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1

If $x^3+ax+b = 0$, then the point $(x,0)$ lies on each of $E$ and $E^d$, for a total of $2$ solutions.


1

I think I solved it now. For future reference, I'll post it here. Let $E$ be an elliptic curve with basepoint $O$, defined over a field $k$. Let $x, y \in k(E)$ be Weierstrass coordinates, i.e. rational functions such that $$\phi = [x, y, 1] : E \rightarrow C$$ is an isomorphism with a curve $C$ given by a Weierstrass equation and $\phi(O) = (0 : 1 : ...


2

Note that $x^3+x=x(x^2+1)$, so the 2-torsion of this elliptic curve will be defined over any field with a $\sqrt{-1}$. $\mathbb{F}_p$ for $p \equiv 1\ (\text{mod}\ 4)$ is such a field, and as the 2-torsion is a subgroup of order 4 for $E(\mathbb{F}_p)$, the claim follows.


4

In the case $p = 4 n + 1$ the number $-1$ is a square $\mod p$. So we can partition the points of $E(\mathbb{F}_p)$ as follows: One point $\infty$ at infinity. One point with $x = 0, y= 0$. Two points $(\epsilon_1, 0)$, $(\epsilon_2, 0)$ with an $x$ that maps to $0$ under $x^3 + x$. (It is $\epsilon_i^2 + 1 = 0$). There remain $4 n - 2$ different $x$ ...


3

The following are some bits and pieces from Section 2.6 of my book. Here I will demonstrate what the OP is trying to do, but with another elliptic curve. Hopefully the OP can replicate this for the elliptic curve above. Let $E/\mathbb{Q}$ be the elliptic curve $y^2=x^3+3$. Its minimal discriminant is $\Delta_E=-3888=-2^4\cdot 3^5$. Thus, the only primes of ...


1

Searching for curves of the form $y^2=x^3+x+k$ gave me the following hit. With $k=6$, the value of $f(x)=x^3+x+6$ is zero for $x=6$ (1 point), and $f(x)$ is a non-zero square of $\Bbb{F}_{19}$ for eight choices $x=0,2,3,4,10,12,14,18$ (16 points). Including the point at infinity gives us a total of 18 points. The Mathematica snippet counter = 1; For[y = ...


1

Let $E: y^2=f(x)$ and $\psi_3(x) = 2f(x)f''(x)-(f'(x))^2$, where $f(x)$ is a monic cubic polynomial with three distinct roots (because $E$ is non-singular!), as above. This can be shown using the following hint: A quartic polynomial $p(x)=a_4x^4+\cdots+a_0$, with $a_4>0$, has exactly two real roots if $p(x)$ takes negative values at all the zeros of ...


0

I think I got a solution so I will share it so you can give me your opinion: We know that this quartic polynomial will have at least two real roots , suppose we have 4 roots $\{ \alpha_1,\beta, \gamma,\alpha_2 \}$ conveniently orderded $\alpha_1< \beta <\gamma< \alpha_2$. Note that $f''(x)=6x+2a$ and $\psi_3'=12f(x)$ Since $\psi_3(\alpha_1)=0 ...


1

Let $L$ be a lattice. Then the explicit equation of the associated elliptic curve is given by $$ E_L\colon y^2=4x^3-g_2(L)x-g_3(L). $$ Here the Eisenstein series are given by $$ g_2(L)=60\sum_{L^{\times}}\frac{1}{\omega^4},\quad g_3(L)=140\sum_{L^{\times}}\frac{1}{\omega^6}. $$ If we can compute these series explicitly, then we have an explicit equation of ...


2

I think that, in general, this is a tough question. If $\Lambda$ is a lattice in $\mathbb C$, then the elliptic curve $\mathbb C/\Lambda$ has an equation of the form $y^2=x^3+g_2(\Lambda)x+g_3(\Lambda)$ where $$g_2(\Lambda)=60\sum_{0\neq \omega\in \Lambda}\frac{1}{\omega^4}$$ and $$g_3(\Lambda)=140\sum_{0\neq \omega\in \Lambda}\frac{1}{\omega^6}$$ If ...


