Download On the exact number of solutions of certain linearized equations

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On the exact number of solutions of certain
linearized equations
Ferruh Özbudak · Zülfükar Saygı
Received: date / Accepted: date
Abstract In this note we have revisited some of of the results of Trachtenberg
[9], which are directly related with the number of solutions of some special
linearized polynomials over finite fields. In some cases we give improvements.
Also, we give some results on the exact number of solutions of certain linearized
equations depending on the coefficients of that equation.
Keywords Linearized equations · Finite fields
1 Introduction
In this note we have revisited some of of the results of Trachtenberg [9], which
are directly related with the number of solutions of some special linearized
polynomials over finite fields. In some cases we give improvements.
It is known that, finding the number of solutions of linearized polynomials is
important in many application in the literature. For example, one can see such
equations in some recents works on the correlations of some special sequences
[2–8].
Let p be an odd prime and n, k be positive integers. Throughout this note,
we will use the following notations. We set q = pk , e = gcd(n, k), n1 = n/e,
q1 = pe and q2 = pnk/e .
We have the following extensions of Fq1 :
Ferruh Özbudak
Department of Mathematics and Institute of Applied Mathematics, Middle East Technical
University, İnönü Bulvarı, 06531, Ankara, Turkey
Tel.: +90-312-210 2991
E-mail: [email protected]
Zülfükar Saygı
Department of Mathematics, TOBB University of Economics and Technology, Söğütözü
06530, Ankara, Turkey
E-mail: [email protected]
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Ferruh Özbudak, Zülfükar Saygı
F
nk
pe
= Fq 2
n
e
F pn
F pk = F q
n
e
F pe = F q1
Fig. 1 Extensions of Fq1
Lemma 1 Let B ⊆ Fpn be a non-empty set. If B is linearly independent over
Fq1 , then B is also linearly independent over Fq .
Proof Here, we give a sketch of the proof only. The proof follows from the
observation that n1 = [Fq2 : Fq ] is equal to [Fpn : Fq1 ] and some arguments
from field extensions.
t
u
Here we note that this lemma is a stronger version of [9, Lemma 4]. Using
this lemma one can easily decide the possible values for number of solutions
of certain linearized equations. But it is not easy to find the exact number of
solutions of that equation. In the following section we will give some results
concerning this problem. In many cases we can easily find the exact number of
solutions of linearized equations depending on the coefficients of that equation.
2 Main Results
Recall that n1 = n/e. Throughout this section we assume that n1 ≥ 2 and we
will use the notation Norm for the relative norm map from Fpn to Fq1 (that
pn −1
is, Norm(x) = x q1 −1 for any x ∈ Fpn ). We start with a useful proposition.
Proposition 1 Let α ∈ Fpn \ {0} and N (α) denote the number of z ∈ Fpn
such that
z q − αz = 0.
Let ψα be the map on Fpn given by
ψα : Fpn → Fpn
x 7→ xq − αx.
Then we have
N (α) =
1, if Norm(α) 6= 1,
q1 , if Norm(α) = 1.
On the exact number of solutions of certain linearized equations
3
Let Aα (T ) ∈ Fpn [T ] be the Fq1 -linearized polynomial given by
n1 −1
Aα (T ) = T q1
n1 −1
+ α q1
+··· + α
n1 −2
T q1
n1 −1
n −2
+q1 1
+ α q1
n −1
n −2
q1 1 +q1 1 +···+q12
T q1 + α
n1 −3
T q1
n −1
n −2
q1 1 +q1 1 +···+q1
(1)
T.
If Norm(α) = 1, then we also have the followings:
1. Kerψα is the roots of the polynomial T q1 − αT over Fpn . This polynomial
is separable and splits over Fpn .
2. Imψα is the roots of the polynomial Aα (T ). This polynomial is separable
and splits over Fpn .
Using Proposition 1 we obtain the following result.
