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I have a question which is probably caused by some confusion I have with extensions of local fields. Let us fix a finite extension of number fields $L/K$ and fix further $\mathfrak p$ denote a non trivial prime ideal of $\mathcal O _K$. Now, for every prime $\mathfrak P$ of $\mathcal O _L$ dividing $\mathfrak p$ we get an embedding $\sigma_\mathfrak P : K_\mathfrak p \to L_\mathfrak P$.

Is it now true or false that for every $i\geq 1$ we have $$\bigcap_{\mathfrak P | \mathfrak p} \ \sigma_\mathfrak P ^{-1}(\mathfrak P ^ i \mathcal O _{L_\mathfrak P}) \subset \mathfrak p ^i \mathcal O _{K_\mathfrak p} \ \ \ ?$$

The question occurred during my struggles to understand relations between ray class fields of different number fields.

Thank you very much in advance!

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You should replace $\mathfrak{P}$ and $\mathfrak{p}$ by the ideals generated by these in the completions you consider. Also, you can remove "$1+$" on both sides of the inclusion. –  Plop May 3 '11 at 18:26
    
You're right, this is what I meant. –  Amateuresque May 4 '11 at 8:51
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1 Answer 1

up vote 2 down vote accepted

It is false in general: because of ramification, you can have that $\sigma_{\mathfrak{P}}^{-1} \left( \mathfrak{P}^2 \mathcal{O}_{L_{\mathfrak{P}}} \right) = \mathfrak{p} \mathcal{O}_{K_{\mathfrak{p}}}$.

However, if $\mathfrak{p} = \mathfrak{P}_1^{e_1} \ldots \mathfrak{P}_k^{e_k}$, then for any $r$, $\sigma_{\mathfrak{P}_r}^{-1} \left( \mathfrak{P}_r^n \mathcal{O}_{L_{\mathfrak{P}_r}} \right) = \mathfrak{p}^m \mathcal{O}_{K_{\mathfrak{p}}}$ where $m$ is the floor of $n/e_r$, simply by comparing valuations (and there is no need to intersect for the different $\mathfrak{P}$).

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Thank you very much for your answer! –  Amateuresque May 4 '11 at 8:55
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