Let f(X) = X³ − 2 ∈ ℚ[X] and let K ⊂ ℂ be the splitting field of f(X) over ℚ. Let ω = e^{2πi/3}. Which of the following statements are true?

A, D. K = ℚ(∛2, ω) has degree 6 over ℚ with non-abelian Galois group S₃; over ℚ(ω) the degree is 3, so the group is ℤ/3.

Part CCSIR NET December 2023

Galois theory essentials

Let $f(X) = X^{3} - 2 \in \mathbb{Q}[X]$ and let $K \subset \mathbb{C}$ be the splitting field of f(X) over $\mathbb{Q}$ . Let $\omega =$ $e^{2\pi{}i/3}$ . Which of the following statements are true?

A.The Galois group of K over $\mathbb{Q}$ is the symmetric group $S_{3}$ .✓ Correct
B.The Galois group of K over $\mathbb{Q}(\omega)$ is the symmetric group $S_{3}$ .
C.The Galois group of K over $\mathbb{Q}$ is $\mathbb{Z}/3\mathbb{Z}$ .
D.The Galois group of K over $\mathbb{Q}(\omega)$ is $\mathbb{Z}/3\mathbb{Z}$ .✓ Correct

The idea, in plain words

The symmetry group depends on what you start from. Beginning with the rationals you need both a cube root and a rotation to reach all three roots, giving a group of size six. Hand over the rotation for free at the start and only the cube root is left to permute, giving a much smaller group.

Solution

$K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ has degree 6 over $\mathbb{Q}$ with non-abelian Galois group $S_{3}$ ; over $\mathbb{Q}(\omega)$ the degree is 3

so the group is $\mathbb{Z}/3$ .

Why each option is right or wrong

Checked against 3 symbolic computations

A.Correct
True.

The roots are $\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^{2}\sqrt[3]{2}$

so $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ .

The degrees multiply: $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3 (X^{3} - 2$ is Eisenstein at 2) and $[\mathbb{Q}(\omega) : \mathbb{Q}] = 2$

giving $[K : \mathbb{Q}] = 6 ($ verified).

The Galois group embeds in $S_{3}$ as it permutes three roots, and | $S_{3}$ | = 6 (verified)

so it is all of $S_{3}$ .
B.WrongBase field or ring
False on order grounds.

Once $\omega$ is in the base field, $[K : \mathbb{Q}(\omega)] = 6/2 = 3$

so the Galois group has order 3 — it cannot be $S_{3}$

which has order 6.

Enlarging the base field can only shrink the Galois group.
C.WrongBase field or ring
False.

|Gal $(K/\mathbb{Q})$ | $= [K : \mathbb{Q}] = 6$

so it cannot be $\mathbb{Z}/3\mathbb{Z}$ .

It is also non-abelian: complex conjugation and the 3-cycle on the roots do not commute.

This option and (B) are the same confusion in mirror image — each attaches the right group to the wrong base field.
D.Correct
True.

$\mathbb{Q}(\omega)$ already contains the cube roots of unity (SymPy confirms $\omega^{3} = 1)$

so $K = \mathbb{Q}(\omega)(\sqrt[3]{2})$ is a degree-3 Kummer extension, cyclic with group $\mathbb{Z}/3\mathbb{Z}$ .

The trigger: adjoining the relevant roots of unity to the base is what turns a radical extension cyclic.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Galois theory essentials

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