Let $\{\alpha_i\}_{i ∈ [n]} ⊆ \bar{\Bbb Q}$ be a set of $n ≥ 1$ algebraic numbers. Suppose that for all $i,j ∈ [n]$, there is a $\Bbb Q$-isomorphism $\phi_{ij} : \Bbb Q(\alpha_i) \tilde{\rightarrow} \Bbb Q(\alpha_j)$ with $\phi_{ij}(\alpha_i) = \alpha_j$. Let $f := \prod_{i=1}^n (X - \alpha_i)$.
To demonstrate: $f ∈ \Bbb Q[X] \Rightarrow$ $f$ is irreducible over $\Bbb Q$.
By assumption, the minimal polynomials of the $\alpha_i$ over $\Bbb Q$ are all equal. I also know that if $f$ is separable, and Gal$(f)$ acts transitively on the roots of $f$, then $f$ is irreducible. As $\Bbb Q$ is perfect, all minimal polynomials are indeed separable, so if $f$ were the minimal polynomial of some element, transitivity of the aforementioned action will imply the result. So my question is: how do we show this? And how does the fact that $f ∈ \Bbb Q[X]$ come into the story?
Edit
Yikes, if $f$ were the minimal polynomial of some element, it would be irreducible regardless, of course... But of course $f$ is separable, as the $\alpha_i$ are all different. This leaves only transitivity.
Note that each $\phi_{ij}$ can be extended to an isomorphism of the splitting fields of $f$ and $\phi_{ij}(f)$ $$\psi : Ω^f_{\Bbb Q}\tilde{\rightarrow} Ω^{\phi_{ij}(f)}_{\Bbb Q}, \,\,\,\,\,\,\,\, \psi|_{\Bbb Q(\alpha_i)} = \phi_{ij}.$$
Here $Ω^f_{\Bbb Q} = \Bbb Q(\alpha_1, \alpha_2, \ldots, \alpha_n)$, and by assumption, $f ∈ ℚ[X]$, so $\phi_{ij}(f) = f$.
That means that for every two roots $\alpha_i, \alpha_j$ we have our $\phi_{ij}$ satisfying $$\psi(\alpha_i) = \phi_{ij}(\alpha_i) = \alpha_j. $$ Thus each root can be mapped onto any other root under $\psi ∈ \text{Aut}_{\Bbb Q}(\Bbb Q(\alpha_1, \alpha_2, \ldots, \alpha_n)) = \text{Gal}(f)$, so $\text{Gal}(f)$ acts transitively on the set of roots of $f$.