# Tag Info

0

A shorter solution depending on a bit more machinery is the following. As others have said, an abelian group of order $9$ is either cyclic or isomorphic to $C_3\times C_3$. It is known (see e.g. Silverman) that the $3$-torsion of this curve (over an algebraic closure of $\Bbb{F}_{11}$ is isomorphic to $C_3\times C_3$. Furthermore, the Weil pairing on that ...

2

You are almost there. Show that $f((1-\omega)/3) = f(\omega(1-\omega)/3)$, because the difference $(1-\omega)/3 - \omega(1-\omega)/3\in \Lambda$.

3

I believe you are looking for the Shioda-Tate formula. See Corollary VII.2.4 in page 70 of Miranda's "The basic theory of elliptic surfaces", or here. Note: let $k$ be a field and let $C/k$ be a smooth projective curve defined over the field $k$ and has genus $g$. The function field of $C/k$ will be denoted by $K=k(C)$. An elliptic surface $\mathcal{E}$ ...

2

Answer to Question 1. Denote $s=x+y$. Then consider equation $$(s-y)^3+s^3+(s+y)^3=z^3,\tag{1}$$ for positive $s,y,z$. First, consider any such integer solutions: for $s>y$ and for $s<y$. Eq. $(1)$ is equivalent to $$3s^3+6sy^2=z^3.\tag{2}$$ Denote $z=3c$, then $(2)$ is equivalent to $$s^3+2sy^2=9c^3.\tag{3}$$ To check all up-to-6-digital ...

2

That depends on how you visualize $\mathbb{P}_k^2$. One way to think of it is that it is an affine plane, i.e. $k^2$, together with an extra (projective) "line at infinity". To make this precise, one identifies (or maps) a point $(x, y) \in k^2$ with a point $[x, y, 1] \in \mathbb{P}_k^2$. Then the rest of the points of the projective plane are of the form ...

0

So we have to prove that $j:\mathbb{H} \longrightarrow \mathbb{C}$ is surjective. By the given q series of $j(\tau)$it is clear that $j(\tau)$ $$j(\tau ) = \frac{1}{q} +744+ 196884q+ 21493760q^2.+ c_2q^3......\qquad where \quad q = e^{2\pi i \tau}$$only goes to infinty when q=0 or the imaginary part of $\tau$ goes to $\infty$. and the pole is a order of ...

2

You can check that $E: f(x,y)=0$ is singular in characteristic 2, using the definition of singular point: A point $P=(x_0,y_0)$ on $E$ is singular if $\partial f/\partial x = \partial f/\partial y = 0$ at $P$. Now take $f(x,y) = y^2 - (x^3+ax+b)$, and suppose for simplicity $K=\mathbb{F}_2$. Then, $P=(a,b)$ is a point on the curve, and you can check ...

1

Depending on how cryptography heavy/math heavy the resulting paper is supposed to be you might want to include theory relating to attacks on curves. This would be weil-tate pairing and possibly the background needed for the weil-descent attack. Who is supposed to be the target audience of the paper? Is this a math paper for academic cryptographers? Or more a ...

2

The correct transformations are the following (assuming the characteristic of the field of definition is not $2$ or $3$). First change $y\longrightarrow y-(a_1x+a_3)/2$, so the new equation has the form $$y^2=x^3+Ax^2+Bx+C.$$ And now change $x\longrightarrow x-A/3$, so that the new equation has the form $$y^2=x^3+ax+b.$$ Clearly, both changes of variables ...

0

You can go from the "long" form to the short form by the transformation $$z=y+\frac12 a_1 x$$ Since you will be interested in rational solutions, and since $a_1$ is rational, studying the short form tells you what you need to know about the long form as well.

1

The slope of the tangent line to $E:y^2=4x^3-g$ at $P=(x_1,y_1)$ is $$m = \frac{6x_1^2}{y_1}$$ as long as $y_1\neq 0$ (i.e., as long as $P$ is not a point of order $2$). Now, let $P=(0,\sqrt{-g})$. Then, the tangent line $L$ to $P$ is a line of slope $0$ passing through $P$, i.e., $y=\sqrt{-g}$. Thus, $L$ intersects $E$ at three points, which can be found ...

1

In general, if $F$ is a field, and $V$ is a variety defined by polynomials defined over $F$, it is very common notation to write $V/F$ to indicate that "$V$ is defined over $F$", and simply read $V/F$ as "$V$ over $F$" or "$V$ defined over $F$" (so you interpret / as "over", and not as any type of quotient space).

