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? 



Elimination of noetherian hypothesis for abelian schemes
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. 


