5
$\begingroup$

Consider $\mathbb Q_p^{\text{ur}}$ the maximal unramified extension of the p-adic numbers. Suppose that on $\mathbb Q_p$ we have the usual absolute value that extends $|\frac{a}{b}|_p=\frac{1}{p^{v_p(a)-v_p(b)}}$ to $\mathbb Q_p$. Now it is known that $|\cdot|_p$ extends uniquely to $\mathbb Q_p^{\text{ur}}$. Is this absolute value discrete w.r.t. $\mathbb Q_p^{\text{ur}}$ ?

The relevant definitions can be found here (pages 105-124).

Since the extension of $|\cdot|_p$ is not discrete w.r.t. $\mathbb Q_p^{\text{al}}$, I had a first naive guess that this is also the case for $\mathbb Q_p^{\text{ur}}$. However, it is known that $\mathbb Q_p^{\text{ur}}$ is obtained by adjoining to $\mathbb Q_p^{\text{ur}}$ the roots of $1$ of order coprime to $p$, and obviously on each of these generators the absolute value must be $1$, which makes me believe the opposite, but how would one compute the absolute value for an arbitrary element in $\mathbb Q_p^{\text{ur}}$?

$\endgroup$
  • 1
    $\begingroup$ I'm not sure what you want, but how about: given an element $\alpha$ in the unramified closure, take any finite unramified extension $E$ of $k=\mathbb Q$ containing $\alpha$, compute the $k$ norm of $N^K_k(\alpha)$, and take $n$-th root of that, where $n=[K:k]$? $\endgroup$ – paul garrett Mar 13 '17 at 19:57
  • 3
    $\begingroup$ Let $K\supset k$ be an extension of local fields. If you arrange it so that $V$ on $K$ restricts to the given $v$ of $k$, then the value group of $V$ is $1/e$ times the value group of $v$ (as subgroups of $\Bbb Q^\times$). Here, $e$ is the ramification index, which is $1$ for unramified extensions. So the value group of the maximal (algebraic) unramified extension of $\Bbb Q_p$ is just $\Bbb Z$. When you complete, the value group doesn’t change either. $\endgroup$ – Lubin Mar 15 '17 at 4:12
  • $\begingroup$ I might say in addition that it is an Unfriendly Act to give a reference to the title page of a book or set of notes, but without a page number. $\endgroup$ – Lubin Mar 15 '17 at 23:58
  • $\begingroup$ Thank you very much for the suggestion. You are absolutely right about the remark. $\endgroup$ – user223794 Mar 16 '17 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.