1. Introduction

Let be a finite field of elements, , is the characteristic of . Let

denotes the n-th Bernoulli-Goss number ^[5] which is a special value of the zeta function of Goss and is in . In the following we give a characterization of monic irreducible polynomials dividing by introducing the numbers , for .

2. Definitions and Notations

In this section, we introduce some definitions and notation that will be used throughout the paper .

• is a finite field of elements , is a power of a prime , ;

•  ,, ;

• ;

• Let , we say that is prime if and is irreducible;

• is the adic valuation where is a prime ;

• ;

• ;

• .

3. Carlitz Module

Let be the Carlitz module which is a morphism of -algebras from into the- endomorphisms of the additive group given by , for

and

for

3.1.1. Lemma ([5], Proposition 3.3.10)

Let , then

Where for

3.1.2. Lemma

Let of degree n, then

1).

2).

Proof

The proof is very easy and can be done with the following

Hints:

1). By induction on i

2). This is obvious.r

3.1.3. Lemma

Let P be a prime of degree d and let , then

Proof

The proof can be done by induction on k. r

4. A remarkable Congruence

4.1.1. Definition

Let , we set

1.

2.

We have : , and using Carlitz's theorem (^[5], Theorem 3.1.5),

Now, we present our first theorem which generalizes a result of Goss appeared in ^[5], page 325, line 19 for i=1.

4.1.2. Theorem

Let , then

Proof

We have

On the other hand, we have :

So the logarithmic derivative of is:

Thus

Therefore:

On the other hand, we have :

Since

We deduce that

By identification, we obtain:

Therefore :

This terminates the proof.r

4.1.3. Definition

We define the i-th Bernoulli-Goss numbers as follows:

and

4.1.4. Theorem ([11], Theorem10)

Let be a prime of degree d,

then

Proof

We have

Therefore

For , we denote

According to Sheats (^[9]), we have if therefore , thus

for .

Hence, it follows that:

So according to Theorem 4.1.2, we have :

This terminates the proof. r

4.1.5. Lemma

Let P be a premier of degree d, then

Proof

This can be shown by a combination of an induction on k, and lemma 3.1r

Now, we present the following remarkable congruence:

4.1.6. Theorem([11],Theorem 11)

Let P be a premier of degree d, then

Proof

We have

Since for, we have by Theorem 4.1.4

, and

5. The Numbers M(d)

We note that :

5.1. Definition

For, we set

and r

According to theorem 4.1.6 if P is a prime of degree d, then

5.2. The Number M(2)
5.2.1. Lemma

M(2) is the product of distinct monic irreducible polynomials (prime) of A of degree p.

These polynomials are the divisors of the - th Bernoulli-Goss number r

Proof

We have:

Let be a irreductible of degree d such that

Let is the smallest integer such that

Because

This proves that : P divides

or

But r

The previous lemma answers the question: What are the primes of degree 2 dividing the - th Bernoulli-Goss number .

i. e

Conclusion

•  If , there is exactly primes of degree 2 satisfying the equation

•  If there is no prime of degree 2 satisfying the equation.

5.3. Number M(3)

Let P be a prime of degree 3 which divides M(3), P is a divisor of the -th Bernoulli-Goss number

Let and

Let: , we have :

There is two possible cases:

Case 1 if , then

therefore is a root of the polynomial with . We have

Then

Moreover :

if

Since

Let F an irreducible of degree d which divides is of degree 3, because if is a root of F, then

This proves that: F divide

But r

Therefore there is irreducible polynomial of degree 3 which divides and if F divide

Conclusion:

For

there is : irreducible polynomials of degree 3 dividing M(3)

Indeed, in this case, and therefore the equation

(1)

has two solutions in

For each there is irreducible polynomials of degree 3 which divide , and thus divide (M)3.

Thus, if P is an irreducible of degree d which divides is a root of P, then .

Therefore

But:

Since is a root of (1).

This proves that :

Case 2 if , then

Therefore:

We set

Since and Tr is linear, then

Because:

So we have:

From :

Let , has degree 3

and because

Now we are looking for ,

We look for F(T) of degree 3 such that

We have :

And then

We set

and we want to get

we have

So we set

Therefore the polynomial is as follows

Thus Q(T) is an irreducible of degree 3 with constant term , because we have in the other case. r

Before concluding we will answer the following question: for is there infinitely many primes such that :

(2)

5.3.1. Proposition

Let , there is at least one prime of degree d such that

Proof

We can assume .

According to (^[7], Proposition 5.5) , we have :

where is the number of irreducible polynomials of degree , is the smallest prime factor of d

If we had

we would have :

i.e

which is impossible if .

On the other hand:

Therefore

Thus, there is at least

prime of degree d which satisfy

Conclusion

In this paper , we showed that there are infinitely many primes such that

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