Let $\alpha$ and $\beta$ be two linearly independent algebraic numbers. Is it always true that $\mathbb{Q}(\alpha,\beta) \cong \mathbb{Q}(\alpha + \beta)$?
Certainly $\mathbb{Q}(\alpha + \beta) \subset\mathbb{Q}(\alpha, \beta)$. The tricky thing is showing the reverse inclusion.
Let $\deg(\alpha) = m$ and $\deg(\beta) = n$ then $\mathbb{Q}(\alpha,\beta)$ has dimension $mn$ over $\mathbb{Q}$. My thought here is that you can take successive powers of $\alpha + \beta$ and generate a system of equations. For some sufficient power, the matrix of coefficients will be solvable. With that, you can isolate $\alpha$ and $\beta$ which proves the reverse inclusion.
Does anyone have any insights?
Let $\alpha=\sqrt{3}-\sqrt{2}$ and $\beta=\sqrt{2}$. Then $$\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\sqrt{3})\neq \mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\alpha,\beta).$$
Of course the primitive element theorem tells you that for all but finitely many $c\in \mathbb{Q}$, $$\mathbb{Q}(\alpha+c\beta)=\mathbb{Q}(\alpha,\beta).$$