Tag Info

New answers tagged

0

@Zilin J.:I have a method that sends all binomial addition $a+b=c$ (so, for example, $a^4+b^4=c^4$) to an elliptical curve, noted $V_A$, of the shape $$X^3+Y^3=AZ^3$$, where $A$ is a cube-free number, and this curve is birational equivalent to a Wierstrass one, the curve $$y^2z=x^3-432A^2z^2$$ (one can obviously work with affine curve making $z=1$), via ...


0

Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write $$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag 1$$ Next, taking the logarithmic derivative of $(1)$ yields $$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$ And then, taking the derivative of ...


1

Another way (using properties of polygamma): $$A\equiv\sum_{m=0}^\infty\frac{1}{(z/\pi+m)^2}=\Psi^{(1)}(1,\frac{z}{\pi})$$ $$B\equiv\sum_{m=-\infty}^0\frac{1}{(z/\pi+m)^2}=\sum_{m=0}^\infty\frac{1}{(-z/\pi+m)^2}=\Psi^{(1)}(1,-\frac{z}{\pi})$$$$=-\Psi^{(1)}(1,\frac{z}{\pi})+\frac{\pi^2}{z^2}+\pi^2(1+\cot(z)^2)$$ Then ...


5

Show that $f(z)=\sum_{m=-\infty}^\infty (z+\pi m)^{-2} -(\sin z)^ {-2}$ is a bounded holomorphic function in $\Bbb C$, whence it is constant. Since it is $0$ in one point (for example at $i\infty$), it is identically $0$. To bound $f$ consider vertical strips (motivated by the series and the periodicity of the sine. To show it is holomorphic note the poles ...


0

I don't have a solution, but I have a simplification. I'd start with: $$\frac{d}{dx} -\cot(z) = \sin^{-2}z$$ $$\int \frac{1}{(z + \pi m)^2}dz = -\frac{1}{z + \pi m}$$ So then you would just need to prove that $\cot(z)$ is equal to $\sum_m \frac{1}{1+m\pi}$.


4

Hint: Differentiate twice the natural logarithm of Euler's infinite product expression for the sine function.


0

Am not sure what you mean by discreet parameters, degree and rank? That will not be enough. As you said, take $C$ to be a smooth cubic curve. Then $\mathcal{O}_C$ and a line bundle $L$ of degree zero but $L\neq \mathcal{O}_C$ have the same discreet parameters. But ...


1

I hope you can find this useful. All elliptic curve $C(\mathbb {Q})$ is canonically write as $y^2=4x^3-g_2x-g_3$... (1) in which the three roots of the right side are distinct; by a birational transformation $(x,y)\to (\frac x4, \frac y4)$ you have $y^2=x^3-h_2x-h_3$... (2) where $h_2$ and $h_3$ can be supposed rational integers. Being $e_1, e_2, e_3$ ...


0

The proof of that result, usually called (an explicit version of) the Weak Mordell-Weil theorem, can be found in Silverman's Arithmetic of Elliptic Curves book. The proof uses Galois cohomology and some minor arithmetic that can be followed with the knowledge of a few of the main theorems of global class field theory.


4

Try to take mod $3 \rightarrow y^3 \equiv 1 \pmod {3} \rightarrow y \equiv 1 \pmod {3} \implies y^3 \equiv 1 \pmod {9}$ Therefore $3x^2 + 3x + 7 \equiv 1 \pmod {9} \rightarrow 3x^2 + 3x \equiv 3 \pmod {9} \rightarrow x^2 + x \equiv 1 \pmod {3}$ This is impossible. If $x \equiv 1 \pmod {3}, x^2 + x \equiv 2 \pmod {3}$ If $x \equiv 2 \pmod{3}, x^2 + x ...


1

David Eisenbud, Mark Green and Joe Harris: Cayley-Bacharach theorems and conjectures, Bulletin American Math. Society 33 (1996) 295-324.


1

Comments under the question suggest that possibly the difficulty was in doing arithmetic modulo a prime. How does one find $-23/4$ in $\mathbb F_5$? $-23$ reduces to $2$, and to divide by $4$ you multiply by the mod $5$ multiplicative inverse of $4$, which you need to find. Trial and error will find the multiplicative inverse when the modulus is $5$, but ...


1

Claim: Let $E: y^2=x^3+D$ and $p>3$ be a prime. Then, there is no point of order $p$ in $E(\mathbb{Q})$. Here are some hints. Let $p>3$ be a prime as in the statement of the claim: If $q$ is a prime such that $q\equiv 2 \bmod 3$, and $q$ does not divide $6D$, then $E(\mathbb{F}_q)=q+1$. A prime $q$ that does not divide $6D$ is a prime of good ...


2

The short answer is given by the Grothendieck-Lefschetz trace formula $$ \#X(\mathbf F_p) = \sum (-1)^i \mathrm{Tr}(\mathrm{Frob}_p \mid H^i_c(X;\mathbf Q_\ell)).$$ For $X$ a compact modular curve associated with a congruence subgroup $\Gamma$, this means that you need to understand the Galois representation $H^1(X,\mathbf Q_\ell)$, as the trace on $H^0$ and ...


