Discussion
The degree of the splitting field of the polynomial \(x^{3} - 2 \in \mathbf{Q}[x]\) is :
Answer: B
The degree is 6.
The roots of \(x^{3} - 2\) are \(\sqrt[3]{2}\), \(\sqrt[3]{2}\,\omega\) and \(\sqrt[3]{2}\,\omega^{2}\), where \(\omega = e^{2\pi i/3}\). The splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).
\(x^{3} - 2\) is irreducible by Eisenstein at 2, so \([\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] = 3\). That field is real, so it does not contain \(\omega\), which satisfies \(x^{2} + x + 1\); adjoining \(\omega\) doubles the degree:
\[ \begin{aligned} [\mathbf{Q}(\sqrt[3]{2}, \omega) : \mathbf{Q}] \\ &= 3 \times 2 = 6 \end{aligned} \]
3 is the degree of one real root only. The Galois group is \(S_{3}\), of order 6, which also rules out 4 and 5.
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