If $[K(\alpha):K]=p\neq q=[K(\beta):K]$ then $[K(\alpha+\beta):K]=pq$ I am having some trouble with the following problem:

Let $K\subseteq L$ be fields. Suppose $\alpha,\beta\in L$ are algebraic elements over $K$ of degrees $p,q$ respectively, where $p$ and $q$ are distinct primes. Show that $\alpha+\beta$ is algebraic over $K$ with degree $pq$.

I have shown that $\alpha+\beta$ is algebraic over $K$, but I am having trouble showing that the degree is $pq$. I have previously shown that $[K(\alpha,\beta):K]=pq$, so I have been trying to use this result. I thought that if I could show that $K(\alpha,\beta)=K(\alpha+\beta)$ then I'd be done. To show this I first noted that clearly $K(\alpha+\beta)\subseteq K(\alpha,\beta)$. Then we have
$$pq=[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha+\beta)][K(\alpha+\beta):K]$$
so that $[K(\alpha+\beta):K]=1,p,q,pq$. I tried looking at the cases when it is equal to $1,p,q$ and deriving a contradiction, but have been unsuccessful. 
Another way I thought of solving this is to show that $\alpha,\beta\in K(\alpha+\beta)$, which would allow me to conclude that $K(\alpha+\beta)=K(\alpha,\beta)$, but I am not sure how to complete this either.

I am looking for some assistance to show that $[K(\alpha+\beta):K]\neq 1,p,q$, or that $\alpha,\beta\in K(\alpha,\beta)$. If you have any other solutions to this problem I would like to see those too.
 A: As pointed out in my comment above, this does not hold in general. I will provide a solution for characteristic $0$ case. Assume $p<q$.

Let $K_1$ be the Galois closure of $K(\alpha)/K$, $K_2$ be the Galois closure of $K(\beta)/K$. Denote $\beta:= \beta_0, \beta_1, \cdots, \beta_{q-1}$ be all the conjugates of $\beta$ over $K$. Becuase $|\text{Gal}(K_2/ K)|$ is divisible by $q$, there exists $\sigma\in \text{Gal}(K_2/ K)$ which is a cyclic permutation. Let $\sigma(\beta_0) = \beta_1 , \sigma(\beta_1) = \beta_2 , \cdots, \sigma(\beta_{q-1}) = \beta_0$. Since the extension $K_1K_2/K_2$ is normal, we can extend $\sigma$ to an element in $G:= \text{Gal}(K_1K_2/ K)$. The number $[K(\alpha+\beta):K]$ is the orbit size of $\alpha+\beta_0$ under $G$. 
Now, consider the numbers
$$\alpha,\sigma(\alpha), \cdots, \sigma^{q-1}(\alpha)$$
they're roots of minimal polynomial of $\alpha$ over $K$, since $p<q$, two of them must coincide, so we have an integer $1\leq k \leq q-1$ such that $\alpha = \sigma^k (\alpha)$. Now we apply $\sigma^k$ to $\alpha + \beta_0$ sucessively:
$$\alpha + \beta_0 \mapsto  \alpha + \beta_k \mapsto \alpha + \beta_{2k}   \mapsto \cdots \mapsto \alpha + \beta_{(q-1)k} $$
where the subscripts have to be interpreted modulo $q$. Since $(k,q) = 1$, we see that all $\alpha + \beta_0 , \cdots ,\alpha + \beta_{q-1}$ are in the orbit, and they're all distinct. Hence $[K(\alpha+\beta):K] \geq q$
If $[K(\alpha+\beta):K] = q$, then $\alpha + \beta_0 , \cdots ,\alpha + \beta_{q-1}$ are all the roots of the minimal polynomial of $\alpha + \beta$ over $K$, hence
$$(\alpha + \beta_0) + (\alpha + \beta_1) + \cdots + (\alpha + \beta_{q-1}) \in K \implies q\alpha \in K$$
where we have used the fact that $\beta_0 + \cdots + \beta_{q-1} \in K$. Since $K$ has characteristic $0$, we have $\alpha \in K$, contradiction. Therefore $[K(\alpha+\beta):K] = pq$, the proof is completed.
