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Let $K\supset F$ a finite extension of local fields. It means that the valuation $v_K$ extends the valuation $v_F$. We denote with $\pi_K$ and $\pi_F$ the uniformizer parameters and with $\mathcal O_K$, $\mathcal O_F$ the rings of integeters.

Clearly we have the following inclusion of open, compact subgroups:

$$K\supset\ldots\supset\pi_K^r\mathcal O_K\supset\pi_K^{r+1}\mathcal O_K\supset\ldots\supset \{0\}$$

$$F\supset\ldots\supset\pi_F^r\mathcal O_F\supset\pi_F^{r+1}\mathcal O_F\supset\ldots\supset \{0\}$$

Is it true that $\operatorname{Tr}_{K|F}\left(\pi_K^r\mathcal O_K\right)\subset\pi_F^s\mathcal O_F$ for some $s\in\mathbb Z$? Can we say something about such $s$? in other words I wanted to know if the trace map preseves the structure of power of ideals in the local fields.

Thanks in advance

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    $\begingroup$ You can show that $v(\alpha) = v(\alpha^\sigma)$ extends the valuation to finite extensions, or that $v(\alpha) = v(\alpha^\sigma)$ in the normal closure, or start with $\pi_F = \pi_K^n u, u \in a+\pi_K O_K$ to find a basis of $O_K/O_F$, the conjugates of $\pi_F$, the traces $\endgroup$ – reuns Feb 16 at 12:00

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