Let $f$ be a polynomial over a field $K$ of degree $n$ such that $f(x^2)$ is separable. Assume that the Galois group $G$ of (a splitting field of ) $f(x^2)$ is maximal, that is to say, the wreath product $G= C\_2 {\rm wr} S\_n$. (In particular, the Galois group of $f$ is $S\_n$.)


The action of the group $G$ on the roots of $f(x^2)$ coincides with the imprimitive action of the wreath product. 


My question is whether there is a natural set with Galois action which corresponds to the primitive action of the wreath product. 



Realization of the primitive action of a wreath product in a Galois group
Yes, there is such a set. First, identify $C\_{2}$ with $\{ \pm 1 \}$, and note that the primitive action of $C\_{2}~{\rm wr}~S\_{n} = C\_{2}^{n} \rtimes S\_{n}$ is on the set of functions $\theta : \{ 1, 2, \ldots, n \} \to C\_{2}$. An element of $C\_{2}^{n}$ acts component-wise and an element of $h \in S\_{n}$ acts by sending $\theta$ to the function $h \cdot \theta(s) = \theta(h^{-1}(s))$. (I'm following the notation from [here](https://groupprops.subwiki.org/wiki/Action_of_wreath_product_on_function_space).)


Let $\beta\_{1}$, $\beta\_{2}$, $\ldots$, $\beta\_{n}$ be the roots of $f$, and define $S = \{ \pm \sqrt{\beta\_{1}} \pm \sqrt{\beta\_{2}} \pm \cdots \pm \sqrt{\beta\_{n}} \}$. Given an element $s = \sum\_{i=1}^{n} \epsilon\_{i} \sqrt{\beta\_{i}}$, we can identify $s$ with the function $\theta : \{ 1, 2, \ldots, n \} \to C\_{2}$, where $\theta(i) = \epsilon\_{i}$. (It's easy to see that $\sqrt{\beta\_{1}}$, $\sqrt{\beta\_{2}}$, $\ldots$, $\sqrt{\beta\_{n}}$ are linearly independent over $K$, and so this is well-defined.)


Let $L$ be the splitting field of $f(x)$ and $M$ be the splitting field of $f(x^{2})$. The elements of $C\_{2}^{n}$ correspond to those elements in ${\rm Gal}(M/L)$. For $\sigma \in {\rm Gal}(M/L)$, define $\epsilon\_{i} = \sigma(\sqrt{\beta\_{i}})/\sqrt{\beta\_{i}}$. Then, $\sigma$ corresponds to the function $\theta(i) = \epsilon\_{i}$.


If $h \in S\_{n}$, the element of ${\rm Gal}(M/K)$ corresponding to $h$ is the one for which $\sigma(\sqrt{\beta\_{i}}) = \sqrt{\beta\_{h(i)}}$. We then have that if $s = \sum \epsilon\_{i} \sqrt{\beta\_{i}} \in S$, then
$$ \sigma(s) = \sum \epsilon\_{i} \sqrt{\beta\_{\sigma(i)}} = \sum\_{i} \epsilon\_{\sigma^{-1}(i)} \sqrt{\beta\_{i}}. $$


This shows the actions of $C\_{2}^{n}$ and $S\_{n}$ on the set of functions corresponds to the actions of $C\_{2}^{n}$ and $S\_{n}$ on $S$.


Addendum for those who are might be confused about the terminology of "primitive" action: In the literature on wreath products, the action of $A~{\rm wr}~B$ on the set of functions is sometimes called the product action. This product action is primitive if and only if $B$ is transitive and $A$ is primitive but not regular. (This is apparently Lemma 2.7A from Dixon and Mortimer's book "Permutation groups".) The action of $C\_{2}~{\rm wr}~S\_{n}$ is not primitive because $C\_{2}$ is primitive *and* regular. 


