I have a problem with the proof of Hazewinkel in his "Local Class Field" article.
First I don't understand why is $F"$ well defined ? I think it needs $L\cap K_r=K$ but I don't see why it is true.
And secondly, why is it so obvious that $L'/K$ is totally ramified
I need some days to find what happened.
First; indeed $K_L=K_{nr}\cap L$ ($\subset$ is obvious) because $K_{nr}\cap L$ contains $K$, is in $L$ and is not ramified because in $K_{nr}$; so we have the reversed inclusion. The same reason worth for any extension of $K$.
For $F''$ to be well defined we just have to see why $l_1k_1=l_2k_2\in LK_r$ implies $F''(l_1k_1)=F''(l_2k_2)$. $l_1k_1=l_2k_2\Rightarrow {k_1\over k_2}={l_2\over l_1}=k\in K_r\cap L=K_L$ then $k_1=kk_2$ and $l_2=kl_1$ and consequently $$F''(k_1g_1)=Frob(k_1)F'(l_1)=Frob(k)Frob(k_2)\frac1{F'(k)}F'(l_2)=Frob(k_2)F'(l_2)=F''(k_2l_2)$$ because $Frob=F'$ on $K_L$.
For the second point.
The good question is : What is $K_{L'}$ ?
Because of the above $K_{L'}=K_{nr}\cap L'=K_s$ for a certain $s\leq r$ because $L'\subset K_rL$. But then $F''|_{K_{L'}}=id$ because $K_{L'}\subset L'=Fix(F'')$ but too $F''|_{K_{L'}}=Frob|_{K_s}$ because we are in $K_r$. The only way to have $Frob|_{K_s}=id$ is $K_s=K$ therefore there is no unramified field between $K$ et $L'$.
Now I just hope i didn't write somewhere an enormity ! So $L'$ is indeed totally ramified.