1

The idea is, $E$ has lots of points in $\overline{\mathbb F_p}$, not all of which are in $\mathbb F_p$. However, if I have an $\overline{\mathbb F_p}$-point, then it is actually an $\mathbb F_p$-point iff it is fixed by the $p^{th}$ power map. This follows because the solutions to $x^p=x$ in $\overline{\mathbb{F_p}}$ (of which there are $p$) are canonically ...

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I would highly suggest Barry Mazur's "Finding meaning in error terms", Bulletin (New Series) of the American Mathematical Society, Volume 45, Number 2, April 2008, Pages 185–228.

2

Setting $x+y=3t$ and $xy=s$, we obtain that $$9t^2-s = (t+1)^3 \implies s = -t^3+6t^2-3t-1$$ $x$ and $y$ satisfying the quadratic $a^2 -3ta + s =0$. This means $9t^2-4s = k^2$. Eliminating $s$, we obtain $$4t^3-15t^2+12t+4 = k^2 \implies 64t^3 - 240t^2 + 192t + 64 = (4k)^2$$ $$(4t-5)^3 - 108t + 189 = (8k)^2 \implies (4t-5)^3 - 27(4t-5) + 54 = (4k)^2$$ Hence, ...

1

Hint $1$ : Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes: $$y^2-3ty+(3t)^2-(t+1)^3=0$$ a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square. Hint $2$: $\Delta_y=(t-2)^2(4t+1)$ Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$ And it's your turn to do some work! Edit Because my ...

0

Yes - if they are $m$-isogenous, then the modular polynomial $\Phi_{m} (j, j')$ is zero. (This is actually the definition of $\Phi_m$). If they are not $m$-isogenous, then the best I can say is that the corresponding lattices will be (rationally) commensurable, so that any representatives $\tau, \tau'$ in the Poincaré half-plane will be related by a rational ...

0

I have definitely forgotten to count the projective points. But we will get to that later. Firstly, let's prove $E(\mathbb{Z}/N\mathbb{Z})=E(\mathbb{F}_p)\times E(\mathbb{F}_q)$. Consider the map $$f:E(\mathbb{Z}/N\mathbb{Z})\to E(\mathbb{F}_p)\times E(\mathbb{F}_q)$$ given by ...

1

Your computations look fine. I haven't seen elliptic curves modulo a composite number in cryptography before, but I am not an expert in this. On the other hand, if you were going to start with the two curves over $\mathbb{F}_p$ and $\mathbb{F}_q$ you'd need the condition $4a^3+27 b^2 \neq 0$ for nonsingularity of $$y^2=x^3+ax+b.$$ This would fail for your ...

1

You are correct that $[0:1:0]$ is not on the curve $y^2=x^3+Axz^4+Bz^6$ in those modified projective coordinates $[x:y:z]$, and $[1:1:0]$ is indeed on the curve. I am not an expert though in cryptography... the literature in cryptography seems to insist in calling the point at infinity by $[0:1:0]$ even though this point is not on the curve in those ...

0

No, this is not necessarily the case. Here is a silly example. Using the same argument one can come up with more exotic examples. An elliptic curve is a pair $(E,\mathcal{O})$ consisting of a smooth projective curve $E$ of genus $1$, and a point $\mathcal{O}$ that will serve as identity of the group. Suppose $V=(E,[0:1:0])$, where $E: y^2=x^3-2$, with ...

0

For what it's worth, I used Magma to compute the rank and torsion subgroup of the jacobian, as follows: P<x>:=PolynomialRing(Rationals()); C:=HyperellipticCurve(x^5-x,-1); J:=Jacobian(C); RankBounds(J); C:=HyperellipticCurve(4*(x^5-x-1/4),0); TorsionSubgroup(Jacobian(C)); For the torsion, Magma requires the curve to be of the form $y^2=f(x)$, so I ...

3

This is not true. For instance $j(\tau)=j(\tau+1)=j(-1/\tau)$ for every $\tau\in\mathbb{H}$. More generally, $$j(\tau) = j\left(\frac{a\tau+b}{c\tau+d}\right)$$ for any matrix $\left(\begin{array}[cc] aa & b\\ c& d\end{array}\right)$ in $\Gamma(1)=\operatorname{SL}(2,\mathbb{Z})$. It is true, however, that $j:\mathbb{H}/\Gamma(1)\to \mathbb{C}$ is a ...

