# Tag Info

## New answers tagged elliptic-curves

2

I am assuming you mean strictly doing everything in $\Bbb Q$, including eventual transmission. It is not sensible to do computations in $\Bbb Q$ and reducing to $\mathbb F_p$ before sending out since that is harder to do and you do not gain anything. The summary of the following discussion is that you cannot use any generated points to keep secrets. ...

2

You can use Magma. The code E:=EllipticCurve("35a2"); E2:=IsogenousCurves(E)[1]; A,B:=IsIsogenous(E,E2); A; B; returns true Elliptic curve isogeny from: CrvEll: E to CrvEll: E2 taking (x : y : 1) to ((1/81*x^9 + 2/9*x^8 + 13*x^7 + 3392/27*x^6 - 30325/9*x^5 - 85999*x^4 - 24206654/27*x^3 - 50989888/9*x^2 - ...

1

You could try reading (the relevant parts of) Qing Liu's book on Algebraic geometry or the book on Neron models by Bosch-Lutkebohmert-Raynaud to get a feeling for elliptic curves over one-dimensional schemes. You could also try reading some papers where abelian schemes are used, e.g., Szpiro's asterisque (1985) on the Mordell conjecture, or Jinbi Jin's ...

4

There are various ways to do this, but I will use the method you show. We are given the elliptic curve $$x^3+17x+5 \pmod{59}$$ We are asked to find $8P$ for the point $P = (4,14)$. I will do one and you can continue. We have: $$\lambda = \dfrac{3 x_1^2 + A}{2 y_1} = \dfrac{3 \times 4^2 + 17}{2 \times 14} = \dfrac{65}{28} = 65 \times 28^{-1} ... 4 Instead of following blindly Wikipedia's formulas, it is best to understand how to calculate P+Q, or P+P, given an elliptic curve. Let us assume for simplicity that the curve is given by E:y^2=x^3+Ax+B, and P,Q\in E. In order to find P+Q, first find the equation of the line L through P and Q, find the third point R of intersection of L ... 1 Perhaps others can find an elementary solution. This is an elliptic curve, of rank 1 and trivial torsion subgroup. So the group of all rational points is generated by one single rational point, namely P=(1,1). Hence, every rational point on the curve is of the form nP for some n\in\mathbb{Z}. Here are some multiples:$$2P=(2 , -3 ),\ 3P=(13 , 47),\ ...

5

As Robert Auffarth said in a comment, a $\mathbb{Q}$-isogeny of degree $p$ is a non-constant morphism of elliptic curves $E\to E'$ defined over $\mathbb{Q}$, that sends zero to zero, i.e., $\mathcal{O}_E\mapsto \mathcal{O}_{E'}$, and has degree $p$. Since the degree of the map equals the size of the kernel, the degree $p$ condition means that the kernel has ...

7

Try the first chapters of Arithmetic Moduli of Elliptic Curves, by N. Katz and B. Mazur, Annals of Math Studies 108

2

If $f_P(S)$ is a function with divisor $m[P]-m[\mathcal{O}]$, then $f_P$ has a zero at $P$ and a pole at $\mathcal{O}$. Then, for a fixed constant $T$, the function $h_{P,T}(S)=f_P(T+S)$ is simply a translation of $f_P$ by $T$. Thus, $h_{P,T}$ has a zero when $T+S=P$, i.e., at $S=P-T$ and a pole at $T+S=\mathcal{O}$, i.e., at $S=-T$. Moreover, the ...

1

There is exactly one way in which the tangent line at $P$ can meet $E$ at a point $[-2]P$ with the same $x$-coordinate as $P$ which I missed. This is so when $P$ is in fact a flex. That totally makes sense because there are 9 flexes on an elliptic curve and $|E(3)|=9$.

3

Look at the curve $y^2=x^3+1$. The point $P=(0,1)$ has order three. Draw the curve, find $2P$, and everything should be a lot clearer.

1

Suppose that (i) $a^2+b^2=c^2$ and (ii) $b^2+c^2=d^2$. Then, if we divide $(i)$ by $b^2$ we obtain $u^2+1=v^2$ and dividing (ii) by $b^2$ obtain $1+v^2=w^2$, where $u=a/b$, $v=c/b$ and $w=d/b$. In particular $u^2+1=v^2$, $u^2+2=w^2$, and $u^2$ are all squares. Hence, $(u^2,uvw)$ is a point in the elliptic curve $$E:y^2=x(x+1)(x+2).$$ After a change of ...

6

Let $A,B$ be integers, and let $E_{A,B}: y^2=x^3+Ax^2+Bx$ be an elliptic curve. Then, the rank of the Mordell-Weil group $E_{A,B}(\mathbb{Q})$ is less or equal to $\nu(A^2-4B)+\nu(B)-1$, where $\nu(N)$ is the number of positive prime divisors of $N$. This is shown in Milne's "Elliptic Curves", Proposition 5.6, or here (Proposition 1.1). In particular, ...

