Elimination of noetherian hypothesis for abelian schemes 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?   
 A: 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.  
