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By Mazur's theorems (Reference:- Modular curves and the Eisenstein ideal, 1977), we know that the only rational points of X_0(N) for N any prime > 163 are the two cusps (o) and (oo) (|X_0(N)(Q)| = 2 ).

Is there any bound for |X_0(N)(Q)|, where N is an arbitrary +ve integer?.

More Generally,

Is there a bound on the number of rational points on the modular curve i.e. for |X_0(N)(K)|, where K is some number field. We know that |X_0(N)(K)| is finite.

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Isn't the modular curve rational for some values of $N$? –  Will Sawin Feb 17 '12 at 6:56
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You might wish to look at Merel's paper (Bornes pour la torsion des courbes elliptiques sur les corps de nombres. Invent. Math. 124 (1996), no. 1-3, 437–449) and the corresponding Bourbaki talk by Edixhoven (Rational torsion points on elliptic curves over number fields. Séminaire Bourbaki, 36 (1993-1994), Exposé No. 782, 19 p., numdam.org/numdam-bin/fitem?id=SB_1993-1994__36__209_0). –  Chandan Singh Dalawat Feb 17 '12 at 8:54
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The $N$ is your question is prime. –  Chris Wuthrich Feb 17 '12 at 9:26
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Either $N$ is prime or there are more than two cusps. –  stankewicz Feb 17 '12 at 9:34
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Mazur proves that if $N$ is prime and $>13$, then $X_0(N)(\mathbf{Q})$ is finite (Theorem 6). Right before Theorem 7, he says that one "expects" that if $N>163$ ("all $N$"), then $X_0(N)(\mathbf{Q})=\{0,\infty\}$. The "all $N$" clause seems to indicate that he is including composite $N$, but for non-prime $N$ there are more than two cusps; maybe he just meant that there are only cusps. He refers to a paper of Ogg, where the latter says that the then-known list up to $N=163$ "may very well be a complete list" of non-cusp points. Nowhere, does he prove what you claim in the first sentence. –  Rob Harron Feb 17 '12 at 22:00
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1 Answer

up vote 12 down vote accepted

Let me restrict throughout to prime $N \geq 23$. This ensures that, indeed, $|X_0(N)(K)|$ is finite, for any number field $K$.

If your question is "For given $N$ and $K$, how big is $|X_0(N)(K)|$?", then I don't really know. Certainly you'll have the two cusps, and possibly some "CM points" coming from CM elliptic curves; e.g., if $K = \mathbb{Q}(i)$, and $N$ splits in $K$. But these cusps and CM points are somehow "trivial". Remember Ogg in "Diophantine Equations and modular forms", 1975; "The conclusion toward which we are tending seems to be that modular curves only have rational points for which there is a reason".

If your question is "Given a number field $K$, is there a bound $C(K)$ such that, if $N > C(K)$, then $X_0(N)(K)$ is trivial?", then this question was studied by Momose in 1995 ("Isogenies of Prime Degree over number fields", Compositio Mathematica, 97). He proved (Theorem A in loc.cit.) that there is a bound $C(K)$ such that, if $N > C(K)$, then any noncuspidal point in $X_0(N)(K)$ is one of three kinds (which he calls 'Type 1', 'Type 2', 'Type 3'). He then asks "Under what conditions on $K$ are there only finitely many points of these three types?" For instance, he shows that, for $K$ quadratic and not imaginary quadratic of class number one, then $X_0(N)(K)$ has noncuspidal points for only finitely many $N$.

Making this $C(K)$ effective has been done in a paper of Agnès David ("Caractère d’isogénie et critères d’irréductibilité", Théorème II, available on the arXiV). This paper also explains effectively how Momose's Type 1 and 2 points can only occur for finitely many primes (the type II case might require GRH for a really strong Effective Cheboratev Density Theorem, I'm not sure). Type 3 points don't occur if you assume your $K$ does not contain an imaginary quadratic field and its Hilbert Class Field.

I was curious how large these bounds were in a specific example like $K=\mathbb{Q}(\sqrt{5})$; using the formulae in David's paper, I got that, for $N > 8 \times 10^{119}$, $X_0(N)(K)$ has only the two cusps.

Dmitry Vaintrob's answer to this question may also be useful. I think their Preprint is now on the arXiv.

EDIT. Regarding your new question "Is there any bound for $|X_0(N)(\mathbb{Q})|$, where $N$ is an arbitrary positive integer?", the answer is yes.

As a rough rule, $X_0(N)(\mathbb{Q})$ contains only the cusps. How many cusps are there? Page 107 of Diamond And Shurman's book contains a nice table, which tells you that the number of cusps is $\sum_{d|N}\phi(gcd(d,\frac{N}{d}))$.

If $N \leq 10$, or $N = 12,13,16,18,25$, then the genus of $X_0(N)$ is zero, and hence will have infinitely many rational points. If $N = 11,14,15,17,19,21,27$, then the genus is one; it turns out that these curves have rank zero, and in addition to the cusps, have $3,2,4,2,1,4,1$ more rational points, respectively. If $N = 37,43,67,163$, then, in addition to the cusps, there are $2,1,1,1$ more rational points, respectively. The results in this paragraph were certainly known before Mazur (I'm not quite sure exactly when).

Fact. For any other integer $N$, there are only the cusps.

This fact was proved for prime $N$ by Mazur. In the introduction to his "Rational Isogenies of Prime Degree" paper, he reduced the fact (for all $N$) to dealing with the cases $169,91,65,39,125$. I believe these cases were subsequently dealt with by work of Kenku and Mestre.

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My question is "For given N and K, how big is |X_0(N)(K)|? Thanks for the related literature and for describing your work. –  Srilakshmi Feb 17 '12 at 14:06
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