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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 ...

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 ...

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In Silverman's book "The Arithmetic of Elliptic Curves" the classical results on the Mordell-Weil rank of these curves are summarised in chapter $X$, section $6$, which is called "The Curve $E: Y^2=X^3+DX$" for fourth-power free integers $D$. Proposition $6.2$ gives us the formula, for $D=p$ prime, $${\rm rank} \; E(\mathbb{Q})+\dim_2 Ш ... 2 You could try with Sage, which is open-source and also available online. It has a vast library of functions related to elliptic curves, although I don't know if it can compute E(\Bbb{Q})/2E(\Bbb{Q}) directly. It shouldn't be too difficult to pick up, especially if you know a bit of Python. You can start looking at the tutorial and at the section of the ... 1 Let (A,\mathfrak{m}) denote the local ring of C at P, with uniformizing parameter s\in\mathfrak{m}. Let f be a function on C, which is regular at P. Let Spec(A') be an open affine neighborhood of P which does not meet the locus which maps to \infty. Then, we can view f in two ways: (1) f is an element of A' and hence defines an ... 1 Assuming your question is "why is the statement in your title true?", then this is a situation where the language of varieties gets a bit confusing. The key question to ask yourself is "rational over what field?". The point is that over an algebraically closed field \overline{K}, every curve admits (infinitely many) rational points. You can see this ... 1 Here is a very computational way of creating isogenies. Coming from someone who has digested approximately the first half of Silverman's AEC but not much more, so I am unable to use any of the high-powered tools. We can create an isogeny from a given elliptic curve E_1 to some other by specifying a finite subgroup A of the additive group of E_1. At ... 1 If a point P=(x,y) has order three, then 2P=-P, so that the duplication formula for the x-coordinate gives, with y^2=f(x)=x^3+ax+b, and 16(4a^3+27b^2)\neq 0,$$ 0=\Psi_3(x)=2f(x)f''(x)-f'(x)^2=3x^4 + 4ax^3 + 6ax^2 + 12x -a^2. 

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.

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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 ...

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

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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 ...

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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, ...

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