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The modular curve $X_0(15)$ has a canonical model over $\mathbf{Q}$, and it has genus $1$. As the cusp $\infty$ is rational, it is an elliptic curve. Roughly, my question is whether we can find all the rational points, and properties of the elliptic curves corresponding to the non-cuspidal ones, after a reasonable computation.

Let me elaborate: To find the rank, one could compute the unique newform of weight $2$ and level $\Gamma_0(15)$ using Sage (or by hand if one is keen) to find $$q - q^2 - q^3 - q^4 + q^5 + q^6 + 3q^8 + q^9 - q^{10} - 4q^{11} + \ldots$$

Now this information (maybe compute some more terms) can be used to approximate the value at $1$ of its $L$-function and observe it is non-zero. A result by Kolyvagin shows the rank is $0$. Moreover, we notice from inspection of the above Fourier coefficients that the order of the torsion group has to be a divisor of $8$. The cusps are $0,\infty, \frac{1}{3}$ and $\frac{1}{5}$, and hence they correspond to rational points on the canonical model as they all have different denominators.

In conclusion: To prove that the rational torsion is of order $8$, it would suffice to find a single rational elliptic curve with a rational subgroup of order $15$. It turns out there are four such curves, an they form an isogeny class of conductor $50$.

1) Could we somehow have expected that there are indeed $4$ non-cuspidal rational points? How could one have guessed that the corresponding curves have conductor $50$?

Note that these points are not Heegner as their associated curves are not CM, so you can't "cheat". Actually, in this particular case I am mainly concerned about the curves being semi-stable, so finding an actual model for them is quite a lot stronger.

2) Is there an argument that shows they are not semi-stable, without computing the Néron model? Could we have predicted some of their properties/invariants?

Remark: Note that I consider the computation of the $q$-expansions of a certain space of newforms to be 'reasonable', but not the computation of the canonical model followed by a $2$-descent (say).

Remark: This question originated as follows. In one of the arguments leading up to a proof of modularity of semi-stable elliptic curves over $\mathbb{Q}$, one argues that the canonical model $X_0(15)_{/\mathbf{Q}}$ has exactly $8$ rational points, and the non-cuspidal points are not semi-stable. This led me to wonder how much of this we could actually deduce without a large scale search, and how much of this could be anticipated without computing anything.

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This computation isn't that hard; it was already known 40+ years ago, since the result is reported in the introductory notes to the "Antwerp Table" (Modular Forms in One Variable = LNM 476). The curve $X_0(15)$ can be computed via $q$-expansions or otherwise, and it happens to have a $2$-torsion point so a $2$-descent is easy and turns out to yield rank $0$. Finding the full torsion group is then routine, and $q$-expansions (or other methods) can be used to find the $j$-invariants etc. of the 15-isogenous curves parametrized by the non-cusp rational points. –  Noam D. Elkies Jul 2 '13 at 16:41
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I must confess I got the information on the conductors being 50 from the Antwerp volume you cite. Apparently I overestimated the 2-descent, and it's very reassuring to know one can always just do that, thanks! I should probably rephrase the bit about "reasonable computations"... That said, I would still be very excited if there were a good answer to either of the questions above. –  Bertie Wooster Jul 2 '13 at 17:36

1 Answer 1

The number of cusps on a modular curve is given here: http://math.stackexchange.com/questions/430912/number-of-cusps-of-an-modular-curve-x-0n

There are ways to guess things about semi-stability wrt torsion structure. For instance $2^8$ can be the 2-local conductor of a curve with a 13-isogeny, but $2^k$ cannot for $k\in\{3,5,7\}$. I thought a reference for this removal of odd valuation at 2 was Brumer and/or Kramer from a paper in the 70s, but I can't seem to find it stated.

Here is a list I found somewhere, no idea about its provenance. Maybe it just comes from parametrising the modular curve (genus 0 case) and staring at $p$-valuation of resulting $(c_4,c_6)$ coefficients. But as above, I think there is some result from the representation side that can be applied.

Ignoring all exotic isogenies, working on minimal quadratic twist
2-isog need f3=0,1,2
3-isogs need f2=0,1,2
6-isogs need f2=0,1,2 f3=0,1,2 vp=0,1 for p>3
4-isog need vp=0,1 for p!=2 and f2=0,1,3,5 [so 12-isog are ss]
8-isog need vp=0,1 for p!=2 and f2=0,1,3
16-isog need vp=0,1 for all p
9-isogs need vp=0,1 for all p!=3 where could have f3=3 [so 18-isog are ss]
5-isogs need f2=0,1,8 and f3=0,1     vp=0,1 for p=3(4) [same for 10,25]
7-isogs need f2=0,1 and f3=0,1,4     vp=0,1 for p=2(3)
13-isogs need f2=0,1,8 and f3=0,1,4  vp=0,1 for p=11(12)

For 5,7,13 isogs:
 The ec minimal twist must also be minimal for modforms with character
 which means that all primes with vp>=2 are "minus" primes,
 that is,        p=1(12), p=5(12) and vp(c4)=1 and vp(c6)>=2
 for p>3:                 p=7(12) and (vp(c6)=1 or vp(c4)>=2)
 except possibly for p when there is a p-isogeny, in which case
 the notion of minimal twist is not isogeny-invariant (this is
 handled for p=2,3 by the finding-method, but not for 5,7).
 For p=2,3 the notion of "minus" is included in the above.
For 10,25 isogs: here 5 cannot be a plus prime
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Do you know what such a representation theoretic argument would look like? Thanks for your list. It still seems a bit of a mystery where conductor 50 comes from, though. Then again, there might not be a nice explanation, in which case this might be a bad question. –  Bertie Wooster Jul 3 '13 at 9:44
    
The list is useless for your case. On the first line, it ignores all exotic isogenies (of which 15 is an example). –  v08ltu Jul 3 '13 at 12:09

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