Clarification on internal sets I am trying to better understand a definition given by Robinson on page 823 of the paper "Nonstandard Arithmetic".
Fix a field $F$ and Galois extension $\Phi$ of $F$ (possibly of infinite degree). One then has enlargements $F^*, \Phi^*$. One can find a certain field $\Psi$, a subfield of $\Phi^*$ containing $F^*$ and of starfinite degree over $F^*$, which is also normal over $F^*$ (basically it is obtained as bound for all the finite normal extensions of $\Phi$ over $F$; I don't think it's particular construction matters too much).
The author then remarks that $\Psi$ has a Galois group $H$ over $F^*$, and writes that it "consists of all internal automorphisms of $\Psi$ fixing $F^*$, and writes that "$H$ consists of all the internal automorphisms of $\Psi$ which leave the elements of $F^*$ invariant".
My concern is that if $H$ is not internal, then it seems like it could not have any internal automorphisms. I think I can resolve this as follows: given the field $K$, the set $S$ of fields containing $K$ in some (fixed) larger field can be defined in a first-order fashion. If we apply this to $F$ and $\Phi$ we see that the set of fields in $\Phi^*$ containing $F^*$ is an internal set, and every element of an internal set is internal. Is that the right way to go about this?
On the other hand, I don't think that the set of elements in $H$ is internal except in the case where $\Phi$ has finite degree over $F$, otherwise $\Phi^* - H$ would be internal (and nonempty because of the degree), and then one could descend to $\Phi/F$ and conclude that no element of $\Phi$ is in any finite extension of $F$, which is absurd since the extension is algebraic.
Is my understanding of these points correct? My intuition (and probably my understanding) of internal-external sets is pretty cloudy but this would help to clarify things.
 A: Here $\Psi$ is meant to be internal.  As Robinson mentions on page 820, all the bounds of concurrent relations in the enlargement are supposed to be internal, and that is the manner in which $\Psi$ is defined.  Indeed, it would not make sense to talk about $\Psi$ being "starfinite" if it were not internal, since the notion of "degree" (which would be an element of ${}^*\mathbb{N}$) is only defined for internal field extensions.
It's not clear to me exactly what argument you have to "descend to $\Phi/F$ and conclude that no element of $\Phi$ is in any finite extension of $F$".  I think what you mean is something like the following.  Since ${}^*\Phi\setminus \Psi$ is nonempty,
$$(*)\text{ there is an element of ${}^*\Phi$ that is not in any finite extension of $F$.}$$  Then, since the enlargement is an elementary extension, the corresponding statement must be true of $\Phi$.  That is, there is an element of $\Phi$ that is not in any finite extension of $F$, which is a contradiction since $\Phi$ is algebraic over $F$.
The problem with this argument is that $(*)$ is not true in the language of the enlargement (and so cannot be transferred back down to $\Phi$).  In that language, the word "finite" refers to the internal notion of finite (i.e., "starfinite"), so "finite extension of $F$" really means "starfinite extension of $F$".  While $\Psi$ contains every subfield of ${}^*\Phi$ that is actually a finite extension of $F$, it does not contain every starfinite extension of $F$.
