# Tag Info

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Using the Weierstrass's function is the simplest way to a closed form. Nevertheless, the ODE can be solved in terms of Elliptic integrals, which are more commonly used.

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Hints: The translation $x \mapsto x+1$ transforms the cubic polynomial on the RHS to a depressed cubic. An appropriate scaling should produce the coefficient '4' for $x^3$.

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Typo: the equation of the first curve should be $y^2 = x^3 + x^2 - 2$. Hint: $y=y'$, so you need only find a transformation from $x$ to $x'$ that kills the quadratic term of the cubic on the right-hand side.

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Well, you must have the wrong equations, because they are not isomorphic. The $j$-invariant classifies elliptic curves up to isomorphism (over $\mathbb{C}$), and the $j$-invariants of these curves are $432/7$ and $-64/25$, respectively. Since they are distinct, they are not isomorphic. In light of Noam Elkies' answer, the $j$-invariant of $y^2=x^3+x^2-2$ is ...

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I'll write something, probably incomplete, and hopefully someone more capable than I will fill in the gaps. Let's start at the beginning: number fields are finite extensions of the rational field $\mathbb Q$. For every such extension $K/\mathbb Q$, we can associate a group, called the class group, which more or less measure how close the ring of integers in ...

3

Yes, $E(\Bbb C)\cong\Bbb C/L$ for a lattice $L$. And $\Bbb C/L$ is isomorphic to...? (You may be aware that $\Bbb Q/\Bbb Z$ is the torsion subgroup of ____)

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There is already an elementary proof here, but I just wanted to say that this equation defines an elliptic curve $E:y^2=x^3+3x^2+x$. The Mordell-Weil theorem states that the set of rational points is a finitely generated abelian group, so $$E(\mathbb{Q})\cong T_E \times \mathbb{Z}^{R_E},$$ where $T_E$ is a finite subgroup, and $R_E\geq 0$ is called the rank ...

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Well $a^3+3a^2+a=a(a^2+3a+1)$ and you can see that $\gcd(a,a^2+3a+1)=1$. So if it's a square, then $a^2+3a+1$ has to be a square too. But that's not possible, because it's strictly between $(a+1)^2=a^2+2a+1$ and $(a+2)^2=a^2+4a+4$.

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The reason why you are getting this result is that if $E/K$ is a curve given by a singular Weierstrass equation, and $E$ has a cusp at $(0,0)$, with "tangent" line $y=0$ at $(0,0)$, then $E_{\text{ns}}(K)\cong K$ are isomorphic as abelian groups, where $E_\text{ns}(K)$ is the group formed by non-singular points on $E$ (i.e., $E(K)$ except $(0,0)$). The ...

2

Am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$? No, using a free basis $F_2 = \langle a,b \rangle$, the map given by $f(a)=a$, $f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1}$ is not an inner automorphism but it induces the identity on $\mathbb{Z}^2$, as does the identity automorphism ...

1

You have to interpret expressions like $\frac{5}{4}$ in the finite field $\mathbb{F}_3$ of $3$ elements: this fraction actually means that you multiply the element $5=1+1+1+1+1=2$ with the multiplicative inverse of $4=1$. So which element $a$ in $\mathbb{F}_3$ has the property, that $a*1=1$? It is the element $a=1$ itself. Consequently $\frac{5}{4}=2$

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The congruence class of $10$ is a square in $\mathbb{F}_{13}$. Indeed, $$6^2\equiv 36\equiv 10 \bmod 13$$ and similarly $(-6)^2\equiv 7^2\equiv 49\equiv 10\bmod 13$. Thus, for $x\equiv 3$ and $-3\equiv 10\bmod 13$, there are two possibilities for $y$, namely $6$ and $7\bmod 13$. Notice that your algorithm could have run into trouble if $1+5x^2\equiv 0 \bmod ... 1 This is a property of abelian groups, and the fact that the group is coming from elliptic curves has little to do with it. Let$G$be a finite abelian group and let$\phi:G\to G$be a homomorphism. Then, $$|\ker(\phi)|=|\operatorname{coker}(\phi)|.$$ This follows from the first isomorphism theorem, in particular from the fact that$G/\ker(\phi) \cong ...

