most recent 30 from http://mathoverflow.net 2013-05-20T17:13:08Z http://mathoverflow.net/feeds/user/23501 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123126/do-isogenies-with-rational-kernels-tend-to-be-surjective/123228#123228 Maarten Derickx 2013-02-28T15:44:42Z 2013-03-02T16:38:12Z

<p>I recycled my code from the other thread to test this. There are 559 elliptic curves of conductor &lt; 300000 that have rank 1 and a rational 5 torsion point. Of these 559 curves there are 452 for which the map $\mathbb Z \to \mathbb Z$ induced by $\eta$ is surjective (this is about 81%).</p> <p>I did the same computation for rational 7 torsion point. The problem now becomes that the dataset is very small because there are only 31 elliptic curves of conductor &lt; 300000 of rank 1 with a rational 7 torsion point. But the remarkable thing is that for 30 of these 31 cases the map $\mathbb Z \to \mathbb Z$ is surjective.</p> <p><strong>Edit:</strong> I updated the results after discovering a small bug in my code that caused some curves in the database to be skipped in the test.</p>

http://mathoverflow.net/questions/114504/the-boundedness-of-the-rank-of-twists-of-a-fixed-curve Maarten Derickx 2012-11-26T09:09:05Z 2012-11-26T20:02:39Z

<p>It is conjectured that there are elliptic curves over $\mathbb Q$ of arbitrarily high rank. I was wondering wether someone made a similar conjecture if one restricts to a fixed $j$-invariant. If there are specific values of $j$ known such that the following question has a negative answer would also be nice to know.</p> <p>Let $j \in \mathbb Q$ be fixed. Do there exist elliptic curves $E/\mathbb Q$ with $j(E)=j$ of arbitrary high rank over $\mathbb Q$?</p> <p>I'm specifically interested in the value $j=0$.</p>

http://mathoverflow.net/questions/113968/surjectivity-of-reduction-maps-of-elliptic-curves-over-q/114003#114003 Maarten Derickx 2012-11-20T23:51:47Z 2012-11-24T01:23:25Z

<p><strong>Part 1</strong> is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.</p> <p>Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.</p> <p>A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at <a href="https://sage.mderickx.nl/home/pub/9" rel="nofollow">https://sage.mderickx.nl/home/pub/9</a></p> <p><strong>Update on part 1:</strong></p> <p>I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor &lt; 100000.</p> <p>ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.</p> <p><strong>Part 3:</strong> Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.</p> <p>In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.</p>

http://mathoverflow.net/questions/96158/extending-birational-isomorphisms-between-planar-curves-to-the-p2 Maarten Derickx 2012-05-06T21:27:49Z 2012-06-24T20:31:24Z

<p>Let $k$ be a field and let $C,D$ be two integral curves in $\mathbb{P}^2_k$. Now let $f:C \to D$ be a birational isomorphism. Can $f$ be extended to $\mathbb{P}^2_k$. To be precise, does there exist a birational isomorphism $F:\mathbb{P}^2_k \to \mathbb{P}^2_k$ such that $F$ and $f$ agree on a (non empty) open subset of $C$?</p> <p>I am mainly interested in the case when $k=\mathbb Q$. </p> <p><strong>[Edit]</strong> Since it is apparently easy to read over I will state it here explicitly. The map $F$ is allowed to be a <strong>birational</strong> isomorphism. It is clear that my statement is false when you want $F$ to be an isomorphism since it has to send curves of the same degree to curves of the same degree.</p> <p>If we take for example $C$ to be the curve $x=0$ and $D$ to be the curve $y^2z=x^3+x^2z$ then $C$ and $D$ are birationally equivalent but clearly no automorphism of $\mathbb{P}^2_k$ sends C to D. However it is possible with a birational isomorphism.</p>

http://mathoverflow.net/questions/129818/elliptic-curves-over-qq-with-identical-13-isogeny Maarten Derickx 2013-05-16T11:00:01Z 2013-05-16T11:00:01Z

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=&lt;P&gt; and E'=&lt;P'&gt; 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) .

http://mathoverflow.net/questions/123126/do-isogenies-with-rational-kernels-tend-to-be-surjective/123228#123228 Maarten Derickx 2013-03-02T16:39:08Z 2013-03-02T16:39:08Z

