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 ∈ 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 ) ∈ ℕ 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 ∈ 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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