### Motivation


I learned about this question from a wonderful article [Rational points on curves](https://www.math.mcgill.ca/darmon/pub/Articles/Expository/12.Clay/paper.pdf) by Henri Darmon. He gives a list of statements (some are theorems, some conjectures) of the form


* the set $\{$ objects $\dots$ over field $K$ with good reduction everywhere except set $S$ $\}$ is finite/empty


One interesting thing he mentions is about [abelian schemes](https://en.wikipedia.org/wiki/Abelian_variety#Abelian_scheme) in the most natural case $K = \mathbb Q$, $S$ empty. I think according to the definition we have a trivial example of relative dimension 0.


### Question



> 
> Why is the set of non-trivial abelian schemes over $\mathop{\text{Spec}}\mathbb Z$ empty?
> 
> 
> 


### Reference


This is proven in [Il n'y a pas de variété abélienne sur Z](https://doi.org/10.1007/BF01388584 "Fontaine, JM. Invent Math 81, 515–538 (1985). zbMATH review at https://zbmath.org/?q=an:0612.14043") by Fontaine, but I'm asking because: (1) Springer requires subscription, (2) there could be new ideas after 25 years, (3) the text is French and could be hard to read (4) this knowledge is worth disseminating.



Why no abelian varieties over Z?
It's a result related in spirit to Minkowski's theorem that $\mathbb Q$ admits no non-trivial unramified extensions. If $A$ is an abelian variety over $\mathbb Q$ with everywhere good reduction, then for any integer $n$ the $n$-torsion scheme $A[n]$ is a finite flat group scheme over $\mathbb Z$. Although this group scheme will be ramified at primes $p$ dividing $n$, Fontaine's theory shows that the ramification is of a rather mild type: so mild, that 
a non-trivial such family of $A[n]$ can't exist.


In the last 25 years, there has been much research on related questions, including by 
Brumer--Kramer, Schoof, and F. Calegari, among others. (One particularly interesting recent variation is a joint paper of F. Calegari and Dunfield in which they use related ideas to construct a tower of closed hyperbolic 3-manifolds that are rational homology spheres, but whose injectivity radii grow without bound.)


EDIT: I should add that the case of elliptic curves is older, due to Tate I believe,
and uses a different argument: he considers the equation computing the discriminant of 
a cubic polynomial $f(x)$ (corresponding to the elliptic curve $y^2 = f(x)$) and shows
that this solution equation has no integral solutions giving a discriminant of $\pm 1$.


This direction of argument generalizes in different ways, but is related to a result of Shafarevic (I think) proving that there are only finitely many elliptic curves with good reduction outside a finite set of primes. (A result which was generalized by Faltings to abelian varieties as part of his proof of Mordell's conjecture.)


Finally, one could add that in Faltings's argument, he also relied crucially on ramification results for $p$-divisible groups, due also to Tate, I think, results which Fontaine's theory generalizes. So one sees that the study of ramification of finite flat groups schemes and $p$-divisible groups (and more generally Fontaine's $p$-adic Hodge theory) plays a crucial role in these sorts of Diophantine questions. A colleague describes it as the ``black magic'' that makes all Diophantine arguments (including Wiles' proof of FLT as well) work.


P.S. It might be useful to give a toy illustrative example of how finite flat group schemes give rise to mildly ramified extensions: consider all the quadratic extensions of $\mathbb Q$ ramified only at $2$: they are ${\mathbb Q}(\sqrt{-1}),$ ${\mathbb Q}(\sqrt{2})$, and ${\mathbb Q}(\sqrt{-2})$, with discriminants $-4$, $8$, and $-8$ respectively. Thus ${\mathbb Q}(\sqrt{-1})$ is the least ramified, and not coincidentally, it is the splitting field of the finite flat group scheme $\mu\_4$ of 4th roots of unity.


