Exercise 9.19

Suppose that γ = A + Bω is primary and that A = 3 M − 1 and B = 3 N . Prove that χ γ ( ω ) = ω M + N and that χ γ ( λ ) = ω 2 M .

Answers

Proof. We verify first that if γ = − γ 1 γ 2 , with

γ = A + Bω , A = 3 M − 1 , B = 3 N , γ 1 = A 1 + B 1 ω , A 1 = 3 M 1 − 1 , B 1 = 3 N 1 , γ 2 = A 2 + B 2 ω , A 2 = 3 M 2 − 1 , B 2 = 3 N 2 ,

then M ≡ M 1 + M 2 ( mod 3 ) , N ≡ N 1 + N 2 ( mod 3 ) .

− γ 1 γ 2 = − A 1 A 2 + B 1 B 2 + ( − A 1 B 2 − A 2 B 1 + B 1 B 2 ) ω = A + Bω ,

therefore

3 M − 1 = A = − A 1 A 2 + B 1 B 2 ≡ 3 ( M 1 + M 2 ) − 1 ( mod 9 ) ,

thus M ≡ M 1 + M 2 ( mod 3 ) .

3 N = B = − A 1 B 2 − A 2 B 1 + B 1 B 2 ≡ 3 ( N 1 + N 2 ) ( mod 9 ) ,

thus N ≡ N 1 + N 2 ( mod 3 ) .

By induction, if γ = ± γ 1 γ 2 ⋯ γ t = ( − 1 ) t − 1 γ 1 γ 2 ⋯ γ t , where γ i = A i + B i ω , A i = 3 M i − 1 , B i = 3 N i , then

M ≡ M 1 + ⋯ + M t ( mod 3 ) , N ≡ N 1 + ⋯ + N t ( mod 3 ) .

By Exercise 9.3,

χ γ ( ω ) = χ γ 1 ( ω ) ⋯ χ γ t ( ω ) = ω M 1 + N 1 ⋯ ω M t + N t = ω ( M 1 + ⋯ + M t ) + ( N 1 + ⋯ + N t ) = ω M + N ,

and by Eisenstein’s result,

χ γ ( λ ) = χ γ 1 ( λ ) ⋯ χ γ t ( λ ) = ω 2 M 1 ⋯ ω 2 M t = ω 2 ( M 1 + ⋯ + M t ) = ω 2 M .

Conclusion: if γ = 3 M − 1 + 3 Nω , then

χ γ ( ω ) = ω M + N , χ γ ( λ ) = ω 2 M .

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2022-07-19 00:00
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