Exercise 2.24

Supply the details to the proof of Theorem 3.

Answers

As Adam Michalik, I suppose that there is a misprint, we must prove Theorem 4 :

Let k a finite field with q elements.

q deg p ( x ) diverges, where the sum is over all monic irreducible p ( x ) in k [ x ] .

Proof. Notations :

P : set of all monic polynomials p in k [ x ] .

P n : set of all monic polynomials p in k [ x ] with deg ( p ) n .

M : set of all monic irreducible polynomials p in k [ x ] .

M n : set of all monic irreducible polynomials p in k [ x ] with deg ( p ) n .

We must prove that p M q deg p ( x ) diverges.

p P q deg p ( x ) diverges :

f P n 1 q deg f = d = 0 n deg ( f ) = d 1 q d = d = 0 n 1 q d Card { f P | deg ( f ) = d } = d = 0 n 1 q d q d = n + 1 .

So f P q deg f diverges.

f P q 2 deg f converges :

f P n q 2 deg ( f ) = d = 0 n deg ( f ) = d 1 q 2 d = d = 0 n 1 q 2 d Card { f P | deg ( f ) = d } = d = 0 n 1 q d 1 1 1 q

As any finite subset of P is included in some P n , f P q 2 deg f converges.

p M q deg p ( x ) diverges :

Let M n = { p 1 , p 2 , , p l ( n ) } the set of all monic irreducible polynomials such that deg p i n . Let

λ ( n ) = i = 1 l ( n ) 1 1 1 q deg ( p i ) .

For simplicity, we write l = l ( n ) for a fixed n . Then

λ ( n ) = i = 1 l a i = 0 1 q a i deg p i = ( 1 + 1 q deg p 1 + 1 q deg p 1 2 + ) × × ( 1 + 1 q deg p l + 1 q deg p l 2 + ) = ( a 1 , , a j ) l 1 q deg ( p 1 a 1 p l a l )

Since the monic prime factors of any polynomial p P n are in P n , the decomposition of p is p = p 1 a 1 p l a l , so

λ ( n ) p P n 1 q deg p = n + 1 .

So lim n λ ( n ) = : this is another proof that there exist infinitely many monic irreducible polynomials in k [ x ] (cf Ex. 2.1).

log λ ( n ) = i = 1 l ( n ) log ( 1 1 q deg p i ) = i = 1 l ( n ) m = 1 1 m q m deg p i = 1 q deg p 1 + + 1 q deg p l ( n ) + i = 1 l ( n ) m = 2 1 m q m deg p i

Yet

m = 2 1 m q m deg p i m = 2 1 q m deg p i = 1 q 2 deg p i 1 1 1 q deg p i = 1 q 2 deg p i q deg p i 2 q 2 deg p i

(the last inequality is equivalent to 2 q deg p i ). So

log λ ( n ) 1 q deg p 1 + + 1 q deg p l ( n ) + 2 ( 1 q 2 deg p 1 + + 1 q 2 deg p l ( n ) ) .

As 1 q 2 deg p 1 + + 1 q 2 deg p l ( n ) is less than the constant f P q 2 deg f , if p M q deg p ( x ) converges, then log λ ( n ) C , where C is a constant, so λ ( n ) e C for all n , in contradiction with lim n λ ( n ) = .

Conclusion : p M q deg p ( x ) diverges. □

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2022-07-19 00:00
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