If $\phi$ is an F-map from $K$ to $E$, both field extensions of $F$, then $\alpha \in K$ and $\phi (\alpha)$ have the same minimum polynomial

Question

Definition of an F-map:

If $K$ and $E$ are field extensions of $F$, an F-map is a homomorphism, $\phi: K \rightarrow E$ such that $F$ is fixed.

I'm reading over a proof of why the number of distinct F-maps from $K$ to $E$ is less than or equal to $[K:F]$ (assuming finite) and the fact that $\alpha$ and $\phi (\alpha)$ have the same minimum polynomial is one of the facts stated, though I can't see why.

Answer 1

Let $p(X)=\sum a_kX^k\in F[X]$. Then $$ \phi(p(\alpha))=\sum \phi(a_k)\phi(\alpha)^k=\sum a_k\phi(\alpha)^k=p(\phi(\alpha))$$ Conclude that $$p(\alpha)=0\iff p(\phi(\alpha))=0.$$