Take a quadratic extension of a field of characteristic 0. Is it Galois? Take a degree 2 extension on top of that. Does it have to be Galois over the base field? What statement in group theory can you think of that reflects this?
Solution
Let \(K/F\) be quadratic with \(\operatorname{char}F=0\). If \(K=F(\alpha)\) and the minimal polynomial of \(\alpha\) is \[ x^2+bx+c, \] then its other root is \(-b-\alpha\in K\). Hence the polynomial splits in \(K\), and characteristic \(0\) gives separability. Thus every quadratic extension in characteristic \(0\) is Galois.
A tower of quadratic extensions need not be Galois over the bottom field. For example, \[ \mathbb Q\subset \mathbb Q(\sqrt2)\subset L:=\mathbb Q\bigl(\sqrt{1+\sqrt2}\bigr). \] The second step is quadratic because \(1+\sqrt2\) is not a square in \(\mathbb Q(\sqrt2)\): its norm is \(-1\), whereas the norm of a square is a square in \(\mathbb Q\). Let \[ \alpha=\sqrt{1+\sqrt2}. \] Then \(\alpha\) has polynomial \[ x^4-2x^2-1, \] whose other roots include \[ \pm\sqrt{1-\sqrt2}, \] which are nonreal. Since \(L\subset\mathbb R\), those roots are not in \(L\). Thus \(L/\mathbb Q\) is not normal and hence not Galois.
The group-theoretic analogue is that normality is not transitive: \[ H\trianglelefteq K,\quad K\trianglelefteq G \centernot\implies H\trianglelefteq G. \] For example, \[ \langle(12)(34)\rangle\trianglelefteq V_4\trianglelefteq A_4, \] but \(\langle(12)(34)\rangle\) is not normal in \(A_4\).