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 $\nu$.



Henselization of valued field
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 $O\_{v\_x}\cap L$.


