## Return to Question

2 minor stylistic changes, and corrected "unramified" to "totally ramified" in 2nd sentence

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