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This is somewhat related to the question and the answers here:

Is completeness of a field an algebraic property?

My question is (to which I believe the answer must have been known), does every extension of of a non-trivial non-archimedean valuation on $\mathbf{Q}$ to $\mathbf{R}$ fail to make $\mathbf{R}$ into a complete field? (Valuations considered here are general Krull valuations.)

I guess the notion of Cauchy sequence should be generalized a little bit, considering that $\mathrm{cf}(2^{\aleph_0}) > \aleph_0$, i.e. one should allow a Cauchy sequence of any well order type.

Thanks!


Edit: I apologize that I did not make the question as clear as it should have been. Let me explain it furthermore here.

  1. All the valuations here are Krull's general valuations, which could be of any rank, not necessarily absolute values.

  2. For a valued field with higher-rank value group $\Gamma$, the definition of Cauchy sequences (or completeness) depends on the value group, in the sense that a Cauchy sequence must be of length $\mathrm{cf}(\Gamma)$. (hence my second paragraph above). The definition could be found in Engler and Prestel's book Valued Fields. More specifically, let $\kappa=\mathrm{cf}(\Gamma)$, and a sequence $(a_\nu)_{\nu<\kappa}$ is said to be a Cauchy sequence if for every $\gamma\in \Gamma$, there exists some $\delta <\kappa$ such that for all $\mu, \nu \in (\delta, \kappa)$, one has

    $v(a_\mu-a_\nu)>\gamma$.

  3. And the valued field is said to be complete if every Cauchy sequence has a limit, meaning there exist some $a$ in the field, such that $v(a_\nu-a)>\gamma$ for any $\gamma$ and sufficiently big $\nu$, which says that $a$ is a limit point of the sequence in the topological sense.

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Based on your comments to Yuri Zarhin's answer, I think you mean to ask: "Does every extension of a nontrivial Krull valuation...". –  Pete L. Clark Nov 14 '10 at 4:29
    
@Jizhan Hong: could you explain what you mean by the completion of a field equipped with a Krull valuation? As far as I know, the completion concept is standard and useful only for rank one valuations. For instance, in the introduction of Engler and Prestel's Valued Fields they write: "In the theory of valuations, the notion of a completion had to be replaced by that of the so-called henselization." –  Pete L. Clark Nov 14 '10 at 8:33
    
@Pete Clark, I have edited it. I am familiar with the book you said and the sentence you quoted. This is just a question that came up when I was reading the other question mention in the link of my question. It's not obvious to me, and I thought just out of curiosity, why not asked it(it's probably not important)? Maybe someone would have convenient ways to see it. –  Jizhan Hong Nov 14 '10 at 13:57

1 Answer 1

Any nontrivial non-archimedean valuation on the field of rational numbers $Q$ is essentially $p$-adic for some prime $p$, i.e., there exists a prime $p$ such that $|p|<1$ and one may check that such $p$ is unique: $|q|=1$ for all other primes $q$. (This is a classical theorem of Ostrowsky, see first pages of Koblitz's p-adic numbers, p-adic analysis and zeta-functions.) So, if R is complete wrt some extension of the valuation then it must contain the $p$-adic completion of $Q$, i.e., the field $Q_p$ of $p$-adic numbers. More precisely, the field of real numbers $R$ must contain a subfield isomorphic to $Q_p$. In particular, all rational numbers that are squares in $Q_p$ are squares in $R$. But this is not the case: for example, $1-p^3$ is a square in $Q_p$ but not in $R$, since it is negative. This proves that $R$ does not contain a subfield isomorphic to $Q_p$. This implies that in the case of usual valuations the answer to your question is positive in the following sense: any extension of of a nontrivial non-archimedean valuation on $Q$ to $R$ fails to make $R$ a complete field.

Of course, if the original non-archimedean valuation on $Q$ is trivial. i.e., every nonzero rational number has norm 1 then one may extend it to the trivial valuation on $R$; every field is complete wrt the trivial valuation, including $R$.

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Oh, yeah, I forgot to mention that I was not considering the trivial valuation. Yuri, I thought about what you wrote here too. But I don't believe that R has to contain an isomorphic image of Q_p. Cauchy sequences of Q_p may fail to be Cauchy sequences of R, simply because the value group is not of rank 1. –  Jizhan Hong Nov 14 '10 at 1:23
    
@Jizhan If $R$ is complete, won't the closure of $Q$ in $R$ be $p$-adic completion of $Q$, hence isomorphic to $Q_p$? –  Keenan Kidwell Nov 14 '10 at 1:33
    
No, the topology could be changed. For example, consider the extension of Z_(p) on Q to Q(X) with v(X)>Z, then there is a neighborhood of zero not containing any of p^n. –  Jizhan Hong Nov 14 '10 at 2:10
    
Jizhan, I added a couple of words to my answer, in order to make it clear that I deal only with usual valuations. –  Yuri Zarhin Nov 14 '10 at 4:43
    
Sure, Yuri, sorry for the confusion. –  Jizhan Hong Nov 14 '10 at 13:53

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