# What is the Galois group of one ultrapower over another ultrapower?

Let $$F$$ be a field, let $$E$$ be a field extension of $$F$$, and let $$U$$ be an ultrafilter. Then my question is, what is the relationship between the Galois groups $$Gal(\Pi_U E/\Pi_U F)$$ and $$Gal(E/F)$$?

Or if that's too general, is it at least possible to say something in the case when $$E$$ is a number field and $$F=\mathbb{Q}$$?

• Try $E=\mathbb{Q}[\sqrt{2}].$ – dodd Jan 24 at 22:11
• I think that if for every $n$ there are only finitely many degree extensions of $E$ in $F$ then the two Galois groups are isomorphic. – Erik Walsberg Jan 25 at 4:58

$$\newcommand{\Gal}{\operatorname{Gal}}$$If $$E/F$$ is a finite Galois extension, then $$\Gal(\prod_UE/\prod_UF)$$ is canonically isomorphic to $$\Gal(E/F)$$. Indeed, by the primitive element theorem, $$E=F(\alpha)$$ for some $$\alpha\in E$$. This means every element of $$E$$ can be written as a polynomial in $$\alpha$$ with coefficients in $$F$$ of degree smaller than $$d=\deg\alpha$$.
Let $$\alpha^*$$ be the image of $$\alpha$$ in the ultrapower. Then we have $$\prod_UE=(\prod_UF)(\alpha^*)$$, as is straightforward from the description using polynomials above. Therefore any element automorpmism of $$\prod_UE/\prod_UF$$ is determined by the image of $$\alpha^*$$, which on $$U$$-most indices must coincide with some conjugate of $$\alpha$$, and hence is induced by an automorphism of $$E/F$$. Conversely, any automorphism of $$E/F$$ induces an automorphism of the ultrapowers.
If $$E/F$$ is Galois but not finite, then $$\prod_UE/\prod_UF$$ need not a Galois extension, in fact it can fail to even be algebraic extension. This happens whenever $$U$$ is a nonprincipal ultrafilter on $$\mathbb N$$ (or more generally for any $$U$$ which is not countably complete). Indeed, in that case for any $$n\in\mathbb N$$ we can pick $$\alpha_n\in E$$ which has degree at least $$n$$ over $$F$$. Then $$[(\alpha_n)]\in\prod_UE$$ will not satisfy a polynomial equation of degree $$\leq n$$ over $$\prod_UF$$ for any $$n$$ by Łoś, so will be transcendental. You can still ask for the automorphism group of this extension, but I'm not sure it will have a nice explicit description. A natural guess would be that the automorphism group is $$\prod_U\Gal(E/F)$$, but I have a suspicion it will be larger than that.
If $$E/F$$ is infinite and $$U$$ is countably complete, then at the very least you can get that $$\prod_UE/\prod_UF$$ is algebraic and Galois. At least in the case of $$E,F$$ countable you should be able to get an isomorphism like in the first part of my answer, but I have no idea about general case.
Edit: as Andreas Blass points out in a comment below, if $$U$$ is countably complete, then $$\prod_U E\cong E$$ if $$E$$ is countable. More generally, if $$U$$ is $$\kappa$$-complete, then $$\prod_UE\cong E$$ whenever $$|E|<\kappa$$. This in particular will hold if $$U$$ is countably complete and $$|E|$$ is below the first measurable.
• About the last sentence of the answer: If $U$ is countably complete, and $E,F$ are countable, then the ultrapowers by $U$ are canonically isomorphic to $E,F$ themselves, so the Galois group doesn't change. The same is true if the cardinalities of $E,F$ are uncountable but smaller than the first measurable cardinal. – Andreas Blass Jan 24 at 23:53