TITLE: Prove that $k(\alpha+\beta)=k(\alpha,\beta)$
QUESTION [6 upvotes]: I am trying to solve the following problem:

Let $k$ be a finite field and let $k(\alpha,\beta)/k$ be finite. If $k(\alpha)\cap k(\beta)=k$, prove that $k(\alpha,\beta)=k(\alpha+\beta)$.

What I already know:

$k(\alpha+\beta)\subset k(\alpha,\beta)$
$k(\alpha,\beta)/k$ is a Galois extension.
$[k(\alpha,\beta):k]=[k(\alpha):k][k(\beta):k]$.

Some thoughts in mind:

Prove that $\alpha\in k(\alpha+\beta)$ or $\beta\in k(\alpha+\beta)$.
Prove that $[k(\alpha,\beta):k(\alpha+\beta)]=1$
Try to express $[k(\alpha+\beta):k]$ in terms of $[k(\alpha):k]$ and $[k(\beta):k]$.
Try to use some Fundamental Theorem of Galois Theory.

However, I still have no idea how to move on. Any help?

REPLY [2 votes]: Since they're Galois, you know that $$\text{Gal}(k(\alpha,\beta)/k)\cong\text{Gal}(k(\alpha)/k)\times\text{Gal}(k(\beta)/k).$$ But then, after the identification you can see that you can produce $[k(\alpha):k][k(\beta):k]$ different automorphisms of $\alpha+\beta$, by acting just on $\alpha$ while fixing $\beta$ from the subgroup
$$\{1\}\times\text{Gal}(k(\beta)/k)$$
and then applying automorphisms to all of those $[k(\alpha):k]$ distinct numbers from the 
$$\text{Gal}(k(\alpha)/k)\times\{1\}$$
factor. Hence the degree of the extension it generates is at least this size--i.e. $[k(\alpha):k][k(\beta):k]$--but then
$$[k(\alpha):k][k(\beta):k]\le [k(\alpha+\beta):k]\le [k(\alpha,\beta):k]\le [k(\alpha):k][k(\beta):k]$$
so equality is everywhere.