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1

Changing the sign of the y will indeed result in another point. Let (x, y) be a point on the curve, then y^2=x^3+ax+b=(-y)^2. Thus the point (x, -y) is also on the curve. Finding 5 more points can be done by adding and doubling the given points until you have five more points. There is no 'mod' in this case, since the field over which the curve is defined ...


1

They're using $\lambda$ as the slope of the line $\vec{PQ}$. Namely, $\lambda = \frac{y_Q - y_P}{x_Q - x_P}$. Now the idea is: take the line $y - y_P = \lambda (x - x_P)$ and find a third point (other than $P$ and $Q$) on this line intersected with the curve. To do this, plug in $x = x_P + t$ and $y = y_P + \lambda t$: $(y_P + \lambda t)^2 = (x_P + t)^3 ...


1

There are formulae for these operations: http://en.m.wikipedia.org/wiki/Elliptic_curve_point_multiplication. Simply fill in the formulae to get the answers.


1

You should let a computer try all possibilities for nP, until nP=O. Then n* is the order of P. EDIT: To do this using a computer or by hand, use the formulae found on wikipedia: http://en.m.wikipedia.org/wiki/Elliptic_curve_point_multiplication Instead of division, you should multiply by the modular multiplicative inverse, since we are working in a finite ...


2

See Harald Hanche-Olsen for most of the solution. When the two points are the same, there are lots of lines through them. You need the line that is tangent to the curve at that point. Differentiate the equation to find the slope of the tangent $$ 2y\frac{dy}{dx}=3x^2+2x-1$$ and, oddly enough, the slope of the line $dy/dx$ is zero again, and the line is ...


3

For the first one, find the equation of the line through the two given points: It is $y=1$. Find the three intersection between this line and the curve. Substituting in $y=1$ in the curve equation and simplifying gives $x^3-x=0$, with the three solutions $x=0$, $x=1$, and $x=-1$. (If the equation had been any harder, you could have simplified it by dividing ...


2

You don't need a text on cryptography, these are elementary facts that you find on every elliptic curves text. For example, check J.Silverman's "The arithmetic of elliptic curves" or Washington's "Elliptic curves: number theory and cryptography".


2

The degree of the line bundle is $3$. This means that a generic section of $\mathcal O(3p)$ has $3$ zeroes. This means that a generic line in $\mathbb P^2$ intersects the image of the curve in $3$ points, which means the image has degree $3$. This kind of computation is part of the basic package related to using line bundles to embed varieties into ...


1

A nonsingular curve $C\subset\mathbb{P}^2$ of degree $d$ always has genus $g = \frac{(d-1)(d-2)}{2}$ (this can be computed explicitly), so $g=1$ implies $d=3$.


1

Let $f(x,y) = x^3+ax+b - y^2$. Then our equation is $f(x,y) = 0$. At a point $(p,q)$ on this curve, the tangent line is: $\frac{\partial f}{\partial x}(p,q) (x-p) + \frac{\partial f}{\partial y}(p,q) (y-q) = 0$ Parametrize this with $x(t) = p - t\frac{\partial f}{\partial y}(p,q)$ and $y(t) = q + t\frac{\partial f}{\partial x}(p,q)$. Then we are to show ...


2

Careful! The elliptic curve is defined over a field $\mathbb F$, but is itself a group $G$. These are two completely different structures. If $(x,y)$ is a point on the elliptic curve (and therefore an element of $G$), then $x$ and $y$ are elements of the field $\mathbb F$. The "characteristic of the group" makes no sense; presumably you mean the ...


4

You should begin by carefully understanding, in the case that $E = \mathbb C/\Lambda$ is an elliptic curve over the complex numbers, the canonical isomorphism between the $\ell$-adic Tate module and $\mathbb Z_{\ell} \otimes_{\mathbb Z} \Lambda$ (or, what is the same, the inverse limit $\varprojlim_n \Lambda/\ell^n\Lambda$). Once you've understood that, ...


