Show $h(x) in F[x]$

Question

$Q)$ 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 $, $alpha_1alpha_2 + alpha_2alpha_3+alpha_1alpha_3 in F$ and  $alpha_1alpha_2alpha_3 in F$,
So the $phi(alpha_1 + alpha_2 + alpha_3), phi(alpha_1alpha_2 + alpha_2alpha_3+alpha_1alpha_3), phi(alpha_1alpha_2alpha_3) in F$ (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.

Kenny Wong · Answer

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 $F$.

Show $h(x) in F[x]$

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