New answers tagged

1

$P$ 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 ...


1

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 ...


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

Not a coincidence, definitely. $70$ is a Pell number, so $2\cdot 70^2+1=99^2$, and some solutions of $$ 1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = q^2 $$ can be derived by imposing that both $2n+1$ and $\frac{n(n+1)}{6}$ are squares: that leads to a Pell equation.


5

There is only one solution(except $1=1$). There is a proof in Mordell's book on Diophantine Equations. The problem is attributed to Lucas: with N > 1 is when N = 24 and M = 70. This is known as the cannonball problem, since it can be visualized as the problem of taking a square arrangement of cannonballs on the ground and building a square pyramid ...


2

The notation $\tfrac{1}{p}P$ is a little bit ambiguous. I understand (b) as saying that there is a point $Q$ such that $pQ=P$ and $Q$ is defined over $K_{\lambda}=L_w$ for any $w \mid \lambda$. This is clearly equivalent to saying that $L_w(Q)=L_w$, i.e. that all $w$ split in $L(Q)$. Because all the $p$-torsion points are defined over $L$, all the points $Q'...


1

Well, use part a of the question, where you have proved that $(\phi_q-p^m)^2=0$. Suppose that $\phi_q-p^m\neq0$. Then $\phi_q-p^m$ is surjective on $E(\bar{\mathbf F}_q)$ by Theorem 2.22. Then $(\phi_q-p^m)^2=(\phi_q-p^m)\circ(\phi_q-p^m)$ is surjective too. But this endomorphism was shown to be $0$ on $E(\bar{\mathbf F}_q)$. Hence $E(\bar{\mathbf F}_q)=\{0\}...


0

As we are in the case of algebraically closed fields, statements about $E(\bar K)$ can be proved using (algebraic) geometry. The image of $\alpha$ (when it is non-constant) would be an irreducible subvariety of the 1-dimensional $E(\bar K)$. The theorem is: In an $n$-dimensional irreducible variety any proper subvariety is strictly lower dimensional. (This ...


2

No, such $n$ does not exist in general. Take for instance $E: y^2=x^3+x+2$ defined over $\mathbb{F}_5$. Then, $E(\mathbb{F}_5)$ has four points $$(0 : 1 : 0), (1 : 2 : 1), (1 : 3 : 1), (4 : 0 : 1)$$ and the group $E(\mathbb{F}_5)$ is cyclic of order $4$. Now take $Q=(1:2:1)$, which is a point of order $4$, and $P=2Q=(4 : 0 : 1)$. Then, there is no $n$ such ...


1

I had the same question. I've solved the problem partially, so I share my idea. I extend the set $S$ to $S' = S \cup \{v \in M_K : E' \mbox{ has bad reduction at } v\}$. I use the notation same as Silverman's AEC. Let $v \in S'$, and consider the localization at $v$. A commutative diagram of exact sequences of $G_{\bar{k_v}/k_v} = G_v/I_v$-module \begin{...


0

As Nefertiti pointed out, since $i$ is a unit in whatever ring it sits in, it’s less useful to ask about $i$-torsion points. But let’s take the curve $E:\,y^2=x^3-x$, which does have an automorphism $[i]$, namely $[i](x,y)=(-x,-iy)$. Then you might ask for $E[1+i]$, and that’s just the set of points $P$ on $E$ for which $P+[i](P)=\Bbb O$


0

I've straighten out this problem. I appreciate Ferra's informative comments. Let $f \in Hom(G_{\bar{K}/K},M;S)$ and $K_f$ the fixed field of $\ker(f)$. Then $K_f$ is an abelian extension of $K$ having exponent $m$ that is unramified outside of $S$. $\because$ $K_f$ is abelian because $M$ is abelian, $K_f$ is unramified outside of $S$ because $f$ is ...



Top 50 recent answers are included