# Algebraically maximal valued fields

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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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) –  Hurkyl Nov 23 '11 at 17:29
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. –  Conrad Nov 24 '11 at 11:43