Let $E$ and $F$ two finite extensions of a field $K$ of degree $[E:K]=m$ and $[F:K]=n$ such that $\gcd(m,n)=1$. Show that if $\alpha\in F$ has degree $r$ on $K$ therefore $\alpha$ has degree $r$ on $E$.
What I did is, there is an irreducible polynomial $f(x)=a_0+a_1x+...+a_rx^r \in K[x]$ such that $f(\alpha)=0$. But how can I continue ? The problem is $\alpha\notin E$ by hypothesis, no ?
It's just a matter of divisibility.
We know that $[K[\alpha]:K]=r$, but $K\subset K[\alpha]\subset F$, so $r$ divides $m$, since $m=[F:K]=[F:K[\alpha]][K[\alpha]:K]=[F:K[\alpha]]*r$.
Therefore $(r,n)=1$.
Let's assume $r>1$, since $r=1$ means $r\in K$, and the problem becomes very easy.
As you said, $\alpha\not\in E$ since $r$ doesn't divide $n$, and $K\subset E\subset E[\alpha]$ is a tower of finite extensions, but also $K\subset K[\alpha]\subset E[\alpha]$ is a tower of extensions, so $r$ and $n$ divide $[E[\alpha]:K]$. But $[E:K]=n$, $[E[\alpha]:E]\le r$ lead to $$ rn|[E[\alpha]:K]=[E[\alpha]:E][E:K]\le rn \implies [E[\alpha]:K]=rn \implies [E[\alpha]:E]=r $$