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Let $G$ be a $k$-group-functor. Then the following conditions are equivalent:
$G$ is the Cartier dual of a constant group.
$G$ is an affine $k$-group and the $k$-ring $O(G)$ is generated by the morphisms $G\to \mu_k$ (these are called characters of $G$).
A $k$-group satisfying the conditions of the previous remark is called diagnalizable k-group.
Let $G$ be a $k$-group. Then the following conditions are equivalent:
$G\otimes_k k_s$ is diagonalizable.
$G\otimes_k K$ is diagonalizable for a field $K\in M_k$.
$G$ is the Cartier dual of an étale $k$-group.
$\hat D(G)$ is an étale? $k$-formal group.
$Gr_k(G,\alpha_k)=0$
(If $p\neq 0)$, $V_G$ is an epimorphism
(If $p\neq0)$, $V_G$ is an isomorphism
A $k$-group satisfying the conditions of the previous theorem is called multiplicative k-group.
Multiplicative $k$-groups correspond by duality to étale formal $k$-groups.
The category $ACm_k$ of multiplicative $k$-groups forma a subcategory of the category $AC_k$ of affine commutative $k$-groups which is stable under forming subgroups, quatients, extensions (the set of these properties says that the subcategory is thick) and limits.
$ACm_k$ is (contravariant) equivalent to the category of Galois modules: To $G$ corresponds the Galois module $\hat D(G\otimes_k k_s)(k_s)=Gr_{k_s}(G\otimes_k k_s,\mu_{k_s})$.
If $E$ is an étale $k$-group, then $D(E)$ is multiplicative and $\hat D(D(E))=E$. And we have $D(D(E))=E$. The duality is hence given by $E\to D(E)$ , $G\to D(G)$ without reference to formal groups.
Last revised on July 19, 2012 at 00:07:30. See the history of this page for a list of all contributions to it.