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## Hot answers tagged elliptic-curves

6

Let $\infty_+$ and $\infty_-$ be the (real) points at infinity of the curve (corresponding to $y/x^2 = +1$ and $y/x^2 = -1$) In general (unless $p^2=1$), a parabola $y = px^2+qx+r$ intersects the elliptic curve at $4$ points. More precisely, $Div(y-px^2-qx-r) = [P_1]+[P_2]+[P_3]+[P_4]-2[\infty_+]-2[\infty_-]$. From a point $P$, you find the parabola that ...

3

That is false in general. For example if $E\colon y^2=x^3+x+1$, then $E(\mathbb F_5)=9$. It is true that if $p>2$ is a prime of good supersingular reduction, then the trace of the Frobenius at $p$ is $0$, so that $E(\mathbb F_p)=p+1$.

3

It seems to be possible to use the fact that $x^3+2y^3+6xyz = 3z^3$ is an elliptic curve with Weierstrass form $y^2 + 6 x y + 16 y = x^{3} - 105 x^{2}$ and that it has rank $1$ to produce integer solutions. I used the following Sage script to do so experimentally. R.<x, y, z> = QQ[] eq = x^3+2*y^3+6*x*y*z-3*z^3 P = [1,1,-1] E = ...

3

Setting $x=X-\frac{A}{3}$ should transform the Montgomery form to short Weierstrass form. Then you can use the known Weierstrass formulae. However, to save work, helpful people have prepared an explicit formulas database which might interest you, particularly this entry. It should give you $$y_3 = ... 2 The only relevant aspect is that \left(\frac{B}{p}\right)=+1. Then you can find a scale factor s\in\mathbb{F}_p^\times for the substitution y=sY that makes the new B' = Bs^2\equiv1\pmod{p}. 2 Consider the multiplication by a map on E. You have, by Theorem III.6.2,$$ [a] = [1] - \left( ([1] - \phi) \circ ([1] - \hat{\phi}) \right ) + \phi \circ \hat{\phi}. $$Using that an isogeny is a homomorphism, you can expand it out to obtain$$ [a] = [1] - [1] + \phi + \hat{\phi} - \phi \circ \hat{\phi} + \phi \circ \hat{\phi}, $$which simplifies to ... 2 If a formal group is defined over a p-adic ring \mathscr O, with residue field k of characteristic p, then there certainly is a pretty standard definition of the height of the formal group. It’s just the height of the k-series. All of this is in the very old text of Fröhlich, I think. As you expect, the height of the formal group of an elliptic ... 2 Answering the question as clarified in a comment. Remember the process for passing between homogeneous and affine coordinates: a pair (x, y) corresponds to [x: y: 1]. Remember also the process for reducing homogeneous coordinates: we scale by a power of the uniformizer (which I'll call \pi, which is mildly in conflict with your notation) until all of ... 2 The points [x:y:z] \in \ker(\pi) are those mapped in [0:1:0] so they must satisfy v(x),v(z)>0 and v(y)=0. 2 Let P=(x,y) be a point on the elliptic curve which is not the identity. We want to be determine whether nP=O, so for this we need to go through the arduous process of explicitly calculating what nP is. This gets very messy considering you need to determine when you are adding a point to itself. Fortunately, there are these things called division ... 2 Applying Hasse's theorem, we find that h\in[h_{\text{min}},h_{\text{max}}] with$$\begin{align} h_{\text{min}} &= \frac{p+1 - 2\sqrt{p}}{n} & h_{\text{max}} &= \frac{p+1 + 2\sqrt{p}}{n} & \therefore\quad h_{\text{max}}-h_{\text{min}} &= \frac{4\sqrt{p}}{n} \end{align}$$Consequently, if n>4\sqrt{p} (it usually is), the uncertainty ... 2 We’re looking in the local, complete situation above \ell at the p^m-torsion points of E for all m. What does it mean to say that T_p(E)^{G_v}\ne0, where G_v=G_{K_v}, the Galois group of an algebraic closure of K_v over K_v? It would mean that there was a consistent sequence of p^m-torsion points of E, in particularly infinitely many of ... 2 The problem here is that all of the operations you are doing are occurring modulo 65537, so we have to do them step-by-step. First of all, when working modulo 65537, "division by 3" really means "multiplication by the inverse of 3", which is 21846 because 3 \cdot 21846 \equiv 1 \bmod 65537. Therefore,$$ \frac{(6117)^3 + 5(6117)^2 + (6117)}{3} "=" ...

2

Both signs of $s$ are equally usable, but one has to make the choice first and then stick with that for consistency. Doing Montgomery elliptic curve arithmetic with the signs of all incoming $Y$ coordinates flipped produces correspondingly flipped outputs, so that is just an automorphism. Accordingly, both candidates for $G$ that you have considered are ...

2

Expanding on Ferra's answer. This is a true statement for $p > 3$ a prime of supersingular reduction, i.e. one for which the endomorphism ring is an order in a quaternion algebra. This is because we have that if $\phi_p$ is frobenius then we know that: $[|{E(\mathbb{F}_p)}|] = [\text{deg}(1 - \phi_p)] = (1 - \phi_p)(1 - \widehat{\phi}_p) = [1] + ... 2 You don't need the Lutz-Nagell Theorem to see this. Let$P=(x,y)$have order 2. Then$P=-P$but we know$-P=(x,-y)$and therefore we have$(x,y)=(x,-y)$hence$y=0$. Conversely, if$P$is a point such that its$y$-coordinate is zero, then we again see that$P=-P$so$P$has order 2. 1 The$x$coordinate is a branched map$E \to P^1$and makes$P^1$into the quotient of$E$by its automorphism$(P \mapsto -P)$. If you know the$x$coordinate of$P$, this is the same thing as knowing$\{P, -P\}$. Let$Q = (\alpha,0)$. Since$Q= -Q$,$\{P+Q,-P+Q\} = \{P+Q,-(P+Q)\}$so the addition by$Q$induces a map$f : P^1 \to P^1$taking the$x$... 1 It's a homogeneous equation, so if$(x,y,z)$is a solution then$(kx,ky,kz)$is also a solution: \begin{eqnarray*} (kx)^3+2(ky)^3-3(kz)^3 &=& 0 \\ \\ k^3x^3 + 2k^3y^3 - 3k^3z^3 &=& 0 \\ \\ k^3(x^3 + 2y^3-3y^3) &=& 0 \end{eqnarray*} If it has one, non-zero solution then it will have infinitely many solutions. Clearly ... 1 If we start with $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = n$$ and set $$x = -X^2, y = 2(nX+4-Y), z = 4X$$ then we get $$-\frac{X^{2} n^{2} + X^{3} - Y^{2} + 8 \, X n + 16}{2 \, {\left(X n - Y + 4\right)} X} = 0$$ So we want the numerator of this expression to equal$0$which we can rewrite as $$X^{3} + X^{2} n^{2} + 8 \, X n + 16 = Y^2$$ If ... 1 Take$E: y^2+y=x^3-x^2$, aka 11a3 in Cremona's tables. Then, the image of the representation mod$5$is precisely given by all the matrices of the form $$\left(\begin{array}{cc} 1 & b\\ 0 & c \end{array}\right)$$ where$b\in \mathbb{Z}/5\mathbb{Z}$and$c\in (\mathbb{Z}/5\mathbb{Z})^\times$. I know this, because the type of image is listed here. ... 1 The two fields you are considering are the same, because of Krasner's lemma, which has as a corollary that separable closure and$p\$-adic completion commute for global fields. See for example prop. 8.1.5 of Neukirch's "Cohomology of number fields".

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