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Let $(K, \nu)$ be a valued field and $x$ is transcendental over $K$. Is there exist a henselian extension of $(K, \nu)$ in between $(K, \nu)$ and $(K(x), \nu^{'})$ where $\nu^{'}$ is an extension (arbitary) of $\nu$.

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up vote 1 down vote accepted

In general the answer is no: let $(K,v)$ be non-henselian and let $v_x$ be the Gauss extension of $v$ to the rational function field $K(x)$:

$v_x(a_nx^n+\ldots +a_0):=\min (va_i : i=0,\ldots ,n)$.

Then $(L,v_x)$ is non-henselian for every subfield $L\subseteq K(x)$.

Proof: by assumption there exists a monic polynomial $f\in O_v[z]$ such that the reduced polynomial $\overline{f}\in k[z]$, $k$ the residue field of $O_v$, has a simple root $\bar{\alpha}$ in $k$ but $f$ has no root in $K$ that reduces to $\bar{\alpha}$. This property of $f$ does not change when considering $f$ as a polynomial in $(O_{v_x}\cap L)[z]$: $K$ is algebraically closed in $K(x)$ and thus in $L$. The residue field of $O_{v_x}$ is $k(x)$, hence $k$ is algebraically closed in the residue field of $O_{v_x}\cap L$.

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Thank you Hagen –  Rajnish Jan 15 '13 at 3:41
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