Given $f = x^3 - 3x + 1$, I want to find the set of all elements of the splitting field of $f$ over $\mathbb{Q}$ such that the orbit under the action of G has at most 2 elements where $G$ is the group of automorphisms of $E$. This set of elements should be an extension of $\mathbb{Q}$.
I have already shown that such a set is a subfield of the splitting field in a previous problem. I attempted to find $Gal_\mathbb{Q}(f)$ and I got that it has 3 distinct real nonrational roots and has discriminant 81, which is a square and is therefore isomorphic to $\mathbb{Z}_3$. Now I don't know how to find the subfield of elements whose order under the action of $G$ is at most 2. Any suggestions?
Obviouly the elements whose orbit is $1$ is $\mathbb{Q}$, as this elements are precisely the fixed field of $G$, which is $\mathbb{Q}$, as the extension is Galois.
Now we consider element whose orbit is of size $2$. Let $\alpha \in E$ and assume that the orbit of $\alpha$ under action of $G = \text{Gal}(E/\mathbb{Q})$ is of size $2$. Let $\alpha_1$ be the other element in the orbit. Then we have that $\sigma(\alpha) = \alpha_1$ for some $\sigma \in G$. We'll prove that $g(x) = (x-\alpha)(x-\alpha_1)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. This is true as $g$ is fixed by $G$ (can you see why?) and so $g \in \mathbb{Q}[x]$ and the irreducibility follows trivialy, as we can't have $\alpha \in \mathbb{Q}$. (In fact $\sigma(\alpha)$ is a root of the minimal polynomial of $\alpha$ over $\mathbb{Q}$, so we conclude that $\deg \min(\alpha,\mathbb{Q}) \ge 2$ and as $\min(\alpha,\mathbb{Q}) \mid g$ we get $\min(\alpha,\mathbb{Q}) = g$)
Now consider the subfield $L = \mathbb{Q}(\alpha)$. We have that $[L:\mathbb{Q}] = \deg g = 2$. However then we have that:
$$3 = [E:\mathbb{Q}] = [E:L][L:\mathbb{Q}]= 2[E:L]$$
But this leads to an obvious contradiction. Hence there are no elements whose orbit under the action of $G$ is of size exactly $2$. From here we deduce that the field of such elements is $\mathbb{Q}$
In fact we know that $\sqrt{\Delta}$ is an element whose orbit is at most $2$, as the only possibilty for mapping it is to $-\sqrt{\Delta}$ and itself. So we know that $\mathbb{Q}(\sqrt{\Delta})$ is a subfield of the wanted field. In fact one can prove that this is exactly the wanted subfield.
Let $L$ be the wanted field then we prove that $[L:\mathbb{Q}] \le 2$. Indeed if it wasn't then there exists a primitive element $\alpha$ who has at least three conjugates in the splitting field of $f$. (This elements might not be in $L$, but are certainly in $E$, as the extension is Galois). However the Galois group acts trasitively on the roots of an ireducible polynomial and so we have that the orbit of $\alpha$ has at least three element, which is a contradiction. Therefore we can conclude that $[L:\mathbb{Q}] \le 2$ and as $\mathbb{Q}(\sqrt{\Delta}) \subseteq L$ we have that $\mathbb{Q}(\sqrt{\Delta}) = L$
However note that this doesn't work when the polynomial is of degree more than $3$. For example take $x^4+1$. Then it's splitting field is $E=\mathbb{Q}(\sqrt{2},i)$. Now $\sqrt{2}$ and $i$ both have an orbit of $2$ under the action of the $\text{Gal}(E/\mathbb{Q})$, but $\sqrt{2} + i$ has an orbit of $4$ elements, so those elements don't constitute a field. In deed if that was the case then $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(i) \subseteq L$, which means that $L=E$, which is impossible.