Does the sum generate the field compositum when degrees are prime?

Question

Inspired by Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

Suppose $k$ is a field and $\alpha,\beta$ are algebraic over $k$ such that $[k(\alpha):k]=p$ and $[k(\beta):k]=q$ with distinct primes $p,q$. Then $$ 1<[k(\alpha,\beta):k(\alpha)]\leq q \quad\text{so}\quad 1<[k(\alpha,\beta):k(\alpha)]\cdot[k(\alpha):k]=[k(\alpha,\beta):k]\leq pq $$ But $[k(\alpha):k]=p$ and $[k(\beta):k]=q$ are divisors of $[k(\alpha,\beta):k]$, so $[k(\alpha,\beta):k]=pq$

Now $k(\alpha+\beta)\subset k(\alpha,\beta)$ is a subfield, so $[k(\alpha+\beta):k]=:n$ divides $pq$.

Is it true that $n\neq p$ and $n\neq q$ and therefore $k(\alpha+\beta)= k(\alpha,\beta)$ or is there a counterexample?

EDIT: The proof for $k=\mathbb Q$ is quite long, maybe in this situation there is a direct proof? And I somehow cannot find a good counterexample for $k$ a finite field.

Answer 1

Well, the answer (for $\operatorname{char}k=0$) can be found here

https://mathoverflow.net/questions/26832/degree-of-sum-of-algebraic-numbers

and here (cited there)

http://alpha.math.uga.edu/~pete/Isaacs70.pdf

The result is quite nice: If $\operatorname{char}k=0$, then $$ [k(a):k][k(b):k] \text{rel. prime} \Rightarrow [k(a+b):k] = [k(a):k][k(b):k] \Rightarrow k(a+b)=k(a,b) $$