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This is a computer assisted answer that shows explicitly the corresponding field extensions of $\Bbb F_{13}$ that contain all $3$-torsion, respectively all $9$-torsion points. (A comment would not fit with all delivered information.) Sometimes it is good to be able to verify and construct explicitly objects in the framework of a given elliptic curve. ...


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It's easier to see this using the short Weierstrass form. Write $C : y^2 = x^3 + ax + b$. Then, by calculus, $$\frac{dy}{dx} = \frac{3x^2+a}{2y}$$ By definition, $$\operatorname{div} dx = \sum_{P \in C} \operatorname{ord}_P(dx)(P)$$ and furthermore, by definition $$\operatorname{ord}_P(dx) = \operatorname{ord}_P\left(\frac{dx}{dt}\right)$$ for any choice of ...


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The Handbook of Elliptic Curves by Ian Connell is a wonderful reference. http://webs.ucm.es/BUCM/mat/doc8354.pdf (...after a quick search.) In section [1.4 Cubic to Weierstrass: Nagell's algorithm] we have the steps of the algorithm that can be applied for any (non-singular) cubic to put it in Weierstraß form. (Characteristic not two, three.) Then ...


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I can answer your very first question on why $\tau_\ell(P,P')$ is an $\ell$th root of unity. If I'm not mistaken, they are assuming in the paper that $E(\mathbb F_q)[\ell]$ (the $\ell$-torsion points in $E(\mathbb F_q)$) is a cyclic group. (In general, if $\ell$ and $q$ are coprime, $E[\ell]\cong (\mathbb Z/\ell\mathbb Z)^2$ has order $\ell^2$, and all its ...


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Whenever you have two identity elements they coincide. Let $e$ and $f$ be two identity elements. Then we get $$e = e \cdot f = f,$$ where the first equality holds as $f$ is an identity element and the second one as $e$ is an identity element. Any yes, for elliptic curves the point at infinity is the identity element.


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I'm going to approach your questions in reverse order. For an abelian variety of dimension $g$, it is well known that the determinant representation of the Tate module is $\chi^g$, i.e. the $g$-th power of the cyclotomic character (the argument I'll give below involves matrices so won't be nice to give in the general case but hopefully it will convince you ...


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I think this is straightforward after noting that for a finite-dimensional vector space $V$, $End(V)$ is the Lie algebra of $Aut(V)$. One has to be a bit careful in the ultrametric case because the exponential map does not converge on the entire Lie algebra, but it does on an open additive subgroup, and its image is an open subgroup (here: of $Aut(V)$); ...


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Finding the preimage of an arbitrary point is equivalent to evaluating the dual isogeny on the point. Your question shows that if you can compute dual isogenies then you can compute preimages. Conversely, if you can evaluate preimages, then you can multiply by the degree of the isogeny to evaluate the dual isogeny. Therefore one should not expect to be able ...


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I will take for granted that $$ E (\mathbb Q)\overset\phi\longrightarrow E'(\mathbb Q)\overset q\longrightarrow \mathbb Q^\times/(\mathbb Q ^\times)^2 $$ is an exact sequence. I will also take the lemma for granted. (a) Then the image in $\mathbb Q^\times/(\mathbb Q ^\times)^2 $ of the positive divisors $1,2,4,5,10,20$ of $20$ has representatives $1,...


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If $D$ has negative degree, then $l(D)=0$. So start an induction on the degree starting with $\deg D=-1$. In general, write $D=D'+(P)$ for some point $P$. It suffices to prove $l(D)\le l(D')+1$. Otherwise one must have $L(D)$ properly containing $l(D')$. Let $f$, $g\in L(D)-L(D')$. Then $f$ and $g$ have the same order at $P$, and so $f-ag$ has degree at one ...


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Let $K = \mathbb{F}_q$. Then the $j$-invariant of a montgomery curve is given by $$j = \frac{256(A^2-3)^3}{A^2 - 4}$$ and therefore the isomorphism class of $M_{A,B}$ over $\mathbb{F}_q^{\text{alg}}$ is completely determined by the number $A^2$. Remark: $q$ is a power of some odd prime, i.e. $256 \neq 0$. One can show that not every element arises as the ...


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Using sage in a simpler manner we obtain the following detailed information: sage: F = GF(5) sage: curves = [] sage: for a, b in cartesian_product([F, F]): ....: try: ....: curves.append(EllipticCurve(F, [a, b])) ....: except: ....: print "Singular curve for a=%s b=%s" % (a, b) ....: Singular curve for a=0 b=0 Singular curve for a=2 ...


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If I understand you correctly you have two groups of order three which means they are isomorphic as there only is one group of order $3$ up to isomorphism.


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$$g_2(w_1,w_2) =60\sum_{u \ne 0 \in\ \Bbb{Z}w_1+\Bbb{Z}w_2} u^{-4}$$ For $a,b,c,d\in \Bbb{Z},ad-bc=\pm 1$ then $$\Bbb{Z}(aw_1+bw_2)+\Bbb{Z}(cw_1+dw_2) = \Bbb{Z}w_1+\Bbb{Z}w_2$$ thus $$g_2(a w_1 +bw_2,cw_1+dw_2)=g_2(w_1 ,w_2)$$ Finally $g_2(r w_1,rw_2) = r^{-4} g_2(w_1 ,w_2)$.


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The upvoted comment above solves it for me. All credit goes to user10354138.


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This is proved in Silverman's book "The Arithmetic of Elliptic Curves" on page 311. The rank $r(E)$ of $E:y^2=x^3+px$ for $p$ prime is given as follows: $$ r(E)= \begin{cases} 0, \text{ if } p\equiv 7,11 \bmod 16 \\ 0 \text{ or }1, \text{ if } p\equiv 3,5,13,15 \bmod 16 \\ 0 \text{ or }1 \text{ or }2, \text{ if } p\equiv 1,9 \bmod 16 \end{cases} $$ You ...


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Let $X$ a compact Riemann surface and $\omega$ a non-zero holomorphic one-form having no zeros (that is in every local chart $\omega = f(z)dz$ with $f$ locally analytic with no zero). Let $$F(p) = \int_{p_0}^p \omega$$ (in local chart $F(p) =F(a)+ \int_a^p f(z)dz$) Let $U$ be the universal cover of $X$ : that is $U$ is the Riemann surface obtained from ...


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Your reasoning is correct. Substitute $Y = \dfrac{3x^2}{2y} X - \dfrac{x^2 - 2c}{2y}$ into $Y^2 = X^3 + c$ to obtain a cubic for $X$. Just don't brute force it. The key point is to observe that two of its roots are known, and equal to $x$ (do you see why?). Use a sum of roots theorem to get the third one. BTW, to use the sum of roots, yo don't need the ...


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