Keywords: BernoulliGoss, Carlitz Module, congruence, irreducible polynomials.
Turkish Journal of Analysis and Number Theory, 2014 2 (1),
pp 1318.
DOI: 10.12691/tjant214
Copyright © 2013 Science and Education Publishing. All Rights Reserved.
1. Introduction
Let be a finite field of elements, , is the characteristic of . Let
denotes the nth BernoulliGoss 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. LemmaLet 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. LemmaLet 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. DefinitionLet , 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. TheoremLet , 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. DefinitionWe define the ith BernoulliGoss 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. LemmaLet 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. DefinitionFor, 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. LemmaM(2) is the product of distinct monic irreducible polynomials (prime) of A of degree p.
These polynomials are the divisors of the  th BernoulliGoss 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 BernoulliGoss 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 BernoulliGoss 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. PropositionLet , 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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