3

In the classical theory, you see that the set-theoretic points of the complex manifolds $Y(N)$, $Y_1(N)$, etc., are in bijection with certain complex tori plus some additional data (level structure) (up to isomorphism). But they are in fact moduli spaces in the more precise sense (involving Yoneda's lemma) for complex analytic elliptic curves over more ...


1

As can be easily seen, there exist pairs of curves over $\mathbb{F}_q$ with the same trace of Frobenius, but whose Frobenius actions on $E[\ell]$ are different, more or less because the $\ell$-torsion lies in different extensions of $\mathbb{F}_q$. No, this can not happen. If two curves have the same trace of Frobenius, then, by Tate's isogeny theorem, ...


1

The simplest way is to use existing methods in computer algebra systems, e.g. if you use the online Magma calculator here there are now awfully sophisticated algorithms there for this sort of thing. To learn more you could read the relevant section in the Magma handbook here In the case of the first of your curves, if I put in the following ...


1

Consider the equation you wrote down: $$\frac{z}{y} = \left(\frac{x}{y}\right)^3 + a\frac{x}{y}\left(\frac{z}{y}\right)^2+b\left(\frac{z}{y}\right)^3,$$ and for simplicity give names $X=z/y$ and $Z=z/y$, so you have: $$Z = X^3+aXZ^2+bZ^3.$$ You are trying to find the $p$-adic valuation of $Z$, i.e., $\nu_p(Z)$, knowing that $\nu_p(X)\geq n$. You need to use ...


1

Yes, there are methods to calculate points on elliptic curves. There are books dedicated to this topic... I'd recommend Silverman and Tate's "Rational Points on Elliptic Curves", for instance.


1

If you take a non branched point $z$, you can find a global section that vanishes simply at $z$ but will also vanish at another point $z' \neq z$ (because $\phi^{-1}(\phi(z)) = \{z,z'\}$). Then $2p \sim z + z'$. There is no absurdity there.


2

Here is an alternative approach. Suppose $E(\mathbb{Q})$ contains a point $P$ of order $4$. Since there is a single point of order $2$ over $\mathbb{Q}$, namely $Q=(0,0)$, we must have $2P=Q=(0,0)$. Now we can use the formulas for multiplication by $2$ (Silverman's "Arithmetic of Elliptic Curves", page 54): $$x([2]P) = ...


1

I'm pretty much ignorant about the higher powered theory of elliptic curves, but given that you know $|E(\Bbb{Q})_{tors}|$ to be a factor of $4$, the problem is reduced to a simple calculation. We have $$E(\Bbb{C})[2]=\{P_\infty,(0,0),(i,0),(-i,0)\}.$$ So if your torsion group has order four, it cannot be the Klein 4-group. Consequently it has to be cyclic ...


0

For abelian varieties, there is no a priori defined Weil pairing of $A[n]$ with itself. Rather, there is a dual abelian variety $\widehat{A},$ and a perfect pairing $A[n] \times \widehat{A}[n] \to \mu_n.$ A polarization on $A$ is an isogeny $A \to \widehat{A}$, satisfying some additional conditions. (See wikipedia, or these notes of Brian Conrad.) ...


1

Let $A$, $A'$ be abelian varieties over $\mathbb{R}$, and let $f:A\to A'$ be a morphism. If $f(0_A)\neq 0_{A'}$, then we define $\widehat{f}:A\to A'$ as $\tau_P\circ f'$ where $\tau_P(Q)=Q-P$, and $P=f(0_A)$. Then, $\widehat{f}$ is a morphism and $\widehat{f}(0_A)=0_{A'}$. In addition, we require here that $f$ is surjective with finite fibres ...


2

Note that if you change variables $x=u-v$, $y=v$, then you obtain $$(u-v)^3+3(u-v)+v^3+3v=n$$ which yields $$u^3-3u^2v+3uv^2+3u=n.$$ If there is a solution with $x,y\in\mathbb{Z}_3$ to the original equation, then there is a solution with $u=x+y$, and $v=y\in\mathbb{Z}_3$. Now, if $3|n$, then $3|u^3$ and therefore $3|u$. However, the equation can also be ...


2

Magma says that $\# H(\mathbb{F}_{103})=104$ and $\# J(\mathbb{F}_{103})=10610$. Here is the code I used: P<x> := PolynomialRing(GF(103)); C := HyperellipticCurve(x^5+1); #C; J:=Jacobian(C); #J;


2

My guess is that the answer to your question is "no", i.e., there is no elliptic curve without CM with the specified property on the discriminant, and I was very close to prove it... Hopefully you can finish it up somehow. However, there is another CM curve with the same property, namely $E: y^2=x^3-1/16x$! Let $E: y^2=f(x)=x^3+Ax+B$, for some ...


1

Let $\Lambda$ be a lattice, generated by $w_1$ and $w_2$. Then, $$\Lambda = \{aw_1+bw_2 : a,b\in\mathbb{Z}\}.$$ If $\alpha\in\Lambda$ there is no reason, in general, to expect that $\alpha\Lambda \subseteq \Lambda$. In other words, $\alpha w_1$ and/or $\alpha w_2$ may not be elements of $\Lambda$. Since $\mathbb{C}/\Lambda \cong \mathbb{C}/\lambda\Lambda$, ...



Top 50 recent answers are included