1

If you make a standard change of variables $X=u^2x$ and $Y=u^3y$, the shape of the Weierstrass equation is preserved, and the coefficients of the new equation are $$Y^2+a_1uXY+a_3u^3Y =X^3+a_2u^2X^2+a_4u^4X+a_6u^6.$$ In other words, the new coefficients replacing the $a_i$ are of the form $a_iu^i$, for $i=1,2,3,4,6$. Since no coefficient changes by a power ...

1

This is because one thinks of each symbol having a weight. If $y$ has weight 3, $x$ has weight 2, and each $a_i$ has weight $i$, then the total weight of each term is exactly 6. But where do these numbers come from? To answer that question, one has to look at the order of the pole at $\infty$. Consider the homogeneous form of the Weierstrass equation: $$v^2 ... 1 Note that the point at infinity, together with the points (-1,0),(0,0),(1,0) form a subgroup of  E(\Bbb Q) isomorphic to (\Bbb Z/2\Bbb Z)^2, you will find this subgroup in E(\Bbb F_p) for any p > 2. Moreover, your curve has complex multiplication (by i : (x,y) \mapsto (-x,iy)), so a lot can be said via the explicit class field theory for ... 0 For variety, we can compute E(\mathbf{F}_3) by using Hasse's theorem, which says it has between 4 - 2\sqrt{3} and 4 + 2\sqrt{3} points: so it's somewhere from 1 to 7 points. Because the curve has three points of order 2, \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} must be a subgroup. And thus it must be the entire group, since the ... 1 The elliptic curve E:y^2=x^3-x has discriminant \Delta=64, so it only has bad reduction at p=2. For any other p>2, the curve has good reduction and E/\mathbb{F}_p is an elliptic curve, in particular, E(\mathbb{F}_p) is a finite abelian group. Take for instance p=5. You can easily find all the points of E in \mathbb{P}^2(\mathbb{F}_5), ... 1 The Wikipedia entry is correct. The sum of two points is supposed to be the reflection across the x-axis of the third point of intersection of E with the line through P and Q. The equation of the line through P and Q is y = \lambda (x - x_1) + y_1. So we need to plug in x_3 and then take the negative, giving$$y_3 = \lambda(x_1 - x_3) - ...

3

Since the isogeny is defined in projective coordinates, you can write it as well in this form: $$[x:y:1] \mapsto [x(x^3+2x+1):x^3y-2xy-x-2y-1:x^3].$$ Now it is clear that $[0:0:1]$, $[0:1:1]$, and $[0:-1:1]$, map to $[0:1:0]$, the point at infinity in $E_1$. Indeed, they map to $[0:-1:0]=[0:-3:0]=[0:1:0]$, respectively, which are all the same point in ...

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If you want to compute the rank yourself, for curves like this with full 2-torsion it is often not too hard by hand: see Silverman's book "The Arithmetic of elliptic curves". Or you can use the implementations available in (for example) Sage: sage: E = EllipticCurve([0,2,0,-3,0]) sage: E.label() '24a1' sage: E.rank() 0 sage: E.torsion_points() [(-3 : 0 : ...

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You can search for points on this sort of equation (of any degree) using Sage-s interface to Michael Stoll's ratpoints C program, but it is hidden: sage: from sage.libs.ratpoints import ratpoints sage: ratpoints([1,0,3,0,1],1000) [(1, 1, 0), (1, -1, 0), (0, 1, 1), (0, -1, 1)] Now this is the equation of a curve of genus 1, so if it has any rational points ...