3

Oh, well. I cannot find this claim in Wikipedia... one case is easy, the other, not. Any primitive Pythagorean triple is $2xy, x^2 - y^2, x^2 + y^2$ with one of $x,y$ odd and the other even, and $\gcd(x,y) =1.$ We cannot have a Pythagorean triple with legs $x^2 - y^2, x^2 + y^2,$ because we would be solving $$2 x^4 + 2 y^4 = z^2$$ with one of $x,y$ odd. ...

1

In positive characteristics, isogenies may ramify. In general, see Silverman's "The Arithmetic of Elliptic Curves", Theorem 4.10 in Chapter III, where he shows that if $\phi:E_1\to E_2$ is a non-constant isogeny then (1) For every $Q\in E_2$, we have $\# \phi^{-1}(Q)$ equals the separable degree of $\phi$. (2) For every $P\in E_1$, the ramification index ...

1

As far as I know there is no code to solve this problem distributed with Magma as standard. Your choices are: use the Sage implementation; implement the algorithms in Magma for yourself; or find someone who has appropriate Magma code and persuade them to share it with you. EDIT. This answer is totally wrong, as ccorn's comment above shows. I had apparently ...

2

Let $E$ and $E'$ be elliptic curves, and let $\phi:E\to E'$ be a $p$-isogeny (i.e., $\phi$ is an isogeny of degree $p$), where $p$ is prime. In particular, $\phi$ is a group homomorphism from $E$ to $E'$ and its kernel $\ker(\phi)$ is a group of size $p$. Prove that every $P$ in $\ker(\phi)$ has order dividing $p$. Prove that the prime-to-$p$ torsion ...

3

In general, if a curve is given by $Y^2=X^3+AX^2+BX+C$, a change of variables $Y=y$ and $X=x-A/3$ will provide a model for the curve of the form $y^2=x^3+A'x+B'$. The reason is that $$X^3+AX^2+\cdots = (x-A/3)^3 + A(x-A/3)^2+\cdots$$ and the coefficient in $x^2$ is given by $3(-A/3)+A=0$.

3

Let $E$ be given by a Weierstrass equation $$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$ If the characteristic of the field is not $2$, then we can complete the square in $y$ to obtain $$(y+(a_1x+a_3)/2)^2 = x^3+a_2x^2+a_4x+a_6+(a_1x+a_3)^2/4.$$ You can write the last equation as $E':Y^2=f(x)$. In this model, it is clear that if $P'=(x_0,Y_0)$, then $-P'$ (the ...

2

First, let me point out that $E:x^3-2y^3=1$, together with the point $[1,0,1]$, is an elliptic curve. Its Mordell-Weil group can be calculated by finding a Weierstrass model, which can be chosen to be $y^2=x^3-27$, and then verify that $E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$. Thus, there are only two rational points on $E$, namely $[1,0,1]$ and ...

0

You can calculate the size of $E_\text{ns}(\mathbb{F}_\ell)$. See Chapter III, Exercise 3.5 of Silverman's "The Arithmetic of Elliptic Curves".

2

Let us begin with the quartic $$m^4-4m^3n-6m^2n^2+4mn^3+n^4=d^2.$$ Dividing through by $n^4$, we obtain $$m'^4-4m'^3-6m'^2+4m'+1=d'^2,$$ with $m'=m/n$ and $d'=d/n^2$. Now we can homogenize this equation to obtain $$E:M^4-4M^3N-6M^2N^2+4MN^3+N^4=D^2N^2,$$ which has a rational point $P=[M,N,D]=[0,0,1]$. The curve $E$ is in fact non-singular, and of genus $1$, ...

1

This is a great observation! Let $E/\mathbb{Q}$ be a curve given by a model $v^3=q(u)$, for some cubic polynomial $q\in\mathbb{Q}[u]$, and assume that the projective closure of this model, i.e., $V^3=W^3q(U/W)$, is smooth, and it has a rational point $P$. Then, $(E,P)$ is an elliptic curve defined over $\mathbb{Q}$. Moreover, $E$ admits an endomorphism ...

2

The group structure of the torsion subgroup may be the same, but the group law may look very different! I find the following example to be interesting and related to your question. Let $E:y^2=x^3+1$ with zero at $[0,1,0]$, and consider $E':y^2=x^3+1$ where we now declare zero to be $[2,3,1]$. Then, $E$ and $E'$ are clearly birationally equivalent via the ...

3

The statement: "The number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a,b)∈ \mathbb{Z}^{2}$ such that $D=−16(4a^{3}+27b^{2})$." is certainly false as stated. For example, the discriminant of the model for the curve $$E:y^2 + y = x^3 - x^2 - 10x - 20$$ is $-11^5$. Any model of the form $y^2=x^3+Ax+B$ with ...

3

Hint: the existence of the Weil pairing.

1

We can use $\mathbf{R}$ in place of $\mathbf{Q}$, if we like. Draw a picture of a curve that has three real $2$-torsion points in Weierstrass form. How many of the $2$-torsion points can be the double of some other point? Recall that every line intersects an elliptic curve in exactly three points, counting multiplicity and the point at infinity.

