Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 2.2.17* (Coefficients of a power series divisible by $p$)

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 ) [ [ x ] ] , and more precisely f ( x ) is in the subring A = [ [ x ] ] of formal series with integer coefficients, because 1 ( 1 x ) = i = 0 x i A so that c i (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 [ [ x ] ]

1 x p = ( 1 x ) p i = 0 c i x i .

This comes down to apply the ring homomorphism

φ { [ [ 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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