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## Hot answers tagged elliptic-curves

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 ...

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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 ...

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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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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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