- Characterize the splitting field and all intermediate fields of the polynomial \(\left(x^{2}-2\right)\left(x^{2}-3\right)\left(x^{2}-5\right)\) over \(\QQ\). Using this characterization, find a primitive element of the splitting field.
Solution
Let \[ f(x)=(x^2-2)(x^2-3)(x^2-5). \]
1. The splitting field is\[ L=\QQ(\sqrt2,\sqrt3,\sqrt5). \]
The roots of \(f\) are \[ \pm\sqrt2,\qquad \pm\sqrt3,\qquad \pm\sqrt5. \] Thus every splitting field contains \(\sqrt2,\sqrt3,\sqrt5\), while the field generated by those three elements contains all six roots. Hence it is exactly the splitting field.
2. One has\[ [L:\QQ]=8 \qquad\text{and}\qquad \Gal(L/\QQ)\cong (C_2)^3. \]
First \(\QQ(\sqrt2,\sqrt3)/\QQ\) has degree \(4\): \(\sqrt3\notin\QQ(\sqrt2)\), since an equality \((a+b\sqrt2)^2=3\) with \(a,b\in\QQ\) forces \(2ab=0\) and then gives either \(a^2=3\) or \(2b^2=3\), both impossible in \(\QQ\).
Set \(E=\QQ(\sqrt2,\sqrt3)\). The extension \(E/\QQ\) is biquadratic with Galois group \((C_2)^2\), so its only quadratic subfields are \[ \QQ(\sqrt2),\qquad \QQ(\sqrt3),\qquad \QQ(\sqrt6). \] The field \(\QQ(\sqrt5)\) is distinct from all three, so \(\sqrt5\notin E\). Hence \[ [L:\QQ]=[L:E][E:\QQ]=2\cdot4=8. \]
Each independent choice of signs on \(\sqrt2,\sqrt3,\sqrt5\) defines a \(\QQ\)-automorphism of \(L\), giving eight automorphisms. Thus \[ \Gal(L/\QQ)\cong (C_2)^3. \]
3. The seven quadratic intermediate fields are\[ \QQ(\sqrt2),\ \QQ(\sqrt3),\ \QQ(\sqrt5),\ \QQ(\sqrt6),\ \QQ(\sqrt{10}),\ \QQ(\sqrt{15}),\ \QQ(\sqrt{30}). \]
The index-\(2\) subgroups of \((C_2)^3\) correspond to the nonzero linear functionals on the \(3\)-dimensional \(\FF_2\)-vector space, hence there are seven. Equivalently, every nontrivial squareclass generated by \(2,3,5\) gives one quadratic field. The seven nonzero products are \[ 2,3,5,6,10,15,30, \] which gives exactly the displayed list.
4. The seven quartic intermediate fields are\[ \begin{aligned} &\QQ(\sqrt2,\sqrt3),\qquad \QQ(\sqrt2,\sqrt5),\qquad \QQ(\sqrt3,\sqrt5),\\ &\QQ(\sqrt2,\sqrt{15}),\qquad \QQ(\sqrt3,\sqrt{10}),\qquad \QQ(\sqrt5,\sqrt6),\\ &\QQ(\sqrt6,\sqrt{10}). \end{aligned} \]
Quartic intermediate fields correspond to subgroups of order \(2\) in \(G=(C_2)^3\). For a nonempty subset \(S\subseteq\{2,3,5\}\), let \(\sigma_S\) change the sign of \(\sqrt d\) exactly for \(d\in S\). The seven nontrivial order-\(2\) subgroups are \(\langle\sigma_S\rangle\), and their fixed fields are respectively \[ \begin{array}{c|c} S & L^{\langle\sigma_S\rangle}\\ \hline \{2\} & \QQ(\sqrt3,\sqrt5)\\ \{3\} & \QQ(\sqrt2,\sqrt5)\\ \{5\} & \QQ(\sqrt2,\sqrt3)\\ \{2,3\} & \QQ(\sqrt6,\sqrt5)\\ \{2,5\} & \QQ(\sqrt{10},\sqrt3)\\ \{3,5\} & \QQ(\sqrt{15},\sqrt2)\\ \{2,3,5\} & \QQ(\sqrt6,\sqrt{10}). \end{array} \] Each displayed field has degree \(4\) over \(\QQ\), so these are exactly the fixed fields.
5. Together with \(\QQ\) and \(L\), the fields in 3 and 4 are all intermediate fields.
The group \((C_2)^3\) has \[ 1+7+7+1=16 \] subgroups: one of dimension \(0\), seven of dimension \(1\), seven of dimension \(2\), and one of dimension \(3\) over \(\FF_2\). By the fundamental theorem of Galois theory, these correspond bijectively to intermediate fields. The fields already listed account for all sixteen.
6. A primitive element is\[ \alpha=\sqrt2+\sqrt3+\sqrt5. \]
Every \(\sigma\in G\) changes signs independently. If \(\sigma\neq1\), it changes at least one sign, and therefore \[ \alpha-\sigma(\alpha) =2\sum_{d\in S}\sqrt d \] for some nonempty \(S\subseteq\{2,3,5\}\). Under the standard real embedding every term in this sum is positive, so the difference is nonzero. Hence the stabilizer of \(\alpha\) in \(G\) is trivial.
By Galois theory, \[ [\QQ(\alpha):\QQ]=|G|=8. \] Since \(\QQ(\alpha)\subseteq L\) and \([L:\QQ]=8\), one gets \[ L=\QQ(\alpha). \] Thus \(\alpha\) is a primitive element of the splitting field.