1

Your question is not a mathematics question but a Sage question! The output for E.integral_points() is a list of points, in your example a list with one element. If you extract that point like this: sage: E = EllipticCurve([0,1,2,8,15]) sage: E Elliptic Curve defined by y^2 + 2*y = x^3 + x^2 + 8*x + 15 over Rational Field sage: E.integral_points() [(0 : 3 ...

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What you want is the general 3-division polynomial (whose roots are the x-coordinates of the points of order 3). Here's how you can compute it in characteristic 2 in Sage: sage: F2 = GF(2) sage: R.<a1,a2,a3,a4,a6,x> = F2[] sage: E = EllipticCurve([a1,a2,a3,a4,a6]) sage: f3 = E.division_polynomial(3,x) sage: f3.factor() a1^2*x^3 + a1*a3*x^2 + x^4 + ...

2

You can use the Hasse-Weil bounds to prove that the Hasse-Weil L-function $L(C,s)$ of a curve $C$ converges for $\Re(s)>3/2$. See page 312 in Husemoller's "Elliptic Curves".

1

This is done in Silverman and Tate's "Rational points on elliptic curves" in Section 5 of Chapter III (see in particular the discussion at the top of page 85). Notice however that even though the proof does not mention Selmer groups explicitly, they are lurking in the background. The authors try hard to keep the proof elementary for undergraduates, but once ...

4

I describe a coding theoretical application as promised. Assume that $q=2^m$, and let $$tr:\Bbb{F}_q\to\Bbb{F}_2, x\mapsto \sum_{i=0}^{m-1}x^{2^i}$$ be the trace function. It is relatively easy to show that (ask if you haven't seen it) for a fixed element $a\in\Bbb{F}_q$ the equation $$y^2+y=a$$ has two solution $y\in\Bbb{F}_q$, if $tr(a)=0$ and no ...

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One application is to the Ramanujan--Petersson conjecture for weight two modular forms: the Eichler--Shimura congruence relation relates the $p$th Hecke correspondences on the modular curve to the Frobenius correspondence in char. $p$, and RH (applied to the reduction mod $p$ of the modular curve) then gives what you want. As well as being of interest in ...

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This might be slightly out there, but you can calculate the Betti numbers of a toric variety using the Weil conjectures. See page 94 of Fulton's "Introduction to Toric Varieties." edit: Also, you can use it to define the Hasse-Weil L-function. This will help you do things like state Taniyama-Weil, modularity, and work on the Langlands program. For these last ...

4

Number theorists (arithmetic geometers) are interested in finding rational points on all diophantine equations. The case of elliptic curves is the first one that we do not know how to solve: Diophantine equations in one variable, i.e., polynomials $p(x)$ in one variable with integer coefficients: it is easy to find all the rational roots of a given ...

1

The intuition behind this rests on two facts: A morphism $\phi:E_1\to E_2$ between elliptic curves (that sends $O_E$ to $0_{E'}$) sends collinear points to collinear points. The addition on an elliptic curve is defined so that if $P,Q,R$ are the three intersection points of a line with $E$, then $P+Q+R=0$. Thus, if $\phi:E_1\to E_2$ is a morphism as ...

3

$\newcommand\O{\mathcal O}$Let $C$ be an elliptic curve and $\ast:C\times C\to C$ be the operation that takes $P,Q\in C$ to the third point $P\ast Q$ on the line through $P$ and $Q$. If we choose a zero element $\O$, we obtain a group $(C,+)$ with $$P+Q = \O \ast (P\ast Q).$$ Picking a different zero $\O'$ gives a group $(C,\oplus)$ with \begin{align} ...

0

Holden Lee had the right idea in the comments. Here is how it can be implemented. Let $E/\mathbb{Q}:y^2=x^3+x+1$. The discriminant is $-2^4\cdot 31$, so it only has bad reduction at $2$ and $31$. Moreover, When $p=5$, the curve $E/\mathbb{F}_5$ has $9$ points. When $p=7$, the curve $E/\mathbb{F}_7$ has $5$ points. Now let \$E_{A,B}/\mathbb{Q}: ...

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