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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 ...


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 ...


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 $\mathcal{E}$ ...


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 $$L(E,s) = \sum_{n\geq 1} \frac{a(n)}{n^{s}}$$ ...


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 ...


3

This is not true. For instance $j(\tau)=j(\tau+1)=j(-1/\tau)$ for every $\tau\in\mathbb{H}$. More generally, $$j(\tau) = j\left(\frac{a\tau+b}{c\tau+d}\right)$$ for any matrix $\left(\begin{array}[cc] aa & b\\ c& d\end{array}\right)$ in $\Gamma(1)=\operatorname{SL}(2,\mathbb{Z})$. It is true, however, that $j:\mathbb{H}/\Gamma(1)\to \mathbb{C}$ is a ...


3

Setting $x+y=3t$ and $xy=s$, we obtain that $$9t^2-s = (t+1)^3 \implies s = -t^3+6t^2-3t-1$$ $x$ and $y$ satisfying the quadratic $a^2 -3ta + s =0$. This means $9t^2-4s = k^2$. Eliminating $s$, we obtain $$4t^3-15t^2+12t+4 = k^2 \implies 64t^3 - 240t^2 + 192t + 64 = (4k)^2$$ $$(4t-5)^3 - 108t + 189 = (8k)^2 \implies (4t-5)^3 - 27(4t-5) + 54 = (4k)^2$$ Hence, ...


3

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


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 ...


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 ...


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 ...


2

Hint $1$ : Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes: $$y^2-3ty+(3t)^2-(t+1)^3=0 $$ a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square. Hint $2$: $\Delta_y=(t-2)^2(4t+1)$ Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$ And it's your turn to do some work! Edit Because my ...


2

Yes, this is possible. In general, your curve $E/\mathbb{C}(t)$ satisfies the Mordell-Weil Theorem (see Chapter III, Theorem 6.1, of "Advanced Topics in the Theory of Elliptic Curves" by Silverman), and the Mordell-Weil group can be trivial, so there would be no points. For instance, in this article, Cox shows that the curve $y^2=4x^3-3x-t$ has trivial ...


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 ...


2

The idea is, $E$ has lots of points in $\overline{\mathbb F_p}$, not all of which are in $\mathbb F_p$. However, if I have an $\overline{\mathbb F_p}$-point, then it is actually an $\mathbb F_p$-point iff it is fixed by the $p^{th}$ power map. This follows because the solutions to $x^p=x$ in $\overline{\mathbb{F_p}}$ (of which there are $p$) are canonically ...


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.


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 ...


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

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 ...


2

Another good reference is Diamond and Shurman's "A First Course in Modular Forms", or William Stein's "Modular Forms".


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.


2

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


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

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 ...


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 ...


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).


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 ...



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