Let \(p(x)\) be a polynomial over \(\mathbb{Q}\) with Galois group \(\mathbb{Z}_2 \times \mathbb{Z}_2\). What can be said about the solvability of \(p(x)\) by radicals? What if the Galois group is \(S_3\)?
Solution
1. A polynomial over \(\mathbb{Q}\) is solvable by radicals if its Galois group is a solvable group.
Over a field of characteristic \(0\), the Galois-theoretic criterion for solvability by radicals says that a polynomial is solvable by radicals if and only if the Galois group of its splitting field is solvable.
2. If the Galois group is \(\mathbb{Z}_2\times\mathbb{Z}_2\), then \(p(x)\) is solvable by radicals.
The group \(\mathbb{Z}_2\times\mathbb{Z}_2\) is abelian, hence solvable. For example, if \(H\) is any subgroup of order \(2\), then \[ 1\triangleleft H\triangleleft \mathbb{Z}_2\times\mathbb{Z}_2 \] has abelian quotients. By 1, \(p(x)\) is solvable by radicals.
3. If the Galois group is \(S_3\), then \(p(x)\) is also solvable by radicals.
The alternating subgroup \(A_3\) is normal in \(S_3\), and \[ 1\triangleleft A_3\triangleleft S_3 \] has quotients \[ A_3\cong C_3, \qquad S_3/A_3\cong C_2. \] Thus \(S_3\) is solvable. By 1, \(p(x)\) is solvable by radicals in this case as well.