Exercise 9.3

Let γ a primary prime. To evaluate χ γ ( μ ) we see, by Exercise 2, that it is enough to evaluate χ γ ( 1 ) , χ γ ( ω ) , χ γ ( λ ) , and χ γ ( π ) , where π is a primary prime. Since 1 = ( 1 ) 3 we have χ γ ( 1 ) = 1 . We now consider χ γ ( ω ) . Let γ = a + and set a = 3 m 1 and b = 3 n . Show that χ γ ( ω ) = ω m + n .

Answers

Proof. Let γ = a + = 3 m 1 + 3 . Then χ γ ( ω ) = ω N ( γ ) 1 3 (remark (b) of Theorem 1).

N ( γ ) 1 = ( 3 m 1 ) 2 + ( 3 n ) 2 3 n ( 3 m 1 ) 1 = 9 m 2 6 m + 9 n 2 9 nm + 3 n N ( γ ) 1 3 = 3 m 2 2 m + 3 n 2 3 nm + n n + m [ 3 ]

Thus, for γ = a + = 3 m 1 + 3 ,

χ γ ( ω ) = ω N ( γ ) 1 3 = ω n + m .

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2022-07-19 00:00
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