Let \(K\) be the splitting field of \(f(x) = x^4 - 4x^2 - 3\) over \(\mathbb{Q}\).
- What is the dimension \([K : \mathbb{Q}]\)?
- What is the Galois group \(\operatorname{Gal}(K/\mathbb{Q})\)? (Identify its isomorphism type and describe how its elements act as permutations on the roots of \(f\)).
- How many subfields does \(K\) have of each dimension over \(\mathbb{Q}\)?
- How many subfields of \(K\) of each dimension are Galois over \(\mathbb{Q}\)?
Solution
Let \(r_1=\sqrt{2+\sqrt7}>0\) and \(r_2=i\sqrt{\sqrt7-2}\), so that \(r_2^2=2-\sqrt7\).
1. The roots of \(f\) are \(\pm r_1,\pm r_2\), and \(K=\QQ(r_1,i\sqrt3)\).
With \(u=x^2\), the equation \(u^2-4u-3=0\) has roots \(u=2\pm\sqrt7\), so the roots of \(f\) are \(r_1,r_2,-r_1,-r_2\). Moreover \(r_1r_2=i\sqrt{(\sqrt7+2)(\sqrt7-2)}=i\sqrt3\). Thus \(K=\QQ(r_1,r_2)\) contains \(i\sqrt3\), and conversely \(r_2=i\sqrt3/r_1\in\QQ(r_1,i\sqrt3)\).
2. The polynomial \(f\) is irreducible over \(\QQ\), and \([K:\QQ]=\boxed{8}\).
By the rational root theorem the only candidate rational roots are \(\pm1,\pm3\), and \(f(\pm1)=-6\), \(f(\pm3)=42\). A factorization into two monic quadratics over \(\QQ\) would pair the four roots. The pairing \(\{r_1,-r_1\},\{r_2,-r_2\}\) gives \(x^2-(2\pm\sqrt7)\notin\QQ[x]\). The other two pairings give constant terms \(\pm r_1r_2=\pm i\sqrt3\notin\QQ\). Hence \(f\) is irreducible and \([\QQ(r_1):\QQ]=4\). The field \(\QQ(r_1)\) is real and \(i\sqrt3\) is not, so \([K:\QQ(r_1)]=2\) by step 1, and the tower law gives \([K:\QQ]=8\).
3. Number the roots \((r_1,r_2,-r_1,-r_2)\) as \(1,2,3,4\). Then \(\operatorname{Gal}(K/\QQ)\) is the dihedral group \(D_4\) of order \(8\) generated by \(\sigma=(1\,2\,3\,4)\) and complex conjugation \(\tau=(2\,4)\), with \(\tau\sigma\tau=\sigma^{-1}\).
Each automorphism sends a root to a root and commutes with negation, so it preserves the partition \(\{\{1,3\},\{2,4\}\}\). The permutations of \(\{1,2,3,4\}\) preserving this partition form a dihedral subgroup of \(S_4\) of order \(8\), and \(\abs{\operatorname{Gal}(K/\QQ)}=[K:\QQ]=8\) by step 2, so the Galois group is this whole subgroup. It contains \(\sigma\) and \(\tau\); complex conjugation fixes \(\pm r_1\) and exchanges \(r_2\) and \(-r_2=\overline{r_2}\).
4. The quadratic subfields are \(\QQ(\sqrt7)\), \(\QQ(i\sqrt3)\), and \(\QQ(i\sqrt{21})\), fixed respectively by \(\langle\sigma^2,\tau\rangle\), \(\langle\sigma^2,\sigma\tau\rangle\), and \(\langle\sigma\rangle\).
We have \(\sqrt7=r_1^2-2\) and \(i\sqrt3=r_1r_2\). The automorphism \(\sigma\) sends \(r_1\mapsto r_2\) and \(r_2\mapsto-r_1\), so \(\sigma(\sqrt7)=r_2^2-2=-\sqrt7\), \(\sigma(i\sqrt3)=-r_2r_1=-i\sqrt3\), and \(\sigma(i\sqrt{21})=\sigma(\sqrt7\cdot i\sqrt3)=i\sqrt{21}\). The element \(\sigma^2\) negates every root, so it fixes \(\sqrt7\) and \(i\sqrt3\); \(\tau\) fixes \(r_1\) and negates \(r_2\), so it fixes \(\sqrt7\) and negates \(i\sqrt3\); hence \(\sigma\tau\) negates \(\sqrt7\) and fixes \(i\sqrt3\). The three subgroups of order \(4\) in \(D_4\) are \(\langle\sigma\rangle\), \(\langle\sigma^2,\tau\rangle\), and \(\langle\sigma^2,\sigma\tau\rangle\), and each fixes the listed element of degree \(2\); by the Galois correspondence it is the whole fixed field [@DF04].
5. The subfields of degree \(4\) are the fixed fields of the five subgroups of order \(2\): \(\QQ(\sqrt7,i\sqrt3)\) for the center \(\langle\sigma^2\rangle\), and four fields for \(\langle\tau\rangle,\langle\sigma^2\tau\rangle,\langle\sigma\tau\rangle,\langle\sigma^3\tau\rangle\), among them \(\QQ(r_1)\) for \(\langle\tau\rangle\) and \(\QQ(r_2)\) for \(\langle\sigma^2\tau\rangle\).
\(D_4\) has five elements of order \(2\): \(\sigma^2\) and the four reflections \(\sigma^j\tau\). The field \(\QQ(\sqrt7,i\sqrt3)\) has degree \(4\) and is fixed by \(\sigma^2\). The involution \(\tau=(2\,4)\) fixes \(r_1\), and \(\sigma^2\tau=(1\,3)\) fixes \(r_2\); each of \(\QQ(r_1)\), \(\QQ(r_2)\) has degree \(4\) by step 2.
6. A subfield is Galois over \(\QQ\) exactly when its subgroup is normal in \(D_4\).The normal subgroups are \(D_4\), the three subgroups of order \(4\), the center \(\langle\sigma^2\rangle\), and \(\{1\}\); the four reflection subgroups are not normal.
The first sentence is the Galois correspondence [@DF04]. Subgroups of index \(2\) are normal, and the center is normal. Conjugation by \(\sigma\) sends \(\sigma^j\tau\) to \(\sigma^{j+2}\tau\), so no reflection subgroup is normal.
7. Q.E.D.
By steps 2 through 6: (a) \([K:\QQ]=8\); (b) \(\operatorname{Gal}(K/\QQ)\cong D_4\), acting on the roots as in step 3; (c) \(K\) has one subfield of degree \(1\), three of degree \(2\), five of degree \(4\), and one of degree \(8\); (d) of these, one of degree \(1\), three of degree \(2\), one of degree \(4\), and one of degree \(8\) are Galois over \(\QQ\).