Let \(F/K\) be a field extension, and let \(\alpha_1,\dots,\alpha_n\in F\) be algebraic over \(K\). Show that \[ K[\alpha_1,\dots,\alpha_n]=K(\alpha_1,\dots,\alpha_n). \]
Solution
It is enough to prove that adjoining one algebraic element to a field already gives a field.
Let \(L\) be a field and let \(\alpha\) be algebraic over \(L\), with minimal polynomial \(m_\alpha(x)\in L[x]\). Evaluation induces an isomorphism \[ L[x]/(m_\alpha)\cong L[\alpha]. \] Since \(m_\alpha\) is irreducible, \((m_\alpha)\) is maximal, so \(L[\alpha]\) is a field. Hence \[ L[\alpha]=L(\alpha). \]
Apply this inductively. Starting with \(K_0=K\), define \[ K_i=K_{i-1}[\alpha_i]. \] Each \(\alpha_i\) is algebraic over \(K\), hence also algebraic over the larger field \(K_{i-1}\). Therefore every \(K_i\) is a field, and \[ K_i=K_{i-1}(\alpha_i). \] After \(n\) steps, \[ K[\alpha_1,\ldots,\alpha_n] =K_n =K(\alpha_1,\ldots,\alpha_n). \]