# Tag Info

## New answers tagged elliptic-curves

2

There is Elliptic Curves: Number Theory and Criptography by Lawrence, where on chapter 8 the author treats this subject.

2

There are fairly detailed notes by Douglas Ulmer available online. They cover the important case of function fields over finite fields. I couldn't tell if this is the case that interest you.

0

This might not be worth an answer but since you haven't gotten one yet, here a few references. One can get an elliptic surface with multiple fibers from a surface without multiple fibers by a process called "logarithmic transformation". It changes a single fiber into a fiber of the same type but it will be a multiple fiber. References for this process are ...

1

You are almost right: since $\phi\widehat{\phi}=2$ by definition of dual and $\phi^2=-2$, you have $\widehat{\phi}=-\phi$. Now you should get the right result in (b), your argument is correct!

5

Take two linearly independent rational differential forms $\omega_1, \omega_2\in L(K)=\Omega^1(C)$ with divisors $\operatorname {div} (\omega_1)=P_1+P'_1$ and $\operatorname {div} (\omega_2)=P_2+P'_2$ . The rational function $f=\frac {\omega_1}{\omega_2}\in \operatorname {Rat}(C)$ then has a divisor of the required form $$\operatorname ... 3 The question reduces to show that if \mathfrak a is a radical ideal, then its extension is also radical. Equivalently, if K[X]/\mathfrak a is reduced, then K[X]/\mathfrak a\otimes_K\overline K is also reduced. If K is a perfect field (and this is an assumption on page 1 of Silverman's book), then this holds as it is pointed out in this answer and ... 1 If E is an elliptic curve over \mathbb{F}_q, then the Weil pairing E(\mathbb{F}_q)\times E(\mathbb{F}_q)\rightarrow \mathbb{F}_q^* shows that there exist positive integers m_1,m_2 such that$$ E(\mathbb{F}_q)\cong \mathbb{Z}/m_1 \mathbb{Z} \times \mathbb{Z}/m_2 \mathbb{Z}, $$with m_1\mid gcd(m_2,q-1), see Chapter III, Corollary 8.1.1 in ... 0 After thinking a bit, I think I've solved this: set \Delta(f) = [\deg f]:E\rightarrow E. Then it's easy to see that \zeta: = f\circ \hat{\phi} - \hat{\psi}\circ \Delta(f) is a morphism E\rightarrow E'' whose image is contained in the kernel of \psi. Now \psi is a finite étale morphism so \ker\psi is a finite subgroup of E'' of order \deg\psi. ... 1 If you start with x = X, y = Y, i.e. simply renaming the variables, then trivially$$ X^3+Y^3−X^2+Y−1 = 0 $$Now you multiply each term by either Z^3, Z^2, Z^1 or Z^0 = 1 so that the sum of the exponents of each term is 3. So X^3 and Y^3 gets multiplied by 1, X^2 gets multiplied by Z and so on. You end up with$$ X^3+Y^3−ZX^2+Z^2Y−Z^3 ...

1

I found a reference which has the required result with proof (which I've copied here). I claim no originality. Lemma (38.9.9): Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d]:A\to A$ is étale if and only if $d$ is invertible in $k$. Proof: Observe that $[d](x+y)=[d](x)+[d](y)$. Since translation by a point is an automorphism of $A$, ...

3

The equations which define $P_3$ have coefficients in $k$, so $Gal(\bar k/k)$ preserves the locus of these equations i.e it fixes the coordinates of $P_3$ since it is the unique point of this locus.

0

After some insight courtesy of Achille Hui, I was able to answer my own question. It is well known (see also this) that $x^3+y^3=N$ is birationally equivalent to the elliptic curve $u^3-432N^2=v^2$ using the transformation $x=\frac{36N+v}{6u}$, $y=\frac{36N-v}{6u}$. In this comment, Hui suggested the command, Q := PolynomialRing(Rationals()); E00 := ...

