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6

The Galois representation is reducible iff there is a one-dimensional Galois-stable $\mathbb{F}_{\ell}$-subspace, say $C$. Then $E \rightarrow E/C$ is a $\mathbb{Q}$-rational isogeny. Conversely, if $E \rightarrow E'$ is a degree $\ell$ isogeny, its kernel $C$ is a Galois-stable subgroup of $E(\overline{\mathbb{Q}})$ of order $\ell$ so gives a ...


4

If $p$ is $\equiv 1 \bmod 3,$ then we may write $p = a^2 + 3 b^2$, where $a$ is pinned down by requiring $a \equiv 1 \bmod 3$ (and I won't bother to pin down $b$). Then, writing $a_p = 1 + p - | C(\mathbb F_p)|$, as usual, one has that $$a_p = 2a,$$ or equivalently, $$| C(\mathbb F_p)| = 1 + p - 2 a .$$ In terms of Qiaochu's answer, the relevant modular ...


2

At a prime $p$ for which $E$ has good reduction, the coefficient $a(p)$ is just $p - 1 + \# E(\mathbb{F}_p)$, where $E(\mathbb{F}_p)$ is the set of points of $E$ modulo $p$. At the remaining primes, the definition is slightly more complicated. If $N$ is the conductor of $E$ and $p$ divides $N$ but $p^2$ does not divide $N$, then $E$ has multiplicative ...


1

The Hasse principle automatically holds for an elliptic curve $E$ over $\mathbf{Q}$. By definition, $E$ has a marked point defined over $\mathbf{Q}$. The correct statement is that the Hasse principle holds for a genus one curve $C$ defined over $\mathbf{Q}$ if and only if $C$ represents the trivial class in the Tate-Shafarevich group of its Jacobian. In ...


1

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 ...


1

The rational points of an elliptic curve are an abelian group, i.e. a $\mathbf Z$-module and the points of finite order its torsion subgroup. The last equality is part of the distributive laws for modules.



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