5

Let's start over the complex numbers, so that $E$ is an elliptic curve over $\mathbb C$. Then the natural map $\mathrm{End}(E) \to \mathrm{End}(H^1(X,\mathbb C))$ is injective. Suppose now that you have an elliptic curve over $\mathbb F_p$. What could the analogue of this statement be? If $E$ is an elliptic curve over an arbitrary field $k$, and $\ell$ ...


5

There are many, many ways in which the knowledge of the various Tate modules gives you information about the elliptic curve. For example: Theorem(Neron-Ogg-Shafarevich): Let $E$ be an elliptic curve over a number field $K$. Let $p$ be a prime of $\mathbb{Q}$. Then, $E$ has good reduction at $\mathfrak{p}$, for $p\nmid\mathfrak{p}$, if and only if the ...


1

You are asking why $f=x^2 - 2x_1x+x_1^2$ divides the difference of the right and left of your equation. Equivalently, why $\,f$ divides $2my_1(x-x_1) +y_1^2-x^3-Ax-B$. But $2my_1=3x_1^2+A$, and $y_1^2 = x_1^3+Ax_1+B$. I think you can do the rest.


0

The long Weierstrass form $y^2+xy+y=x^3$ is transformed into the short Weierstrass form, namely to $$ y^2=x^3+621x+9774. $$ The formulas for the necessary substitutions are given here.


7

Definition 2 is wrong. Any curve has dimension 1. Strictly speaking, definition 1 is also wrong. You need also a base point on the curve to have an elliptic curve; otherwise you just have a genus $1$ curve. Alternatively, an elliptic curve is an abelian variety of dimension $1$, i.e. an abelian variety which is also a curve. Such a curve necessarily has ...


3

We have: $$\tag 1 y^2 = x^3 + ax + b$$ To add a point, we have: $\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}$ $v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}x$ $x_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}$ $y_3 = -(\lambda x + v)$ (Calculate this out) Next, we can substitute $(1)$ ...


0

Now try a substitution $x=u+k$ with $k$ a constant, multiply out and choose $k$ so there is no $u^2$ term. With $k=-1/12$ the cubic in $x$ then in terms of $u$ is $$u^3+\frac{23}{48}u+\frac{181}{864},$$ if my calculations are OK. Anyway that's the idea to finish from where you are.


2

Do you know the group law (of an elliptic curve)? Assuming we're on a field of characteristic $\;\neq 2,3\;$ ,we can define: $$t:=\frac{3\cdot 2^2}{2\cdot 1}=6$$ $$x_1:=t^2-2\cdot2=32\\y_1:=1+6(32-2)=181$$ and we get a new solution $\;(32\,,\,\,181)\;$


2

One approach (see Example $15.5$ at Elliptic Curve Cryptography), is the following. Take $x = 0 \ldots 16$ and for each $x$ solve: $$y^2 = x^3 + 2x + 2 \pmod {17}$$ This yields the following sets of points: $x = 0, 7, 10, y = 6, 11$ $x = 3, 5, 9, y = 1, 16$ $x = 6, y = 3, 14$ $x = 13, y = 7, 10$ $x = 16, y = 4, 13$ You can use Hasse's Theorem to ...


2

You can find quite comprehensive list of books that deal with elliptic curves on the official site of Andrej Dujella: http://web.math.pmf.unizg.hr/~duje/literatura.html#EK


3

In addition to Dietrich's complete answer, using Sage you can compute $E(\mathbb F_{27})$: E = EllipticCurve(GF(27,'a'),[0,0,1,2,0]); E.abelian_group() Additive abelian group isomorphic to Z/14 + Z/2 embedded in Abelian group of points on Elliptic Curve defined by y^2 + y = x^3 + 2*x over Finite Field in a of size 3^3 to find that $E(\mathbb F_{27}) ...


6

The Hasse bound holds for all elliptic curves over finite fields. Since the discriminant is $\Delta=-539\equiv 1(3^k)$, we really have an elliptic curve over $\mathbb{F}_{3^k}$ for all $k\ge 1$. I count $\# E(\mathbb{F}_3)=7$, and this satisfies the Hasse bound, because $3\le 2\sqrt{3}$. The computation for $\mathbb{F}_9$ and $\mathbb{F}_{27}$ is probably ...



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