# Maarten Derickx

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 Name Maarten Derickx Member for 1 year Seen yesterday Website Location Leiden, Netherlands Age 26
I'm an Algant PhD student at Leiden, Bordeaux en Milan.
 2d comment Elliptic curves over QQ with identical 13-isogenyNo 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=

and E'= 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) . May7 awarded ● Yearling Mar2 comment Do isogenies with rational kernels tend to be surjective?Apparently there was a bug in my code that explains why my results where different. Mar2 revised Do isogenies with rational kernels tend to be surjective?deleted 16 characters in body Mar2 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 Feb28 comment 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. Feb28 answered Do isogenies with rational kernels tend to be surjective? Dec6 comment 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 $End(V)$. Dec3 awarded ● Commentator Dec3 comment transform a polynomial into another one upto a constantIn 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. Nov28 comment A question of line bundle for finite etale coveringWhy did you make this a comment and not an answer? Nov26 comment 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. Nov26 comment 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? Nov26 comment 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. Nov26 revised The boundedness of the rank of twists of a fixed curve.added 17 characters in body Nov26 asked The boundedness of the rank of twists of a fixed curve. Nov25 comment Surjectivity of reduction maps of elliptic curves over QNo, 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. Nov25 comment Surjectivity of reduction maps of elliptic curves over QI'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 $(0 : 121176 : 1)$. Nov24 revised Surjectivity of reduction maps of elliptic curves over Qdeleted 1 characters in body; added 6 characters in body Nov23 accepted Surjectivity of reduction maps of elliptic curves over Q Nov23 revised Surjectivity of reduction maps of elliptic curves over Qadded 1472 characters in body; deleted 1 characters in body Nov21 revised Surjectivity of reduction maps of elliptic curves over QUpdated the answer with new found counter examples. Nov21 awarded ● Supporter Nov21 revised Surjectivity of reduction maps of elliptic curves over Qadded 5 characters in body Nov21 awarded ● Teacher Nov20 answered Surjectivity of reduction maps of elliptic curves over Q