The usual way to define the Henselisation $A^h$ of a local ring $(A, \mathfrak{m})$ is to take direct limit $\varinjlim (B, q)$ over all etale neighborhoods of $A$ (i.e. pairs $(B,q)$ where $B$ is an etale $A$-algebra, prime ideal $q$ lies over $\mathfrak{m}$ such that the induced map $A/\mathfrak{m} \to \kappa(q)= B_q/q$ is an isomorphism).
James Milne gave in his book Etale Cohomology at page 37 in Example 4.10 an alternative construction for Henselisation for case $A$ beeing normal as follows:
Let $K$ be field of fractions of $A$, and $K_s$ be a separable closure of $K$. The Galois group $G$ of $K_s$ over $K$ acts on the integral closure $B$ of $A$ in $K_s$. Let $n$ be a maximal ideal of $B$ lying over $m$, and let $D \subset G$ be the decomposition group of it, that is, $D = \{\sigma \in G \ \vert \ \sigma(n)=n \}$. Let $A^h$ be the localization at $n^D$ of the integral closure $B^D$ of $A$ in $K^D_s$. (Here
$$ B^D= \{b \in B \ \vert \ \sigma(b) = b \text{ for all } \sigma \in D \} $$
Claim: $A^h$ is the Henselization of $A$.
If $A^h$ were not Henselian, there would exist a monic polynomial $f(T) \in A^h[T]$ that is irreducible over $A^h$ but whose reduction $\overline{f}(T) \in (A^h/n^D)[T]$ factors into relatively prime factors. But from such an $f$ one can construct a finite Galois extension $L$ of $K_s^D$ such that the integral closure $A'$ of $A^h$ in $L$ is not local. This is a contradiction since the Galois group of $L$ over $K^D_s$ permutes the prime ideals of $A$ lying over $n^D$ and hence cannot be a quotient of $D$. To see that $A^h$ is the Henselization, one only has to show that it is a union of etale neighborhoods of $A$, but this is easy using (3.21).
Theorem 3.21: Let $X$ be a connected normal scheme, and let $K =R(X)$ the fraction field.
Let $L$ be a finite separable field extension of $K$ , let $X'$ be the normalization of $X$ in $L$, and let $U$ be any open subscheme of $X'$ that is disjoint from
the support of $\Omega^1 _{X'/X}$. (sheaf of relative Kähler differentials) Then $U \to X$ is etale, and conversely any separated
etale morphism $Y \to X$ of finite-type can be written $Y = \coprod U_i \to X$
where each $U_i \to X$ is of this form.
QUESTIONS
Question 1: Above is stated that if $A^h$ would not be Henselian and therefore there would exist an irreducible $f(T)$ over $A^h$ whose reduction $\overline{f}(T)$ factors, then one could construct a finite Galois extension $L$ of $K_s^D$ such that the integral closure $A'$ of $A^h$ in $L$ is not local. How one could construct such non local $A'$ from this $f$ splitting over residue field?
My idea was the following one: I choose an arbitrary zero $a \in K_s$ of $f$ and consider the extension $L:= K_s^D(a)$. Then the integral closure $A'$ of $A^h$ in $L$ contains $A^h[a]= A^h[T]/(f)$. Since $A' $ is finite over $A^h$, it is finite over $A^h[a]$ and in order to show that $A'$ is not local it suffice to show that $A^h[a]$ is not local. $A^h[a]$ is finite over $A^h$ too and therefore every maximal ideal of $A^h[a]$ lies over $n^D$.
Denote by $k^h$ the residue field $A^h/n^D$. Then the fiber over $n^D$ is $k^h \otimes A^h[a]= k^h[T]/(\overline{f})$ factors by assumption that $overline{f}$ splits and Chinese remainder. So $A^h[a]$ has more then one maximal ideal and therefore not local, therefore $A'$ not local too.
Is the approach ok so far? I'm pretty sure that this argumentation should work, but find it too laborious. Is there an easier or say more natural way to construct such non local $A'$ as claimed above?
Question 2: At the end of the paragraph is claimed that quoted Theorem 3.21 helps to realize $A^h$ as union of etale neighbourhoods of $A$. How should (3.21) be applied here reach the claimed result?
Idea: $\Omega^1 _{X'/X}$ has closed support. Therefore we could try to exhaust $A_h$ by union of localization $B^D_s$ for $s \in B^D- n^D$ with $D(s) \cap \operatorname{Supp}(\Omega^1 _{X'/X}) = \emptyset$.
Due to Thm 3.21 these are etale over $A$. Problem: Are these also etale neighbourhoods of $A$? Note that these shold contain prime ideals lying over $m$ with $A/\mathfrak{m} =\kappa(q)$. I don't know how to control this condition.