Apparently there was a bug in my code that explains why my results where different.

http://mathoverflow.net/questions/123126/do-isogenies-with-rational-kernels-tend-to-be-surjective Maarten Derickx 2013-02-28T16:58:25Z 2013-02-28T16:58:25Z

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.

http://mathoverflow.net/questions/37021/is-fourier-analysis-a-special-case-of-representation-theory-or-an-analogue/37071#37071 Maarten Derickx 2012-12-06T01:08:13Z 2012-12-06T01:08:13Z

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 $End(V)$.

http://mathoverflow.net/questions/115212/transform-a-polynomial-into-another-one-upto-a-constant Maarten Derickx 2012-12-03T00:50:17Z 2012-12-03T00:50:17Z

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.

http://mathoverflow.net/questions/114518/a-question-of-line-bundle-for-finite-etale-covering Maarten Derickx 2012-11-28T23:43:11Z 2012-11-28T23:43:11Z

Why did you make this a comment and not an answer?

http://mathoverflow.net/questions/114504/the-boundedness-of-the-rank-of-twists-of-a-fixed-curve/114520#114520 Maarten Derickx 2012-11-26T12:57:04Z 2012-11-26T12:57:04Z

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.

http://mathoverflow.net/questions/114504/the-boundedness-of-the-rank-of-twists-of-a-fixed-curve/114519#114519 Maarten Derickx 2012-11-26T12:38:18Z 2012-11-26T12:38:18Z

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?

http://mathoverflow.net/questions/114504/the-boundedness-of-the-rank-of-twists-of-a-fixed-curve Maarten Derickx 2012-11-26T10:54:24Z 2012-11-26T10:54:24Z

@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.

http://mathoverflow.net/questions/113968/surjectivity-of-reduction-maps-of-elliptic-curves-over-q/114003#114003 Maarten Derickx 2012-11-25T21:57:59Z 2012-11-25T21:57:59Z

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.

http://mathoverflow.net/questions/113968/surjectivity-of-reduction-maps-of-elliptic-curves-over-q/114003#114003 Maarten Derickx 2012-11-25T16:22:15Z 2012-11-25T16:22:15Z

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 $(0 : 121176 : 1)$.

http://mathoverflow.net/questions/96158/extending-birational-isomorphisms-between-planar-curves-to-the-p2/96166#96166 Maarten Derickx 2012-05-11T21:15:23Z 2012-05-11T21:15:23Z

Now $E(\Gamma(3))/\mathbb{P}^1$ doesn't have a section of infinite order. But by choosing a suitable line $L \in \P^2$ to parameterize the elliptic elliptic curves in the Hesse pensil by one can find a different elliptic curve $E \to \P^1$ wich has a section of infite order, and this $E$ will also be birational equivalent to $\P^2$. Now the translation by this point of infinite order is an isomorphism from $E$ to itself and hence gives a birational map from $\P^2$ to itself. This birational automorphism will be the translation by a point of infinite order in most of its fibers over $L$.

http://mathoverflow.net/questions/96158/extending-birational-isomorphisms-between-planar-curves-to-the-p2/96166#96166 Maarten Derickx 2012-05-11T21:02:52Z 2012-05-11T21:02:52Z

Yesterday during lunch at Univeriteit Leiden I also heard another argument (by bas Edixhoven) why the &quot;negative result for elliptic curves&quot; argument cannot work in the birational case. The argument is by considering the Hesse pencil $u(x3+y3+z3)+vxyz$ . Now by blowing up in the 9 points where rational map $\mathbb{P}^2 \to \mathbb{P}^1$ given by $(x:y:z)\mapsto(x^3+y^3+z^3:xyz)$ is not defined we get $E(\Gamma(3))\mapsto \mathbb{P}^2$. $E(\Gamma(3))$ together with the map to $\mathbb{P}^1$ (via $\mathbb{P}^2$) will be the universal elliptic curve with full level 3 structure.

http://mathoverflow.net/questions/96158/extending-birational-isomorphisms-between-planar-curves-to-the-p2/96166#96166 Maarten Derickx 2012-05-07T09:38:54Z 2012-05-07T09:38:54Z

Thanks for the answer. Altough it was not the answer to my question it stil was instructive.