Exercise 8.23

Let f ( x 1 , x 2 , , x n ) 𝔽 p [ x 1 , x 2 , , x n ] . Let N be the number of zeros of f in 𝔽 p . Show that N = p n 1 + p 1 a 0 ( x 1 , , x n ζ af ( x 1 , , x n ) ) .

Answers

Proof. Let A r = { ( x 1 , x 2 , , x n ) 𝔽 p n | f ( x 1 , x 2 , , x n ) = r } . Then 𝔽 p n = r 𝔽 p A r , so, for all a 𝔽 p ,

( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) = r 𝔽 p ( x 1 , x 2 , , x n ) A r ζ ar = r 𝔽 p | A r | ζ ar .

Let m ( r ) = | A r | = N ( f ( x 1 , x 2 , , x n ) = r ) . Then

a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) = a 𝔽 p r 𝔽 p m ( r ) ζ ar = r 𝔽 p m ( r ) a 𝔽 p ζ ar

As a 𝔽 p ζ ar = 0 if r 0 , and a 𝔽 p ζ ar = p if r = 0 , we obtain

a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) = m ( 0 ) p = pN .

Moreover

a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) = p n + a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) ,

so

pN = p n + a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) .

In conclusion,

N = p n 1 + p 1 a 𝔽 p ( x 1 , x 2 , , x n ) 𝔽 p n ζ af ( x 1 , x 2 , , x n ) .

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2022-07-19 00:00
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