20

This isn't really a full answer, but it's too long for a comment, and perhaps it's informative all the same.
Your sum $S_k[\mathcal{O}]$ can be written as the value at $s = k$ of the sum
$$\sum_{0 \ne \lambda \in \mathcal{O}} \frac{\lambda^k}{Nm(\lambda)^s} = \sum_{n \ge 1} a^{(k)}_n n^{-s},$$
where $a^{(k)}_n := \sum_{N(\lambda) = n} \lambda^k$.
Now, I ...

answered Sep 27 '20 at 13:26

David Loeffler

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15

(Collecting comments into a community wiki answer.)
There is a standard method for transforming a smooth cubic into Weierstrass form. See for example Section 1.3 or Appendix B of Silverman and Tate's book Rational Points on Elliptic Curves. It is also implemented in Sage as $\mathtt{EllipticCurve\_from\_cubic}$.
For $n=6$ your curve is isomorphic to $Y^2 + ...

15

To get a model with good reduction at $2$, take $y = 2Y + x^3 + x^2 + 1$,
subtract $(x^3+x^2+1)^2$ from both sides, and divide by $4$ to get
$$ Y^2 + (x^3+x^2+1) \, Y = -x^5-x^3+x^2-x. $$
(A similar tactic of un-completing the square
is well-known for elliptic curves.)

answered Feb 26 at 22:51

Noam D. Elkies

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14

### How to prove there are exactly $8$ integer points on the elliptic curve $y^2 = x^3 + 17$ [duplicate]

Using the mathematical programming language Sage, we can run the E.integral_points() command to get a (proof verified) confirmation that there are only $8$ integral points.
The points are $(-2,3)$, $(-1,4)$, $(2,5)$, $(4,9)$, $(8,23)$, $(43, 282)$, $(52, 375)$, and $(5234,378661)$. Note that some would call this $16$ points, since all of these points remain ...

13

The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian
of $X_0(35)$) has rank 0 (as shown for example by a 2-descent
computation in Magma); it is isomorphic to
${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$,
with generators the difference of the two points at infinity
on $X_0(35)$ and the 2-torsion point corresponding to the
...

13

There is a subtle problem with this idea, that causes serious problems. You observed that $\Lambda_\ell \otimes \mathbb Z_\ell = T_\ell$ but didn't find any other information for it. There is a reason for that.
Let $K$ be the field generated by the coordinates of the $\ell$-power torsion points of $E$. Given an $\ell$-power torsion point defined over $F$, to ...

12

Technically speaking, an elliptic curve is a genus 1 curve with a choice of rational point. The automorphism group of an elliptic curve is the subgroup of automorphisms of the genus 1 curve that fix that rational point. So there is no contradiction with the facts you looked up, it just means that no point is fixed by all of $S_3$, and indeed (1:1:1) is not ...

11

Good question! I wished I understood what this conjecture really means, concretely. First, I should say that in my paper with Bhatt on prismatic cohomology, much of the content of these conjectures has been proved, although the connection is not made very clear, and we only work after $p$-adic completion for some prime $p$; it would be possible to go back ...

11

(The statements you quote are only true if you are working over a field of characteristic zero or $p > d$. I will continue to make that assumption)
The formula is not $I(3)=3d(d-2)$ but rather $\sum_{s>2} (s-2)I(s) = 3d(d-2)$ (changing notation slightly so $I(s)$ counts the points with contact exactly $s$ instead of at least $s$).
The typical curve (...

10

Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $E = \{ y^2 = x^3+x \}$ over $\mathbb{Z}[i]$, and let $p$ be a prime which is $3 \bmod 4$. Let $E/p$ be the reduction of $E$ modulo $p$ (which remains prime in $\mathbb{Z}[i]$). Then $E/p$ has the following endomorphisms:
The $p$-power ...

answered Sep 22 '20 at 14:52

David E Speyer

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9

For $k=1$, the surface $y^2=x^3+z^4$ is a rational surface, so it has lots of rational points. The substitution $x=zu$ and $y=z^2v$ leads to $z=u^3/(v^2-1)$, so for almost all $u,v\in\mathbb Q$, the point
$$
\left( \frac{u^4}{v^2-1},\; \frac{u^6v}{(v^2-1)^2},\; \frac{u^3}{v^2-1} \right)
$$
satisfies $y^2=x^3+z^4$. (The same thing should work in general, take ...

answered Aug 26 '20 at 17:59

Joe Silverman

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9

Generalizing Noam's comment and expanding a bit on anon's comment, if you put your curve in Weierstrass form
$$ y^2 = P(x) = x^3+ax^2+bx+c\quad\text{with $a,b,c\in\mathbb Z$,} $$
then the Lutz-Nagell theorem says two things:
Any torsion point $(x,y)\in E(\mathbb Q)$ satisfies $x,y\in\mathbb Z$.
For such a point, either $y=0$, which case the point has order ...

8

Let's discuss the $\mathbb G_m$ question first. For $p$ to split completely in the field generated by the $d$-torsion of $\mathbb G_m$, i.e. the field generated by the $d$th roots of unity, a necessary and sufficient condition is that $\mathbb F_p$ contains the $d$th roots of unity, i.e. $p \equiv 1\mod d$. This requires $p>d$ so I guess the analogous ...

8

Yes, $\omega$ is an invariant differential, also characterized by being nowhere vanishing (including the point at infinity).
At every point, $dx$ and $dy$ together span the cotangent space (which is $1$-dimensional). From the second equation, you see that $dx$ vanishes precisely when $y$ vanishes, and $dy$ vanishes precisely when $P'$ vanishes. This also ...

