Exercise 9.16

Is x 3 ≡ 2 − 3 ω ( mod 11 ) solvable ? Since D ∕ 11 D has 121 elements this is hard to resolve by straightforward checking. Fill in the details of the following proof that it is not solvable. χ π ( 2 − 3 ω ) = χ 2 − 3 ω ( 11 ) and so we shall have a solution iff x 3 ≡ 11 ( mod 2 − 3 ω ) is solvable. This congruence is solvable iff x 3 = 11 ( mod 7 ) is solvable in ℤ . However, x 3 ≡ a ( mod 7 ) is solvable in ℤ iff a ≡ 1 or 6 ( mod 7 ) .

Answers

Warning: false sentence, since

N ( 2 − 3 ω ) = ( 2 − 3 ω ) ( 2 − 3 ω 2 ) = 4 + 9 − 6 ( ω + ω 2 ) = 4 + 9 + 6 = 1 9 ( a n d n o t 7 ! ) .

Proof. Since 1 9 is a rational prime, and since π = 2 − 3 ω and 1 1 are primary primes, by the Law of Cubic Reciprocity, and by Exercise 9.15 (with p = 1 9 ≡ 1 ( m o d 3 ) ),

∃ ⁡ x ∈ D , 2 − 3 ω ≡ x 3 [ 1 1 ] ⟺ χ 1 1 ( 2 − 3 ω ) = 1 ⟺ χ 2 − 3 ω ( 1 1 ) = 1 ⟺ ∃ ⁡ x ∈ D , x 3 ≡ 1 1 [ 2 − 3 ω ] ⟺ ∃ ⁡ x ∈ ℤ , x 3 ≡ 1 1 [ 1 9 ]

Moreover, by Proposition 7.1.2 (with p = 1 9 , d = ( p − 1 ) ∧ 3 = 3 , ( p − 1 ) ∕ d = 6 ),

∃ ⁡ x ∈ ℤ , x 3 ≡ 1 1 [ 1 9 ] ⟺ 1 1 6 ≡ 1 ( m o d 1 9 ) ,

which is true : 1 1 6 = 1 2 1 3 = ( 1 9 × 6 + 7 ) 3 ≡ 4 9 × 7 ≡ 1 1 × 7 ≡ 7 7 ≡ 1 [ 1 9 ] .

Conclusion: there exists x ∈ D such that 2 − 3 ω ≡ x 3 ( m o d 1 1 ) .

With some computer code, we find a solution x = 1 + 8 ω (and its associates ω 2 x = 7 − ω , ω x = − 8 − 7 ω ≡ 3 + 4 ω ( m o d 1 1 ) ) :

x 3 = ( 1 + 8 ω ) 3 = 3 2 1 − 1 6 8 ω ≡ 2 − 3 ω ( m o d 1 1 ) .

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Note: The sentence becomes true if we replace 2 − 3 ω by the primary prime 2 + 3 ω . Since N ( 2 + 3 ω ) = 7 , with the same reasoning,

∃ ⁡ x ∈ D , 2 + 3 ω ≡ x 3 [ 1 1 ] ⟺ χ 2 + 3 ω ( 1 1 ) = 1 ⟺ ∃ ⁡ x ∈ D , x 3 ≡ 1 1 [ 2 + 3 ω ] ⟺ ∃ ⁡ x ∈ ℤ , x 3 ≡ 1 1 ≡ 4 [ 7 ] ⟺ 4 2 ≡ 1 ( m o d 7 )

but 4 2 ≡ 2 ≢ 1 ( m o d 7 ) , so the equation x 3 ≡ 2 + 3 ω ( m o d 1 1 ) is not solvable.

( x 3 ≡ a ( m o d 1 1 ) is solvable in ℤ iff a

7-13 = a^2 ≡ 1 ( m o d 7 ) iff a ≡ ± 1 ( m o d 7 ) .)

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2022-07-19 00:00
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