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In general, given two vector bundles $E,F$ of ranks $e,f$ on any variety $C$ we have the formula: $$\operatorname {det} (E\otimes F)= (\operatorname {det} E)^{\otimes f} \otimes (\operatorname {det} F)^{\otimes e} $$ where $\operatorname {det} E=\wedge^eE$ etc. Applying this to the case at hand, we write $$ \operatorname {det}V \stackrel ...


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I think this should follow from the fact that there exists a perfect pairing $$\langle\cdot ,\cdot\rangle \colon H_1(X_0(N),\mathbb R)\times H^0(X_0(N),\Omega^1_{\mathbb C})\to \mathbb C$$ $$\langle\{\alpha,\beta\},f\rangle=\int_{\alpha}^{\beta}f$$ Here $\{\alpha,\beta\}$ is the real homology class on $X_0(N)$ of any path from $\alpha$ to $\beta$ in ...


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The discussion only rests on the sign of the discriminant, $-(2-c)(3-c)$. When negative, there are no asymptotes, hence ellipse (or circle). When positive, two asymptotes, hyperbola.


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I have since worked out the answer to this. So in case anybody comes across the question, I will sketch it below. Observe $\rho$ has the minimal polynomial $x^2+x+1=0$, so if $\Lambda$ is the lattice $\mathbb{Z}+\rho\mathbb{Z}$, we have $\rho \Lambda = \rho\mathbb{Z}+\rho^2\mathbb{Z}=\rho\mathbb{Z}-\mathbb{Z}-\rho\mathbb{Z}=\Lambda$. But clearly $g_2(a ...


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My copy of Silverman is in my office, so I cannot recall/check all the key bits. I just have this hunch that this is related to the existence of an order four automorphism of elliptic curves of this type. This is too long to fit into a comment, so an answer it is. To move the identity element to the origin, $(z,w)=(0,0)$ we do the usual change of ...


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I agree with the OP and Bruno Joyal that the statement of this exercise is faulty. As you say, condition (ii) had better hold for any rational point $O$ given that we've defined the group law in such a way to make $O$ the origin. Unfortunately I could not remember what I had in mind when I wrote this, so I uploaded a new copy in which condition (ii) is ...


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The Diophantine equation $Ax^n+By^n=Cz^n$ for coprime integers $(A,B,C)$ is closely related to FLT, and has been studied, too. There are choices for $A,B,C$ and $n=p$, such that there are indeed some nontrivial solutions, e.g., $$ x^5+13y^5=8z^5 $$ has the nontrivial solution $(x,y,z)=(3,1,2)$. But usually there are very few solutions. For an overview see ...


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This equation was solved in a paper of Henri Darmon and Loic Merel in Crelle's journal in 1997. The techniques involved are similar to those in Wiles' proof, with additional complications due to the fact that the trivial solution $x=y=z=1$ actually corresponds to a modular form of weight $2$ and level $32$.


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As it has been pointed out, a sufficient condition is taking the ground field to be infinite. This can be traced back to the following very general considerations: if $f\in A[(X_i)_{i\in I}]$ is a polynomial ($A$ a commutative ring), then we get an induced map $\tilde{f}:A^I\rightarrow A$ obtained by substiution. It turns out that the map $f\mapsto\tilde{f}$ ...



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