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In my first post, I asked if $\mathbb{Q}_p$ is the completion of $\mathbb{Q}$ over some valuation of an prime ideal (and it seems to be true, according to an answer). Now I am asking if this can be generalized for finite extensions of $\mathbb{Q}_p$.

To put it in other words: Let $K$ be a finite extension of $\mathbb{Q}_p$. Question: Is there a valuation $v$ of some prime ideal in $\mathcal{O}_k$ on a global field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that the completion $k_v$ of $k$ over $v$ is equal to $K$?

ANT Learner
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    Does this answer your question? Is every finite Galois extension of $\mathbb{Q}_p$ the completion of a number field at some valuation? This is true, but not completely straightforward. It follows from Krasner's lemma, which roughly says that the roots of nearby polynomials in $\mathbb Q_p[x]$ generate the same extension. So take the polynomial $f$ that generates $K$ and choose a nearby rational polynomial. – Mathmo123 Jan 14 '21 at 18:03
  • @Mathmo123 Thank you for the link to the thread. Although the title seems to be really helpful, I cannot see in the linked post where exactly the valuation comes into play. It seems to not be included in the poster's description at all. – ANT Learner Jan 14 '21 at 18:12
  • I forgot to add that I was interested in valuations wrt. prime ideals - I hope this does not confuse anyone. – ANT Learner Jan 14 '21 at 18:25
  • There are also some linked mathoverflow posts in the comments? Completion with respect to valuations is completely standard terminology, so don't worry! – Mathmo123 Jan 14 '21 at 18:36
  • Did anyone try to show it from that $p$-adic Galois extensions are solvable ? Solvable means that once enough roots of unity are added the extension is contained in a radical extension and since $K^/K^{n}$ is easily shown finite depending solely on reduction data we get that these radical extensions are completions of number fields. For the subextensions we use the fundamental theorem of Galois theory. – reuns Jan 14 '21 at 18:57
  • @reuns Is there a way to prove that $K^{\times}/K^{\times n}$ is finite without using Krasner's lemma? Without knowing that it's finite, I don't say why any given element should have a rational representative. – Mathmo123 Jan 14 '21 at 20:26
  • @Mathmo123 For $K$ a $p$-adic field, $1+p^2 O_K$ has finite index in $O_K^\times$ and $(1+p^2 O_K)^n = 1+p^{k+2} O_K$ where $p^k| n$ so $(1+p^2 O_K)/(1+p^2 O_K)^n$ is finite and so are $O_K^{\times}/O_K^{\times n}$ and $K^{\times}/K^{\times n}$. If $K$ is the completion of a number field $F$ then $K^{\times}/K^{\times n}$ has representatives in $F$. – reuns Jan 14 '21 at 20:32

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Yes. Let $K$ be a finite extension of $\mathbb Q_p$. We want to show that there is a finite extension $F$ of $\mathbb Q$, and a prime ideal $\mathfrak p$ in the ring of integers $\mathcal O_F$ of $F$, such that $K$ is the completion of $F$ with respect to the $\mathfrak p$-adic valuation on $F$.

Krasner's lemma tells us that there is an irreducible polynomial $g(t)$ (irreducible in the ring $\mathbb Q_p[t]$, that is) with rational coefficients, and a root $\beta$ of $g(t)$, such that $K = \mathbb Q_p(\beta)$. Then $g(t)$ is also irreducible in the ring $\mathbb Q[t]$, so if we set $F = \mathbb Q(\beta)$, then $[F : \mathbb Q] = [K : \mathbb Q_p] = \operatorname{deg} g$.

Now let $| - |$ be the unique extension of the $p$-adic absolute value to $K$. Restrict $|-|$ to an absolute value on $F$. A version of Ostrowski's theorem for number fields states that this absolute value is either archimedean, or arises from a prime ideal $\mathfrak p$ in the ring of integers $\mathcal O_F$ of $F$. Specifically, there is a real number $0 < \rho < 1$ and a prime ideal $\mathfrak p$ such that

$$|x| = \rho^{\operatorname{ord}_{\mathfrak p}(x)}.$$

Obviously this absolute value is not archimedean on $F$ (its restriction to the $p$-adic absolute value on $\mathbb Q$ isn't, after all). Anyway, the completion of $F$ with respect to $\mathfrak p$ is the same as the metric space completion of $F$ with respect to $|-|$. Finally, just as the metric space completion of $\mathbb Q$ with respect to $|-|$ gets you $\mathbb Q_p$, so does the metric space completion of $F = \mathbb Q(\beta)$ with respect to $|-|$ get you $\mathbb Q_p(\beta) = K$.

D_S
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