Give an example of an extension of degree 3 over \(\mathbb{Q}\) which is not Galois over \(\mathbb{Q}\).
Suppose \(K\) is a Galois extension of \(\mathbb{Q}\) with \([K:\mathbb{Q}] = 105\), and that \(L\) is a subfield of \(K\) with \([L:\mathbb{Q}] = 3\). Show that \(L\) is Galois over \(\mathbb{Q}\).
Solution
1. For part (a), \(\mathbb Q(\sqrt[3]{2})\) is a non-Galoisextension of degree \(3\).
Let \(a=\sqrt[3]{2}>0\). Eisenstein’s criterion at \(2\) makes \(x^3-2\) irreducible over \(\mathbb Q\), so \([\mathbb Q(a):\mathbb Q]=3\) [@DF04]. The other two roots are \(a\omega\) and \(a\omega^2\) for a primitive cube root of unity \(\omega\). They are nonreal, whereas \(\mathbb Q(a)\) is contained in \(\mathbb R\). Thus an irreducible polynomial over the base field has a root in the extension but does not split there. The extension is not normal and hence not Galois.
2. For part (b), \(H=\operatorname{Gal}(K/L)\) is a normalsubgroup of \(G=\operatorname{Gal}(K/\mathbb Q)\).
The finite Galois correspondence gives \(|G|=105\) and \([G:H]=[L:\mathbb Q]=3\), hence \(|H|=35\) [@DF04]. Let \(G\) act by left multiplication on the three left cosets of \(H\), giving \(\rho:G\to S_3\). The action is transitive: \(yx^{-1}\) sends \(xH\) to \(yH\).
The order of the image divides \(105\), by the first isomorphism theorem, and divides \(6=|S_3|\), by Lagrange’s theorem. It is therefore \(1\) or \(3\). Transitivity on three points excludes order \(1\), so \(|\rho(G)|=3\) and \(|\ker\rho|=35\). Every element of the kernel fixes the coset \(H\), so belongs to \(H\). Since both subgroups have order \(35\), \(H=\ker\rho\). Kernels are normal, so \(H\lhd G\).
3. The field \(L\) in part (b) is Galois over \(\mathbb Q\).
An intermediate field of a finite Galois extension is Galois over the base exactly when its corresponding subgroup is normal [@DF04]. Apply this criterion to the subgroup in step 2.