Keywords: Bernoulli-Goss, Carlitz Module, congruence, irreducible polynomials.
Turkish Journal of Analysis and Number Theory, 2014 2 (1),
pp 13-18.
DOI: 10.12691/tjant-2-1-4
Received January 12, 2014; Revised February 12, 2014; Accepted February 20, 2014
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 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. 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 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. 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 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. 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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