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Does anyone know of an elementary proof that an algebraically maximal field is Henselian (ie one that does not assume knowledge of henselizations)?

Definitions: We say a valued field $(K,v)$ is algebraically maximal if it has no proper algebraic intemediate extensions, that is any algebraic valued field extension of $(K,v)$ either increases the value group or the residue field.
We say a valued field is Henselian if if satisfies Hensels lemma, for instance any polynomial $X^n+X^{n-1}+a_{n-2}X^{n-2}+...+a_0$ with $v(a_{n-2}),...,v(a_0)>0$ has a root $b$ in $K$ with $v(b)\geq 0$.

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  • $\begingroup$ Does the "obvious" approach not work? Assume the field is not Henselian, choose a counterexample polynomial $f(X)$, then look at the algebraic field extension $K[X]/f(X)$? (I suppose you need to show that if $f$ is reducible, that one if its factors is a counterexample polynomial) $\endgroup$
    – user14972
    Nov 23 '11 at 17:29
  • $\begingroup$ The problem here is that you somehow need to put a valuation on $K[X]/f(X)$ which does not increase the value group or the residue field. $\endgroup$
    – Conrad
    Nov 24 '11 at 11:43
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Probably this is not a 'direct proof', but 'abstractly' it is: If it were not Henselian, then one could consider its Henselization that, by definition, preserves value group and residue field, and it is therefore a non-trivial extension; this contradicts the assumption of algebraic maximality.

If I might add a short remark: The two notions are equivalent in characteristic 0. Thus, it makes sense to talk about the positive characteristic case. For the latter, the converse is true as far as you consider the field to be 'defectless'; moreover, if you add the extra-assumption of 'perfectness' (so independently from the characteristic):

A defectless, perfect valued field is Henselian if and only if algebraically maximal.

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  • $\begingroup$ Sorry, I realized you wrote "not assuming Henselization" only after I wrote my answer! I apologise. You can thus ignore it. $\endgroup$
    – Pietro
    Aug 16 '12 at 21:11
  • $\begingroup$ Thanks for the answer. At the time I was trying to find a quick and dirty way of showing the existence of henselisations for valued fields of characteristic zero. I didn't come up with one, but I don't really need it anymore. Tho' I would be interested if you happened to know of such a proof. $\endgroup$
    – Conrad
    Aug 18 '12 at 23:49

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