Fontaine’s theorem
Definition 1. An abelian scheme $A \to S$ is a smooth proper scheme with connected fibers, which is a group object.
This implies that all fibers are abelian varieties, and hence it is automatically commutative. For any $n$, the map $n \colon A \to A$ is finite locally free, and then $A[n]/S$ is finite locally free over $S$ of degree $n^{2g}$, where $g = \dim A_s$. This is automatically étale over $S$ away from points of characteristic dividing $n$.
Theorem 2 (Fontaine). There is no abelian scheme $A \to \operatorname{Spec} \mathbb{Z}$ of positive dimension.
The way to approach this is by analyzing $A[n] \to \operatorname{Spec} \mathbb{Z}$.
Corollary 3. The only relative curve $\mathcal{C} \to \operatorname{Spec} \mathbb{Z}$ (more precisely, relative dimension one smooth proper with connected geometric fibers) over $\mathbb{Z}$ is $\mathbb{P}_ \mathbb{Z}^1$.
This is because we can look at the Jacobian $\operatorname{Pic}^0(\mathcal{C})$. It is proper because of the valuative criterion and smooth because the obstruction in $H^2$ vanishes. So if $g \gt 0$ we get a contradiction. When $g = 0$ then this is classified by $H^1(\operatorname{Spec} \mathbb{Z}, \mathrm{PGL}_ 2)$ and this is trivial because a quaternion algebra over $\mathbb{Q}$ is split unless it is ramified at at least two places (one of which has to be finite).
No elliptic curves over ℤ
For an elliptic curve $E/K$ over a field $K$, we have a distinguished point $P_0 \in A(K)$. Now using $\mathscr{O}(3P_0)$, we get an embedding $E \hookrightarrow \mathbb{P}_ K^2$, and moreover there is a line that intersects $E$ at $P_0$ of order $3$. Setting this tangent line to be $z = 0$ and $P_0 = [0:1:0]$ we get an equation of the form $$ zy^2 + a_1 zxy + a_3 z^2y = x^3 + a_2 z x^2 + a_4 z^2 x + a_6 z^3. $$
Here, if we use the change of variables $x \mapsto u^2 x$ and $y \mapsto u^3 y$ then we get the equation but modified by $a_i \mapsto u^i a_i$. So over a base scheme $B$ and a line bundle $\mathscr{L}/B$, we can look at $$ \mathbb{P}_ B(\mathscr{E}), \quad \mathscr{E} = \mathscr{O} z \oplus \mathscr{L}^{-2}x \oplus \mathscr{L}^{-3}y, $$ where we think of variables as $$ z \in H^0(B, \mathscr{E}), \quad x \in H^0(B, \mathscr{E} \otimes \mathscr{L}^2), \quad y \in H^0(B, \mathscr{E} \otimes \mathscr{L}^3). $$ Now given these coefficients $a_i \in H^0(B, \mathscr{L}^i)$, we get the abstract Weierstrass model $$ \mathcal{W} \subseteq \mathbb{P}(\mathscr{E}) $$ cut out by the above equation.
Proposition 4. If we have an isomorphism $\mathcal{W} \cong \mathcal{W}^\prime$ then this comes from
- an isomorphism $\varphi \colon \mathscr{L} \cong \mathscr{L}^\prime$ and
- an isomorphism $$ \psi \colon \mathscr{O} \oplus \mathscr{L}^{-2} \oplus \mathscr{L}^{-3} \cong \mathscr{O} \oplus \mathscr{L}^{\prime-2} \oplus \mathscr{L}^{\prime-3} $$ that preserves the flat, with the additional property that the map on its associated graded recovers the powers of $\varphi$.
What is the way to think about this? This $\mathscr{L}$ will be the Hodge bundle $\omega_{\mathcal{W}/B}$, because the invariant differential $dx/y$ has degree $1$ as $dx/y \mapsto u^{-1} dx/y$.
It turns out there is a section $$ \Delta \in H^0(B, \mathscr{L}^{12}) $$ whose zeros are the singular fibers, and it is an integral polynomial in $a_i$. For $R$ a discrete valuation ring and $K = \operatorname{Frac}(R)$ and $E/K$, we may consider all integral Weierstrass models over $R$. We can ask for the minimal value of $\operatorname{val}(\Delta)$, and it turns out that this is zero if and only if the elliptic curve has good reduction.
Theorem 5. If $C$ is a Dedekind scheme, then there exists a unique $\mathcal{W} \to C$ whose completion at each $x \in C$ is minimal Weierstrass.
Note that here $\mathscr{L}$ does not have to be trivial. But over $\mathbb{Z}$, if there is an elliptic curve then $\Delta = \pm 1$. We can actually rule this out using arithmetic.