Let $F$ be a field, $\alpha$ and algebraic number over $F$ with minimal polynomial $P$. Let $\phi_0 : F \to E$ be a morphism from $F$ to another field $E$. I would like to say that there are as many extensions of $\phi_0$ to $\phi : F[\alpha] \to E$ as there are roots for $\phi_0(P)$. I see how the argument works in the case where $\phi_0$ is trivial, however I have issues generalising it.
In the case of a trivial $\phi_0$, we show that such $F$-morphisms $F[\alpha] \to E$ are in bijection with the roots of $P$ in $E$. Indeed, since $\phi$ is a $F$-morphism, $\phi(\alpha)$ is also a root of $P$ so that $\phi$ permutes the roots. Conversely, if $\beta$ is another root of $P$ in $E$, it defines a morphism $F[\alpha] \to E$ by $F[\alpha] \simeq F[X]/(P) \to E$ where the arrow is just the evaluation at $\beta$.
I would like to do the same for a general morphism $\phi_0$, and show that the sought morphism extensions are in bijection with roots of $\phi_0(P)$. If $\alpha$ is a root of $P$, it still holds that $\phi(\alpha)$ is a root of $\phi_0(P)$. Conversely, let $\beta$ be a root of $\phi_0(P)$. There is at the evaluation morphism $F[X] \to E$ that factorises by the minimal polynomial of $\beta$. However, here, I am not sure if we can conclude that this one is $P$ (not even $\phi_0(P)$, what about irreducibility ?), so that I don't see the algorithm $F[X]/(\mathrm{minpol}(\beta)) \simeq F[\alpha]$ as before.
Is there something I missed in this construction?