# Tag Info

## New answers tagged elliptic-curves

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$.$\becauseK_f$is abelian because$M$is abelian,$K_f$is unramified outside of$S$because$f\$ is ...

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