1

If instead of looking at $x=\frac{a^2}{b^2}$ you look at $x=-\frac{a^2}{b^2}$ you end up with $25b^4-a^4=n^2$ which has as an easy solution $a=2,b=1$. In fact the point $P=(-4,6)$ lies on the curve, and you can easily check that $2P$ has non-integral coordinates.


0

Most of this post is a quilt of examples and comments others have offered me when asked about formal schemes, which I've found very helpful. Hopefully, they'll help you. This is just following up on Sebastian's answer to explain why your intuition is right: the formal completion of E along the "0"-section is like the Taylor series expansion of E about the ...


-2

Rewrite as $56u^2+12u+(1-w^3)=0$ This has roots $u=\frac {-12 \pm \sqrt{144-224(1-w^3)}} {112}$ $u=\frac {-12 \pm \sqrt{144-224+224w^3}} {112}$ $u=\frac {-12 \pm \sqrt{224w^3-80}} {112}$ $u=\frac {-12 \pm 4\sqrt{14w^3-5}} {112}$ $u=\frac {-3 \pm \sqrt{14w^3-5}} {28}$


4

Multiplying by $14^3$, we obtain $$14^4\cdot 2^2 u^2 + 14^3 \cdot 12u + 14^3 = (14w)^3 \implies (14(28u+3))^2+980 = (14w)^3$$ This is a Mordell equation of the form $Y^2 = X^3-980$, which has $5$ solutions given by $(14,42)$, $(21,91)$, $(29,153)$, $(126,1414)$ and $(326,5886)$. Of this only $(14,42)$ has $Y$ of the form $14(28u+3)$. Hence, the only ...


2

modulo my unreliable arithmetic, the semi-group operation is: $$ \theta_a \circ \theta_b = (\theta_a+\theta_b)\frac{\left(1+\sqrt{1+\frac{4\theta_a\theta_b}{(\theta_a+\theta_b)^2}} \right)}2 $$ which does not look too promising on the associativity front


2

Unless I have misunderstood OP, this seems not to be the case. I have drawn the following in GeoGebra, on which it is clearly false: For clarity I didn't draw A+(B+C) nor (A+B)+C. These points would be reflections of A(BC) and (AB)C, respectively, in $y$ axis. From the picture it's clear that these will be distinct points.


0

If I hoped for a group law, I would begin by looking for the neutral element. In the case of elliptic curves, one works in projective coordinates and the neutral element is the point at infinity. In your case, the hyperbola is $xy=1$ so when you projectivize it you get $xy-z^2=0$. The trouble is that $z$ may now introduce a $-$ sign, which completely your ...


1

No, you don't just "add $z$'s" until the equation is homogeneous, and no, you don't substitute $z=1$, as the comments above may suggest. This works, but that's not the reason, and not what is formally happening. Let's work backwards. Let's start with a curve given in projective coordinates, let's say $$C: f(X,Y,Z)=ZY^2 - X^3 - XZ^2-Z^3=0,$$ i.e., ...


2

For what it's worth, I took a generator $P$ of the Mordell-Weil group of $y^2=x^3-20$ and computed the preimage of $k\cdot P$ for $|k| \leq 300$ back to the original curve, and the only instance where the point was integral was for $k=0$.


2

$$ \gcd(y, 7y^2 + 6 y + 2) = 1,2 $$ The first case is odd $y,$ so that $7y^2 + 6y+2$ is odd and $\gcd(y, 7y^2 + 6 y + 2) = 1.$ Both $y$ and $7y^2 + 6 y + 2$ must be cubes. Take $y = n^3.$ We want $7n^6 + 6 n^3 + 2$ to be a cube. Cubes are $1,0,-1 \pmod 9.$ If $n \equiv 0 \pmod 3,$ then $7n^6 + 6 n^3 + 2 \equiv 2 \pmod 9$ and is not a cube. If $n^3 \equiv 1 ...


0

The "defined over $\mathbb{Q}$" part is nicely explained in Section 1.2, page 44, of Chapter III (by David Rohrlich) in the compilation of articles "Modular Forms and Fermat's Last Theorem" (Editors: Cornell, Silverman, and Stevens).


