The answer is certainly "Yes", but this is the problem I met in Alegebraic Algebraic Number Theory by Neukirch. I guess that I must make be doing something wrong, since otherwise I will get the statement "There are no totally ramified extensions except the trival ones".
Let $K$ be Henselian field, $L/K$ be a finite, totally ramified extension. Let $\lambda$ and $\kappa$ be the residue field of $L$ and $K$ respectively. Because $L/K$ is unramifiedtotally ramified, and $K$ is the maximal unramified subextension, so we have $\lambda=\kappa$. If $L\ne K$, let $a \in L-K$, since . Since the valuation is non-trival, by multiple certain multiplying by an element in $K$, we can suppose $a \in O_L$, the valuation ring of $L$. By
Because "The the valuation ring of $L$ is the integral closure of the valuation ring of $K$ in $L$ "(P144, Chapt 2 Theorem (6.2) of Neukirch), let $f(x) \in O_K[x]$ be the minimal polynomial of $a$ in $O_K$, where $O_K$ is the valuation ring of $K$. (one One can prove $f(x)$ is monic, and I guess it may different differ from the minimal polynomial in $K$), where $O_K$ is the valuation ring of over $K$. K$.) Let $\bar{f}(x)$ be the corresponding polynomial in over the residue field $\kappa$. It must be the minimal polynomial of $\bar{a} \in \lambda$, because , otherwise, by Hensel's lemma, $\bar{f}$ admits a factorization in $\kappa[x]$ implies $f$ admits a factorization in $O_K[x]$. But since $\lambda=\kappa$, we get $deg(\bar{f})=deg(f)=1$ ($f$ is monic). This means $a\in K$, a contradiction. This means $L=K$ !
