Exercise: Let $a$ and $b$ be algebraic over $\mathbb Q$ with minimal polynomials $g$ and $h$, let the other roots of $g$ and $h$ be $a = a_1, \dots, a_n$ and $b = b_1, \dots, b_m$ respectively (in the splitting field of $gh$). Suppose that the elements $a_i + b_j$ are different for each index. Show that $\mathbb Q(a+b) = \mathbb Q(a,b)$
My progress: Since $a + b \in \mathbb Q(a,b)$, it is obvious that $\mathbb Q (a + b) \subseteq \mathbb Q (a,b)$. I am stuck on how to prove it the other way. I can see that proving $a \in \mathbb Q(a + b)$ would be sufficient (because then $ b \in \mathbb Q(a+b)$ as well)
It's also true that in the splitting field of $g \cdot h$, we can write $g$ as $(x - a_1) \dots (x - a_n)$ and $h$ as $(x - b_1) \dots (x - b_m)$. I am struggling to find a way to use this information to prove $\mathbb Q(a + b) \supseteq \mathbb Q(a,b)$
Let $K$ be the normal closure of $\mathbb Q(a,b)$, i.e. the splitting field of $g,h$. If $a\not\in \mathbb Q(a+b)$, then there is an element $\sigma\in \text{Gal}(K/\mathbb Q)$ such that $\sigma$ fixes $a+b$ but not $a$, therefore $\sigma(a+b) = \sigma(a) + \sigma(b) = a_i + b_j = a + b$ and $a_i\not=a$.