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Exercise 2.2.17* (Coefficients of a power series divisible by $p$)

Let the number c i be defined by the power series identity

( 1 + x + ⋯ + x p − 1 ) ∕ ( 1 − x ) p − 1 = 1 + c 1 x + c 2 x 2 + ⋯ .

Show that c i ≡ 0 ( mod p ) for all i ≥ 1 .

Answers

Proof. Consider

f ( x ) = ( 1 + x + ⋯ + x p − 1 ) 1 ( 1 − x ) p − 1 = 1 − x p ( 1 − x ) p = ∑ i = 0 ∞ c i x i = 1 + c 1 x + c 2 x 2 + ⋯

( c 0 = f ( 0 ) = 1 ).

We can consider these equalities in the real (or complex) field, and the series as convergent series for | x | < 1 . But it is more convenient here to consider f ( x ) as a formal series, so that f ( x ) ∈ ℝ [ [ x ] ] , and more precisely f ( x ) is in the subring A = ℤ [ [ x ] ] of formal series with integer coefficients, because 1 ∕ ( 1 − x ) = ∑ ⁡ i = 0 ∞ x i ∈ A so that c i ∈ ℤ (we can compute c i using Exercise 1.4.7).

Here we denote [ c ] p the class modulo p of some integer c (and 1 = [ 1 ] p , 0 = [ 0 ] p ).

Then we reduce modulo p the equality in ℤ [ [ x ] ]

1 − x p = ( 1 − x ) p ∑ i = 0 ∞ c i x i .

This comes down to apply the ring homomorphism

φ { ℤ [ [ x ] ] → 𝔽 p [ [ x ] ] ∑ i = 0 ∞ a i x i ↦ ∑ i = 0 ∞ [ a i ] p x i .

We obtain

1 − x p = ( 1 − x ) p ∑ i = 0 ∞ [ c i ] p x i ∈ 𝔽 p [ [ x ] ] .

Using ( p k ) ≡ 0 ( 1 ≤ k ≤ p − 1 ) , we obtain the equaliy in 𝔽 p [ [ x ] ]

( 1 − x ) p = 1 − x p .

(This is true is p is an odd prime, and also true if p = 2 , since 1 = − 1 in 𝔽 2 .) Thus

1 − x p = ( 1 − x p ) ∑ i = 0 ∞ [ c i ] p x i ∈ 𝔽 p [ [ x ] ] .

Moreover 1 − x p is invertible in 𝔽 p [ [ x ] ] , since

( 1 − x p ) ∑ i = 0 ∞ x pi = 1 .

Therefore, simplifying by ( 1 − x p ) ,

1 = 1 + [ c 1 ] p x + [ c 2 ] p x 2 + ⋯ ,

so

0 ≡ c 1 ≡ c 2 ≡ ⋯ ( mod p ) .

This shows that c i ≡ 0 ( mod p ) for all i ≥ 1 . □

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2024-08-10 11:24
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