Discussion
What is the degree of the splitting field \(x^{3} - 2\) over Q ?
Answer: A
The splitting field of \(x^{3} - 2\) over Q has degree 6.
The roots are \(\sqrt[3]{2}\), \(\omega\sqrt[3]{2}\) and \(\omega^{2}\sqrt[3]{2}\), where \(\omega\) is a primitive cube root of unity, so the splitting field is \(\mathbf{Q}(\sqrt[3]{2}, \omega)\).
\[ \begin{aligned} [\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] &= 3 \\ [\mathbf{Q}(\sqrt[3]{2}, \omega) : \mathbf{Q}(\sqrt[3]{2})] &= 2 \\ \text{total} &= 3 \times 2 = 6 \end{aligned} \]
The first degree is 3 because \(x^{3} - 2\) is irreducible (Eisenstein at 2). The second is 2 because \(\omega\) satisfies \(x^{2} + x + 1\) and is not real, while \(\mathbf{Q}(\sqrt[3]{2})\) is real. The Galois group is \(S_{3}\), of order 6. The answer 3 stops at the real root, and 2 and 4 do not divide in the right way.
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