Let $\alpha, \beta \in \mathbb{C}$, such that $\alpha + \beta$, and $\alpha\beta$ are algebraic.
Show that $\alpha$ and $\beta$ are algebraic.
attempt: Suppose $\alpha, \beta \in \mathbb{C}$ such that $\alpha + \beta$, and $\alpha\beta$ are algebraic.
Let $\alpha + \beta$, $\alpha\beta$ are algebraic be algebraic over $F$. And let $\alpha, \beta \in K$, where $K$ is an extension of $F$. then $\alpha + \beta $ and $\alpha\beta$ lie in the extension $K = F(\alpha,\beta)$, which is finite over $F$.
so $[F(\alpha, \beta) : F] = [F(\alpha, \beta): F(\alpha + \beta, \alpha\beta)][F(\alpha + \beta, \alpha\beta) : F] \leq 2[F(\alpha + \beta, \alpha \beta) : F] $, since $\alpha + \beta$ and $\alpha \beta$ are roots .
So $[F(\alpha + \beta) : F]$ is finite and in the extension $F(\alpha,\beta)/F$ the elements must be algebraic over $F$, then $\alpha , \beta$ are algebraic.
Can someone please verify this? Or help me give a better approach. Thank you!
Best Answer
Hint: $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$.