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Let $E$ be an unramified quadratic extension of a local field $F$, with $p$ odd. Let $E^1$ denote the set of norm $1$ elements of $E$. What can be said about the following index: $$ {\rm 1.}\ \ \ \ [ (E^1 \cap (1 + P_E )) : E^1 \cap (1 + P_E^M )] $$ where $M$ is a positive integer.

We know that in this case $N_{E/F}(1+P_E )=1+P_F$. If we consider the short exact sequence $$ 1\longrightarrow E^1\longrightarrow R_{E}^{\times}\longrightarrow R_{F}^{\times}\longrightarrow 1 $$ and intersect this with $1+P_{E}$, we get $$ 1 \longrightarrow E^1\cap (1+P_E)\longrightarrow 1+P_E\longrightarrow 1+P_{F}\longrightarrow $$

but I don't think this helps in computing the index in 1.

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    $\begingroup$ Could you please rewrite this into TeX, etc? $\endgroup$ Commented Nov 24, 2023 at 23:05

1 Answer 1

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You can reduce your problem to computing the index $[E^1 \cap (1+P_E^k ):E^1 \cap (1+P_E^{k+1})]$, for all $k\geqslant 1$. Hint : prove and use the fact that for all $x\in P_E^{k}$, $N_{E/F} (1+x)\equiv 1+{\rm Tr}_{E/F}(x)$ mod $1+P_E^{k+1}$.

Remark : you do not need to assume $p$ odd.

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  • $\begingroup$ I was able to prove that $N_{E/F}(1+x)\equiv 1+Tr_{E/F}(x) \mod (P_{E}^{k+1})$, but I couldn't figure out how that helps in computing the index. $\endgroup$
    – Ekta
    Commented Nov 25, 2023 at 22:50
  • $\begingroup$ Also thank you for fixing the latex problem $\endgroup$
    – Ekta
    Commented Nov 25, 2023 at 22:50

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