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Chris Wuthrich
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For $n$ large enough the Galois group of $F_n/F_{n-1}$ identifies with the kernel $G$ of $\operatorname{GL}_2(\mathbb{Z}/p^{n+1}\mathbb{Z}) \to \operatorname{GL}_2(\mathbb{Z}/p^{n}\mathbb{Z})$. Let $P$ and $Q$ be a basis of $E_{p^{n+1}}$. The norm of $P$ is written as a sum over $a$, $b$, $c$, $d$ modulo $p$ as \begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ &= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ &= p^4P+p^n(\sum_{a=0}^{p-1} a)\, P + p^n (\sum_{c=0}^{p-1} c)\, Q \\ &= p^4 P\end{align*}\begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ &= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ &= p^4P+p^{n+3}(\sum_{a=0}^{p-1} a)\, P + p^{n+3} (\sum_{c=0}^{p-1} c)\, Q \\ &= p^4 P\end{align*} and similar for $Q$ so it is multiplication by $p^4$ on all of $E_{p^{n+1}}$.

For $n$ large enough the Galois group of $F_n/F_{n-1}$ identifies with the kernel $G$ of $\operatorname{GL}_2(\mathbb{Z}/p^{n+1}\mathbb{Z}) \to \operatorname{GL}_2(\mathbb{Z}/p^{n}\mathbb{Z})$. Let $P$ and $Q$ be a basis of $E_{p^{n+1}}$. The norm of $P$ is written as a sum over $a$, $b$, $c$, $d$ modulo $p$ as \begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ &= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ &= p^4P+p^n(\sum_{a=0}^{p-1} a)\, P + p^n (\sum_{c=0}^{p-1} c)\, Q \\ &= p^4 P\end{align*} and similar for $Q$ so it is multiplication by $p^4$ on all of $E_{p^{n+1}}$.

For $n$ large enough the Galois group of $F_n/F_{n-1}$ identifies with the kernel $G$ of $\operatorname{GL}_2(\mathbb{Z}/p^{n+1}\mathbb{Z}) \to \operatorname{GL}_2(\mathbb{Z}/p^{n}\mathbb{Z})$. Let $P$ and $Q$ be a basis of $E_{p^{n+1}}$. The norm of $P$ is written as a sum over $a$, $b$, $c$, $d$ modulo $p$ as \begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ &= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ &= p^4P+p^{n+3}(\sum_{a=0}^{p-1} a)\, P + p^{n+3} (\sum_{c=0}^{p-1} c)\, Q \\ &= p^4 P\end{align*} and similar for $Q$ so it is multiplication by $p^4$ on all of $E_{p^{n+1}}$.

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Chris Wuthrich
  • 8.9k
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  • 47

For $n$ large enough the Galois group of $F_n/F_{n-1}$ identifies with the kernel $G$ of $\operatorname{GL}_2(\mathbb{Z}/p^{n+1}\mathbb{Z}) \to \operatorname{GL}_2(\mathbb{Z}/p^{n}\mathbb{Z})$. Let $P$ and $Q$ be a basis of $E_{p^{n+1}}$. The norm of $P$ is written as a sum over $a$, $b$, $c$, $d$ modulo $p$ as \begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ &= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ &= p^4P+p^n(\sum_{a=0}^{p-1} a)\, P + p^n (\sum_{c=0}^{p-1} c)\, Q \\ &= p^4 P\end{align*} and similar for $Q$ so it is multiplication by $p^4$ on all of $E_{p^{n+1}}$.