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Let $p$ be a fixed prime number.
Let $k$ be a ring. Then the assignation
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 $O_k^\mathbb{N}$.
There is an exact sequence $0\to \Lambda_k^{(n+1)}\to \Lambda_k^{(n)}\to \alpha_k\to 0$
We have $\Lambda_k=lim_n \Lambda_k/\Lambda_k^{(n+1)}$.
Each $\Lambda_k /\Lambda_k^{(n+1)}$ is is an $n$-fold extension of the additive group $\alpha_k$.
If $k$ is a field then $\Lambda_k$ is a unipotent group.
For $F=(1-t+\cdots )\in \Lambda(k)$ there is an isomorphism of $k$-schemes
If $k=\mathbb{Q}$ and $F(t)=exp(-t)$ we have $F(at)F(bt)=F(A(a+b)t)$ and $\phi_F$ is an isomorphism $\phi_F:\alpha_k^{\mathbb{N}_{\gt 0}}\to \Lambda_k$.
If $k$ is a field with characteristic $p$ it is not possible an $F$ such that $F(at)F(bt)=F(ct)$. However there is always a formula $F(at)F(bt)=\Pi_{i\gt 0} F(\lambda_i(a,b)t^i$ where $\lambda_i(X,Y)\in k[X, Y]$. This is verified by Möbius inversion.
(Möbius inversion) Let $\mu$ be the Möbius function. Then Möbius inversion gives
The Artin-Hasse exponential is defined by the morphism
We have $E((a_i),t)E((b_i),t)=E((S_i(a_0,\dots,a_i,b_0,\dots,b_i),t)$ where $s_i\in \mathbb{Z}_{(p)}[X_0,\dots, X_i,Y_0,\dots, Y_i]$
$P\in \Lambda(R)$, $R\in M_{\mathbb{Z}_{(p)}}$ can be uniquely written as
where $(a_n)\in R^\mathbb{N}$
The $\mathbb{Z}_{(p)}$-group $\Lambda_{\mathbb{Z}_{(p)}}$ is isomorphic to the n/(n,p)=1-power of the subgroup image of $E$.
By base-change a similar statement applies to $\Lambda_{\mathbb{F}_p}$. This shows that the Artin-Hasse exponential plays over $\mathbb{F}_p$ a similar role as the usual exponential over $\mathbb{Q}$.
Last revised on May 27, 2012 at 13:39:07. See the history of this page for a list of all contributions to it.