If $[K(\alpha):K]=p\neq q=[K(\beta):K]$ then $[K(\alpha+\beta):K]=pq$

Solution 1:

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.