$[k(\alpha):k]=p, [k(\beta):k]=q$, $p>q$ are primes, then $k(\alpha,\beta)=k(\alpha+\beta)$ Let $p>q$ be primes. Suppose $L\mid_{k}$ is an algebraic extension and $\alpha,\beta\in L$ are such that $[k(\alpha):k]=p$, $[k(\beta):k]=q$ and characteristic of $k$ is coprime with $p$. Show that $k(\alpha,\beta)=k(\alpha+\beta)$.
From the given condition what we know is: $[k(\alpha,\beta):k]=pq$ and $\alpha$ is separable over $k$. Hence we can conclude that $k(\alpha,\beta)=k(\alpha+\lambda\beta)$ for some $\lambda\in k^{*}$. But I'm stuck at showing that $\lambda=1$
 A: I EDIT my post to answer to your objection.
The  extensions $k(a)$ and $k(b)$ are linearly disjoint since $p, q$ are 2 distinct primes. It follows e.g. that $b$ is not in $k(a)$, hence $a + b$ is not in $k(a)$ either. Similarly, $a+b$ is not in $k(b)$, hence the field $K = k(a+b)$ is distinct from $k(a)$ and $k(b)$. If $K$ were distinct from $L = k(a,b)$, its degree over $k$ would be $p$ or $q$. Suppose that it is $p$. Then $L$ would contain the compositum of $k(a)$ and $K$, which has degree $p^2$: impossible. Same argument when replacing $p$ by $q$.
A: In view of the questions raised by Starfall and user26857 about a certain degree equal or not  to $p^2$ , my previous solution works only in the Galois setting, hence it must be modified in the general case. The proof will be more complicated and will follow part I, §12 of Kaplansky’s book  « Fields and Rings ». The natural starting point is to assume separability all along and introduce the normal closures $M$ and $N$ of $k(a)$ and $k(b)$ respectively. Let us proceed in 3 steps :
1°. Let us show that if $p$ is not the characteristic of $F$, then $M$ is also a normal closure of $ F(a_1 – a_2)$, where the $a_i$’s are the distinct conjugates of $a$ : the Galois group of $M/F$ contains an automorphism $\tau$ of order $p$ which permutes the $a_i$’s, and we can  number these so that $\tau$ is the $p$-cycle $(1, 2, …, p)$. Since $\tau$ stabilizes the normal closure $M’$ of  $ F(a_1 – a_2)$, it follows that $a_1 – a_2, a_2 – a_3$, etc. all belong to $M’$, and hence also all the $a_1 – a_i$’s. The sum of the latter elements is $pa_1$, which implies that $a_1 \in M’$. Similarly, all the $a_i$’s are in $M’$, and thus $M’ = M$. 
2°. Working in a normal closure $L$ of $F(a,b)$, with obvious notations,  suppose that $a_i – a_j = b_h – b_k$ for a certain 4-uple $(i,j,h,k)$. Then $a_i – a_j \in N$. But by the previous result, this implies $M\subset N$, which is impossible because $[N :F]$ divides $q !$ and hence is not divisible by $p$ > $q$. Thus no difference of two conjugates of $a$ equals the difference of two conjugates of $b$.
3°. We know that $[F(a,b) : F] = pq$ . Let us show that for any $i = 1,…, p$ and any  $h = 1,…, q$, there exists an automorphism of $L/F$ which sends $a$ to $a_i$ and $b$ to $b_h$. This follows easily from the existence of an automorphism fixing $a$ and sending some $b_k$ to $b$. But by hypothesis, the conjugates of $b$ over $F(a)$ are still $b_1,…,b_q$, and the normality of $L$ /$F(a)$ ensures the existence of the desired automorphism.The property just shown means that all the elements $a_i + b_h$ are conjugates of $a + b$. The property in 2° shows that all the $pq$ sums $a_i + b_h$ are distinct. Hence $a + b$ has degree $pq$ over $F$, and we are done.
If the additional conditions (in italics) are not met, I have no clue.