0

For $P \in E(k)/pE(k)$, consider the set $\{Q \in E(\overline{k}) :pQ = P\}$. This is principal homogeneous under $E[p](\overline{k})$, so it is a good candidate for the $\overline{k}$-points of the torsor we are looking for. In fact, getting the actual torsor is just a matter of reformulating the above. Identify the point $P$ with $\mathrm{Spec}(k)$. ...


2

Have a look at Silverman and Tate's "Rational Points on Elliptic Curves". There, in page 22, they tell you how to transform any non-singular cubic into a Weierstrass form. The reason why you don't see much work on curves of the form $y^3=x^3+\cdots$ is that we first bring it to a Weierstrass form and then work there.


0

This has been noted before. See "Why is this a characterization of isogenies of elliptic curves? (From Silverman)". The statement $\phi(x_1,y_1)\in E_2(K(x_1,y_1))$ indeed just says that $\phi$ is a morphism landing on $E_2$ (which is true since $\phi$ is an isogeny by assumption).


1

Such equations are called Cubic plane curves; references are given here. The projective version is given by $F(x,y,z)=0$ where $F$ is a non-zero linear combination of the third-degree monomials $$ x^3, y^3, z^3, x^2y, x^2z, y^2x, y^2z, z^2x, z^2y, xyz. $$ For $z=1$ we obtain the affine version. Any non-singular cubic curve can be transformed into the ...


1

This can be done in $4$ steps: Prove it directly for $p=2,3$, i.e., find a non-trivial solution modulo $2$ and one modulo $3$. Prove that $C: 3x^3+4y^3+5z^3=0$ is a non-singular curve over $\mathbb{F}_p$, where $p>3$ is a prime. Show the following: if $C:F(x,y,z)=0$ is a non-singular curve given by a homogeneous polynomial of degree $d\geq 1$, then ...


3

There are some significant problems with the content of that page. First of all, the $L$-series is usually defined as an Euler product, but the product appearing at the bottom of that page is not correct (when $p$ is good, the Euler factor should be $(1-a(p)p^{-s}+p^{1-2s})^{-1}$. With this definition, then $$L(E,s) = \sum_{n\geq 1} \frac{a(n)}{n^{s}}$$ ...


5

Using the Weierstrass Normal form, let's try to understand the elliptic integral $$ \int \frac{dy}{\sqrt{4y^3 - g_2 y - g_3}} $$ In order to reduce the number of parameters, Fricke and Klein do a change of variables $y = \frac{g_2}{g_3}z$. The elliptic curve has two basic periods $$ \Omega = \int \frac{dz}{\sqrt{4z^3 - g(z+1)}} \text{ and } -H = \int ...


2

$\newcommand{\Cpx}{\mathbf{C}}$Here's a fairly detailed sketch of the underlying framework, together with hints for applying the machinery to your situation. Generalities: Let $X$ be a connected holomorphic manifold, $\phi:X \to X$ a biholomorphism, and $A$ the cyclic group generated by $\phi$. To get a manifold quotient, we'll assume $A$ acts properly ...


3

This is shown in Theorem 2.3.1.(b), Chapter V, of Silverman's "The Arithmetic of Elliptic Curves".


3

Your are making some confusion between morphisms of groups and morphisms of curves. A morphism of groups $A,B$ is a map $A\to B$ which respects the operations, a morphism of (algebraic) curves $X,Y$ is a map $X\to Y$ whose components can be written as polynomials. An isogeny of elliptic curves $E,E'$ over a field $k$ is a morphism of algebraic curves $E\to ...


0

A shorter solution depending on a bit more machinery is the following. As others have said, an abelian group of order $9$ is either cyclic or isomorphic to $C_3\times C_3$. It is known (see e.g. Silverman) that the $3$-torsion of this curve (over an algebraic closure of $\Bbb{F}_{11}$ is isomorphic to $C_3\times C_3$. Furthermore, the Weil pairing on that ...


