I am Pierre MATSUMI. Could you please teach me what is effective Mordell.

Assume that f(X,Y) = 0 which defines smooth affine curve with genus > 1, and that there will be the solution X=n/m, Y=n'/m' in rational number Q. Then,

Theorem(?)(Effective Mordell): max(|n|,|m|,|n'|,|m'|) < const_f for some constant const_f with some constant in positive real number.

Is this statement right?

I found some article where, if C denotes the proper smooth curve defined by f(X,Y) after compactification, effective Mordell is equivalent to the fact that there is some non-trivial function F:C ---> P^1 and the height of F(n/m,n'/m') is bounded.

I am NOT sure whether this definition is equivalent to the above Theorem(?).... Please just teach me.

Sincerely yours, Pierre MATSUMI

  • 1
    $\begingroup$ I think effective simply means that we know an algorithm to determine const_f from the coefficients of f. At the moment we don't know such an algorithm, we only know that const_f exists. $\endgroup$
    – GH from MO
    Mar 1, 2013 at 18:10
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    $\begingroup$ So your "Theorem(?)" is a theorem (of Faltings), but it is not the effective version. An effective version is not known at the moment. For certain families of curves there is an effective version, but that is another issue. $\endgroup$
    – GH from MO
    Mar 1, 2013 at 18:15

1 Answer 1


If one had an effective version of the abc conjecture, then Elkies showed how to use it to obtain an effective version of the Mordell conjecture (using Belyi maps).

In your formulation, the theorem would be effective if there was an algorithm to compute the constant const_f for any given f.

For example, assume that $f$ has coefficients in $\mathbb Z$, and let $H(f)$ be the maximum of the absolute values of its coefficients. Then the following would be an effective version of the Mordell conjecture: $$ \max(|n|,|m|,|n'|,|m'|) \le 10^{10^{10^{H(f)+\deg(f)+1000}}}. $$ NOTE: I'm not saying that this statement is known; it's not. (Although I'd be surprised if it isn't true.) But it illustrates what is meant by an "effective bound".

  • $\begingroup$ Thanks a lot! I finally got the meaning of ``effective"!! Pierre $\endgroup$ Mar 2, 2013 at 8:42

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