Show $h(x) \in F[x]$ $Q)$ There are fields $\mathbb{Q} \subset F \subset K$ (Here the $\mathbb{Q}\subset $F and $\mathbb{Q}\subset $ K are Galois extension)
Say the $f(x) \in  F[x]$ and $g(x) \in \mathbb{Q}[x]$ with $f(\alpha_1) = g(\alpha_1)$ for some $\alpha_1 \in K$. Plus $f$ has roots $\alpha_1, \alpha_2$ and $\alpha_3$ in $K$
(Here the $f$ is a irreducible over $F$ and the $g$ is a irreducible over $\mathbb Q$
)
There is a $\phi \in G(K/\mathbb{Q}) s.t. h(x) = (x-\phi(\alpha_1))(x-\phi(\alpha_2))(x-\phi(\alpha_3))$ with $h \neq f$
Show $h(x)$ is a irreducible in $F[x]$

I've already shown under hypothesis, $h(x) \in F[x]$ (I.e. $irr(\phi(\alpha_1), F) = h(x) $ by using $G(K/F) \lhd G(K/\mathbb{Q})$)
But the problem is I couldn't show $h(x) \in F[x]$ the hypothesis that I used.
So I checked the answer sheet but it said
$f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$  and $h = \phi(f)$
Hence $\alpha_1 + \alpha_2 + \alpha_3 \in F $, $\alpha_1\alpha_2 + \alpha_2\alpha_3+\alpha_1\alpha_3 \in F$ and  $\alpha_1\alpha_2\alpha_3 \in F$,
So the $\phi(\alpha_1 + \alpha_2 + \alpha_3), \phi(\alpha_1\alpha_2 + \alpha_2\alpha_3+\alpha_1\alpha_3), \phi(\alpha_1\alpha_2\alpha_3) \in F$ (I.e. $h \in F[x]$).
Why does those results happening? I can't understand Why does the $\phi$'s image of those are element in $F$ at all. Any help always welcome. Thanks.
 A: I think what you're missing is that if $\mathbb Q \subset F \subset K$, with $F$ Galois over $\mathbb Q$ and $K$ Galois over $\mathbb Q$, and if $\phi \in {\rm Gal}(K / \mathbb Q)$, then for every $\alpha \in F$, $\phi(\alpha)$ is also in $F$.
Why is this true? Take an $\alpha \in F$, and consider its minimal polynomial $m(X)$ over $\mathbb Q$. Since $\alpha$ is a root of $m(X)$ and since $\phi$ fixes $\mathbb Q$, $\phi(\alpha)$ must also be a root of $m(X)$. But since $F$ is Galois over $\mathbb Q$, any irreducible polynomial in $\mathbb Q[X]$ with at least one root in $F$ splits completely in $F$. This applies to $m(X)$, which is irreducible and has at least one root in $F$ (namely $\alpha$). Hence $m(X)$ splits completely in $F$, i.e. all roots of $m(X)$ are contained in $F$, including $\phi(\alpha)$.

By the way, once you realise that $\phi$ maps $F$ into $F$, it's not too hard to convince yourself that restriction $\phi|_F$ of $\phi$ to $F$ is an automorphism of $F$ (which fixes $\mathbb Q$). Once you realise this, and once you realise that your $h$ is simply $\phi(f)$, it's not hard to see that $f$ being irreducible over $F$ implies that $h$ is irreducible over $F$.
