I've been trying to understant this problem from Gallian's Abstract algebra:
Suppose that $p(x) \in F[x]$ and $E$ is a finite extension of $F$. If $p(x)$ is irreducible over $F$ and deg $p(x)$ and $[E:F]$ are relatively prime, show that $p(x)$ is irreducible over $E$.
Answer in the books says: Let $\alpha$ be a zero of $p(x)$ in some extension of $F$. First note $[E(\alpha):E] \leq [F(\alpha):F] = $ deg $p(x)$. I don't understand how can we use that extension degree od $F(\alpha)$ over $F$ is equal to deg $p(x)$ when we don't have that polynomial is monic so we only know that it's irreducible and that $\alpha$ is his zero but we don't know that it's minimal polynomial. I'm obviously missing something but I can't see what it is.
If the leading coefficient of $p(x)$ is $c$, then $\frac 1cp(x)$ is the minimal polynomial of $\alpha$.