Let \(L\) be the splitting field of \(x^5 - 3\) over \(\mathbb{Q}\).
Let \(\zeta_5\) be a primitive 5th root of unity. Prove that \(\mathbb{Q}(\zeta_5) \subset L\).
What is the Galois group of \(L\) over \(\mathbb{Q}(\zeta_5)\)?
What is the dimension of \(L\) over \(\mathbb{Q}\)?
Solution
Write \(\alpha=\sqrt[5]{3}\) for the positive real root, \(\zeta=\zeta_5\), and \(F=\mathbb Q(\zeta)\). The answers are \[ F\subset L=F(\alpha),\qquad \operatorname{Gal}(L/F)\cong C_5,\qquad [L:\mathbb Q]=20. \]
1. The splitting field contains \(\zeta\) and equals\(\mathbb Q(\alpha,\zeta)\).
The five roots are \(\alpha\zeta^j\) for \(0\leq j<5\). The field \(L\) contains both \(\alpha\) and \(\alpha\zeta\), so it contains their ratio \(\zeta\); here \(\alpha\ne0\). Conversely, \(\mathbb Q(\alpha,\zeta)\) contains every root. These two containments prove the assertion.
2. The degrees are \([L:F]=5\) and \([L:\mathbb Q]=20\).
Eisenstein’s criterion at \(3\) makes \(x^5-3\) irreducible over \(\mathbb Q\) [@DF04], so \([\mathbb Q(\alpha):\mathbb Q]=5\). Also \(\zeta\) is a root of \[ \Phi_5(x)=x^4+x^3+x^2+x+1. \] Its translate is \[ \Phi_5(t+1)=t^4+5t^3+10t^2+10t+5, \] which is Eisenstein at \(5\). Translation of the variable is an automorphism of \(\mathbb Q[x]\), so \(\Phi_5\) is irreducible and \([F:\mathbb Q]=4\).
Set \(d=[L:F]\). Since \(L=F(\alpha)\) and \(\alpha\) satisfies a degree-\(5\) polynomial over \(F\), we have \(1\leq d\leq5\). The tower law gives \([L:\mathbb Q]=4d\). The subfield \(\mathbb Q(\alpha)\subseteq L\) has degree \(5\), so the same law implies \(5\mid4d\). As \(\gcd(4,5)=1\), we obtain \(5\mid d\), hence \(d=5\) and \([L:\mathbb Q]=20\).
3. The Galois group over \(F\) is cyclic of order \(5\).
The extension \(L/F\) is a splitting field of \(x^5-3\) in characteristic zero, so it is finite Galois. Its automorphism group has order \([L:F]=5\) [@DF04]. An \(F\)-automorphism must send \(\alpha\) to a root \(\alpha\zeta^j\), and this image determines it because \(L=F(\alpha)\). Thus the map \[ \operatorname{Gal}(L/F)\longrightarrow\mathbb Z/5\mathbb Z, \qquad \sigma\longmapsto j\quad \text{when }\sigma(\alpha)=\alpha\zeta^j \] is injective. It is a homomorphism: every such automorphism fixes \(\zeta\), so composition adds the exponents modulo \(5\). Both groups have five elements, making the map an isomorphism. In particular the automorphism fixing \(F\) and sending \(\alpha\) to \(\zeta\alpha\) generates the group.