A question on an answer on Math Overflow about Artin approximation I have a question on an answer of this Math Overflow question.
Let $(A,I)$ be a commutative excellent normal local domain. The completion
$$
\hat A=\underset{\longleftarrow}{\operatorname{lim}} A/(I^nA)
$$
has $A$ as a subring by the canonical map. An answer to the referred question on MO states that the henselization $A^h$ of $A$ can be defined as the separable closure $S$ of $A$ in $\hat A$. More precisely, an element $a\in \hat A$ is in $A^h$ iff there if a separable polynomial $f\in A[X]$ with $f(a)=0$ (is this interpretation correct?). According to the answers, this should follow from Artin approximation.

How does this description of $A^h$ follows from Artin approximation? Where do I need the properties ''excellent'' and ''normal''? Where ''separability''?

At least for the (probably easy?) direction $S\subseteq A^h$ there should be an other argument.
To cite a lecture of Popescu, 

a Noetherian local ring $(A,I)$ has the property of approximation if
  every finite system of polynomial equations $f$ over $A$ in
  $X_1,\ldots,X_n$ has its solutions in $A$ dense with respect to the
  $I$-adic topology in the set of its solutions in the completion $\hat A$
  of $A$; that is, for every solution $\hat x$ of $f$ in $\hat A$ and
  every positive integer $c$ there exists a solution $x$ of $f$ in $A$
  such that $$x\equiv \hat x\mod I^c\hat A.$$

The Artin approximation theorem is the statement that some rings have this property of approximation. This should be relevant for the question with a single polynomial equation instead of a system. I think one needs ''normality'' and ''excellence'' for the Artin approximation theorem to hold but I can't find it in the literature formulated with these properties. Moreover, I am curious where the separability comes into the game.
 A: If $(R, \mathfrak m)$ is a Noetherian local ring with henselization $R^h$, then the map on completions $R^\wedge \to (R^h)^\wedge$ is an isomorphism, see Lemma Tag 06LJ. Since also $R^h$ is Noetherian (ibid.) we may think of $R^h$ as a subring of its completion (because the completion is faithfully flat in the Noetherian case). In this way we see that we may identify $R^h$ with a subring of $R^\wedge$.
Now we can start to think about which elements of $R^\wedge$ are in $R^h$. For simplicity we assume $R$ is a domain with fraction field $K$. Clearly, every element $f$ of $R^h$ is algebraic over $R$, in the sense that there exists an equation of the form $a_n f^n + \ldots + a_1 f + a_0 = 0$ for some $a_i \in R$ with $n > 0$ and $a_n \not = 0$.
Conversely, assume that $f \in R^\wedge$, $n \in \mathbf{N}$, and $a_0, \ldots, a_n \in R$ with $a_n \not = 0$ such that $a_n f^n + \ldots + a_1 f + a_0 = 0$. OK, now by Artin Approximation if $R$ is a G-ring (slightly weaker than excellent), then, for every $N > 0$ there exists an element $g \in R^h$ with $a_n g^n + \ldots + a_1 g + a_0 = 0$ and $f - g \in \mathfrak m^N R^\wedge$, see Theorem Tag 07QZ. We'd like to conclude that $f = g$ when $N \gg 0$. If this is not true, then we find infinitely many roots $g$ of $P(T)$ in $R^h$. This is impossible because (1) $R^h \subset R^h \otimes_R K$ and (2) $R^h \otimes_R K$ is a finite product of field extensions of $K$. Details omitted.
Conclusion: If $R$ is a local domain with fraction field $K$ and a G-ring, then $R^h \subset R^\wedge$ is the set of all elements which are algebraic over $K$.
