I have a doubt concerning this problem:
Let $K \supset \mathbb Q$ be a field such that $[K:\mathbb Q]=m$ and let $p(x) \in \mathbb Q[x]$ be an irreducible polynomial of degree $n$ over $\mathbb Q$. If $\gcd\{m,n\}=1$, prove that $p(x)$ is an irreducible polynomial over $K$.
My attempt: Let $\alpha \in \mathbb C$ be a root of $p(x)$. Now consider the fields $\mathbb {Q}(\alpha) \subset \mathbb {K}(\alpha)$ and supose that $[K(\alpha):K]=r$ and $[K(\alpha):\mathbb {Q}(\alpha)]=s$. It is true that $ns=mr$ and $\gcd\{m,n\}=1 \Rightarrow n|r$. How to proceed from now on in order to prove that $p(x)$ is also irreducible over $K$? Thank you!
For any field $F$, any polynomial $q\in F[x]$, and any root $\beta$ of $q$, we have $$[F(\beta):F]\leq\deg(q)$$ with equality if and only if $q$ is irreducible. Another way of saying this is that for any $\beta$ that is algebraic over $F$, we have $$[F(\beta):F]=\text{the degree of $m_{\beta,F}\in F[x]$ (the minimal polynomial of $\beta$ over $F$)}$$ Because you've shown that $n$ divides $[K(\alpha):K]$ and it can be at most $n$, we must have $$\deg(m_{\alpha,K})=[K(\alpha):K]=n=\deg(p)$$ But $p\in \mathbb{Q}[x]\subset K[x]$ is a polynomial with $\alpha$ as a root, so the minimal polynomial $m_{\alpha,K}$ must divide it, that is, $$m_{\alpha,K}\mid p$$ Because their degrees are equal, we must have $p=c\cdot m_{\alpha,K}$ for some constant $c$. In particular, $p$ is irreducible because $m_{\alpha,K}$ is by definition.