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Fix a prime $p$ and let $q$ be either a prime distinct from $p$ or $q = \infty$. By my previous question, the absolute value $|.|_q$ on $\Bbb Q$ extends to some absolute value $N_q( \cdot )$ on $\Bbb Q_p$. (I'm not sure that this extension is unique up to equivalence... — By the way, is it true that any extension to $\Bbb Q_p$ of the $p$-adic absolute value on $\Bbb Q$ is equivalent to the usual $p$-adic absolute value on $\Bbb Q_p$?)

My question: is $\Bbb Q_p$ complete with respect to $N_q(.)$ ? If no, what does the completion look like? In particular, is it the same as the completion of $\Bbb Q_q$ with respect to $N_p(.)$ ?

Since these extensions are constructed from Zorn's lemma, I have some trouble in working on these questions.

Alphonse
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    How do you extend the $q$-adic norm to $p$-adic numbers that are not in $\mathbb{Q}$? IMHO the only thing that makes the $q$-adic norm work is the starting values are rational to begin with when $a,b$ are relatively prime with $q$ then $$\left|\frac{aq^\alpha}{b}\right|q=q^{-\alpha}$$ How would you define $$ \left| \sum{k=0}^\infty p^k\right|_q$$ ? – N8tron May 28 '18 at 20:07
  • @N8tron : see the linked question. There is no explicit description, alas, because you have to use the axiom of choice to construct this extension ! – Alphonse May 28 '18 at 20:10
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    nothing in the previous post leads me to believe it can be done – N8tron May 28 '18 at 20:14
  • Let $K_1 = \Bbb Q_p$, let $K_2$ be the completion of $K_1$ w.r.t. $N_q(.)$, and fix an extension $N_{p,2}(.)$ to $K_2$ of the $p$-adic valuation on $K_1$. Then let $K_3$ be the completion of $K_2$ w.r.t. $N_{p,2}$ and fix an extension $N_{q,3}(.)$ to $K_3$ of the $q$-adic valuation on $K_1$. Then let $K_4$ be the completion of $K_3$ w.r.t. $N_{q,3}$ and fix an extension $N_{p,4}(.)$ to $K_4$ of the $p$-adic valuation on $K_2$. ... does this process stop? – Alphonse May 29 '18 at 07:22
  • I don't think that any two extensions of the $q$-adic absolute value are equivalent. Pick two $\Bbb Q$-algebraically independent elements $a,b \in \Bbb Q_p$. We define $$N(a) = 1 = N'(b), N(b)=2 =N'(a).$$ Then I think that $N,N'$ define absolute values on $\Bbb Q(a,b) \subset \Bbb Q_p$ (extending $|.|_q$), and their extensions to $\Bbb Q_p$ should not be equivalent. – Alphonse May 29 '18 at 11:56
  • What is the setup for Zorn's lemma argument? Presumably it will produce a maximal set in the p-adics under which $N_q$ is an absolute value. You have to prove this set is not $\mathbb Q$ and all of $\mathbb Q_p$ – N8tron May 29 '18 at 14:29
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    @N8tron: With Zorn's lemma resp. the axiom of choice one gets field isomorphisms $i_p: \Bbb C_p \simeq \Bbb C$, hence by composing you have an iso $j_{p,q}: \Bbb C_p \simeq \Bbb C_q$. Pull back the $q$-adic value on $\Bbb C_q$ to $\Bbb C_p$ and restrict it to $\Bbb Q_p$. But these constructions are kind of useless because the isomorphisms cannot be made explicit. Compare point 6 of my answer https://math.stackexchange.com/a/2563837/96384 – Torsten Schoeneberg May 31 '18 at 05:40
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    Related: https://math.stackexchange.com/q/338148/96384 – Torsten Schoeneberg May 31 '18 at 06:30
  • Ah thanks for the links it looks like what I'd want to see is the following: For a proof of the fact that C and Cp are isomorphic you can see page 83 of the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - 1978. I've seen that book in our library so I'll go check it out :-) – N8tron May 31 '18 at 12:50

1 Answers1

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There is too much choice in the many places. I can give answers only after find a landscape for each question, so that no choice is involved.

(1) $\Bbb Q_p$ is not closed with respect to any choice of some $N_p$. For this, consider a sequence of the shape $(p/q)^n$.

(2) It is by construction a completion. We have lost the beautiful valuation from $\Bbb Q_p$, at least i cannot see how to make it still work.

(3) In which category do we search for an isomorphism?

(Metric spaces, rings? This structure is common to the two objects. But $\Bbb Q_p$-modules, $\Bbb Q_q$-modules, $\Bbb Q_p\otimes \Bbb Q_q$-modules? I cannot see how to get any of the above structures.)

If there is one, then Zorn is the only solution. As said, there is too much choice in the two used axioms of choice. I know this is not satisfactory, but in my opinion, the question should define at least a map between the two objects that should be the same. All i can say is as follows. Let $\Bbb Q_{p,\text{alg}}$ be the subfield of algebraic numbers of $\Bbb Q_p$. Then we have unique valuations $|\cdot|_p$, $|\cdot|_q$ on some of the following objects:

$\require{AMScd}$ \begin{CD} @. \Bbb Q_p@>>> (?)\\ @. @AAA @AAA \\ @. \Bbb Q_{p,\text{alg}} @>>> \boxed{\ \Bbb Q_{\text{alg}}\ }@>>> (?)\\ @. @AAA @AAA @AAA \\ \Bbb Q @>>> \Bbb Q_{p,\text{alg}}\cap \Bbb Q_{q,\text{alg}} @>>> \Bbb Q_{q,\text{alg}} @>>> \Bbb Q_q \end{CD}

(In order to write the picture i also had to make a choice for each of the two inclusions in $\bar{\Bbb Q}$, here denoted for the notational homogenity with the subscript alg.)

The boxed entry is the last one that has both valuations in a natural way.

I cannot see how to complete "structurally" (not "existentially") the picture. The topological properties are like those when we want to consider "other filters" on $\Bbb Z$, or on one of the algebraic fields below. There are some applications in number theory that need for a short time such filters, but only for a short time. The situation above is much more complicated than the extension of $\Bbb Q_p$ by one transcendent variable $t$, to get $\Bbb Q_p((\, t\, ))$.

dan_fulea
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