0

This is a standard fact about modular functions. For instance, it is explained in Chapter 1, Sections 6 and 7 of Silverman's "Advanced Topic in the Theory of Elliptic Curves".

1

The points on the curve with $y = 0$ correspond precisely to when the tangent line is vertical (and thus not intersecting the curve except at infinity), so that the point is of order $2$. You can see using vertical lines that if $y \ne 0$, then the inverse of a point will be the reflection across the $x$-axis. Thus, the number of such points is determined ...

1

What about the curve $X+Y=0$ defined over $k = \Bbb F_2(Z)$. The point $(Z,Z)$ of the curve is not in the image of $\phi$ even though the ideals of the curve and of its image by $\phi$ are the same.

2

Yes, this is possible. In general, your curve $E/\mathbb{C}(t)$ satisfies the Mordell-Weil Theorem (see Chapter III, Theorem 6.1, of "Advanced Topics in the Theory of Elliptic Curves" by Silverman), and the Mordell-Weil group can be trivial, so there would be no points. For instance, in this article, Cox shows that the curve $y^2=4x^3-3x-t$ has trivial ...

2

Another good reference is Diamond and Shurman's "A First Course in Modular Forms", or William Stein's "Modular Forms".

1

Stevens' 1982 book is a research monograph, aimed at readers who are already experts. There's a big gap between it and Serre's book. To bridge that gap, you might like to try reading some of the articles in the Cornell--Silverman--Stevens volume on Fermat's last theorem (especially Rohrlich's article in that volume on modular curves). That and the ...

1

This equation was solved in a paper of Henri Darmon and Loic Merel in Crelle's journal in 1997. The techniques involved are similar to those in Wiles' proof, with additional complications due to the fact that the trivial solution $x=y=z=1$ actually corresponds to a modular form of weight $2$ and level $32$.

1

The Diophantine equation $Ax^n+By^n=Cz^n$ for coprime integers $(A,B,C)$ is closely related to FLT, and has been studied, too. There are choices for $A,B,C$ and $n=p$, such that there are indeed some nontrivial solutions, e.g., $$x^5+13y^5=8z^5$$ has the nontrivial solution $(x,y,z)=(3,1,2)$. But usually there are very few solutions. For an overview see ...

0

Did you prove that $f$ is holomorphic on $\mathbb{C}\backslash\Gamma$? If you did, then it's easier: it can only have poles in the points of the form $m + in$, and its values in all such points are equal: $f(z) = f(z + 1) = f(z + i)$. Now, if it doesn't have poles in such points, it means that it has no poles at all, so it is bounded on $\mathbb{C}$, ...

1

I have since worked out the answer to this. So in case anybody comes across the question, I will sketch it below. Observe $\rho$ has the minimal polynomial $x^2+x+1=0$, so if $\Lambda$ is the lattice $\mathbb{Z}+\rho\mathbb{Z}$, we have $\rho \Lambda = \rho\mathbb{Z}+\rho^2\mathbb{Z}=\rho\mathbb{Z}-\mathbb{Z}-\rho\mathbb{Z}=\Lambda$. But clearly $g_2(a ... 0 To begin on this topic I highly recommend Lectures on Elliptic Curves, J. W. S. Cassels. London Mathematical Society. Student Texts 24. This book of Professor Cassels is indeed bright and decorated with a truly extraordinary didactic sense. You can get it free by internet. 0 If you are referring to Stein's Complex Analysis (Princeton Lectures in Analysis, vol 2) by Elias M. Stein, Rami Shakarchi, then I'd suggest the following chapters/topics: Chapter 1: Preliminaries to Complex Analysis Chapter 2: Cauchy's Theorem Chapter 3: Meromorphic Functions and the Logarithm Chapter 5: Entire Functions Chapter 6: The Gamma and Zeta ... 2 I agree with the OP and Bruno Joyal that the statement of this exercise is faulty. As you say, condition (ii) had better hold for any rational point$O$given that we've defined the group law in such a way to make$O$the origin. Unfortunately I could not remember what I had in mind when I wrote this, so I uploaded a new copy in which condition (ii) is ... 1 As it has been pointed out, a sufficient condition is taking the ground field to be infinite. This can be traced back to the following very general considerations: if$f\in A[(X_i)_{i\in I}]$is a polynomial ($A$a commutative ring), then we get an induced map$\tilde{f}:A^I\rightarrow A$obtained by substiution. It turns out that the map$f\mapsto\tilde{f}\$ ...

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