Let \(K\) be the splitting field of \(x^3 - 5\) over \(\mathbb{Q}\).
What is the dimension of \(K\) over \(\mathbb{Q}\)?
What is the Galois group of \(K\) over \(\mathbb{Q}\)? (Identify the isomorphism type of the Galois group, and also indicate how its elements act as permutations of the roots of \(f\).)
What are all the subfields of \(K\)? Be sure to explain how you know you have included all of them.
Which of these subfields are Galois over \(\mathbb{Q}\)?
<1>2. The degree is \[ \boxed{[K:\mathbb Q]=6}. \]
<1>3. The Galois group is \[ \boxed{\operatorname{Gal}(K/\mathbb Q)\cong S_3}. \]
<1>4. Define automorphisms \(\sigma,\tau\in\operatorname{Gal}(K/\mathbb Q)\) by \[ \sigma(\alpha)=\zeta\alpha, \qquad \sigma(\zeta)=\zeta, \] \[ \tau(\alpha)=\alpha, \qquad \tau(\zeta)=\zeta^2. \] Then, on the ordered root set \((\alpha,\zeta\alpha,\zeta^2\alpha)\), \[ \sigma=(123), \qquad \tau=(23). \]
<1>5. The complete list of subfields of \(K\) is \[ \mathbb Q, \quad K, \quad \mathbb Q(\zeta), \quad \mathbb Q(\alpha), \quad \mathbb Q(\zeta\alpha), \quad \mathbb Q(\zeta^2\alpha). \]
<1>6. The subfields Galois over \(\mathbb Q\) are exactly \[ \boxed{\mathbb Q,\quad \mathbb Q(\zeta),\quad K.} \]
Solution
1. Let\[ \alpha=\sqrt[3]{5},\qquad \zeta=e^{2\pi i/3}. \]Then the roots of \(x^3-5\) are\[ \alpha,\quad \zeta\alpha,\quad \zeta^2\alpha, \]so\[ K=\mathbb Q(\alpha,\zeta). \]
The polynomial \(x^3-5\) is irreducible over \(\mathbb Q\) by Eisenstein’s criterion at \(5\), so \([\mathbb Q(\alpha):\mathbb Q]=3\). Since \(\alpha\) is real, \(\mathbb Q(\alpha)\subseteq\mathbb R\), while \(\zeta\notin\mathbb R\). Hence \(\zeta\notin\mathbb Q(\alpha)\). The element \(\zeta\) satisfies \(x^2+x+1=0\), so \[ [K:\mathbb Q(\alpha)]=2. \] Therefore the tower law gives \([K:\mathbb Q]=2\cdot3=6\).
Because \(K\) is the splitting field over characteristic \(0\) of the separable polynomial \(x^3-5\), the extension \(K/\mathbb Q\) is Galois. Thus \[ |\operatorname{Gal}(K/\mathbb Q)|=[K:\mathbb Q]=6. \] The Galois group acts faithfully on the three roots, so it embeds in \(S_3\). Since both groups have order \(6\), the embedding is an isomorphism.
The map \(\sigma\) cyclically permutes the three cube roots of \(5\), while \(\tau\) is complex conjugation: it fixes the real root \(\alpha\) and exchanges the two nonreal roots. These permutations generate \(S_3\).
By the fundamental theorem of Galois theory, intermediate fields correspond bijectively to subgroups of \(S_3\). The subgroups of \(S_3\) are exactly \[ S_3,\quad \{e\},\quad A_3,\quad \langle(12)\rangle,\quad \langle(13)\rangle,\quad \langle(23)\rangle. \] Thus there are exactly six intermediate fields. The fixed field of \(A_3=\langle\sigma\rangle\) has degree \(2\) and is \(\mathbb Q(\zeta)\), since \(\sigma\) fixes \(\zeta\). Each order-\(2\) subgroup is the stabilizer of one of the three roots, so its fixed field is respectively one of the three cubic fields generated by that root.
Under the Galois correspondence, an intermediate field is Galois over \(\mathbb Q\) exactly when its corresponding subgroup is normal in \(S_3\). The normal subgroups of \(S_3\) are \[ \{e\},\quad A_3,\quad S_3. \] They correspond respectively to \(K\), \(\mathbb Q(\zeta)\), and \(\mathbb Q\). The three order-\(2\) subgroups are not normal, so the three cubic subfields are not Galois over \(\mathbb Q\).