# nLab Demazure, lectures on p-divisible groups, III.1 the Artin-Hasse exponential series

This entry is about a section of the text

Let $p$ be a fixed prime number.

###### Definition

Let $k$ be a ring. Then the assignation

${\Lambda }_{k}:\left\{\begin{array}{lll}{M}_{k}& \to & R\left[\left[t\right]\right]\\ R& \to & 1+\mathrm{tR}\left[\left[t\right]\right]\end{array}$\Lambda_k:\begin{cases} M_k&\to&R[[t]] \\ R&\to&1+tR[[t]] \end{cases}

sending a k-ring? to the multiplicative group of formal power series with coefficients in $R$ and constant term $1$ is an affine k-group. As a k-functor? ${\Lambda }_{k}$ is equivalent to ${O}_{k}^{ℕ}$.

###### Remark
1. There is an exact sequence $0\to {\Lambda }_{k}^{\left(n+1\right)}\to {\Lambda }_{k}^{\left(n\right)}\to {\alpha }_{k}\to 0$

2. We have ${\Lambda }_{k}={\mathrm{lim}}_{n}{\Lambda }_{k}/{\Lambda }_{k}^{\left(n+1\right)}$.

3. Each ${\Lambda }_{k}/{\Lambda }_{k}^{\left(n+1\right)}$ is is an $n$-fold extension of the additive group ${\alpha }_{k}$.

4. If $k$ is a field then ${\Lambda }_{k}$ is a unipotent group.

###### Remark

For $F=\left(1-t+\cdots \right)\in \Lambda \left(k\right)$ there is an isomorphism of $k$-schemes

${\varphi }_{F}:\left\{\begin{array}{l}{O}_{k}^{{ℕ}_{>0}}\to {\Lambda }_{k}\\ \left({a}_{n}\right)↦{\Pi }_{n}F\left({a}_{n}{t}^{n}\right)\end{array}$\phi_F:\begin{cases} O_k^{\mathbb{N}_{\gt 0}}\to \Lambda_k \\ (a_n)\mapsto \Pi_n F(a_n t^n) \end{cases}

If $k=ℚ$ and $F\left(t\right)=\mathrm{exp}\left(-t\right)$ we have $F\left(\mathrm{at}\right)F\left(\mathrm{bt}\right)=F\left(A\left(a+b\right)t\right)$ and ${\varphi }_{F}$ is an isomorphism ${\varphi }_{F}:{\alpha }_{k}^{{ℕ}_{>0}}\to {\Lambda }_{k}$.

If $k$ is a field with characteristic $p$ it is not possible an $F$ such that $F\left(\mathrm{at}\right)F\left(\mathrm{bt}\right)=F\left(\mathrm{ct}\right)$. However there is always a formula $F\left(\mathrm{at}\right)F\left(\mathrm{bt}\right)={\Pi }_{i>0}F\left({\lambda }_{i}\left(a,b\right){t}^{i}$ where ${\lambda }_{i}\left(X,Y\right)\in k\left[X,Y\right]$. This is verified by Möbius inversion.

###### Remark

(Möbius inversion) Let $\mu$ be the Möbius function. Then Möbius inversion gives

$\mathrm{exp}\left(-t\right)={\Pi }_{n}\left(1-{t}^{n}{\right)}^{\mu \left(n\right)/n}$exp(-t)=\Pi_n (1-t^n)^{\mu(n)/n}
###### Definition and Remark

The Artin-Hasse exponential is defined by the morphism

$E:\left\{\begin{array}{l}{O}_{{ℤ}_{\left(p\right)}}^{ℕ}\to {\Lambda }_{{ℤ}_{\left(p\right)}}\\ \left(\left({a}_{n}\right),t\right)↦{\Pi }_{n\ge 0}F\left({a}_{n}{t}^{{p}^{n}}\right)\end{array}$E:\begin{cases} O^\mathbb{N}_{\mathbb{Z}_{(p)}}\to \Lambda_{\mathbb{Z}_{(p)}} \\ ((a_n),t)\mapsto \Pi_{n\ge 0}F(a_n t^{p^n}) \end{cases}

We have $E\left(\left({a}_{i}\right),t\right)E\left(\left({b}_{i}\right),t\right)=E\left(\left({S}_{i}\left({a}_{0},\dots ,{a}_{i},{b}_{0},\dots ,{b}_{i}\right),t\right)$ where ${s}_{i}\in {ℤ}_{\left(p\right)}\left[{X}_{0},\dots ,{X}_{i},{Y}_{0},\dots ,{Y}_{i}\right]$

$P\in \Lambda \left(R\right)$, $R\in {M}_{{ℤ}_{\left(p\right)}}$ can be uniquely written as

$P\left(t\right)={\Pi }_{\left(n,p\right)=1}E\left(\left({a}_{n}\right),{t}^{n}\right)$P(t)=\Pi_{(n,p)=1} E((a_n),t^n)

where $\left({a}_{n}\right)\in {R}^{ℕ}$

###### Proposition

The ${ℤ}_{\left(p\right)}$-group ${\Lambda }_{{ℤ}_{\left(p\right)}}$ is isomorphic to the n/(n,p)=1-power of the subgroup image of $E$.

By base-change a similar statement applies to ${\Lambda }_{{𝔽}_{p}}$. This shows that the Artin-Hasse exponential plays over ${𝔽}_{p}$ a similar role as the usual exponential over $ℚ$.