For a minimal integral Weierstrass equation, let $\widetilde E$ be the reduced cubic and $\widetilde E_{\rm ns}(k)$ its nonsingular points, with their induced group law. Define the <filtration of elliptic-curve points over a local field> by
$$
E_0(K)=\{P\in E(K):\widetilde P\in\widetilde E_{\rm ns}(k)\},
$$
and
$$
E_1(K)=\ker\left(E_0(K)\longrightarrow\widetilde E_{\rm ns}(k)\right).
$$
The parameter $z=-x/y$ identifies $E_1(K)$ with the <formal group of an elliptic curve> on $\pi\mathcal O_K$. Part (a) therefore gives $E_1(K)[n]=0$. Reduction restricts to the exact sequence
$$
0\longrightarrow E_1(K)\longrightarrow E_0(K)
\longrightarrow\widetilde E_{\rm ns}(k)\longrightarrow0.
$$
Its restriction to $n$-torsion has trivial kernel, yielding the injection
$$
\boxed{E_0(K)[n]\hookrightarrow\widetilde E_{\rm ns}(k).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-125/3/b/solution