All Questions
7 questions
1
vote
0
answers
122
views
Rational points on an elliptic curve the denominator of x is a square
Let $f \in \mathbb Q[x]$ be a squarefree cubic polynomial with nonzero constant coefficient and consider the elliptic curve $E : y^2 = f(x)$.
Define $E(\mathbb Q)' \subseteq E(\mathbb Q)$ as
$$\left \...
- 2,224
2
votes
0
answers
198
views
Finding rational points via birational map
Let $C$ be an affine curve given by $p_C(x,y)=0$ where $$ p_C=2x^3y + 2xy^3 +x^3 + y^3 + 5x^2y + 5xy^2 + 2x^2 + 2y^2 + 2x^2y^2 + 2xy $$
and let $\overline{C}$ denote the projective closure of $C$. For ...
- 3,065
5
votes
1
answer
717
views
rational points of a hyperelliptic curve of genus 3
Let $K=\mathbb{Q}(\sqrt{-1}).$ I have the following hyperelliptic curve of genus 3:
$$ C : y^2 = (x^2-x+1)(x^6+x^5-6x^4 -3x^3+14x^2-7x+1) $$
I want to find $C(K)$. My first attempt was to compute the ...
- 11.3k
3
votes
1
answer
395
views
Finding $K$-rational points on $X_0(35)$
Let $K=\mathbb{Q}(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(35)$?
Recall that $X_0(35)$ is a hyperelliptic curve of genus 3 and has the simplified affine model:
\...
- 11.3k
7
votes
1
answer
218
views
Subfields of Hilbertian fields
This question is about the Remark on the top of page 22 of Serre's Topics in Galois Theory, available here :
http://www.ms.uky.edu/~sohum/ma561/notes/workspace/books/serre_galois_theory.pdf
My ...
- 19.6k
11
votes
5
answers
4k
views
How much do I need to learn algebraic geometry to understand arithmetics over number fields
I am at the stage of learning. Mostly, I am attracted by algebraic number theory. Roughly speaking, I am interested in the rational points of algebraic varieties. I am little bit afraid to start to ...
- 1,499
28
votes
6
answers
2k
views
Patterns among integer-distance points
Mark each point of $\mathbb{N}^2$ ($\mathbb{N}$ the natural numbers) if its
Euclidean distance from the origin is an integer. One obtains a plot like this, symmetric about the $45^\circ$ diagonal.
...
CommunityBot
- 1