Let \(f(x) = x^3 - 7\).
Let \(K\) be the splitting field for \(f\) over \(\mathbb Q\). Describe the Galois group of \(K / \mathbb Q\) and the intermediate fields between \(\mathbb Q\) and \(K\). Which intermediate fields are not Galois over \(\mathbb Q\)?
Let \(L\) be the splitting field for \(f\) over \(\mathbb R\). What is the Galois group \(L/ \mathbb R\)?
Let \(M\) be the splitting field for \(f\) over \(\mathbb F_{13}\), the field with 13 elements. What is the Galois group of \(M / \mathbb F_{13}\)?
Solution
Let \(\alpha=\sqrt[3]{7}\) and \(\omega=e^{2\pi i/3}\), so the roots of \(f\) in \(\CC\) are \(\alpha_1=\alpha\), \(\alpha_2=\omega\alpha\), \(\alpha_3=\omega^2\alpha\).
1. \(K=\QQ(\alpha,\omega)\) has degree \(6\) over \(\QQ\), and \(\Gal(K/\QQ)\cong S_3\), acting as all permutations of \(\alpha_1,\alpha_2,\alpha_3\).
\(f\) is Eisenstein at \(7\), so \([\QQ(\alpha):\QQ]=3\). \(K\) contains \(\alpha\) and \(\omega=\alpha_2/\alpha_1\), and \(\QQ(\alpha,\omega)\) contains all roots, so \(K=\QQ(\alpha,\omega)\). Since \(\QQ(\alpha)\subset\RR\) and \(\omega\notin\RR\), \([K:\QQ(\alpha)]=2\), so \([K:\QQ]=6\). \(\Gal(K/\QQ)\) embeds in the permutation group \(S_3\) of the roots and has order \(6\).
2. The intermediate fields are \(\QQ\), \(\QQ(\omega)=\QQ(\sqrt{-3})\), \(\QQ(\alpha_1)\), \(\QQ(\alpha_2)\), \(\QQ(\alpha_3)\) and \(K\); the three cubic fields \(\QQ(\alpha_i)\) are exactly the ones not Galois over \(\QQ\).
\(S_3\) has subgroups \(1\), \(A_3\), the three subgroups of order \(2\), and \(S_3\). \(A_3\) fixes \(\omega=\alpha_2/\alpha_1\), and \([\QQ(\omega):\QQ]=2=[S_3:A_3]\), so its fixed field is \(\QQ(\omega)\). The subgroup of order \(2\) fixing the root \(\alpha_i\) has fixed field \(\QQ(\alpha_i)\), of degree \(3\). An intermediate field is Galois over \(\QQ\) exactly when its subgroup is normal; \(1\), \(A_3\) and \(S_3\) are normal, and the subgroups of order \(2\) are not.
3. \(L=\CC\) and \(\Gal(L/\RR)\cong\ZZ/2\ZZ\), generated by complex conjugation.
\(\alpha\in\RR\) and \(\omega\alpha\notin\RR\), so \(L=\RR(\omega)=\CC\).
4. \(M=\FF_{13}(\beta)\cong\FF_{13^3}\) for a root \(\beta\) of \(f\), and \(\Gal(M/\FF_{13})\cong\ZZ/3\ZZ\), generated by \(x\mapsto x^{13}\).
Since \(3\mid 13-1\), an element \(a\in\FF_{13}^\times\) is a cube if and only if \(a^4=1\). Here \(7^2=49\equiv10\) and \(7^4\equiv100\equiv9\not\equiv1\pmod{13}\), so \(f\) has no root in \(\FF_{13}\) and, being cubic, is irreducible. Moreover \(3^3=27\equiv1\pmod{13}\), so \(3\) is a primitive cube root of unity in \(\FF_{13}\), and the roots of \(f\) are \(\beta,3\beta,9\beta\in\FF_{13}(\beta)\). Hence \(M=\FF_{13}(\beta)\) has degree \(3\), and the Galois group of a degree-\(3\) extension of finite fields is cyclic of order \(3\), generated by Frobenius.