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It is known that for every abelian scheme $A$ over a ring $R$, there exists a subring $R_0$ of $R$ that is of finite type over $\mathbb{Z}$ and an abelian scheme $A_0$ over $R_0$ such that $A$ is deduced from $A_0$ by base change.

Using the relevant theorems in EGA, I see how one can get $A_0$ over $R_0$ satisfying every hypothesis defining abelian schemes except the "connected geometric fibers" property.

Is this really easy and I'm missing something or there is an argument to be done?

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  • $\begingroup$ So you have a group scheme $A_0$ over $R_0$ which after base-change to $R$ is isomorphic to $A$ over $R$? Certainly the group scheme $A_0$ will have geometrically connected fibres. Indeed, as the map Spec $R \to $ Spec $R_0$ is surjective, the fibre over a (geometric) point $r_0$ in Spec $R_0$ is dominated by the (connected) fibre of some point $r$ in Spec $R$? explicity: choose r in the inverse image of $r_0$. Then $A_r$ dominates $A_{r_0}$. Thus $A_{r_0}$ is connected. $\endgroup$ Jun 29, 2016 at 15:06
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    $\begingroup$ @AriyanJavanpeykar. What if $\text{Spec}(R) \to \text{Spec}(R_0)$ is not surjective? What if the image is not even constructible? You need to show that the locus in the base over which the fibers are geometrically connected is open. This follows, for instance, by considering the open stratum of the flattening stratification of the coherent sheaf $\mathcal{Q}$ coming from EGA III_2 Cor. 7.7.6, p. 201 for the sheaf $\mathcal{F}$ being the structure sheaf of $A_0$. $\endgroup$ Jun 29, 2016 at 15:28
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    $\begingroup$ @JasonStarr: Constructibility is sufficient (don't need openness), and holds much more widely (i.e., without properness hypotheses). See IV$_3$, 9.7.7. $\endgroup$
    – nfdc23
    Jun 29, 2016 at 15:37
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    $\begingroup$ @DocteurCottard: Let me assume that the set is open, $U = D(s_1)\cup \dots \cup D(s_m) \subset \text{Spec}(R_0)$. The image of $\text{Spec}(R)$ is contained in $U$. Thus the images of $s_1,\dots,s_m$ generate the unit ideal of $R$. So there exist elements $t_1,\dots,t_m$ in $R$ such that $1=s_1\cdot t_1 + \dots + s_m\cdot t_m$. After replacing $R_0$ by the subring $R_1=R_0[t_1,\dots,t_m]\subset R$ and replacing $A_0$ by the base change $A_1$ of $A_0$ to $R_1$, we can assume that the geometric fibers are connected. $\endgroup$ Jun 29, 2016 at 16:53
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    $\begingroup$ A different argument from Jason's is here: stacks.math.columbia.edu/tag/05FI $\endgroup$
    – darx
    Jun 29, 2016 at 21:31

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I am just writing my comments as an answer. As nfdc23 explains, there are stronger results that require weaker hypotheses, but let me assume that $R_0$ is a finitely generated algebra contained in $R$, and let $A_0$ be a proper, flat $R_0$-scheme whose geometric fibers are reduced and whose base change $A$ to $R$ has geometrically integral fibers. By Théorème 7.7.6 of EGA $\textrm{III}_2$, there exists a finitely presented $R_0$-module $Q_0$ that represents the covariant functor of $R_0$-modules, $$M\mapsto H^0(A_0,M\otimes_{R_0}\mathcal{O}_{A_0}).$$ Since $A_0$ is $R_0$-flat, the image of $A_0$ in $\text{Spec}(R_0)$ is an open subscheme $V$. Since $A_0$ is $R_0$-proper, this open subset is also closed, hence it is affine and finitely generated. The image of $\text{Spec}(R)$ is contained in $V$. Thus, up to replacing $\text{Spec}(R_0)$ by $V$, assume that $V$ equals $\text{Spec}(R_0)$.

Since $V$ equals $\text{Spec}(R_0)$, the support of $Q_0$ equals all of $\text{Spec}(R_0)$. Let $\text{Fit}_1(Q_0) \subset R_0$ be the Fitting ideal of $Q_0$, cf. Section 20.2 of Eisenbud's "Commutative Algebra with a View Towards Algebraic Geometry". The open complement $U$ of the zero scheme of $\text{Fit}_1(Q_0)$ contains all primes $\mathfrak{p}\subset R_0$ such that $(Q_0)_{\mathfrak{p}}$ is generated by one element as a module over the local ring $(R_0)_{\mathfrak{p}}$. Denoting $\kappa(\mathfrak{p}) = (R_0)_{\mathfrak{p}}/\mathfrak{p}(R_0)_{\mathfrak{p}}$, by Nakayama's Lemma, this is equivalent to the condition that $\text{Hom}_{R_0}(Q_0,\kappa(\mathfrak{p}))$ is $1$-dimensional as a vector space over $\kappa(\mathfrak{p})$, i.e., $H^0(A_0,\mathcal{O}_{A_0}\otimes_{R_0} \kappa(p))$ is $1$-dimensional as a vector space over $\kappa(\mathfrak{p})$. By hypothesis, for every prime of $R$, the corresponding vector space is $1$-dimensional. Thus, the image of $\text{Spec}(R)$ in $\text{Spec}(R_0)$ is contained in the open subset $U$.

The open subset $U$ is a union of basic open subsets $D(s_i)$. Since the image of $\text{Spec}(R)$ is contained in $U$, the elements $s_i$ generate the unit ideal in $R$. Thus, there exist finitely many of these elements, $s_1,\dots,s_m$, and there exist elements of $R$, $t_1,\dots,t_m$, such that $1=s_1\cdot t_1 + \dots + s_m\cdot t_m$ as elements in $R$. Define $R_1$ to be the $R_0$-subalgebra of $R$, $$R_1 = R_0[t_1,\dots,t_m]\subset R.$$ Then $R_1$ is also a finitely generated algebra, and the image of $\text{Spec}(R_1)$ in $\text{Spec}(R_0)$ is contained in $U$. Thus, for the base change $A_1$ of $A_0$ to $R_1$, the geometric fibers of $A_1$ over $R_1$ are integral.

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