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Motivations and Terminology

The term "henselian field" is ambiguous. What I mean when I say that $K$ is a henselian field is that there exists a henselian DVR $R$, such that $K=Frac(R)$. What I mean when I say that $L$ is a complete field is that there exists a complete DVR $S$ such that $L=Frac(S)$.

Note that every complete field is henselian. Examples of complete fields are $\mathbb{Q}((t))$ (the field of formal Laurent series with coefficients in $\mathbb{Q}$), $\mathbb{Q}_p$ (the $p$-adics), and so forth.

When I try to think of henselian fields that are not complete, the ones that immediately come to mind are algebraic over some complete field. For example $\mathbb{Q}((t))(\sqrt{2})$, $\mathbb{Q}_p^{un}$ (the maximal unramified extension of $\mathbb{Q}_p$), and so forth.

Question

Is it true that for every henselian field $K$ there exists a subfield $L\subset K$ such that $L$ is complete and such that $K/L$ is an algebraic extension?

EDIT: I've re-emphasized this in the comments, but I think it is important to put this in the body of the question: both the term "henselian field" and "complete field" are used in many different contexts to mean different things.

Note that under the definitions above $\mathbb{R}$ does not constitute as a complete field. (This is because $\mathbb{R}$ is not the fraction field of a complete DVR.)

Also note that I do not consider $\mathbb{Q}((t^{1/n}))_{n\in\mathbb{N}}$ to be a henselian field. (This is because my definition of a henselian field requires it to be the fraction field of a henselian DVR, not a general henselian ring.)

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  • $\begingroup$ Do you consider $\mathbf R$ to be a complete field? $\endgroup$
    – KConrad
    May 4, 2012 at 2:29
  • $\begingroup$ No. I defined "complete field" to mean the fraction field of a complete DVR. $\endgroup$ May 4, 2012 at 2:35
  • $\begingroup$ Note also that it is not true that the fraction field of a general henselian ring is a "henselian field" under my definition. Only the fraction field of a henselian DVR is a "henselian field". $\endgroup$ May 4, 2012 at 2:38
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    $\begingroup$ How about the henselization of $\mathbb Z_{(p)}$, where $p$ is a prime? $\endgroup$
    – Angelo
    May 4, 2012 at 3:17
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    $\begingroup$ I think you can just argue by cardinality: a complete field has to be uncountable, but there exist countable Henselian fields. There are explicit examples above, but also the theory of Henselian fields is axiomatisable in first order logic, so it follows from basic model theory that there are countable models (with value group $\mathbf{Z}$) $\endgroup$
    – Moshe
    May 6, 2012 at 11:56

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