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Q(ζ3,23,33) is a Kummer extension with quotient (Z/3)2

Example

Let F=Q(ζ3) and

K=F(23,33).

Then K/F is a Kummer extension, and the subgroup generated by the classes of 2 and 3 in F×/(F×)3 is isomorphic to (Z/3)2. Consequently

Gal(K/F)(Z/3)2.

Facts & Assumptions

Given: The field F=Q(ζ3) and the extension K=F(23,33).

[L1]

Kummer theory identifies finite abelian extensions of exponent dividing 3 with subgroups of F×/(F×)3 (Kummer theory classifies finite abelian extensions of exponent dividing n by subgroups between (F×)n and F×).

[L2]

Norm and trace are given by the embedding formulas (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

Verification

technique · direct
1.1

The field F contains μ3, so adjoining cube roots is exactly the Kummer situation of [L1]. To show that the classes of 2 and 3 are independent modulo cubes, suppose 2a3b=c3in F×. Taking norms from F to Q gives 22a32b=NF/Q(c)3. The left side is a rational cube only when 32a and 32b, hence only when 3a and 3b. Therefore the classes of 2 and 3 each have order 3 and generate a subgroup isomorphic to (Z/3)2.

L2algebra
2.1

By [L1], the field generated by the corresponding cube roots is a Kummer extension with Galois group Hom((Z/3)2,μ3)(Z/3)2. Since K is exactly that field, the stated conclusion follows.

L1step 1.1

Depends on

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