Let \(\alpha,\beta\) be algebraic over a field \(F\). Prove that \[ \alpha\pm\beta, \qquad \alpha\beta \] are algebraic over \(F\), and if \(\beta\ne0\), then so is \[ \alpha\beta^{-1}. \]
Solution
Since \(\alpha\) and \(\beta\) are algebraic over \(F\), the extensions \[ F(\alpha)/F, \qquad F(\beta)/F \] are finite. Therefore the compositum \[ F(\alpha,\beta) \] is finite over \(F\).
Every element of a finite field extension is algebraic over the base field. Now \[ \alpha+\beta, \quad \alpha-\beta, \quad \alpha\beta \] all lie in \(F(\alpha,\beta)\). If \(\beta\ne0\), then \(\beta^{-1}\in F(\beta)\subseteq F(\alpha,\beta)\), so \[ \alpha\beta^{-1}\in F(\alpha,\beta) \] as well.
Hence all the displayed elements are algebraic over \(F\).