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I'll try to rephrase to hopefully clarify in some detail how to use the Nagell-Lutz theorem and Mazur's theorem to find the torsion group. So we work with the points $P_i$ that we find from Nagell-Lutz. We add $P_i$ to itself atmost $12$ times, that is for each $P_i$ we calculate the set $\{nP_i \text{ | } 2 \leq n \leq 12\}$. If in this set we find the ...

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The result specifies what are the exact possibilities for the torsion part of the group of rational points on an elliptic curve (over the rationals). By Mordell's theorem you know that $E(\mathbb{Q})_{\text{torsion}}$ is some finite abelian group. Now, you know that it cannot by just whatever finite abelian group but it must be one of a very particular ...

2

I guess your question is about the special case of Mazur's theorem for the curves $y^2=x^3+ax$, as the proof of the full theorem is much involved. Anyway, you could look at Silverman's The Arithmetic of Elliptic Curves, GTM 106, pages 310+311 for an idea how to prove this.

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For this curve the torsion points are $P = (-2, 0)$ of order $2$ and the point at infinity. The reason why your points $P_1, ..., P_4$ are not torsion points are due to the fact that the Nagell-Lutz Theorem only gives you the possible candidates for the torsion points. In other words, Nagell-Lutz is not a if and only if-theorem. When checking the list of ...

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Yes, it is correct that there are no rational points of finite order. And your application of the Nagell-Lutz Theorem seems valid too.

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One has that $O$ is the neutral element with respect to the group law $\oplus$. So $P \oplus Q \oplus O = P \oplus Q$ and $P \oplus Q = O$ is essentially the definition of $Q = - P$ as $-P$ denotes the inverse element of $P$, and $Q$ apparently is the inverse element of $Q$ as $P \oplus Q = 0$. For the second part of the question recall from the ...

0

For any elliptic curve $C$ over any base field there are affine coordinates $(x, y)$ such $C$ can be written in the form $$y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6. \qquad (\ast)$$ If the characteristic of the underlying field if not $2$ or $3$, then we can choose coordinates $(x, y)$ such that the curve is $$y^2 = x^3 + px + q.$$ As an indication of what goes ...

2

Try a change of variables $X = x +\alpha$ and $Y = y + \beta$ for suitable $\alpha$ and $\beta$, to put it in the required form. This will give equations on $\alpha,\beta$, that you will be able to solve because the characteristic of $k$ is not $2$ not $3$.

