18
votes
Accepted
Rational perfect power values of $y(y+1)$
There are no such solutions. Let $x=a/b$ and $y=c/d$ be reduced fractions. Then
$a^n/b^n=(c(c+d))/d^2$ and since both sides are reduced fractions we get that
$a^n=c(c+d)$ and $b^n=d^2$. From the ...
- 4,189
16
votes
rational points of a hyperelliptic curve of genus 3
It turns out that $C(K) = C(\mathbb Q) = \{\infty_+, \infty_-, (0,1), (0,-1), (1,1), (1,-1)\}$.
To see this, consider a point $P \in C(K)$ and write $\bar{P}$ for its image under the nontrivial ...
- 11.3k
14
votes
Accepted
From a physicist: How do I show certain superelliptic curves are also hyperelliptic?
This curve is not hyperelliptic unless $n=2$ or $N=2$.
First, note that it is more convenient to write the curve as
$$ w^n = \frac{ \prod_{\alpha=1}^N (z- u_\alpha )} {\prod_{\alpha=1}^N (z- v_\alpha)}...
- 148k
7
votes
Is it true that every mapping class in $\mathrm{Mod}(\Sigma_3)$ commutes with some hyperelliptic involution?
No. After the Nielsen realization theorem, this follows from the classification of automorphism groups of genus three Riemann surfaces, which can be found in many places. For example, the cyclic group ...
- 40.2k
7
votes
Accepted
If a genus 2 curve has no $k$-rational points, can it have a $k$-rational divisor class of degree 1?
Rummaging a bit through the LMFDB turns up the curve
https://www.lmfdb.org/Genus2Curve/Q/129600/b/129600/1
with equation $y^2 = -(2x^3+3x-2)(2x^3+4x^2+x-2)$
with no rational points (indeed trivial ...
- 79.9k
6
votes
Accepted
Degenerations of hyperelliptic coverings
If $p_1 = p_2 \ne p_3 = p_4 \ne p_5 \ne p_6$ then the normalization of the double cover branched at the divisor $D = \sum_{i=1}^6 p_i$ is a smooth irreducible rational curve. If also $p_5$ and $p_6$ ...
- 39.3k
6
votes
Accepted
Finding the $K=\mathbb{Q}(\sqrt{6})$-rational points on the twist of $X_{0}(26)$
I used Magma to point search on $C/K$ up to a height of $1000$ and it appears that $C(K) = \emptyset$. If that's true, then one can probably use the Mordell-Weil sieve to prove it. Here's a bit more ...
- 20.4k
5
votes
Accepted
Degree of irrationality and hyperelliptic curves
Yes, because then there is a nonconstant map from projective space to the symmetric square of $C$.
- 79.9k
5
votes
Elliptic factors in the Jacobian and zeta function
This is an incomplete answer to the second question (Jackson has indicated the answer to the first question in the comments).
If you ask the same question over a general number field $F$, then you ...
- 2,606
4
votes
Is the Jacobian isogenous over $\mathbb{F}_p$ to the direct product of the elliptic curves?
If $$y^2 = (x^3+b) (x^3-b) = x^6-b^2$$ then $$(y)^2 = (x^2)^3 - (b^2)$$ and $$(y x^{-3} b^{-1} )^2 = (- x^{-2})^3 + (b^{-2}) $$ giving two maps to elliptic curves, but not the elliptic curves you ...
- 148k
4
votes
An isogeny between Jacobians of hyperelliptic curves
In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product
$E_{1,t}\times E_{2,...
- 1,386
4
votes
Accepted
Can I calculate congruent zeta function of given hyperelliptic curve by hand?
Well, I am such an outsider to this field that I hesitate to reveal how fragmentary my knowledge is. So fragmentary, in fact, that I can only tell you the sketchiest facts, without any explanation at ...
- 4,193
4
votes
Accepted
Generalization of torsion points on Jacobian of genus 2 over finite fields (with respect to the theta divisor)
The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C$...
- 984
4
votes
Easiest example where field of definition is not field of moduli
I believe that the other answer to this question is incorrect (I don't have the reputation to comment on it). The curve
$$y^2=(x^2-1)(x^2-2)(x-i)$$
is in fact defined over $\mathbb{R}$. To see this ...
- 41
3
votes
Bounds for the number of points on projective hyperelliptic curves over finite fields
For simplicity, let $q$ be odd. The question is to bound $\mathbb{F}_q$ points on $y^2 = f(x)$ when there are singularities. Write $f(x) = u(x) v(x)^2$ where $u$ is square free. Let $X$ be the affine ...
