    May 20, 2018 at 10:04

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Let me just do the simple thing and calculate the limit. Perhaps this is not what you are looking for, but then we can perhaps refine the question a bit in that case.

Write $\mu_{p^n} = \mathrm{Spec}(k[x_n]/(x_n^{p^n} - 1))$. Then the natural map $\mu_{p^n} \to \mu_{p^{n-1}}$ that I believe you are thinking about is the one induced by the $k$-algebra map \begin{align} \varphi_n \colon k[x_{n-1}]/(x_{n-1}^{p^{n-1}} - 1) & \to k[x_n]/(x_n^{p^n} - 1) \\ x_{n-1} & \mapsto x_n^p. \end{align} For our purposes, note that this is actually a morphism of Hopf $k$-algebras, which amounts to saying that this actually induces a morphism of the corresponding finite group schemes.

Now let's compute: since $\mathrm{Spec}$ yields an anti-equivalence between the category of affine (group) schemes over $k$ and the category of commutative $k$-algebras (perhaps Hopf algebras...), we have \begin{align} \mathbf{Z}_p(1) & := \mathrm{lim}_{\varphi_n}\,\mu_{p^n} \\ & \:= \mathrm{lim}_{\varphi_n}\, \mathrm{Spec}(k[x_n]/(x_n^{p^n} - 1)) \\ & \:= \mathrm{Spec}(\mathrm{colim}_{\varphi_n}\,k[x_n]/(x_n^{p^n} - 1)). \end{align} Now you can verify that $$ \mathrm{colim}_{\varphi_n}\,k[x_n]/(x_n^{p^n} - 1) \cong k[t^{1/p^{\infty}}]/(t - 1) $$ the natural map from the colimit being induced by the maps \begin{align} k[x_n]/(x_n^{p^n} - 1) & \to k[t^{1/p^\infty}]/(t - 1) \\ x_n & \mapsto t^{1/p^n}. \end{align} So $\mathbf{Z}_p(1) = \mathrm{Spec}(k[t^{1/p^\infty}]/(t - 1))$.

It may be useful to describe it's functor of points: given a $k$-algebra $R$, we have \begin{align} \mathbf{Z}_p(1)(R) & = \mathrm{Hom}(\mathrm{Spec}(R),\mathbf{Z}_p(1)) \\ & = \mathrm{Hom}(k[t^{1/p^\infty}]/(t - 1), R) \\ & = \big\{(r_1,r_2,\ldots) \in \prod\nolimits_{i = 1}^\infty R \mid r_1^p = 1 \;\text{and}\;r_{n+1}^p = r_n\;\text{for}\;n\geq 1\big\}, \end{align} i.e. $\mathbf{Z}_p(1)(R)$ consists of sequences of compatible $p$-power roots of unity in $R$, as should be expected.

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    $\begingroup$ This inverse limit group scheme is usually called $\mathbf{Z}_p(1)$. The symbol $\mu_{p^\infty}$ is usually reserved for the direct limit (as sheaves) of $\mu_p \subset \mu_{p^2} \subset ...$ along the usual inclusion maps. $\endgroup$
    – user117273
    May 21, 2018 at 14:21
  • $\begingroup$ @user117273: Ah! Thanks for pointing that out; of course, that notation makes a lot more sense. I have edited the text and made this change! $\endgroup$ May 21, 2018 at 16:07
  • $\begingroup$ Is there any specific reason for using the notation $\mathbb{Z}_p(1)$ what does $1$ stand for here? $\endgroup$
    – grok
    May 23, 2018 at 23:37

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