Show that if \(\alpha, \beta\) are algebraic over \(F\), then \(\alpha\pm \beta, \alpha\beta\), and \(\alpha\beta^{-1}\) (for \(\beta \neq 0\)) are all algebraic over \(F\).
Solution
Since \(\alpha\) and \(\beta\) are algebraic over \(F\), the extension \[ F(\alpha,\beta)/F \] is finite. Indeed, \[ [F(\alpha,\beta):F] =[F(\alpha,\beta):F(\alpha)]\,[F(\alpha):F]<\infty. \] Every element of a finite extension is algebraic over the base field.
Now \(F(\alpha,\beta)\) is a field containing \[ \alpha+\beta,\quad \alpha-\beta,\quad \alpha\beta, \] and, when \(\beta\ne0\), \[ \alpha\beta^{-1}. \] Hence all these elements are algebraic over \(F\).