I have a question about a result on tamely ramified extensions in Neukirch's Algebraic Number Theory.
Proposition 7.7 in chapter II section 7. The question I have is about the proof which starts by proving that if $e = 1$ then $L = K$.
The statement is as follows:
Let $K$ be a Henselian field. A finite extension $L/K$ is tamely ramified if and only if $L = T(\sqrt[m_1]{a_1}, \dots, \sqrt[m_r]{a_r})$
Where $T$ is the maximal unramified subextension of $L/K$, $a_i \in T$ for any $i$ and $m_i \geq 2$ are integers prime to $p$ the characteristic of the residue field.
The statement is clear. The first part of the proof goes as follows:
We may disregard the field $K$ since $L/T$ is tamely ramified if and only if $L/K$ is. Assume that $L/T$ is tamely ramified and the fields $L$ and $T$ have the same valuation group i.e $e = 1$. We want to show $L = T$.
Assume $L \neq T$, then one can find a unit $\epsilon \in L \setminus T $ such that $Tr_{L/T}(e) = 0$. Since $L/T$ is tamely ramified we have $[L:T] = m$ is prime to $p$ and we have
$$ Tr(\epsilon) = \sum_{i=1}^{m} \epsilon_i = 0 $$
Where the $\epsilon_i$'s are the conjugates of $\epsilon$. Since the residue fields of $L$ and $T$ are the same the elements $\epsilon_i$ all reduce to the same element $\overline{\epsilon}$ in the residue field. Hence
$$ \overline{Tr(\epsilon)} = m \overline{\epsilon} = 0 $$
which is impossible since $m$ is prime to the characteristic $p$ and $\overline{\epsilon}$ is not zero.
So $L = T$ when $e = 1$.
The part I don't get is how are we allows to sum the conjugates of $\epsilon$ and reduce to the residue field when we don't know that $L/T$ is a normal extension? (Working on a normal closure may salvage the argument but I'm not sure).
You answered your question : go to the normal closure $F/K$
The true question is why should $\sigma(\varepsilon)$ be in $\varepsilon+m_F$ for each $\sigma\in Aut(F/K)$.
From the definition of Henselian there is that $O_F$ has a unique maximal ideal and any automorphism $\sigma\in Aut(F/K)$ sends $m_F$ to itself.
That $O_L,O_K$ have the same residue field means that $\varepsilon-a\in m_F$ for some $a\in O_K$ thus $\sigma(\varepsilon)-a\in m_F$ and $\sigma(\varepsilon)\in\varepsilon+m_F$.