The resemblance between Mordell's theorem and Dirichlet's unit theorem

Question

The first one states that if $E/\mathbf Q$ is an elliptic curve, then $E(\mathbf Q)$ is a finitely generated abelian group.

If $K/\mathbf Q$ is a number field, Dirichlet's theorem says (among other things) that the group of units $\mathcal O_K^\times$ is finitely generated.

The proof of Mordell's theorem and the proof of Dirichlet's theorem are somewhat similar (a covolume calculation in one case, and what feels to me like bounding a covolume in the other).

How can these two objects be realized as instances of the same construction? Could it be done so well as to reduce the proof of Mordell-Weil and Dirichlet's theorems to a single proof?

In the correspondence between the class number formula and the conjectured formula for the leading term of $L(E/\mathbf Q, s)$ , it appears that $\mathcal O_K^\times$ really does play the role of $E(\mathbf Q)$ (regulator corresponds to regulator, torsion to torsion). From my understanding, the general belief is that these two objets are analogous. But I'm having a hard time putting them on the same footing.

In fact, there is a generalization of the Birch & Swinnerton-Dyer conjecture to any abelian variety over $\mathbf Q$, but in this case the conjectured leading term of the $L$-function is symmetric in $A$ and $\breve{A}$ (where $\breve{A}$ is the dual abelian variety). This conjecture degenerates to the BSD conjecture in the case of an elliptic curve, which is self-dual.

But $\mathcal O^\times$ isn't an abelian variety. At best, $\mathcal O_K^\times$ can be thought as the $\mathcal O_K$-valued points of the group scheme $\mathbf G_m = \text{Spec }(\mathbf Z[x,y]/(xy-1))$. But: (1), $\mathbf G_m$ isn't an abelian variety over any field, and (2), we are looking at its points in the ring of integers of a number field, rather than in a field. So, why should we expect it to be the same as $E(\mathbf Q)$?

Or, perhaps $E(\mathbf Q)$ and $\mathcal O_K^\times$ the wrong objects to be trying to compare?

Answer 1

Some comments.

First notice that the analogue object with $O_K^\star$ should be $E(K)$ (not $E(\mathbb Q)$). If you want to work over $\mathbb Q$, replace $\mathbb G_{m, K}$ with its Weil restriction to $\mathbb Q$ (then you get a torus of dimension $[K: \mathbb Q]$).

As you said, $O_K^\star$ is $\mathbb G_m(O_K)$, the group of sections of the group scheme $\mathbb G_m$ over $O_K$. In fact, the group $E(K)$, despite of the notation, is also the group of sections of some group scheme over $O_K$, namely, the Néron model of $E$ over $O_K$. But we can avoid this rather sophisticate object: let $S$ be a finite set of finite primes of $K$ such that $E$ has good reduction away from $S$ (simply take $S$ containing the set of primes dividing the discriminant of some Weierstrass equation with integral coefficients). This means that $E$ extends to an elliptic curve $\mathcal E$ over the ring $O_S$ of $S$-integers of $K$. Some works have to be done to show that $\mathcal E$ is indeed a group scheme. By the valuative criterion of properness, or just because $\mathcal E$ is projective, the canonical map $\mathcal E(O_S)\to E(K)$ is a group isomorphism bijective. On the other hand, it is easy to show that the finite generation of $O_S^*$ implies the finite generation of $O_K^*$ (the cokernel of $O_K^*\to O_S^*$ is generated as group by the missing primes).

Note that $O_K$ is also finitely generated over $\mathbb Z$.

So a common statement would be :

Let $G$ be a smooth commutative group scheme over $O_S$ with semi-abelian generic fiber $G_K$. Then $G(O_S)$ is a finitely generated abelian group.

Fix an integer $n$ invertible in $O_S$. The standard proof of Mordell starts by proving $E(K)/nE(K)$ is finite (weak Mordell-Weil). The proof extends verbatim to $G$ essentially because the multiplication by $n$ on $G$ is étale. The second part of the proof uses height function for abelian varieties and the finiteness of points of bounded height. For tori, it is similar: using logarithm, one shows that it is discrete. I don't know whether it is possible to unify and generalize these two approachs to show the discreteness in for any commutative algebraic group over $K$.

Of course this point of view doesn't simplify the proof of Mordell-Weil, it just makes complicate the proof of finite generation of $O_K^*$ and of $O_K$.

Answer 2

This is an interesting observation. It has been noticed before, but not much has come out of it as far as I know. The closest to realizing both theorems as instances of a more general theme is this paper of Franz Lemmermeyer "Conics: a poor man's elliptic curves", where he does not only relate units and points on elliptic curves (in fact $S$-units and points on elliptic curves) but also points on conics under a similar view point. See Section 7, for a summary of the comparison.

Answer 3

I write a different answer because the first one is begins to be too long.

Let $S$ be a finite set of primes in $K$. Let G be a smooth (finite type) commutative group scheme over $O_S$ with semi-abelian generic fiber $G_K$. Then $G(O_S)$ is a finitely generated abelian group.

Proof. Let $L$ be any finite extension of $K$ and let $S'$ be any finite set of primes of $L$ containing those dividing a prim of $S$, then $G(O_S)$ is canonically a subgroup of $G(O_{S'})$, and it is enough to show $G(O_{S'})$ is finitely generated. Note that $G(O_{S'})=G_{O_{S'}}(O_{S'})$ and $G_{O_{S'}}$ is again a smooth finite type commutative group scheme over $O_{S'}$.

The generic fiber $G_K$ is extension of an abelian variety $A_K$ by a torus $T_K$. Enlarging $K$ if necessary we can suppose $T_K$ is split ($\simeq \mathbb G_{m, K}^d$). Enlarging $S$ if necessary, we can suppose that $A_K$ has good reduction over $O_S$. Let $A$ be the abelian scheme over $O_S$ extending $A_K$. Then $G_K\to A_K$ extends to $G\to A$ by Néron mapping property. The kernel is a finite type model of $T_K$. Enlarging again $S$, we can suppose it is isomorphic to $\mathbb G_{m, O_S}^d$. So we ended up with an exact sequence $$ 1\to \mathbb G_{m, O_S}^d\to G\to A \to 0.$$ Taking $O_S$-sections, we get $$ 1\to (O_S^*)^d \to G(O_S) \to A(O_S)=A(K). $$ Using Mordell-Weil and the weak Dirichlet, we see that $G(O_S)$ is a finitely generated abelian group.

The above statement is of course well known. A colleague of mine tells me this should be in P. Vojta's papers "Integral Points on Subvarieties of Semiabelian Varieties".