bio | website | |
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location | Nottingham, UK | |
age | ||
visits | member for | 4 years, 10 months |
seen | 3 hours ago | |
stats | profile views | 1,388 |
Jan 21 |
awarded | Nice Answer |
Dec 24 |
comment |
$\zeta$ function for ambiguous class group
It seems to me this is just the Riemann zeta function at $[K:\mathbb{Q}]s$ with finitely many factors modified, no ? |
Dec 21 |
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Can the pre-image of the real points in the complex upper-half plane of a modular elliptic curve under the modular parametrization be identified?
Coordinates on the affine model over $\mathbb{Q}$ are $j(\tau)$ and $j(N\tau)$. These are power-series in $q=e^{2\pi i \tau}$ with real coefficients. Now $\bar{q} = e^{2\pi i(-\bar\tau)}$ confirms that complex conjugation is the reflection on the imaginary axis. |
Dec 21 |
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Can the pre-image of the real points in the complex upper-half plane of a modular elliptic curve under the modular parametrization be identified?
I guess you answer what the preimage of $X_0(N)(\mathbb{R})$ in the upper half plane looks like. Surely the imaginary axis belongs to it as complex conjugation on the curve comes from the reflection through the imaginary axis $\tau\mapsto -\bar\tau$. |
Nov 25 |
comment |
Fermat's last theorem over larger fields
Could a Russian speaker add a summary of the results in this paper to this answer, please ? |
Nov 25 |
answered | On a minimal algebraic number field which satisfies the principal ideal theorem |
Nov 22 |
awarded | Nice Question |
Nov 20 |
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Is elliptic curve point division defined over the field of real numbers?
If you are looking for "procedures" to compute it for general fields, then look at sage's function P.division_points(n). It starts by finding roots of the division polynomial in the field. |
Nov 20 |
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How to explicitly compute lifting of points from an elliptic curve to a modular curve?
If increasing the conductor is computationally hard, maybe you could enlarge the base field to get to rank 2 examples. Find a small conductor curve with large rank over a small quadratic field. |
Nov 17 |
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Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
The Galois group of the equation for $p=3/5$ and $n=7$ is $S_7$, which is not soluble. So don't expect a solution by radicals for $n=7$. |
Nov 12 |
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For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
Yes that is possible for $p=5$. It turns out to be equivalent to be a quadratic twist by 5 of a curve with a rational 5-torsion point. |
Nov 11 |
revised |
For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
added 108 characters in body |
Nov 11 |
comment |
For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
You are right. More generally, If $G$ is the group of all matrices of the form $(\begin{smallmatrix} 1 & * \\ 0 & * \end{smallmatrix})$, then $H^1(G,E[p])= \mathbb{F}_p$ if $p =3$ and it is zero if $p>3$. Your example, and many other curves with a 3-torsion point rational over $\mathbb{Q}$, has indeed this group. |
Nov 11 |
revised |
For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
deleted 4 characters in body |
Nov 11 |
answered | For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial? |
Nov 11 |
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For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
I don't know of an example with $p>2$. Maybe the group theorists can tell us examples of $H^1(H, V)\neq 0$ for subgroups $H$ of $\operatorname{SL}_2(\mathbb{F}_p)$ with $V$ the 2-dimensional vector space over $\mathbb{F}_p$. Say with $p=3$ or $p=5$. |
Nov 11 |
comment |
For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
Over $\mathbb{Q}$, the determinant $G\to \mathbb{F}_p^{\times}$ must be surjective. So your $G$ won't appear as a group for an elliptic curve over $\mathbb{Q}$. |
Nov 9 |
comment |
A number theoretic identity
it is true at least up to $2n+1<1000$. |
Nov 9 |
awarded | Enlightened |
Nov 9 |
awarded | Nice Answer |