- 156k
3
votes
Accepted
Degree of morphisms and isogenies
It's easier if we forget about isogenies: $E_1$ and $E_2$ are isomorphic,
and $X_1$ and $X_2$ are isomorphic, so the cover $X_2\to E_2$
induces a cover $X_1 \to E_1$ of the same degree.
To make this ...
- 879
3
votes
Computing the class group of a quadratic function field
The answer to the precise question is yes. See Theorem 1.2 of:
"Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields," https://arxiv.org/pdf/0912.0325.pdf.
...
- 21.3k
3
votes
An isogeny between Jacobians of hyperelliptic curves
The Tate conjecture is known for function fields (Zarhin). To check whether the Tate modules are isomorphic you need to check that the image of Frobenius match for a finite set of places that can be ...
- 30.5k
3
votes
Accepted
What is the quotient $E \!\times\! E^\prime / G$?
I claim that $E\times E'/G$ is the Weil restriction of $E_{\mathbb{F}_{p^2}}$ w.r.t. $\mathbb{F}_{p^2}/\mathbb{F}_{p}$. (I don't know about the product question, or the Jacobian question; the answers ...
- 19.6k
3
votes
Accepted
Intersection number of divisors on abelian surfaces and its invariance under translation by 2-Torsion points
The answer is yes if $D_2$ is an ample curve. More generally, the following result holds:
Two ample line bundles on an abelian variety
are algebraically equivalent if and only if they differ by a ...
- 66.3k
3
votes
Hyperelliptic curves imply FLT-like results
Let $p$ be an odd prime, and assume $ax^p+by^p+cz^p=0$ with $x\ne0$; then
$$Y^2=X^p+a^2(bc)^{p-1}/4$$
has a nontrivial point with
$$X=-bcyz/x^2\quad\,\quad Y=(-bc)^{(p-1)/2}(by^p-cz^p)/(2x^p)\;.$$
- 13.1k
3
votes
Is it true that every mapping class in $\mathrm{Mod}(\Sigma_3)$ commutes with some hyperelliptic involution?
V.I. Arnold proved that the image of the hyperelliptic group in the mod 2 symplectic group (ie, the image of $HMod_g\to Mod_g\to Sp_{2g}(\mathbb Z)\to Sp_{2g}(\mathbb F_2)$) is the symmetric group on ...
- 8,717
2
votes
Hyperelliptic curves imply FLT-like results
A reference for all quotients of the Fermat curve is Lang's book "Introduction to Algebraic and Abelian Functions". There, you'll find your maps (up to twist) and several others.
If you have a map $X ...
- 30.5k
2
votes
Good lecture notes/books on Jacobian of hyperelliptic curve
An elementary introduction to hyperelliptic curves by Menezes, Wu and Zuccherato gives a good elementary overview, especially if you want to understand the arithmetic on Jacobians.
If you would like ...
- 125
2
votes
Accepted
Identifying elements in the kernel of an explicit endomorphism of a Jacobian variety
I would suggest that you work with the Kummer surface $K$ of $J$
instead of using Mumford coordinates. The advantage is that $K$ is
a quartic surface in $\mathbb P^3$; in the case you are
considering ...
- 11.3k
2
votes
Accepted
Hyperelliptic integrals
In McRae' PhD thesis (W. D. McRae, Riemann theta functions on degenerate surfaces, Phd thesis, 1997, pp. 43), the homology basis $\{\tilde a_j, \tilde b_j\}$ is given as follows. Let $\alpha_j$ be a ...
- 89
1
vote
Accepted
Auto-equivalences of non-trivial components of derived category of $X_{18}$
Let me answer the question by myself. After a intensively literature research, I found that the habilitation of Faenzi,Daniele contains everything I need, here is the link http://dfaenzi.perso.math....
- 1,982
1
vote
Hyperelliptic equation on a function field
There are such bounds, although I don't have a reference handy. I'll try to find one later, unless someone else posts one first. Let $f(X)$ be the polynomial on the RHS, and let $D_f\in\mathbb C[z]$ ...
- 47.4k
1
vote
Accepted
Prime divisors on the Jacobian of a genus 2 curve over $\mathbb{F}_q$ under the $n$ map
If $\iota$ denotes the hyperelliptic involution, then the condition $n[P-\infty] = [Q-\infty]$ is equivalent to $nP+\iota(Q)$ linearly equivalent to $(n+1)\infty$. In a few pathological cases, where ...
- 30.5k
Only top scored, non community-wiki answers of a minimum length are eligible