Let $k$ be a number field, $v$ a discrete non-archimedean valuation on $k$. Let $k_v$ be the completion of $k$ wrt $v$ and $\mathcal O_v$ its valuation ring.

My question is:

If $x\in k_v$ then is there an integer $m\in \mathbb{Z}$ (not merely $x\in \mathbb{Z}_p$) such that $mx\in \mathcal{O}_v?$

Many thanks in advance!


Hint: $v$ corresponds to a prime ${\frak p}$ of $k$. Then ${\frak p}$ lies above $(p)$ for some rational prime $p$. Thus we have an integer with positive valuation...

  • $\begingroup$ I see, so for a large enough positive integer $n$, $v_\mathfrak{p}(p^nx) \ge 0$, hence $p^nx\in \mathcal{O}_v$. Why must a prime ideal of $\mathbb{Z}_p$ be of the form $p\mathbb{Z}_p$ for $p$ a rational prime number? $\endgroup$ – Mark Rodriguez Aug 18 '13 at 16:57
  • 1
    $\begingroup$ @MarkRodriguez While that's true, I didn't say that in my answer. Is that a separate question you wanted to ask? | At any rate, here's a hint: argue that the minimal (normalized) valuation of an element in a prime ideal of ${\Bbb Z}_p$ must be $1$. $\endgroup$ – anon Aug 18 '13 at 17:04
  • $\begingroup$ Sorry I momentarily forgot that $\mathfrak{p}$ was a prime ideal of the ring of integers of $k$ (and not of $\mathcal{O}_v$). Please ignore my question. Regards $\endgroup$ – Mark Rodriguez Aug 18 '13 at 17:11

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