Maarten Derickx
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Registered User
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I'm an Algant PhD student at Leiden, Bordeaux en Milan.
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2d |
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Elliptic curves over QQ with identical 13-isogeny No it really should be X_1(13) x X_1(13) modulo the diagonal in (Z/13Z)^* x (Z/13Z)^* . To see this X_1(13) x X_1(13) classifies pairs (E,E',P,P') where P and P' are points of order 13. Now this data gives in a canonical way an isomorphism phi between the groups E=<P> and E'=<P'> by sending P to P'. To remember the isomorphism but forget the generators P,P' on has to quotient out by the diagonal (Z/13Z)^* x (Z/13Z)^* . Note that (Z/13Z)^* x (Z/13Z)^* doesn't even act on X_0(13) x X_0(13) . |
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May 7 |
awarded | ● Yearling |
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Mar 2 |
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Do isogenies with rational kernels tend to be surjective? Apparently there was a bug in my code that explains why my results where different. |
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Mar 2 |
revised |
Do isogenies with rational kernels tend to be surjective? deleted 16 characters in body |
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Mar 2 |
revised |
Do isogenies with rational kernels tend to be surjective? Edited the results after fixing a bug in the code; added 155 characters in body |
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Feb 28 |
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Do isogenies with rational kernels tend to be surjective? But 2 isogenies will not give new information since the kernel of a 2 isogeny is always rational. Let $phi$ be an isogeny of prime degree between two rank 1 curves. Then either $phi$ or it's dual will be surjective modulo the torsion. So if you look at all isogenies of a fixed prime degree and order them by conductor then the ratio surjective - not surjective will always be 50-50. |
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Feb 28 |
answered | Do isogenies with rational kernels tend to be surjective? |
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Dec 6 |
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Is Fourier analysis a special case of representation theory or an analogue? You should say 'communting invertible matrices' and write $Aut(V)$ instead of $End(V)$. Since multiplication of matrices does not put a group structure on $End(V)$. |
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Dec 3 |
awarded | ● Commentator |
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Dec 3 |
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transform a polynomial into another one upto a constant In the quadratic case a linear transformation is not enough, even over the complex numbers. The reason being that you have only two parameters to chose (a and b), while you have to solve for three equations. To be concrete $p(ax+b)=a_{2} a^{2} x^{2} + (2 a_{2} a b + a_{1} a) x + a_{2} b^{2} + a_{1} b + a_{0}$ so in order for the linear transformation to turn one polynomial into the other you have to have $a_{2} a^{2}=b_2$, $(2 a_{2} a b + a_{1} a)=b_1$ and $ a_{2} b^{2} + a_{1} b + a_{0}=b_0$. These three equations will be rarely solvable at the same time. |
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Nov 28 |
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A question of line bundle for finite etale covering Why did you make this a comment and not an answer? |
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Nov 26 |
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The boundedness of the rank of twists of a fixed curve. Thank you for your reference to the literature with a conjectural answer to my question and for the elaborate discussion of the current status of the conjecture. This was exactly the kind of answer I was hoping for. |
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Nov 26 |
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The boundedness of the rank of twists of a fixed curve. He indeed shows that there are infinitely many by giving a whole family in each case. Is it a coincidence that 2,4 and 6 are exactly the numbers of $\bar k$ automorphisms of any $E$ with those $j$ invariants? |
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Nov 26 |
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The boundedness of the rank of twists of a fixed curve. @Damian: Yes I mean over $\Q$. A positive answer for a single $j$ would indeed prove the conjecture, but it might still be so that there are $j$ for which it is known that my question has negative answer. I also don't understand why I would have to enlarge the field of definition. For example all curves of the form $y^2=x^3+a$ with $a \in \mathbb Q$ have j-invariant zero, and this family of curves contains infinitly many non isomorphic curves. So it makes perfect sense to ask wether this family contains curves of arbitrary high rank. |
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Nov 26 |
revised |
The boundedness of the rank of twists of a fixed curve. added 17 characters in body |
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Nov 26 |
asked | The boundedness of the rank of twists of a fixed curve. |
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Nov 25 |
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Surjectivity of reduction maps of elliptic curves over Q No, in my large search trough the entire Cremona Database for counter exmamples I found counter examples with isogenies of prime degree 2,3,5 and 7. So it seems that there is not much of an obstruction comming from the kind of isogeny. |
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Nov 25 |
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Surjectivity of reduction maps of elliptic curves over Q I'm still continuing my search for more counter examples :). Looking at different sextic twists of y^2=x^3+1 I also found a CM rank 3 counter example to part 1. This counter example is interesting since the Gupta-Murty paper proves that Part 1 holds for CM curves of rank $\geq 6$. So this counter example shows that the Gupta-Murty result at least needs something like rank $\geq 4$ as a condition. In this counter example $E'$ is given by $y^2 =x^3 + 14683622976$ and $\phi$ given by dividing out the group of order $3$ generated by $(0 : 121176 : 1)$. |
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Nov 24 |
revised |
Surjectivity of reduction maps of elliptic curves over Q deleted 1 characters in body; added 6 characters in body |
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Nov 23 |
accepted | Surjectivity of reduction maps of elliptic curves over Q |
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Nov 23 |
revised |
Surjectivity of reduction maps of elliptic curves over Q added 1472 characters in body; deleted 1 characters in body |
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Nov 21 |
revised |
Surjectivity of reduction maps of elliptic curves over Q Updated the answer with new found counter examples. |
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Nov 21 |
awarded | ● Supporter |
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Nov 21 |
revised |
Surjectivity of reduction maps of elliptic curves over Q added 5 characters in body |
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Nov 21 |
awarded | ● Teacher |
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Nov 20 |
answered | Surjectivity of reduction maps of elliptic curves over Q |
