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Let $K$ be a complete field w.r.t discrete absolute value $|\cdot|_K$,

$\mathcal O_K=\{x\in K:|x|_K\leq 1\}$.

$L$ is an extension field with $[L:K]< \infty$ and let $\mathcal O_L$ be the integral closure of $\mathcal O_K$ in $L$.

Also let $\pi$ be the generator of the unique maximal ideal of the DVR $\mathcal O_K$.

If $\pi \mathcal O_L=\mathcal P^e$, with $\mathcal P$ the unique non zero prime in $\mathcal O_L$, then is it true that the absolute value $|\cdot|_{\mathcal P}$ on $L$ extends $|\cdot|_K$ on $K$?

Many thanks for your help.

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  • $\begingroup$ Why do you write $r$ prime ideal factors? The ring $\mathcal O_L$ is a DVR so it has just one nonzero prime ideal. $\endgroup$ – KCd Feb 24 '15 at 12:55
  • $\begingroup$ You are right- thanks for correcting me. $\endgroup$ – mathmo Feb 24 '15 at 13:09
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If depends on how you define $|\pi|_K$ and $|\mathcal{P}|_L$. But if you define $|x|_L = \sqrt[{[L/K]}]{|N_{L/K}(x)|}$ for all $x\in L$ where $N_{L/K}(x)$ is the norm (*) of $x$ over $K$, then yes. Have you already heard about ramification indexes ?

(*) Recall the $N_{L/K}(x)$ is the product of all conjugates of $x$ in $L$ over $K$, which is the same as the determinant of the $K$-endomorphism $y\mapsto xy$ of $L$.

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  • $\begingroup$ Is it true that $N_{L/K}(x)=x^n$ if $x\in K$ where $n=[L:K]$- so your formula will give $|x|_{\mathcal P}=|x|_K$? Sorry for the basic questions but I'm a newcomer to number theory. Btw how do ramification indices help us? regards $\endgroup$ – mathmo Feb 24 '15 at 15:44
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    $\begingroup$ You are right, it is true, so that we deal with an extension of $|\cdot|_K$ indeed. Ramification index gives you basically which "type" of exponents of $|\pi|_K$ absolute values $|x|_L$ for $x\in L$ will have. I advise you "$p$-adic numbers. An introduction" of Fernando Q. Gouvêa for an introduction to ramification theory (basically for finite extensions of $\mathbf{Q}_p$'s) and J.-P. Serre's "Local fields" for a more proper exposition. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Feb 24 '15 at 15:45
  • $\begingroup$ Many thanks! Regards $\endgroup$ – mathmo Feb 24 '15 at 15:55

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