# algebraic geometry – Duality of diagonalizable/constant group scheme?

Let $$M$$ be an arbitrary abstract group and $$S$$ be a scheme. Let $$M_S=coprod_{iin M}S_i$$ where $$S_i=S$$,which can be constructed via glueing data of a scheme see tag 01JA. Then $$M_S$$ is called the constant group scheme over $$S$$ associated to $$M$$. And for any $$S$$-scheme $$T$$, $$M_S(T)={text{locally constant functions }f:|T|to M}$$, where $$|T|$$ has Zariski topology and $$M$$ has discrete topology, and locally constant function is defined to be $$forall tin T,exists Uni ttext{ open}$$ s.t. $$f(U)={f(t)}$$. see tag 03YW

The dual of a group scheme $$G$$ over $$S$$ is a scheme representing the functor $$T/Smapsto mathrm{Hom}_{mathrm{Gr}-T}(G_T,mathbb{G}_m)$$. For a constant group scheme $$M_S$$, its dual is denoted as $$D(M)$$ or $$D_S(M)$$. In particular, it exists and $$DeclareMathOperator{Spec}{Spec}D_S(M)cong Spec mathbb{Z}(M)times_{Spec mathbb{Z}}S$$ where $$mathbb{Z}(M)$$ is the group ring $$bigoplus_{iin M}mathbb{Z}_i$$ with an obvious ring structure. Note that $$D_S(M)$$ is always commutative no matter whether $$M$$ is commutative.

It’s known in a few reference that if $$M$$ is commutative then $$D_SD_S(M)=M_S$$ i.e. $$M_S$$ is reflexive. (e.g. SGA3,Expose VIII,Page 3,Theoreme 1.2 or (Groupe Algeacutebriques by Demazure and Gabriel, Chap II,§2,2.11 Dualite des groupes diagonalisables)). But none of the proof seems convincing. And I seem to find a counterexample in some sense.

From my understanding of the corresponding between $$DeclareMathOperator{itSpec}{mathit{Spec}}D_SD_S(M)(T)=mathrm{Hom}_{mathrm{Gr}-T}(itSpec_{mathcal{O}_T}mathcal{O}_T(M),mathbb{G}_m)$$ and $$M_S(T)$$. A element in $$mathrm{Hom}_{mathrm{Gr}-T}(itSpec_{mathcal{O}_T}mathcal{O}_T(M),mathbb{G}_m)$$ corresponds to a map of $$mathcal{O}_T$$-algebras $$f:mathcal{O}_T(t,t^{-1})to mathcal{O}_T(M)=bigoplus_{iin M}mathcal{O}_T$$ s.t. $$epsilon f(t)=1$$ and $$Delta(f(t))=f(t)otimes f(t)$$.

If we write $$f(t)=sum_{iin M}a_icdot e_iin bigoplus_{iin M}mathcal{O}_T(T)$$ then we have $$a_i cdot a_j=delta_{ij}a_i$$ and $$1=sum_i a_i$$ (so each $$a_i$$ is an idempotent element in $$mathcal{O}_T(T)$$). Clearly $$forall sin T,1|_s=sum_i a_i|_s$$ so at least one of $$a_i$$ is supported on $$s$$. And $$T_{s_i} cap T_{s_j}=T_{s_is_j}=T_0=emptyset$$. So $$(T_{s_i})_{iin M}$$ forms a disjoint open cover of $$T$$ which can be identified with a locally constant function from $$T$$ to $$M$$. But the only problem is, $$sum_{iin M}a_icdot e_iin bigoplus_{iin M}mathcal{O}_T(T)$$ has only finite terms, which means the induced locally constant map has only a finite image. And I believe that there exists locally constant functions $$Tto M$$ with infinite image, which cannot be induced from $$mathcal{O}_T(t,t^{-1})to mathcal{O}_T(M)=bigoplus_{iin M}mathcal{O}_T$$.

So which part of my arugments is wrong? Or do we need some extra conditions on $$M$$ (like finite or finitely generated) or the base scheme $$S$$ (like locally Noetherian, locally connected or connected)? Or the duality only holds in some topology like fppf/etale topology?

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