2

You are almost there. Show that $f((1-\omega)/3) = f(\omega(1-\omega)/3)$, because the difference $(1-\omega)/3 - \omega(1-\omega)/3\in \Lambda$.


4

I believe you are looking for the Shioda-Tate formula. See Corollary VII.2.4 in page 70 of Miranda's "The basic theory of elliptic surfaces", or here. Note: let $k$ be a field and let $C/k$ be a smooth projective curve defined over the field $k$ and has genus $g$. The function field of $C/k$ will be denoted by $K=k(C)$. An elliptic surface $\mathcal{E}$ ...


2

Answer to Question 1. Denote $s=x+y$. Then consider equation $$ (s-y)^3+s^3+(s+y)^3=z^3,\tag{1} $$ for positive $s,y,z$. First, consider any such integer solutions: for $s>y$ and for $s<y$. Eq. $(1)$ is equivalent to $$ 3s^3+6sy^2=z^3.\tag{2} $$ Denote $z=3c$, then $(2)$ is equivalent to $$ s^3+2sy^2=9c^3.\tag{3} $$ To check all up-to-6-digital ...


2

That depends on how you visualize $\mathbb{P}_k^2$. One way to think of it is that it is an affine plane, i.e. $k^2$, together with an extra (projective) "line at infinity". To make this precise, one identifies (or maps) a point $(x, y) \in k^2$ with a point $[x, y, 1] \in \mathbb{P}_k^2$. Then the rest of the points of the projective plane are of the form ...


0

So we have to prove that $j:\mathbb{H} \longrightarrow \mathbb{C}$ is surjective. By the given q series of $j(\tau) $it is clear that $j(\tau)$ $$ j(\tau ) = \frac{1}{q} +744+ 196884q+ 21493760q^2.+ c_2q^3......\qquad where \quad q = e^{2\pi i \tau}$$only goes to infinty when q=0 or the imaginary part of $\tau$ goes to $\infty$. and the pole is a order of ...


2

You can check that $E: f(x,y)=0$ is singular in characteristic 2, using the definition of singular point: A point $P=(x_0,y_0)$ on $E$ is singular if $\partial f/\partial x = \partial f/\partial y = 0$ at $P$. Now take $f(x,y) = y^2 - (x^3+ax+b)$, and suppose for simplicity $K=\mathbb{F}_2$. Then, $P=(a,b)$ is a point on the curve, and you can check ...


1

Depending on how cryptography heavy/math heavy the resulting paper is supposed to be you might want to include theory relating to attacks on curves. This would be weil-tate pairing and possibly the background needed for the weil-descent attack. Who is supposed to be the target audience of the paper? Is this a math paper for academic cryptographers? Or more a ...


2

The correct transformations are the following (assuming the characteristic of the field of definition is not $2$ or $3$). First change $y\longrightarrow y-(a_1x+a_3)/2$, so the new equation has the form $$y^2=x^3+Ax^2+Bx+C.$$ And now change $x\longrightarrow x-A/3$, so that the new equation has the form $$y^2=x^3+ax+b.$$ Clearly, both changes of variables ...


0

You can go from the "long" form to the short form by the transformation $$z=y+\frac12 a_1 x$$ Since you will be interested in rational solutions, and since $a_1$ is rational, studying the short form tells you what you need to know about the long form as well.


1

The slope of the tangent line to $E:y^2=4x^3-g$ at $P=(x_1,y_1)$ is $$m = \frac{6x_1^2}{y_1}$$ as long as $y_1\neq 0$ (i.e., as long as $P$ is not a point of order $2$). Now, let $P=(0,\sqrt{-g})$. Then, the tangent line $L$ to $P$ is a line of slope $0$ passing through $P$, i.e., $y=\sqrt{-g}$. Thus, $L$ intersects $E$ at three points, which can be found ...


1

In general, if $F$ is a field, and $V$ is a variety defined by polynomials defined over $F$, it is very common notation to write $V/F$ to indicate that "$V$ is defined over $F$", and simply read $V/F$ as "$V$ over $F$" or "$V$ defined over $F$" (so you interpret / as "over", and not as any type of quotient space).



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