2

Another approach (see Example $15.5$ at Elliptic Curve Cryptography), is the following. Take $x = 0 \ldots 6$ and for each $x$ solve: $$y^2 \equiv x^3 + 2 \pmod {7}$$ You can set-up two tables (note: each $y$ can produce zero, one or two points): $x = 0 \implies y^2 = 2 \pmod 7$ $x = 1 \implies y^2 = 3 \pmod 7$ $x = 2 \implies y^2 = 3 \pmod 7$ $x = 3 ... 2 Without knowing exactly what properties you would like to test I hope the following will be of some help. The best reference for families of high rank is the tables by Andrej Dujella. The tables are available from his web page here and are very much up-to-date as he is himself a researcher of the subject. If you dig into the very extensive bibliography ... 1 For any group with$9$elements, one can simply pick a nonidentity element$g$and compute the subgroup it generates. Note that there are only two groups of this order$\mathbb{Z}_9$and$\mathbb{Z}_3 \times \mathbb{Z}_3$, and that the former has exactly$2$elements of order$3$. If$g$has order$9$, your group is isomoprhic to$\mathbb{Z}_9$, with ... 3 Well, first you need a Hermitian form$H:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$such that$\Im H(\Lambda\times\Lambda)\subseteq\mathbb{Z}$. Using a little bit of linear algebra, it is easy to see that there is only one$\mbox{mod }\mathbb{Z}$. Explicitly, if$\Lambda=\langle 1,\tau\rangle$for$\tau\in\mathbb{H}$, then every Hermitian form that satisfies ... 1 In a$\textbf{MOV/Frey-Rück}$context; Let$E$be an elliptic curve defined over a finite field$\mathbb{F}_q$, let$n$be a prime dividing$\#E(\mathbb{F}_q)$. The$\textbf{embedding degree}$of$E$with respect to$n$is the smallest integer$k$such that$n$divides$q^k - 1$. Example Let$E: y^2 = x^3 + 1$be an elliptic curve defined over ... 0 Let us write$A*B$for the third point in the intersection of the cubic with the line through$A$and$B$. If we fix a point$O$on the cubic we define$A+B=O*(P*Q)$. Remark For each point$P$of$C$,$-P=(O*O)*P$. Infact$P+X=O$is equivalent to$O*(P*X)=O$, implying that$O*O=P*X$, that gives (applying$*P$to both sides)$(O*O)*P=(P*X)*P$. And the ... 5 This is just a comment on the questions in the updates, but it won't fit nicely in the comment box. About the question in update 1: the answer is that$\mathbf A^1$is the moduli space. The reason the other space isn't is the following: let$C_\lambda$be the curve branched over$\{0,1,\lambda,\infty\}$. Then$C_\lambda \cong C_\mu$iff$\mu$is one of ... 5 a) The moduli space$\mathcal M_g$of smooth projective curves is not isomorphic to$\mathbb A^n_\mathbb C$because its only global holomorphic functions are just the constants:$\mathcal O_\text {hol}(\mathcal M_g)=\mathbb C$. Hence the holomorphic variety underlying$\mathcal M_g$is not even Stein (whereas of course$\mathbb A^n_\mathbb C$is Stein) , a ... 6 Take a look at the book "Moduli of curves" by Harris and Morrison. It has answers to your questions. In particular, for all$g>2$the moduli space contains complete curves, hence cannot be affine. It is a theorem by Diaz that for all$g>22$the moduli space is not even uniratonal. Edit: I misremembered who proved what (should have checked the book ... 2 My answer will probably not be perfect, but I will try to convey my own understanding. When you try to implement the MOV attack (or try to guard from it) what you need to do is transfer the discrete logarithm problem from an EC over$\mathbb{F}_p$to a finite field$\mathbb{F}_{p^k}$. For this it is important to note that you are not actually embedding the ... 2$p$is not unique, but any two possibilities for$p$differ exactly by a 3-torsion point on$E$(so there are exactly 9 possible values for$p$that give the same divisor class). The basics of this are explained in chapter 3 of Silverman's AEC book, for example. The main tool here is Corollary III.3.5 in Silverman's Arithmetic of Elliptic Curves. It says ... 4 Andrew Bremner found the$k$th$m_k$of small height for$k=8,9,10,11$. Given, $$a^4+b^4+c^4 = d^4\tag1$$ $$(p + r)^4 + (p - r)^4 + s^4 = q^4$$ where$R_k = p,q,r,s$, $$R_8 = 6260583580,\; 12558554489,\; -1552770140,\; 11988496761$$ $$R_9 = -1456578618665,\; 2734283895746,\; -639377557145,\; 2452045365504$$ $$R_{10} = -3142543344652846743,\;\, ... 1 The answer is that, "yes", Wiles's argument involves generalized elliptic curves at some points, in so far as (a) it involves arguing on various modular curves X_0(N) and X_1(N), whose cusps have a moduli interpretation in terms of generalized elliptic curves (as noted by [user40276] in comments above), and (b) it involves studying the bad reduction ... 3 The roots of T^2-2T+5 are \rho_{1,2}=1\pm 2{\rm i}, so$$\#E({\bf F}_{p^2})=p^2+1-(\rho_1^2+\rho_2^2)=26-(-6)=32.$$1 Sage gives the following 2P=\left(\frac{13153}{2304} , \frac{1185553}{110592}\right), 6P=\big(\frac{17631797546863867480163645661711294049}{2834578067615933833996300908324147456} ,-\frac{60902529607177336000181399672827762453069546262535228527}{4772353810493036247904139120367622993558177805319376896} \big{)} =(6.220254699738563,-12.761528592718786) 4 Sure there are other ways. For example, 2P=P+P, 3P=2P+P, 6P=3P+3P Also, 2P=P+P, 3P=2P+P, 4P=3P+P, 5P=4P+P, 6P=5P+P and so on. In general, the double and add method you describe is the fastest. It is akin to the square and multiply that you often see in multiplicative groups. 1 Trying to clarify the step that seems to be troubling the OP. We partition the elements x\in\Bbb{F}_p into three subsets A,B and C: Subset A consists of those elements x\in\Bbb{F}_p such that x^3+Ax=0. You should notice that if x\in A, then y=0 is the only solution of y^2=x^3+Ax. Therefore to each x\in A there is a single point (x,0) ... 1 You should send the equation into the congruence modulo 4, for that according to Division Algorithm, (or Complete Residue System modulo 4) we have 4 possibilities for arbitrary number x.$$ \text{If} \ x \equiv 0 \ \text{or} \ 2 \pmod{4} \ \ \text{then} \ \ y^2\equiv 0 - 73 \equiv 3 \equiv -1 \pmod{4}$$So as we know in the congruence modulo$4$, ... 7 Schemes play an enormous role in all the modern theory of elliptic curves, and have done so ever since Mazur and Tate proved their theorem that no elliptic curve over$\mathbb Q$can have a 13-torsion point defined over$\mathbb Q$. For some additional explanation, you could look at this answer. But bear in mind that theorems on the classification of ... 2 I know this question is old (came across it on a google search for something else) but the answer you want is in the (mostly expository) paper: http://arxiv.org/abs/alg-geom/9206008 In short, a genus 2 curve has six branch points for the hyperelliptic map, and a choice of two of them (unordered) is equivalent to a choice of two Weierstrass points, which is ... 2 In fact I wouldn't suggest to compute the degree of that field extension as a way to compute the degree of the isogeny, because it might be difficult. I think it is more reasonable to compute the ramification indexes. So here one would like to compute for example$\sum_{P\in\phi^{-1}(O')} e_{\phi}(P)=\deg\phi$. (Just to be clear, I will call$O$the point at ... 2 In projective coordinates,$[x:y:z]$denotes the same point as$[tx:ty:tz]$(if$t\ne 0$), hence indeed$[0:1:0]$and$[0:2:0]$denote the same point. It is convenient but not mandatry to take the "simplest" representative, which would be$[0:1:0]$. While$[1:1:0]$,$[2:3:0]$,$[1:0:0]$etc. are also infinte points they are different points; they all lie on ... 3 For$x=0$we find the point$(0,0)$. If$x \neq 0$, we have$f(x) = x^3+Ax = -( (-x)^3 + A(-x) ) = -f(-x)$. If$f(x)=0$we find the two points$(x,0)$and$(-x,0)$. If$f(x) \neq 0$then either$f(x)$or$-f(x)$is a square in$\mathbb{F}_p$(using the fact that$p \equiv 3 \mod 4$) so either we find two points$(x,y)$,$(x,-y)$or two points$(-x,y)$, ... 3 Thanks to Jesper Petersen for correcting the equation. Once we have the right coefficients and coordinates, the identity can be verified by writing the$P_i\$ in terms of small generators, avoiding the calculation of huge numbers such as the 50+ digit denominator that arises in the direct verification. According to mwrank the group of rational points has ...

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