Exercise 9.13

Show that π is a cube in D ∕ 5 D iff π ≡ 1 , 2 , 3 , 4 , 1 + 2 ω , 2 + 4 ω , 3 + ω , or 4 + 3 ω ( mod 5 ) .

Answers

Proof. Let π ∈ D , [ π ] ≠ 0 . Then [ π ] is a cube in D ∕ 5 D iff [ π ] ( q 2 − 1 ) ∕ 3 = 1 , with q = 5 , namely [ π ] 8 = 1 (Prop. 7.1.2, where 3 ∣ q 2 − 1 = 24 = | ( D ∕ 5 D ) ∗ | ).

By Exercise 9.12, the class of γ = ωλ has order 8, thus the 8 elements [ γ ] k , 0 ≤ k ≤ 7 are distinct roots of the polynomial x 8 − 1 , which has at most 8 roots. Therefore the subgroup of cubes in ( D ∕ 5 D ) ∗ is

{ 1 , [ γ ] , [ γ ] 2 , … , [ γ ] 7 } .

γ = ω ( 1 − ω ) = ω + 1 + ω = 1 + 2 ω , so

γ 0 = 1 γ 1 = 1 + 2 ω γ 2 ≡ − 3 ≡ 2 [ 5 ] ( Ex . 9.12 ) γ 3 = − 3 − 6 ω ≡ 2 + 4 ω [ 5 ] γ 4 ≡ − 1 ≡ 4 [ 5 ] γ 5 ≡ − 1 − 2 ω ≡ 4 + 3 ω [ 5 ] γ 6 ≡ 3 [ 5 ] γ 7 ≡ 3 + 6 ω ≡ 3 + ω [ 5 ]

Conclusion: If π ≢ 0 ( mod 5 ) , π ≡ α 3 ( mod 5 ) , α ∈ D iff

π ≡ 1 , 2 , 3 , 4 , 1 + 2 ω , 2 + 4 ω , 3 + ω , 4 + 3 ω [ 5 ] .

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2022-07-19 00:00
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