Field extensions and gcd

Question

Let $L|K$ be a field extension and let $u, v \in L$ be algebraic elements over $K$ such that $[K(u):K]=n$ and $[K(v):K]=m$.

1. Show that if $\gcd(m, n)=1$ then $Irr(v, k)$ is irreducible on $K(u)$.
2. Show that if $\gcd(m, n)=1$ then $[K(u, v):K]=mn$.

It's clear that $1$ implies $2$, since $[K(u, v):K]=[K(u, v):K(u)][K(u):K]$, and , by (1), $[K(u, v):K(u)]=m$. I'm stuck on $1$. Can someone help me?

Answer 1

Hint: 2. also implies 1., namely:

$$[K(u,v) :K] = [K(u,v) : K(u)]\cdot [K(u):K] = [K(u,v): K(v)]\cdot [K(v):K]$$

tells you something about the minimal polynomial of $v$ over $K(u)$.

Answer 2

Ok, I think I've made it.

We know that: $$[K(u,v) :K] = [K(u,v) : K(u)]\cdot [K(u):K] = [K(u,v): K(v)]\cdot [K(v):K]$$

Let $[K(u,v) : K(u)]=a$ and $[K(u,v) : K(u)]=b$. The equation above is telling us that $am=bn$. Notice that $n|am$, and, since $mdc(n, m)=1$, this implies that $n|a$. Therefore, $n \leq a$.

On the other hand, $a=Deg(Irred(v, K(u)))\leq Deg(Irred(v, K))=n$. Therefore, $n=a$ and $[K(u,v) :K]=mn$, which proves (2).

Since $Irred(v, K(u))$ divides $Irred(v, K)$ and both are monic and have the same degree, they are the same. Therefore, $Irred(v, K)$ is irreducible over $K(u)$, which proves (1).