Let \(F/K\) be a field extension, and let \(a \in F\) be algebraic over \(K\) of odd degree \([K(a) : K] = 2m + 1\). Prove that \(a^2\) is algebraic over \(K\), that \([K(a^2) : K]\) is odd, and that \(K(a) = K(a^2)\).
Solution
Since \(a\) is algebraic over \(K\), so is \(a^2\), and \[ K\subseteq K(a^2)\subseteq K(a). \] Hence the tower law gives \[ [K(a):K] =[K(a):K(a^2)]\,[K(a^2):K]. \]
Now \(a\) is a root of \[ x^2-a^2\in K(a^2)[x], \] so \[ [K(a):K(a^2)]\le2. \] Therefore this degree is either \(1\) or \(2\). But it divides the odd integer \[ [K(a):K]=2m+1, \] so it cannot be \(2\). Thus \[ [K(a):K(a^2)]=1, \qquad K(a)=K(a^2). \] Consequently \[ [K(a^2):K]=[K(a):K]=2m+1, \] which is odd.