Let $F $ be a subfield of $\mathbb R $ and $f (x) \in F [x]$ an irreducible cubic with discriminant $d\gt 0$ and $A_3$ as Galois group. Let $p $ be a prime and $K=F (r) $, with $r $ real such that $r^p \in F$; show that $K $ cannot contain a splitting field of $F $.
I proceeded in this way: the roots are in $\mathbb R $, because (since $f $ is irreducible) a couple of complex roots would imply that $G_f\cong S_n $. Clearly if $K $, defined as above, contains a splitting field of $f $, then $p=3$; thus, we can write every root as $s= ar^2+br+c$ for some $a,b,c \in F $. Now I would show that the minimal polynomial of every $s$ over $F $ has a root in $\mathbb C $, and this would lead to a contradiction (since $f $ must be the minimal polynomial of all its roots). However I'm not sure at all that this is the right direction.
Moreover the following point asks to show that there is no subfield $E \lt \mathbb R $, with a root tower over $F $, such that $E $ contains a splitting field for $f $. However such a root tower can be only in the form that we just discussed, because if it was like, say, $E:F_1:F$, with $E=F_1 (u) $, $u^3\in F_1$, and $F_1=F(v) $, $v^2\in F $, then the Galois group of $f $ would be $S_n $. However this seems quite too simple, so maybe I shouldn't take for granted that $p=3$; even if it seems reasonable that in a field of the form $F (w) $, $w^p \in F $, $p\gt 3$, there are no elements with minimal polynomial of degree $3$ (condition necessary to be a root of $f $). Can you give me a clarification on this exercise? Thanks a lot
Let $m\ge 1$ be the least integer such that $r^m \in F$.
$r^p \in F,r^m\in F$ implies $r^{\gcd(m,p)}\in F$. If $m< p$ then $r=r^{\gcd(m,p)}\in F$ and the problem is trivial. Thus we assume $m=p$. Since $r^p$ is a root of $x^p-r^p$ and in characteristic $0$ everything is separable, the minimal polynomial of $r$ will be of the form $h(x)=\prod_{j=1}^J (x-r e^{2i\pi n_j/p})$, thus $\pm r^J=|h(0)| \in F$. Overall it means $\deg(h)=J=p$
Since $p$ is prime there is no field in between $F$ and $F(r)$. Thus the problem is to show that $F(r)$ is not the splitting field of your irreducible cubic.
But this is obvious because $F(r)\subset \Bbb{R}\implies e^{2i\pi /p}\not \in F(r)\implies F(r)/F$ is not Galois.