$\QQ(\sqrt[3]{2})/\QQ$ is not Galois

Let \(\sqrt[3]{2}\in\RR\) be the real cube root of \(2\). The extension \(\QQ(\sqrt[3]{2})/\QQ\) has degree \(3\) and trivial automorphism group, so it is not Galois.

The polynomial \(x^3-2\) is irreducible over \(\QQ\) by Eisenstein's criterion at \(p=2\), so \([\QQ(\sqrt[3]{2}):\QQ]=3\). An automorphism \(\sigma\) of \(\QQ(\sqrt[3]{2})\) fixing \(\QQ\) sends \(\sqrt[3]{2}\) to a root of \(x^3-2\) lying in \(\QQ(\sqrt[3]{2})\subseteq\RR\). The only real root of \(x^3-2\) is \(\sqrt[3]{2}\), so \(\sigma=\id\). Hence \(\abs{\Aut(\QQ(\sqrt[3]{2})/\QQ)}=1\ne3\), and \(\QQ(\sqrt[3]{2})/\QQ\) is not Galois by FT-3WWKN.