Determine the Galois groups of the following polynomials over the corresponding fields:
\(x^4- 5\) over \(\mathbb Q, \mathbb Q(\sqrt 5), \mathbb Q(i\sqrt 5)\).
\(x^3 - 2\) over \(\mathbb Q\).
\((x^3-2)(x^2-5)\) over \(\mathbb Q\).
Solution
1. Let \(\alpha=\sqrt[4]{5}\). The splitting field of \(x^4-5\) over\(\QQ\) is\[ L=\QQ(\alpha,i). \]
The roots are \[ \alpha,-\alpha,i\alpha,-i\alpha, \] so \(\QQ(\alpha,i)\) contains all roots and is generated by them.
2. Over \(\QQ\), the Galois group of \(x^4-5\) is the dihedral group\(D_4\) of order \(8\).
The polynomial \(x^4-5\) is Eisenstein at \(5\), hence \[ [\QQ(\alpha):\QQ]=4. \] Since \(\QQ(\alpha)\subseteq\RR\), it does not contain \(i\), so \[ [L:\QQ]=8. \] Define \[ r(\alpha)=i\alpha,\quad r(i)=i, \qquad s(\alpha)=\alpha,\quad s(i)=-i. \] Then \(r\) has order \(4\), \(s\) has order \(2\), and \[ srs=r^{-1}. \] These automorphisms generate \(8\) elements, hence all of \(\operatorname{Gal}(L/\QQ)\). Thus \[ \operatorname{Gal}(x^4-5/\QQ)\cong D_4. \]
3. Over \(\QQ(\sqrt5)\), the Galois group of \(x^4-5\) is\(\ZZ_2\times\ZZ_2\).
Since \(\sqrt5=\alpha^2\), the fixed subgroup is \[ \operatorname{Gal}(L/\QQ(\sqrt5)) =\langle r^2,s\rangle. \] Both \(r^2\) and \(s\) have order \(2\) and they commute. The subgroup has order \(4\), as required because \([\QQ(\sqrt5):\QQ]=2\) and \([L:\QQ]=8\). Hence it is the Klein four group.
4. Over \(\QQ(i\sqrt5)\), the Galois group of \(x^4-5\) is also\(\ZZ_2\times\ZZ_2\).
Put \(\beta=i\alpha^2=i\sqrt5\). Under \(r\) one has \(r(\beta)=-\beta\), while under \(s\) one also has \(s(\beta)=-\beta\). Thus the elements fixing \(\beta\) are \[ 1,\quad r^2,\quad rs,\quad r^3s. \] They form the subgroup \[ \langle r^2,rs\rangle. \] Both generators are involutions and commute, so this subgroup is \(\ZZ_2\times\ZZ_2\). Its order is \(4\), agreeing with \([L:\QQ(i\sqrt5)]=4\).
5. The Galois group of \(x^3-2\) over \(\QQ\) is \(S_3\).
Let \(\gamma=\sqrt[3]{2}\) and let \(\omega\) be a primitive cube root of unity. The splitting field is \[ M=\QQ(\gamma,\omega). \] The polynomial \(x^3-2\) is Eisenstein at \(2\), so \([\QQ(\gamma):\QQ]=3\). Since \(\QQ(\gamma)\subseteq\RR\) and \(\omega\notin\RR\), \[ [M:\QQ]=6. \] The Galois group acts faithfully on the three roots \(\gamma,\omega\gamma,\omega^2\gamma\), hence embeds in \(S_3\). Since its order is \(6\), it is all of \(S_3\).
6. The splitting field of \((x^3-2)(x^2-5)\) over \(\QQ\) is\[ N=M(\sqrt5), \]where \(M=\QQ(\sqrt[3]{2},\omega)\).
The first factor splits in \(M\) and the second in \(\QQ(\sqrt5)\), so their product has splitting field equal to the compositum of these two splitting fields.
7. One has\[ M\cap\QQ(\sqrt5)=\QQ. \]
The intersection is a subfield of the quadratic extension \(\QQ(\sqrt5)/\QQ\), so it is either \(\QQ\) or \(\QQ(\sqrt5)\).
If \(\QQ(\sqrt5)\subseteq M\), it would be a quadratic subfield of the \(S_3\)-Galois extension \(M/\QQ\). Such subfields correspond to index-\(2\) subgroups of \(S_3\). The unique index-\(2\) subgroup is \(A_3\), whose fixed field is the quadratic field generated by the square root of the discriminant of \(x^3-2\). Since \[ \operatorname{disc}(x^3-2)=-108=-3\cdot6^2, \] that field is \(\QQ(\sqrt{-3})=\QQ(\omega)\), not \(\QQ(\sqrt5)\). Therefore the intersection is \(\QQ\).
8. Hence\[ \operatorname{Gal}((x^3-2)(x^2-5)/\QQ) \cong S_3\times\ZZ_2. \]
7. Therefore restriction gives\[ \operatorname{Gal}(N/\QQ) \cong \operatorname{Gal}(M/\QQ)\times \operatorname{Gal}(\QQ(\sqrt5)/\QQ) \cong S_3\times\ZZ_2. \]