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Let $K$ be a local field (locally compact topological field) of characteristic zero. Is it true that $K$ is isomorphic to the completion of a number field under some valuations? If yes, then how to prove it?

I ask this question since in a paper it is said that $k_v^*/(k_v^*)^2$ where $k_v$ is the completion of a number field at some place $v$ has order 1, 2, 4 or 8. From the structure theory of local fields this order argument is incorrect for general local fields of characteristic zero.

If not every local field of characteristic zero comes from the completion of some number field and the above claim about the order of $k_v^*/(k_v^*)^2$ is correct. Can some one prove it or give a reference for it?

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    $\begingroup$ Is $K$ means to be complete with respect to a discrete valuation having finite residue field (i.e., if $K$ meant to be a finite extension of some $\mathbf{Q}_p$)? If so, think about the primitive element theorem and Krasner's Lemma to arrive at an affirmative proof. $\endgroup$
    – user29283
    Commented Mar 28, 2013 at 4:01
  • $\begingroup$ Krasner's Lemma can be used to prove that every local field of characteristic zero comes from completion of a number field. Therefore it answers the question. $\endgroup$
    – ronggang
    Commented Mar 28, 2013 at 8:24

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First of all, it is not correct to say that for a local field $K$, the group $K^\times/K^{\times 2}$ has order $1,2,4$, or $8$. To get a counter-example, think of a sufficiently ramified extension of $\mathbf{Q}_2$, or the local field $\mathbf{F}_2((T))$.

To compute this group for any local field $K$ with (finite) residue field $k$ of cardinality $q=p^f$ ($p$ prime), you have to use the structure of the multiplicative group $K^\times$, which turns out to be isomorphic --- after you choose a uniformiser of $K$ --- to the product $\mathbf{Z}\times k^\times \times U_1$, where $U_1$ is the group of $1$-units (Einseinheiten, the kernel of the map $\mathfrak{o}^\times\to k^\times$, where $\mathfrak{o}$ is the ring of integers of $K$). Now remark that $k^\times$ is cyclic of order $q-1$ and $U_1$ is a $\mathbf{Z}_p$-module (which is finitely generated of rank $[K:\mathbf{Q}_p]$ in the characteristic-$0$ case).

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  • $\begingroup$ It is known that a local field of characteristic zero is a finite extension of $Q_2$. This why I said that $K*/(K^*)^2$ has order 1,2,4 or 8 is incorrect. The question is whether the order argument is true for local fields coming from completions of number field? If every local field comes from the completion of a number field, then of course the order argument is incorrect. $\endgroup$
    – ronggang
    Commented Mar 28, 2013 at 7:32
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    $\begingroup$ @ronggang: read xuhan's comment. Any finite extension $K$ of $Q_p$ is isomorphic to a completion of a finite extension of $Q$. In short, write $K=Q_p(x)$ and write down the min poly of $x$ and then replace its coefficients with rational numbers very very close to those numbers; this works because of Krasner's lemma. $\endgroup$
    – user30035
    Commented Mar 28, 2013 at 7:45
  • $\begingroup$ @ronggang: As my answer shows, the question about the structure of $K^\times/K^{\times2}$ for a local field K has nothing to do with whether K is a completion of a number field or not. $\endgroup$
    – Chandan Singh Dalawat
    Commented Mar 28, 2013 at 8:31
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A nice reference for the relevant proof is http://math.stanford.edu/~conrad/676Page/handouts/localglobal.pdf.

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