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Let $S$ be an integral scheme with function field $K = K(S)$. Let $\mathscr{A}, \mathscr{B}$ be Abelian schemes over $S$. Let $L/K$ be a separable field extension. Given $f_L \in \mathrm{Hom}(\mathscr{A}_L,\mathscr{B}_L)$, why does there exist an étale cover $T \to S$ with function field $L'$, $L/L'/K$ and an extension $f_T \in \mathrm{Hom}(\mathscr{A}_T,\mathscr{B}_T)$ of $f_{L'}$?

What might help: Let $R$ be a local Artinian ring with residue field $k$, and $\mathscr{A}, \mathscr{B}$ Abelian varieties over $R$. Then $\mathrm{Hom}(\mathscr{A}_R,\mathscr{B}_R) \to \mathrm{Hom}(\mathscr{A}_k,\mathscr{B}_k)$ is injective. (This follows from Mumford, GIT, Theorem 6.1 1).)

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    $\begingroup$ There are two elements of the solution. The first is the Weil extension theorem, which implies that the connected components of the Hom scheme $\text{Hom}_S(\mathcal{A},\mathcal{B})$ are proper over $S$. The second is an infinitesimal analysis that implies that $\text{Hom}_S(\mathcal{A},\mathcal{B})\to S$ is an unramified morphism. One proof of this does follow from the Rigidity Lemma, as you suggest. Altogether, $\text{Hom}_S(\mathcal{A},\mathcal{B})$ is a disjoint union of finite unramified $S$-schemes. Every component $T$ that dominates $S$ is finite and 'etale over $S$. $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 8:29
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    $\begingroup$ I just realized, for the argument that I am sketching, you do need to know that $S$ is unibranch, e.g., $S$ is integrally closed in its fraction field. Otherwise there may exist finite and unramified morphisms that are not 'etale, e.g., the normalization of a nodal plane curve (that is irreducible) is finite and unramified, yet it is not 'etale. $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 8:34
  • $\begingroup$ @Jason Starr: Assuming $S$ normal is OK. By the Weil extension theorem, do you mean "If $f$ is a rational map between smooth separated group schemes, defined in codimension $\leq 1$, then it is defined everywhere."? And then apply the valuative criterion for properness? For $\mathrm{Hom}_S(\mathscr{A},\mathscr{B}) \to S$ being unramified: Do you mean I should use the infinitesimal lifting criterion? And why don't you make your comments into an answer? $\endgroup$
    – user19475
    Commented Mar 25, 2017 at 8:57
  • $\begingroup$ The Rigidity Lemma does not give etaleness (and how do you produce accent grave in comments?). Consider a fixed elliptic curve $E$ over a field $k$. Let $S$ be a smooth, connected $k$-curve. Let $\mathcal{B}\to S$ be a family of elliptic curves over $S$ that dominates moduli and such that $\mathcal{B}_s$ is isomorphic to $E$ for some $k$-point $s$ of $S$. Let $\mathcal{A}$ be $E\times_{\text{Spec}\ k} S$. Then the $S$-scheme $\text{Hom}_S(\mathcal{A},\mathcal{B})$ has a closed point over $s$ that does not extend to a section over any etale open of $S$. $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 11:01

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Since the OP suggested it, I am posting my comments as an answer. By the construction of Hilbert and Quot schemes, there is a relative Hom scheme, $\text{Hom}_S(\mathcal{A},\mathcal{B})$ over $S$ whose connected components are quasi-projective over $S$. The claim is that these components are proper over $S$. By the valuative criterion of properness, it suffices to prove this in the special case that $S$ equals $\text{Spec}(R)$ for $R$ a DVR. Denote by $K$ the fraction field of $R$. Then it suffices to prove that every $K$-morphism of $K$-fibers, $\phi_K:\mathcal{A}_K\to \mathcal{B}_K$ extends to an $R$-morphism $\phi:\mathcal{A}\to \mathcal{B}$.

This follows by the extension result of Weil. Here is the basic argument: denote by $\mathcal{A}'$ the closure in $\mathcal{A}\times_S \mathcal{B}$ of the graph of $\phi$. The projection morphism $\text{pr}_{\mathcal{A}}:\mathcal{A}'\to \mathcal{A}$ is projective and birational. Since $\mathcal{A}$ is regular, by Abhyankar, for every irreducible component of the exceptional set, the geometric generic fiber of the component over its image is a ruled variety. By construction, this fiber immerses into a fiber of $\mathcal{B}\to S$. Since these fibers are Abelian varieties, they admit no nonconstant morphism from a genus $0$ curve. Thus, $\mathcal{A}'\to \mathcal{A}$ is an isomorphism. The composition of the projection $\text{pr}_{\mathcal{B}}:\mathcal{A}'\to \mathcal{B}$ and the inverse of this isomorphism is an $S$-morphism $\phi:\mathcal{A}\to \mathcal{B}$ extending $K$. Thus, $\text{Hom}_S(\mathcal{A},\mathcal{B})$ has proper components over $S$.

The Rigidity Lemma from "Geometric Invariant Theory" implies that the morphism $\text{Hom}_S(\mathcal{A},\mathcal{B})\to S$ is unramified. To prove this, it suffices to prove that for every Artin local scheme $\text{Spec}(C)$ over $S$, for every pair of $C$-morphisms of the pullbacks, $$\phi,\psi:\mathcal{A}_C\to \mathcal{B}_C,$$ both of which map the identity section to the identity section, if the restrictions of $\phi$ and $\psi$ over $\text{Spec}(C/\mathfrak{m}_C)$ are equal, then $\phi$ equals $\psi$. In other words, for the difference morphism $\delta = \phi - \psi$, if $\delta_{C/\mathfrak{m}_C}$ equals $0$, then $\delta$ equals $0$. By the Ridigity Lemma, Proposition 6.1(1), p. 115 of "Geometric Invariant Theory", $\delta$ is a composition of the $C$-morphism $\mathcal{A}_C \to \text{Spec}(C)$ and a section $\sigma:\text{Spec}(C)\to \mathcal{B}_C$ of $\mathcal{B}$ over $\text{Spec}(C)$. Since $\delta$ sends the identity section to the identity section, $\sigma$ is the identity section. Thus, $\delta$ is the constant morphism to the identity section, i.e., $\phi$ equals $\psi$. Therefore $\text{Hom}_S(\mathcal{A},\mathcal{B})$ is unramified over $S$.

Since the components are both proper and unramified, then they are finite. For a non-unibranch varieties $S$, there can exist a finite, unramified morphism $S'\to S$ that is not étale. For instance, if $S$ is a nodal plane curve, then the normalization will be finite and unramified, but not étale. However, if $S$ is normal, then every finite, unramified, dominant morphism to $S$ is étale. Therefore, every irreducible component of $\text{Hom}_S(\mathcal{A},\mathcal{B})$ that dominates $S$ is finite and étale over $S$.

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    $\begingroup$ To give a concrete example where $\operatorname{Hom}_S(\mathscr A,\mathscr B)$ is not étale even when $S$ is a smooth curve, consider a family $\mathscr A = \mathscr B = \mathscr E$ of elliptic curves whose general member is not CM but a special member $\mathscr E_s$ is. Then $\operatorname{Hom}_S(\mathscr A,\mathscr B) = \operatorname{End}_S(\mathscr E)$ has components lying only over $s$, corresponding to the extra endomorphisms of $\mathscr E_s$. Similar examples exist in mixed characteristic, e.g. if the reduction is supersingular. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 20, 2017 at 5:57

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