Show that if \(u\in F\) is algebraic of odd degree over \(K\), then so is \(u^2\), and moreover \(K(u) = K(u^2)\).
Solution
Let \[ n=[K(u):K], \] which is odd by hypothesis.
1. The element \(u^2\) is algebraic over \(K\).
Since \(u\) is algebraic over \(K\), the extension \(K(u)/K\) is finite. The element \(u^2\) belongs to the finite extension \(K(u)\), so it is algebraic over \(K\).
2. One has\[ [K(u):K(u^2)]\le2. \]
The element \(u\) is a root of \[ X^2-u^2\in K(u^2)[X]. \] Therefore the minimal polynomial of \(u\) over \(K(u^2)\) has degree at most \(2\), which is exactly the asserted inequality.
3. The degree \([K(u):K(u^2)]\) divides the odd integer \(n\).
By 1, the tower \[ K\subseteq K(u^2)\subseteq K(u) \] consists of finite extensions. The tower law gives \[ n=[K(u):K] =[K(u):K(u^2)]\,[K(u^2):K]. \] Thus \([K(u):K(u^2)]\) divides \(n\).
4. Consequently\[ [K(u):K(u^2)]=1. \]
By 2 this degree is either \(1\) or \(2\). By 3 it divides the odd integer \(n\), so it cannot equal \(2\).
5. Therefore\[ K(u)=K(u^2). \]
The inclusion \(K(u^2)\subseteq K(u)\) is immediate. By 4 the extension has degree \(1\), hence the two fields are equal.