7

It looks like your notion of strong $\mathbb{Q}$-curve over $K$ is what Peter Bruin and Andrea Ferraguti refer to as a $\mathbb{Q}$-curve being completely defined over $K$. Such curves have $L$-function factoring as a product of $L$-series of newforms for $\Gamma_1(N)$. This then seems to coincide with the definition of strongly modular given by Xevi Guitart ...

answered Jun 30 '20 at 17:31

Barinder Banwait

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7

It doesn't have a more standard name than "$M_{1,n}/S_n$".
Belorousski's PhD thesis contains very explicit rational parametrizations of $M_{1,n}$ for $n<11$. I would expect (but haven't checked) that a modification of his constructions would work to show rationality of $M_{1,n}/S_n$, too.
The irrationality of $M_{1,11}$ can be seen from the ...

7

By pts := PointsQI(fourcovers[1], 3*10^12 : OnlyOne := true);
I found the fifth point with the first coordinate
184125172284095573254772251800166095132866268069994053071269775232967258156536458883390620097481326304558948736/556947891689070160911189283508448977897416986918076204554508778044285190076205830897454330201.
A slightly smaller point has the first ...

6

What you're looking for is a section of the sheaf $\omega = \pi_* \Omega^1_{E/B}$, where $\pi: E \to B$ is the structure map, which is a line bundle on $B$. This line bundle $\omega$ has a canonical extension to the compactification $\bar{B}$, and global sections of this line bundle over $\bar{B}$ are exactly weight 1 modular forms (and more generally $H^0(...

answered Oct 23 '20 at 12:03

David Loeffler

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6

This has been alluded to in one of the comments, but if $E(\mathbb Q)$ has an $\ell$-torsion point, then at every prime $p$ of good reduction we have
$$ p+1-a_p = \#E(\mathbb F_p) \equiv 0 \pmod \ell, $$
so the local factor of the $L$-function at $p$ satisfies
$$ L_p(T) = 1 - a_p T + p T^2 \equiv 1 - (p+1) T + p T^2= (1-T)(1-pT) \pmod\ell. $$
Of course, for $...

6

(1) You don't need the full group law, of course. Start with a rational solution to $x^3+y^3=A_0$, use the tangent line to find a new solution, use the tangent line to that solution to find another solution, etc. (I think this idea may be due to Bachet in the 1600s. In any case, it's simple algebra.) Then clear denominators. But maybe you're asking for an ...

6

The Weil bound $a_n \leq d_2(n)\sqrt{n}$ where $n$ is the divisor function gives $ \sum_{n < x} a_n < \sum_{n_1n_2<x} \sqrt{n_1n_2} = O ( x^{3/2} \log x )$. This is the trivial bound in this setting.
To do better than that, the main approach will be to use the modularity theorem, which recognizes $a_n$ as Fourier coefficients of a modular form of ...

5

The mentioned equation has infnitely many solutions. See the paper by A. Bremner and myself: A. Bremner, M. Ulas, On $x^a\pm y^b \pm z^c \pm w^d = 0, 1/a + 1/b + 1/c + 1/d = 1$, Int. J. Number Theory, 7(8) (2011), 2081-2090.

5

I asked Kevin Buzzard to ask John Coates directly, and it's basically as people have surmised: the moniker is due to the fact the curve appears first in Cremona's book as it has the smallest possible conductor, and it has the smallest coefficients. It is not due to historical priority, as Coates knows of 8th/9th century Arabic manuscripts that discuss $y^2 = ...

5

This is an incomplete answer to the second question (Jackson has indicated the answer to the first question in the comments).
If you ask the same question over a general number field $F$, then you will find counterexamples by considering "false elliptic curves" i.e. principally polarized abelian surfaces with quaternionic multiplication. These are ...

5

The (Selmer) ranks of the isogenous twists $y^2=x^3+D$ and $y^2=x^3-27D$ satisfy various relations that often mean the ranks are equal (or differ by 1). I realize these are not the pairs you're studying, but the theory, and especially the techniques, might well be applicable. (Actually, if you work over $\mathbb Q(\sqrt3)$, then $y^2=x^3-27D$ is isomorphic ...

5

Although Magma's MordellWeilShaInformation can't find the fifth generator for the $\mathbb{Z}/8\mathbb{Z}$ curve, it can find the last generator for an isogenous $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$ curve by searching up to height $10^5$ on the $4$-coverings.
SetClassGroupBounds("GRH");
E24 := EllipticCurve([1, 0, 0, -...

5

Dujella's webpage contains relevant information:
Infinite families of elliptic curves with high rank and prescribed torsion
He also has pages describing rank records for individual curves and for curves defined over quadratic fields.

4

My paper
"Unsolvable cases of $P^3+Q^3+cR^3=d PQR$", Rocky Mountain J. of Math. (28),No 3, 1998
gives 7 (infinite) generic classes of unsolvability of the original equation. However
there are still many (generic) cases to prove!
Solutions have been found for all $n$ in the range $0<n\le1000$.

4

The name refers to the concept of an anharmonic ratio, or cross-ratio. Four points $A,B,C,D$ are called equianharmonic if their cross-ratio is a cube root of 1. In that case the 6 cross-ratios obtained by combining the four points in all possible ways are the same ("equal anharmonic ratio" = "equianharmonic").
The curve $y^2=x^3-1=(x-p_1)(...

answered Feb 27 at 22:07

Carlo Beenakker

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4

It's important to be clear that this map on $H^1_{\mathrm{dR}}$ overlies a highly non-trivial map on the base-ring $R$. You can imagine a case where $R$ is something like $\mathbf{Z}_p\langle X \rangle$ and $\phi$ is $X \mapsto X^p$ (this is a bit of an oversimplification, but it might even be literally true for a few small values of $n$ and $p$). So $\phi$ ...

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