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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?

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What do you mean by a covolume-calculation in proving Mordell theorem? IIRC, the proof is about the vanishing of certain cohomology groups, thus obtaining the result. Maybe you are talking about other approaches? –  awllower Feb 23 '13 at 2:36
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One might view this phenomenon this way: We can construct abelian extensions of number fiels(it appearts to work for imaginary number fields) by use of torsion points on $\mathbb E(\mathbb Q)$. And one obtains the unit group in an extension. Moreover, I believe, though I cannot assure it in my recollection, that Dirichlet unit theorem can also be proved by means of cohomology. In any case, this seems to be a nice question. And feel free to inform me of any errors. Thanks. –  awllower Feb 23 '13 at 2:41
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Dear @awllower, thank you for your response. Part of the proof of the Mordell-Weil theorem (the "easy" part) depends on the construction of the height function, and on the descent lemma. This is the part which, to me, has the flavor of a covolume calculation. –  Bruno Joyal Feb 23 '13 at 2:50
    
I see, Thanks for pointing this out. –  awllower Feb 23 '13 at 2:51

3 Answers 3

up vote 8 down vote accepted

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 $O_K$.

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The proposed generalization is false. Take $G$ to be the additive group $\mathbb{G}_a$, take $\mathcal{O}_S$ to be $\mathbb{Z}[1/p]$ for some prime $p$. Then $G(\mathcal{O}_S)$ is $\mathcal{O}_S$ considered as an additive group, which is not finitely generated. –  David Speyer Mar 2 '13 at 19:12
    
@DavidSpeyer: good point. Thanks ! I add a restriction. –  user18119 Mar 2 '13 at 21:20
    
Thank you very much. I will ponder on this. I'll accept your answer in a few days if there is no more input. Again, thank you! –  Bruno Joyal Mar 2 '13 at 23:07
    
Dear @QiL'8: what do you mean by "semi-abelian generic fibre"? –  Bruno Joyal Mar 5 '13 at 1:56
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@Bruno: this means the generic fiber which is an algebraic group over $K$, is an extension of an abelian variety by a torus. For the finite generation, one can always enlarge the base field (the original group of sections is then contained in the new one), so the torus can be supposed to be split, i.e., isomorphic to a $\mathbb G_{m,K}^d$. –  user18119 Mar 5 '13 at 9:22

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.

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One can notice that the conic defined by a Pell-Fermat equation corresponds to a torus of dimension $1$. –  user18119 Mar 1 '13 at 23:56
    
Thank you for the reference! –  Bruno Joyal Mar 3 '13 at 2:31

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".

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