-6

An elliptic curve is not a group so it can't have a subgroup.

3

Two birationally equivalent, projective curves without singularities are isomorphic. Hence two birationally equivalent elliptic curves are isomorphic.

1

$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$. Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$. Since $\gcd(a,b) = 1$, and we know that $2a^2 d^4 \equiv 0 \pmod{p^{2n}}$, thus $d^4 \equiv 0 \pmod{p^{2n}}$ and this implies that $p^n \mid ... 3 Hints: Prove that the curve has genus 1. Find the primes of bad reduction. If$p$is a prime of good reduction, and large enough, then the Hasse-Weil bounds tell you that there is an affine point (not just at infinity) over$\mathbb{F}_p$. Now use Hensel's lemma to lift it up to$\mathbb{Q}_p$. If$p$is a prime of bad reduction, or not large enough, you ... 2 It is not that we can't rescale$X,Y$independently, but rather that we don't want to rescale them by a square, or cubic root (as I suggested in my comment above$X'=\beta^{1/3}, Y'=\alpha^{1/2}Y$), which might not exist in the field$K$over which the curve is defined. 0 Sure you can you do the following: Find the lowest common multiple of the denominators. Multiply the result by a constant so that it becomes a power of$6$. Multiply the result in the equation. Linear transformation. E.g. $$y^2=x^3+\frac{2}{9}x^2+\frac{1}{8}x+2$$ Step 1:$LCM = 72=2^3\cdot3^2$. Step 2: Let$N=2^3\cdot3^4\cdot LCM=46656$. Step 3/4: ... 1 Let me explain why I would not expect this to be true. Suppose that$X$is a$K3$surface with the following properties: First we ask that the automorphism group$G=\operatorname{Aut}(X)$is finite. (Remark: by a theorem of Sterk, this is equivalent to requiring that$X$has fintely many elliptic pencils, though we don't use this.) Note that this gives ... 1 The missing two points are the http://en.wikipedia.org/wiki/Circular_points_at_infinity From the complex projective point of view, a circle is the same as any other conic, and 5 points are needed specify a conic. The reason 3 points in the plane are enough to define a circle is that circles are the conics that pass through the 2 "circular points at ... 1 For Beozout's theorem to work we must work in the complex projective plane, as you noted. Mathematica's Reduce transforms the system $$x^2 + y^2 = z^2 \ \ \text{and}\ \ x^2 + y^2 - x z - y z = 0$$ to $$(y\neq 0\land ((x=0\land y=z)\lor (z=0\land (y=-i x\lor y=i x))))\lor (x\neq 0\land ((y=0\land x=z)\lor (z=0\land (y=-i x\lor y=i x))))$$ We can read off ... 1 Quote from Example section of the Wikipedia entry on Bézout's Theorem (see here): Two circles never intersect in more than two points in the plane, while Bézout's theorem predicts four. The discrepancy comes from the fact that every circle passes through the same two complex points on the line at infinity. Writing the circle$(x-a)^2+(y-b)^2 = r^2$... 1 If you eliminate$Y$from both the equations you get $$-2\,X\,Z^2\,\left(Z-X\right)=0$$ Thus there is a pair of roots at$Z=0$(points at infinity). Set$Z=0$and you get$X^2+Y^2=0$. Setting$X=1$, you get the two other roots $$(1, i, 0)$$ and $$(1, -i, 0)$$ I had a typo in the earlier answer. 1 Theorem: Let #$E(Fq) = 1 - a + q $Write$X_2 − aX + q = (X − α)(X − β)$. Then #$E(F_{q^n}) = 1 − (α^n + β^n) + q^n $for all n ≥ 1. Now for the problem: It is easy. Write$x^2 + 2$=$(x+i \sqrt2)(x-i \sqrt2)$and so #$ E(F_{2^n})=2^n+1 -(x+i \sqrt2)(x-i \sqrt2)$and from there it is then easy to use a phase argument to duduce the answer. Note: ... 3 You are right. The computation is just too inefficient. The best known attack on ECDLP, the pollard rho attack, would be useless against elliptic curves over the rationals. Consider this, if you were to do the computations over a finite field of say 512-bits, you will only have to deal with 512-bit intermediate values along the way. Considering the same ... 4 Intuitively such a thing cannot exist, even with level structure, since the "forget isomorphism class and remember only isogeny class" map from the usual moduli space ought to be algebraic, but an algebraic map of curves has finite fibers but (over$\mathbb{C}$, say) isogeny classes of non-isomorphic curves are infinite. Of course there is the problem that ... 3 When we try to classify elliptic curves up to isogeny, then we use the modular curves$X_0(N)/\mathbb{Q}$, whose non-cuspidal$K$-rational points (for some number field$K$) classify triples$(E/K,E'/K,\phi/K)$of elliptic curves$E$and$E'$defined over$K$, together with an isogeny$\phi:E\to E'$defined over$K$, with cyclic kernel of size$N\$. In other ...

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