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As a general fact, a point is determined by its coordinates. Elliptic curves are usually represented as plane projective curves, in which case they are described by a homogeneous relation between three projective variables $X, Y, Z$. If you are using a Weierstrass model then there is only one point with $Z=0$ (the "point at infinity"), so in this case yes, ...
I think that what you want is described in Proposition 2.5 of Silverman's Arith. of Elliptic Curves: If your elliptic curve $E$ has distinct tangent lines at the singularity, then $E_{\rm ns} \to \bar K^*$ given by $(x,y) \mapsto \frac{y-a_1x-b_1}{ y-a_2x-b_2}$ is an isomorphism of groups. If $E$ has a cusp (the tangent lines are coincide), then $E_{\rm ... 1 Think about the group operation in terms of chords and tangents. So if the chord$L$connecting$P$and$Q$has equation$ax+by+c=0$, then the chord$\phi(L)$connecting$\phi(P)$and$\phi(Q)$has equation$\phi(a)x+\phi(b)y+\phi(c)=0$. If$L$intersects$E$at a third point$R$, then$\phi(L)$intersects$E$at the point$\phi(R)$(here it is essential ... 1 Since there no answers, I will attempt to give one. There may be some small mistakes, but the general theory should be right. Short answer: You will always find primes$p$if$B_2>{p+1+2\sqrt{p}}$, due to Hasse's theorem. Otherwise, you will miss some. The probability of find$p$in stage one is $$\Phi(p,B_1)/x\approx \rho(\ln p / \ln B_1),$$ where ... 2 It is known in general that the group$E(\mathbb{F}_q)$for an elliptic curve$E$over$\mathbb{F}_q$is either cyclic or isomorphic to$\mathbb{Z}/k\mathbb{Z} \times \mathbb{Z}/\ell\mathbb{Z}$for some$k,\ell$. For$y^2=x^3+x+1$over$\mathbb{F}_5$the group is isomorphic to$\mathbb{Z}/9\mathbb{Z}$, so it is cyclic. Indeed,$P=(0,1)$is a generator of ... 2 I believe that the standard reference is Silverman and Brumer, The number of elliptic curves over$\mathbb{Q}$with conductor$N$, Manuscripta Mathematica 91, 1996. They prove that the number of elliptic curves of curves of conductor$N$is bounded above by$N^{\frac{1}{2}+\epsilon}$. I found out that you can read it here: ... 0 You want to find$x_0$such that the$y_0$satisfying $$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$$ is equal to$a.$You should be able to solve this. 1 For any sheaf of$\mathcal O$-modules$\mathcal M$, the sheaf-Hom$Hom_{\mathcal O}(\mathcal O, \mathcal M)$is naturally isomorphic to$\mathcal M$(given by evaluation at the global section$1$of$\mathcal O$). The tensor product$\mathcal M \otimes_{\mathcal O} \mathcal O$is also naturally isomorphic to$\mathcal M$(the isomorphism being given by the ... 0 I don't think this is an answer, but really a comment, but I don't have a high enough reputation to comment: If you're talking about ECIES then, the client Bob, would already have the private key (which is a secret that is kept to himself) to decrypt data that was sent by Alice. Note the that private and public key are related to each other, the public key ... 4 As an example, lets look at Elliptic-Curve-Diffie-Hellman (ECDH). Alice Ephemeral key pair generation Select a private key$n_A \in [1, n-1]$Calculate the public key$Q_A = n_A P$Alice ships$Q_A$to Bob Bob Ephemeral key pair generation Select a private key$n_B \in [1, n-1]$Calculate the public key$Q_B = n_B P$Bob ships$Q_B$to Alice Each ... 2 For simplicity, I’ll write$C_n$for$\mathbb Z/n\mathbb Z$. First, I think you’ll find that$C_2\times C_{36}$and$C_4\times C_{18}$are isomorphic. The best way to write something like this is to make sure the indices divide:$2|36$, but$4$does not divide$18$. Your question about why$C_3\times C_{24}$does not occur is much more interesting, and I’m ... 1 Hint: any ideal$I$of this ring$R=\mathbf{Z}\left[\frac{1+\sqrt{D}}{2}\right]$can be viewed as a lattice sitting inside$\mathbf C$. Moreover, multiplication by elements of$R$preserves this lattice, so it descends to the quotient$\mathbf C/I$. In other words,$\mathbf C/I$has complex multiplication by$R$. Now, try to answer the following questions: ... 2 Your idea is correct. It follows immediately from Lagrange's theorem and the Hasse bound. Could the group be twice as big or bigger? 2 It's easy enough to write down curves with no$p$-adic points. For example $$C: X^3+pY^3=p^2Z^3$$ has no projective solutions over$\mathbb{Q}_p$. Indeed, if it did, then you could assume$X,Y,Z$to be$p$-adic integers with at least one of them a unit. But reducing modulo$p$, you see that$p|X$, and then reducing modulo$p^2$you get$p|Y$and finally ... 3 If two curves are isomorphic then they have the same groups of points. The map$E\mapsto E(K)$is a functor, and functors preserve isomorphisms. For the same reason they also have the same torsion subgroups. I do not really know what you mean by$E/\Lambda$. Are you thinking of period lattices? The notation$E/\Lambda\$ does not really make sense. I am also ...