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I am currently reading Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves", which can be found here.

My question concerns the proof of proposition 1.6.2, page 23 of the book. First, the context is the following.

Let $C/S$ be a smooth commutative group-scheme over a scheme $S$ of relative dimension $1$. Let $A$ be a fixed "abstract" finite abelian group. For any integer $N\geq 1$, we define the $S$-subgroup-scheme of $C/S$ of points of order $N$: for any $S$-scheme $T\rightarrow S$, we have $$C[N](T)=Ker\left(N:C(T)\rightarrow C(T)\right)$$ We fix an isomorphism of abelian groups $$A\simeq \mathbb{Z}/N_1\mathbb{Z} \times\ldots\times \mathbb{Z}/N_r\mathbb{Z}$$ so that the functor on $S$-schemes $Hom_{S-gp}(A,C)$ defined by $$T\mapsto Hom_{gp}(A,C(T))$$ is represented by the $S$-scheme of finite presentation $$C[N_1]\times _SC[N_2]\times _S \ldots \times _SC[N_r]$$

Under this context and these notations, the proposition is the following.

The functor $A$-$Str(C/S)$ on $S$-schemes $$T\mapsto the\space set\space of\space A-structures\space on\space C_T/T$$ is represented by a closed subscheme of $Hom_{S-gp}(A,C)\simeq C[N_1]\times _S \ldots \times _SC[N_r]$ definable locally by $1+\#(A) +\#(A)^2$ equations.

Now, this proof is admittedly easy using the corollary 1.3.7, page 15. It states the existence of a universal closed subscheme for the relation "a given effective proper Cartier divisor in $C/S$ is an $S$-subgroup-scheme".

In order to use this corollary, the "universal" effective Cartier divisor in $C\times _S Hom_{S-gp}(A,C)/Hom_{S-gp}(A,C)$ attached to the "universal" morphism $\phi _{univ}:A\rightarrow C$ is given out of the blue.

This is now the part I do not understand. Where does this universal Cartier divisor come from? How is $\phi _{univ}$ defined? Moreover, in order to be coherent, shouldn't this morphism actually take value in $C\times _S Hom_{S-gp}(A,C)$?

Any help in order to clarify this proof would be highly appreciated. I thank you very much.

NB: for the definition of "$A$-structure", see page 20, 1.5.1. A morphism of abstract groups $\phi: A\rightarrow C(S)$ is an $A$-structure on $C/S$ if the attached effective Cartier divisor $\sum _{a\in A}[\phi(a)]$ is a subgroup of $C/S$.

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    $\begingroup$ You're right that the map is not $A \to C$ but rather $A \to \left(C \times_S \text{Hom}_{S-\text{gp}}(A, C)\right)(\text{Hom}_{S-\text{gp}}(A, C))$. Think about what this means in terms of the fuctor of points: given $T \to \text{Hom}_\text{gp}(A, C(T))$ you want to produce a morphism from $T$ to the fibered product using $a \in A$. $\endgroup$ – RghtHndSd Jun 21 '18 at 1:51
  • $\begingroup$ Thank you for your answer, RghtHndSd. However, I am not sure I understand very well. To build the morphism $\phi _{univ}$, it is enough to build a family of group morphisms between the groups of $T$-points, functorial in $T$ which is any $S$-scheme, right? So namely, I am giving myself a point in $A(T)$ (which, if $T$ is not connected, may be different from $A$, right?), and I need to build an image in $C(T)$, another in $Hom_{gp}(A,C(T))$, which agree on $S(T)$. If I am not mistaken, this should be correct. I do not see how to build this then.. $\endgroup$ – Suzet Jun 21 '18 at 4:26
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You're right that the map is not $A \to C$ but rather $A \to (C×\text{Hom}_{S-\text{gp}}(A,C))(\text{Hom}_{S-\text{gp}}(A,C))$. Think about what this means in terms of the fuctor of points: given $T \to \text{Hom}_{\text{gp}}(A,C(T))$ you want to produce a morphism from $T$ to the fibered product using $a \in A$.

In your comment above, you talk about $A(T)$. However, $A$ is an abstract group (a set with a binary operation, and nothing more!), not a group scheme. In particular, $A$ is not a functor, and so $A(T)$ is meaningless.

Let's fix $a \in A$ and we want to produce a point of $(C \times_S \text{Hom}_{S-\text{gp}}(A, C))(\text{Hom}_{S-\text{gp}}(A, C))$. This means we want to produce a morphism

$$\text{Hom}_{S-\text{gp}}(A, C) \to C \times_S \text{Hom}_{S-\text{gp}}(A, C).$$

Using the identity map $\text{Hom}_{S-\text{gp}}(A, C) \to \text{Hom}_{S-\text{gp}}(A, C)$, by definition of the fibered product, it suffices to produce $\text{Hom}_{S-\text{gp}}(A, C) \to C$ such that the fibered product diagram commutes.

Now we have the universal point $\text{Hom}_{S-\text{gp}}(A, C) \to \text{Hom}_{S-\text{gp}}(A, C)$ given by the identity map, and this gives an element in $\text{Hom}_{\text{gp}}(A, C(\text{Hom}_{S-\text{gp}}(A, C))$. With our fixed $a \in A$, we have now produced the map $\text{Hom}_{S-\text{gp}}(A, C) \to C$.

It remains to check that the outer loop of the fibered product commutes. However since one arrow in the outer loop of our fibered product diagram is the identity, this immediately reduces to checking that the map $\text{Hom}_{S-\text{gp}}(A, C) \to C$ is an $S$-morphism (sorry, I'm too lazy to draw it out here). This is true by definition.

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  • $\begingroup$ Right, I eventually understood it! Thank you very much for your explanations. In fact, I thought of $A$ as the constant groupscheme $A$ over $S$ in my comment. I just realized however that I didn't actually understand the nature of the universal morphism $\phi$. Now I understand how it is built and why it works. Thank you again, your answer is really nice. $\endgroup$ – Suzet Jun 22 '18 at 0:54

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