The Mordell Conjecture/Faltings Theorem says roughly that if $K$ is an algebraic number field and $X$ is an algebraic curve defined over $K$ of genus $g >1$ then the set of $K$-rational points $X(K)$ is finite.

Now in the cases when the genus is $g = 0$ or $g = 1$ we can have an infinite number of $K$-rational points in the curve, but then Mordell's conjecture says that when the genus goes on to the next higher value $g \geq 2$ then this can no longer happen and we can only have a finite number of $K$-rational points lying on the curve (if there are any).

The usual topological interpretation that one sees mentioned in the books when they talk about the genus is as the number of holes in the surface $X(\mathbb{C})$. Thus I was just wondering if there's some way to see intuitively or at least without having to resort to highly technical arguments using scheme theory why this happens?

To be more specific, is there a way to see intuitively how the fact that the surface $X(\mathbb{C})$ has more than one hole (when the genus is greater than one) prevents the corresponding algebraic curve from having an infinite number of $K$-rational points?

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Note: If it helps to ease things at least a little bit, I would be more than happy with an argument for the case when $K = \mathbb{Q}$ the set of rational numbers. I'm aware that this is highly technical stuff so maybe I'm asking for too much, but it would be very interesting if there is an argument for this so I cannot resist asking.

Thank you very much in advance for any help.

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    $\begingroup$ Have you looked at Vojta's proof of the conjecture? It's based on basic ideas of diophantine approximations, generalizing Siegel's Theorem. Here's Silverman's review: ams.org/mathscinet/search/… $\endgroup$ Apr 28, 2011 at 18:31
  • $\begingroup$ @Arturo No I haven't. I'll take a look at it when I am in the university since I can't access that from my apartment. Thank you very much. $\endgroup$ Apr 28, 2011 at 18:38
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    $\begingroup$ I can summarize some of it, but it may not be any more "intuitive". Remember that Mordell objected to it being called "Mordell's conjecture", saying it was more of a question, not even a wild guess, and that he had little or no intuition to base it on. $\endgroup$ Apr 28, 2011 at 18:39
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    $\begingroup$ Related: mathoverflow.net/questions/56011/… $\endgroup$ Apr 28, 2011 at 19:40
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    $\begingroup$ @Adrian: I was trying to write a summary, but in the end there is just too much handwaving beyond me. I can tell you exactly where the genus comes in within Vojta's proof (invoking the arithmetic Riemann-Roch theorem), but I don't really have any intuition on regular Riemann-Roch, let alone the arithmetic version. I suspect Qiaochu's link is a better resource anyway. $\endgroup$ Apr 28, 2011 at 20:22

1 Answer 1


You should certainly look at the answers to the MO question linked to above by Qiaochu. In any case, even over $\mathbb Q$ there is no simple proof of Mordell's conjecture.

Here is one geometric way to think about the conjecture, which may not be emphasized in the linked MO answers:

A projective curve over $\mathbb Z$ admits a projective model over $\mathbb Z$, and by the valuative criterion, a $\mathbb Q$-valued point of this model is the same thing as a $\mathbb Z$-valued point. This leads to the more general problem of thinking about $\mathbb Z$-valued points of curves over $\mathbb Z$.

The reason I write "more general" is because this problem now includes affine curves with coefficients in $\mathbb Z$ (in which case $\mathbb Z$-valued points are not the same as $\mathbb Q$-valued points; it is much more restrictive for a point in affine space to have integral coordinates than rational coordinates).
In this case one has Siegel's theorem mentioned above by Arturo. This says that on an affine curve over $\mathbb Z$ of genus $> 0$ (and by genus here I mean genus of the projectivization) there are only finitely many rational points. Of course, if $g > 1$ this is weaker than Faltings's theorem (which says the same thing even when we pass to the projectivization), but Siegel's theorem was earlier, and includes the case of an elliptic curve with the point at infinity removed. Note that one can also replace $\mathbb Q$ by any number field $K$ and $\mathbb Z$ by its ring of integers, or in fact by any localization of the ring of integers obtained by inverting a finite number of primes, i.e. at any ring of $S$-integers --- see e.g. this formulation of Siegel's theorem.

