# Tag Info

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There is probably a more direct way to prove this, since it is apparently an exercise in Silverman, but one way to see it is: if $C$ has genus $1$, then its Jacobian $E$ is an elliptic curve, and $C$ and $E$ become isomorphic over any field $L$ over which $C$ obtains a rational point.

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I will describe the group law for an elliptic curve in short form, ie $E: y^2=x^3+Ax+B$ for some $A$, $B$. The group law on the elliptic curve can be defined geometrically, if $P$, $Q$ are distinct points, then draw the straight line between the two points. This line will intersect the curve with multiplicity $3$ so let $R$ be the third point. Finally if ...

0

You have slightly misunderstood the situation. There is an additional assumption, namely that the image is non-solvable. Now there is a classification of the subgroups of $\mathrm{PGL}_2(\mathbb{F}_p)$. (Probably the paper of Serre that they cite has it --- did you look at that? Otherwise, one place it is discussed is in Swinnerton-Dyer's article in LNM ...

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Re: question 2, the homogeneity requirement stems from the fact that it is referring to projective coordinates. Elliptic curves live in 2 dimensions so they can be expressed in 3 homogeneous coordinates. The definition of \phi can be homogenised replacing a with az and b with bz^2

1

If a point $P=(x,y)$ has order three, then $2P=-P$, so that the duplication formula for the $x$-coordinate gives, with $y^2=f(x)=x^3+ax+b$, and $16(4a^3+27b^2)\neq 0$, $$0=\Psi_3(x)=2f(x)f''(x)-f'(x)^2=3x^4 + 4ax^3 + 6ax^2 + 12x -a^2.$$

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The isomorphism comes about by looking at lines through the singular point $P$. Indeed, for any point $Q \neq P$ on $C$, the unique line in $\mathbb P^2$ through $P$ and $Q$ does not intersect $C$ anywhere else, because it already intersects with multiplicity $3$ (namely, once at $Q$ and twice at $P$). Thus, we get a map $$f \colon C\setminus\{P\} \to ... 3 The fibers f^*(Q)=Q'+Q'' of your degree-two morphism f:X\to \mathbb P^1 exactly decribe the complete linear system of effective canonical divisors \vert K_X\vert  of X and these fibers are linearly equivalent divisors as Q runs through \mathbb P^1. In particular f^*(P_i)=2P_i\equiv f^*(P_j)=2P_j for 1\leq i,j\leq 6, so that ... 1 Here is a very computational way of creating isogenies. Coming from someone who has digested approximately the first half of Silverman's AEC but not much more, so I am unable to use any of the high-powered tools. We can create an isogeny from a given elliptic curve E_1 to some other by specifying a finite subgroup A of the additive group of E_1. At ... 0 I already found an answer via the comment of mercio here. And I also found similarity of his comment and Theorem 14.16 in the book of D. A. Cox "Primes of Form x^2 + ny^2". The main idea here is the isomorphism that preserve the degree from \text{End}_{\mathbb{C}}(E) to \text{End}_{\overline{\mathbb{F}_p}}(E), where \text{End}_{\mathbb{C}}(E) is an ... 2 You could try with Sage, which is open-source and also available online. It has a vast library of functions related to elliptic curves, although I don't know if it can compute E(\Bbb{Q})/2E(\Bbb{Q}) directly. It shouldn't be too difficult to pick up, especially if you know a bit of Python. You can start looking at the tutorial and at the section of the ... 0 As stated, your problem is slightly wrong. Let's see first what the Hasse bound gives us. Let N=pk for some k\in \mathbb N_{>0} be the number of points of E. Then |pk-p-1|\leq 2\sqrt{p}. Now if k\geq 2 we have that |pk-p-1|\geq p-1. But the inequality p-1\leq 2\sqrt{p} doesn't hold for p\geq 7. Hence, if p\geq 7, then k=1 and N=p. ... 0 Let's start with the long Weierstrass equation for an elliptic curve over a field K:$$ y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6$$If the characteristic of the K is not 2 then we can make the substitution y_1=\dfrac{a_1}{2}(y-a_1x-a_3) which leads us to a medium equation:$$y_1^2 = 4x^3 + b_2x^2 + 2b_4x+b_6 If the characteristic is not 3 then we can ...

1

No, and depending on who you talk to, it is expected that no such curves exist. The Çiperiani-Wiles theorem says that every genus one curve over $\mathbb{Q}$ with semistable Jacobian and local points everywhere must have a point over some solvable extension. As far as I know, both conditions are expected to be able to be removed, so that every genus one ...

2

If $\mathfrak p$ is an integral ideal of $K$, the absolute value of $x\in K$ at $\mathfrak p$ is given by $|x|_{\mathfrak p}=N(\mathfrak p)^{-ord_{\mathfrak p}(x)}$. Let $K=\mathbb Q(\sqrt{-5})$, so that $\mathcal O_K=\mathbb Z[\sqrt{-5}]$ and let $x=\frac{1+\sqrt{-5}}{2}$. Then, as your definition shows, we have that ...

1

The number of points is $1+\sum_{x\in \mathbb F_p}\left(\left(\frac{x^3+ax+b}{p}\right)+1\right)$, where $\left(\frac{x^3+ax+b}{p}\right)$ is the Legendre symbol modulo $p$. If $x$ is not a root of $x^3+ax+b$, the term inside the sum is either $0$ or $2$. Otherwise, it is $1$. If $x^3+ax+b$ has a root, then it has either $1$ or $3$ roots and the sum is odd, ...

1

You can check out the answers to the following (closely) related MSE question: References for elliptic curves

1

The book you are quoting is in fact pretty elementary: a good knowledge of basic algebraic geometry will be more than enough to understand it. You can check for example Hulek's "Elementary algebraic geometry" or M. Reid's "Undergraduate algebraic geometry". In fact Silverman's book is pretty much self-contained by this point of view: the first two chapters ...

1

I think that if $E$ has CM this is false: take an endomorphism $\phi$ such that $\phi^2=[d]$, where $d$ is square-free. Then $\phi\circ\widehat{\phi}=[\deg\phi]=[d]$. Thus $\phi^2=\phi\circ\widehat{\phi}$, which implies $\phi=\widehat{\phi}$, but $\phi$ cannot be multiplication by an integer since it has non-square degree.

1

Assuming your question is "why is the statement in your title true?", then this is a situation where the language of varieties gets a bit confusing. The key question to ask yourself is "rational over what field?". The point is that over an algebraically closed field $\overline{K}$, every curve admits (infinitely many) rational points. You can see this ...

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