# Does the Mordell conjecture imply the Shafarevich conjecture

The base field is a number field.

It is known that the Shafarevich conjecture implies the Mordell conjecture (Kodaira-Parshin).

Is the converse also true?

Note that both conjectures are now theorems (Faltings).

Edit: To be clear, I'm referring to the Shafarevich conjecture for curves. That is, for any number field $K$, finite set of places $S$ of $K$ and integer $g > 1$, the set of curves over $K$ of genus $g$ with good reduction outside $S$ is finite.

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What brings up this question? Since Mordell is about curves and Shaf. has a much broader scope, it doesn't seem natural to think one could prove a converse direction. –  KConrad Mar 15 '12 at 14:39
Well, a strong enough effective version of Mordell (Mordell effectif) implies the abc conjecture. Although it doesnt appear anywhere in the literature to my knowledge, people seem to think that abc should imply Shafarevich. –  Bobby Mar 15 '12 at 15:11
Ow I see what you mean now. I am referring to Shafarevich' original conjecture for curves (not abelian varieties). –  Bobby Mar 15 '12 at 15:21
The Shafarevich conjecture can be interpreted as stating that the set of $S$-integral points (over any number field) on the moduli space of curves of fixed genus (or principally polarized abelian varieties of fixed dimension) is finite. So, in this sense, it is a generalization of Siegel's theorem/Mordell's conjecture. But note that it is a statement about a variety of high dimension, so I very much doubt that you can reduce the Shafarevich conjecture to the Mordell conjecture. Except, of course, in the case of elliptic curves, where the Shafarevich conjecture follows directly from Siegel's theorem as Shafarevich originally remarked.
The moduli space of curves of genus $g$ is not a Weil restriction of a curve, at least when it is non-rational (e.g. for $g \geq 24$). Indeed, if it were, then it would be the Weil restriction of a curve of genus zero, since it is simply connected: geometrically, a Weil restriction is a product. But then it would be rational. –  M P Mar 15 '12 at 18:44
Is $\mathbf{P}^1$ the only simply connected curve? I know elliptic curves have nontrivial etale covers (multiplication by $n$), but why is the etale fundamental group of a hyperbolic curve nontrivial? –  Harized Mar 15 '12 at 20:48
@Harized. Embed the curve in its Jacobian and get covers using multiplication by $n$ on the Jacobian. But there's a lot more. The computation of the fundamental group of a (topological) surface of genus g is a standard exercise in topology. –  Felipe Voloch Mar 15 '12 at 21:18