0

HINT: Two parabolic cylinders cut by a transverse plane. The projections on $x,y,z$ planes are: $$\begin{cases} x^2-2 z+1=0 \\ x^2 - 2 y -1 = 0 \\ y-z +1 =0 \end{cases}$$ The constant has to be set to zero in the last equation in order that it become the generator of a cone passing through origin: $$y=z$$

3

The discriminant of a polynomial $f$ of degree $n$ and leading coefficient $a_n$ is $$\operatorname{disc}f=\frac{(-1)^{\tfrac{n(n-1)}2}}{a_n}\operatorname{Res}(f,f').$$ ($\operatorname{Res}(f,g)$ denotes the resultant of the polynomials $f$ and $g$). Thus it is $0$ if and only if $f$ has a multiple root. For any polynomials $f,g$, it can be shown there ...

1

LONG SOLUTION. The equations defining the parabola can be rewritten as parametric equations as follows: $$\cases{ x=t \cr \displaystyle y={1\over2}t^2-{1\over2}\cr \displaystyle z={1\over2}t^2+{1\over2}\cr }$$ The parametric equation of the line passing through the origin and a point $P(t)=(t,\ t^2/2-{1/2},\ t^2/2+{1/2})$ of the parabola is just ...

1

The problem is reduced to the Euclidean algorithm in $\mathbb C[x]$. This technique works in the case of a nonsingular plane curve given by an equation of the form $y^n = f(x)$, and in particular for an elliptic curve in Weierstrass form. We wish to write $$1 = P\cdot f' + Q\cdot ny^{n-1}$$ for some $P,Q \in \mathbb C[x,y]/(y^n-f)$. As in msteve's answer, ...

1

As is suggested in the question, in order to find a free generator of $\Omega_A^1$, it suffices to find $P,Q \in A$ satisfying $P(3x^2-1) + Qy = 1$. Indeed, a generator is then given by $2P dx + Qdy$. The solution below finds such $P$ and $Q$, but it is rather ad hoc and I would be very interested in seeing a systematic method, if one exists. Let's first ...

3

HINT: You can parametrise in several ways. How about $$x=c_1t^2$$ and $$y=c_2t^3$$ where $c_1^3=c_2^2$ ?

0

2) The passage you need is simply de-homogeneizing with respect to $Z$, which in algebraic terms mean to set $Z=1$. The geometric interpretation of this is choosing as the line at infinity the one given by $Z=0$, and intersecting your projective curve with the copy of the affine plane given by $\{(X:Y:Z)\in \mathbb P^2\colon Z\neq 0\}$. Notice that here the ...

3

It is very easy to find an example. The curve $E\colon y^2=x^3+x+2$ has $j$-invariant $432/7$, which is not an alegbraic integer. Thus $E$ has no CM. In general, if $\Lambda$ is of the form $\mathbb Z+\mathbb Z\tau$ for some $\tau\in \mathbb C$, then $\mathbb C/\Lambda$ has CM if and only if $\tau$ is imaginary quadratic. This makes it easy to construct ...

2

Yes what you are doing seems correct. Take $x=t$ where $t \in \Bbb Q$. Then $y=x^2=t^2 \Rightarrow y \in \Bbb Q$. Also it is not true that '$x^2$ is rational $\Rightarrow x$ is rational'.(Take $x^2=2$, then $x=\pm \sqrt 2 \notin \Bbb Q$) Hence $\{(t,t^2):t \in \Bbb Q\}$ is the set of all rational points on the parabola $y=x^2$.

4

One way to think about this is via the modular curves parametrizing elliptic curves $E$ with either $E[3]$ or $\Delta^{1/3}$ rational. Note that $\Delta^{1/3}$ is rational iff $j^{1/3}$ is rational, because $j = E_4^3 / \Delta$. Assume for simplicity that $K$ contains the cube roots of unity (because $K(E[3])$ contains them in any case thanks to the Weil ...

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