Hot answers tagged

4

A morphism $\phi: \mathbb{P}^m \rightarrow \mathbb{P}^n$ can be given by $\phi =(F_0: \dotsc : F_n)$, where the $F_i$ are homogeneous polynomials of same degree in $m$ variables (See Remark 3.2 on the same chapter of Silverman). The only problem is that the $F_i$ might have common zeros in $\mathbb{P}^m$. Say $P$ is one such point, then $\phi(P) = (0: \...


4

The only integral solutions to your first problem are $(3, \pm 5)$. The general class of equations are known as Mordell's equation. A fairly elaborate discussion and case by case analysis is provided here.


3

Part 2 is easy to see as his algorithm allow us to take square root of "small numbers" modulo p(As the complexity depends is = $\sqrt{|x|} \log^{9}p$). So one can evaluate $\zeta_2,\zeta_4,\zeta_8 $ as all of them involve taking square roots of numbers of small magnitude like $\sqrt{-1},\sqrt{2}$ and as Jyrki mention if $p \neq 1 \mod 16$ then one of them ...


3

The cubic \begin{equation*} u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0 \end{equation*} can be shown to be equivalent to the elliptic curve \begin{equation} y^2=x^3+1296n^2(n^2+n+1)^2 \end{equation} by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient MS-DOS version of Derive. The curve has ...


3

As requested in the comments, I will upgrade my comment to an answer. Keith Conrad's excellent article settles the question. On page $10$, it is noted that for the values $k = -5, -6, 6, 7, -24, 45$, the Mordell curve $y^{2} = x^{3}+k$ has exactly one rational point. Proof of this fact is not given, so the references may be helpful in this regard. He does ...


3

Such a map is called an isogeny, and isogenies over finite fields preserve the number of points (in fact, two elliptic curves over a finite field are isogenous iff they have the same number of points). On the function field side, this corresponds to an embedding of the function field of the original curve into the second one. By the existence of the dual ...


3

The largest value for $f(k)$ that I know of is an example due to Noam Elkies with $k = 509142596247656696242225$, where there are (at least) $125$ pairs of solutions (so $f(k)=250$ in your notation). If ranks of elliptic curves over the rationals are absolutely bounded, then so is $f(k)$ (as long as one restricts to $6$th power free values of $k$ to avoid ...


3

A handful for $n$ odd 13 a: 230153 b: 12792 c: 283945 d: 284233 u: 507 v: 164 15 a: 32625 b: 3472 c: 61455 d: 61553 u: 56 v: 31 23 a: 523367 b: 57072 c: 1413145 d: 1414297 u: 984 v: 667 27 a: 206703 b: 3848 c: 231345 d: 231377 u: 156 v: 37 89 a: 11534489 b: 700920 c: 63439289 d: 63443161 u: 649 v: 540 105 a:...


2

An Attempt You have $a^2+\left(n^2+1\right)b^2=d^2$. All integral solutions $(a,b,d)$ to this equation satisfies (1) $|a|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right)$ for some $r\in\mathbb{Q}$. Now, if $b^2+c^2=d^2$, then (...


2

$P$ isn't an $\ell$th root of unity, but $\tau_\ell(P, P)$ is. The Tate-Lichtenbaum pairing maps into $\mathbf{F}_q^\times/ (\mathbf{F}_q^\times)^\ell$. Since $\ell \mid q - 1$, this group is isomorphic to the group $\mu_\ell$ of $\ell$th roots of unity. Since $\ell$ is prime all nontrivial elements of $\mu_\ell$ are generators, or in other words, as long as ...


2

This is not quite complete in details, and there is some hand waving, some of which I can't justify (see added comment/note), but I obviously hope that all is correct, and that it in the meantime adds something... Setting up (a lot of, and some of it rival) notation: In the following $k$ is a number field. Suppose $V$ is the $A$-torsor corresponding to ...


2

Here is the general picture. Let $f \in \mathcal{M}_k(N,\chi)$ be a modular form of weight $k$, level $N$, and nebentypus $\chi$. Suppose that $f$ has the Fourier expansion \[f(z) = \sum_{m = 0}^{\infty} a_f(m) e(mz).\] Then by replacing $z$ with $nz$, we see that \[f(nz) = \sum_{m = 0}^{\infty} a_f(m) e(mnz) = \sum_{m = 0}^{\infty} a_{f_n}(m) e(mz)\] with \[...


2

Fact : $\mathbb{Z}[\sqrt{-2}]$ is Unique factorization domain. lemma : in every UFD, if the product of two numbers, which are relatively prime is a cube, then each of them must be a cube. There is no any solution $$x^3 = (y+\sqrt{-2})\times(y-\sqrt{-2})$$ the greatest common divisor of these factors will divide 2-times the $\sqrt{-2}$, which is lead to ...


1

Answering my own question. Turns out the $\alpha$ referred to in the notes is exactly the endomorphism which we are trying to reduce mod $\psi_\ell$, although it isn't made very clear. Explanation: Let $\alpha = (\alpha_x(x), \alpha_y(x) y)$ be an endomorphism in $\text{End}(E)$, where $\alpha_x$ and $\alpha_y$ are rational functions. If the denominator of ...


1

If $W$ contains more than one component, then it is the union of projective subvarieties of codimension 1, which necessarily have non-empty intersection. (This is immediate from Bezout but if you prefer a smaller hammer, also follows from the Krull height theorem.) Then, we need to check that $W$ cannot possibly be smooth at a point lying on two or more of ...


1

The reference of this answer is D. Cox 'Primes of the form $x^2+ny^2$'. The formulation 1 is Theorem 10.14 and Corollary 10.20, and 2 is Theorem 7.7 (ii). Then it is more natural that 1 is inferred from 2, than 2 is inferred from 1. The elements in the set of triples described in 2, can be considered as a set of equivalence classes of quadratic forms $C(D)$...



Only top voted, non community-wiki answers of a minimum length are eligible