- Show that any quadratic extension of a field \(F\) with \(\ch(F)\neq 2\) is Galois.
Solution
Let \(K/F\) be quadratic and assume \(\operatorname{char}F\ne2\). Choose \(\alpha\in K\setminus F\). Then \(K=F(\alpha)\) and the minimal polynomial of \(\alpha\) has degree \(2\): \[ m(x)=x^2+bx+c. \] Set \[ \beta=\alpha+\frac b2. \] Then \(K=F(\beta)\) and \[ \beta^2=\frac{b^2-4c}{4}=:d\in F. \] Since \(\beta\notin F\), \(d\) is not a square in \(F\), and the minimal polynomial of \(\beta\) is \[ x^2-d. \] Its two roots are \(\beta\) and \(-\beta\), both of which lie in \(K\). Thus \(K\) is the splitting field of \(x^2-d\) over \(F\), so \(K/F\) is normal.
Because \(\operatorname{char}F\ne2\), the two roots \(\beta\) and \(-\beta\) are distinct; hence the polynomial is separable. Therefore \(K/F\) is finite, normal, and separable, so it is Galois.