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Hi,

I am trying to define an exponential map for finite abelian group schemes. The following looks like it should work, but doesn't (see below). I am putting up this question hoping that someone will know how to fix this.

Let $R$ be a $\mathbb Q$-algebra and let $I\subset R$ be a nilpotent ideal. Let $G$ be a finite free group scheme over $R$. Let $A = \Gamma(G,\mathcal O_G)$. It is an $R$-algebra and $A^\vee = Hom_R(A,R)$ is likewise an $R$-algebra (using the comultiplication corresponding to the group operation of $G$). Let $\epsilon : A\to R$ correspond to the unit element $e\in G(R)$ of $G$. Let $Lie\ G$ be the Lie algebra of $G$: $$ (Lie\ G)(R) = \ker(G(R[t]/t^2)\to G(R)) = [ x\in A^\vee : x(ab) = \epsilon(a)x(b) +\epsilon(b)x(a)]\subset A^\vee. $$ Next we want to define $\exp : I\cdot (Lie\ G)(R)\subset I\cdot A^\vee \to G(R)$. Let $x\in I\cdot (Lie\ G)(R)$. Then we put $$ \exp (x) = \sum_{n=0}^\infty \frac{x^n}{n!} $$ Note that the sum is finite because $I$ is nilpotent. The element $x^n$ corresponds to the multiplication in $A^\vee$ and is defined by $x^n(a) = x(a)^n\in R$.

This all seems like it should work, but unfortunately you will see that $\exp(x):A\to R$ is not a ring homomorphism so it doesn't define an element in $G(R)$.

How to fix this?

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Does "finite free" group scheme mean "structure algebra is locally free of finite rank" as a module over the base ring? If so then over a $\mathbf{Q}$-algebra any such group scheme is finite etale, so the Lie algebra vanishes. What are you trying to do, and why? –  user30180 Apr 18 '13 at 3:14
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