This result in turn is related to the $S$-unit theorem, which says that if you consider the affine curve $\mathbb A^1 \setminus \{0,1\}$, there are only finitely many $\mathcal O$-valued points whenever $\mathcal O$ is the ring of $S$-integers in any number field. (To understand this, note that the description of $\mathbb A^1\setminus \{0,1\}$ as an affine curves is the set of $(x,y,z)$ such that $xy = (x-1)z = 1$. Thus giving an $S$-integral point of $\mathbb A^1\setminus \{0,1\}$ is the same as giving an $S$-unit $x$ so that $x-1$, or equivalently $1-x$, is also an $S$-unit, which in turn is the same as giving a pair of $S$-units satisfying $u+v =1$; this is how the $S$-unit equation is usually phrased.)

What do (affine or projective) curves of genus $g > 1$, affine curves of genus $g = 1$, and the curve $\mathbb A^1\setminus \{0,1\}$ (or indeed $\mathbb P^1$ minus any collection of at least three points) have in common? They are all hyperbolic curves.

So the basic geometric fact underlying Mordell's conjecture, Seigel's theorem, and the $S$-unit theorem, is that there are only finitely many integral (or even $S$-integral, for a given ring of $S$-integers) points on a hyperbolic curve.

There is an analogy (developed in the work of Vojta mentioned by Arturo above) between this situation in arithmetic and a certain phenomenon in complex function theory:

Recall Picard's theorem, which states that a holomorphic function on $\mathbb C$ missing more than one value is constant. Another way to think of this is that it says that a holomorphic function from $\mathbb C$ into $\mathbb A^1\setminus \{0,1\}$ is constant. More generally, the same is true for holomorphic maps of $\mathbb C$ into any hyperbolic Riemann surface. (The usual proof is that one lifts to a map on the universal cover, to get a holomorphic map from $\mathbb C$ to the unit disk, which is constant by Liouville.)

To be more precise, then, there is an analogy between integral points of hyperbolic curves and holomorphic maps from $\mathbb C$ to hyperbolic curves. If you want to see this analogy spelled out in more detail, you can look up some of the references on Vojta's work.

The idea that hyperbolic varieties should not contain many integral points was greatly extended by Lang, in analogy with corresponding results in complex function theory.

  • $\begingroup$ Thank you very much for this wonderful and insightful answer. I'm having some trouble though understanding the description of $\mathbb{A}^1 \setminus \{ 0, 1 \}$ given in the fifth paragraph, as the set of points $(x, y, z)$ such that $xy = (x-1)z = 1$, but maybe that's a question that really does not belong in this one. In any case, thanks a lot. $\endgroup$ Jun 21, 2011 at 5:00
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    $\begingroup$ @Adrian: Dear Adrian, writing $x \neq 0$ (or $x \neq 1$) is not an equation. To get an equation, one instead introduces variables $y$ and $z$, which are metaphors for $x^{-1}$ (which exists if and only if $x \neq 0$) and $(x-1)^{-1}$ (which simiarly exists if and only if $x \neq 1$) and then imposes the equations $xy = 1$ and $(x-1)z = 1$. (This is a special case of the so-called Rabinowitz trick, which shows that Spec $A \setminus V(f)$ is affine for any element $f$ of any ring $A$, by identifying it with Spec $A_f$, where $A_f = A[X]/(1-fX)$.) Regards, $\endgroup$
    – Matt E
    Jun 21, 2011 at 5:19
  • $\begingroup$ Dear Matt, thank you for the kind explanation. I see the point now. $\endgroup$ Jun 21, 2011 at 5:30

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