Theorem 1 Let α, β be nonzero elements of Fpn . Let N (α, β) denote the number of z ∈ Fpn such that
2
(z q − αz) ◦ (z q − βz) = z q − (α + β q ) z q + αβz = 0.
Let Cα,β denote the constant in Fpn defined as
n1 −3
Cα,β
n1 −2
α
αq1 +1
αq1 +···+q1 +1
1
+ αq1 +···+q1 +1 .
= + q1 +1 + q2 +q +1 + · · · + n1 −2
1
q
+···+q
+1
β
β
1
β 1
β 1
Then N (α, β) ∈ {1, q1 , q12 }. Moreover we have the followings:
1. N (α, β) = 1 if and only if Norm(α) 6= 1 and Norm(β) 6= 1.
2. N (α, β) = q1 if and only if one of the followings hold:
(a) Norm(α) = 1 and Norm(β) 6= 1.
(b) Norm(α) 6= 1 and Norm(β) = 1.
(c) Norm(α) = Norm(β) = 1 and Cα,β 6= 0.
3. N (α, β) = q12 if and only if Norm(α) = Norm(β) = 1 and Cα,β = 0.
Proof Here, we give a sketch of the proof only. The proof uses Proposition 1
and the observation that (T q1 − αT ) | Aβ (T ) if and only if Cα,β = 0, where
Aβ (T ) is the Fq1 -linearized polynomial in Fpn [T ] defined in (1).
t
u
The following result is well known if k | n. Here we give a slight extension,
including the case k - n as well. We will use the following proposition later.
Proposition 2 Let m ≥ 2 be an integer. Let
m
A(T ) = T q + Am−1 T q
m−1
+ · · · + A1 T q + A0 T ∈ Fpn [T ]
be an Fq -linearized polynomial with A0 6= 0. If there exists η ∈ Fpn \ {0} such
that A(η) = 0, then there exist β ∈ Fpn \ {0} and Fq -linearized monic and
separable polynomial B(T ) ∈ Fpn [T ] such that
A(T ) = B(T ) ◦ (T q − βT ) .
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Ferruh Özbudak, Zülfükar Saygı
Proof Note that Fq ⊆ Fq2 . Considering A(T ) ∈ Fq2 [T ] we obtain that there
exists B(T ) ∈ Fpn [T ] such that


A(T ) = B(T ) ◦ 
Y
(T − cη) .
(2)
c∈Fq
Here
Y
(T − cη) = T q − βT
(3)
c∈Fq
with β ∈ Fq2 and B(T ) ∈ Fq2 [T ]. It remains to prove that β ∈ Fpn . Indeed
if T q − βT ∈ Fpn [T ], then using (2) we conclude that B(T ) ∈ Fpn [T ], which
completes the proof.
Now we prove that β ∈ Fpn . Comparing the coefficients of degree 1 terms
in both sides of (3) we obtain that
Y
(−cη) = −β and hence β = −η q−1
c∈Fq \{0}
Y
= η q−1 .
c∈Fq \{0}
This shows that β ∈ Fpn \ {0} as η ∈ Fpn \ {0}.
t
u
Let a, b ∈ Fpn \ {0}. Let N denote the number of z ∈ Fpn such that
2
z q + az q + bz = 0.
The main problem of this section is to compute N explicitly. This problem
is now reduced to a “factorization” problem in the following sense:
– If there exist α, β ∈ Fpn \ {0} such that
2
z q + az q + bz = (z q − αz) ◦ (z q − βz) ,
(4)
then N is computed explicitly using Theorem 1 as N = N (α, β).
– If there is no α, β ∈ Fpn \{0} such that (4) holds, then N = 1 by Proposition
2.
The following is a numerical example such that there is no α, β ∈ Fpn \ {0}
satisfying (4).
Example 1 Let p = 3, n = 3, k = 1 and γ be a primitive element in F33 , such
that γ 3 + 2γ + 1 = 0. Then by computer search we see that
z9 + γ7z3 + z
can not be written of the form z 3 − αz ◦ z 3 − βz for all α, β ∈ F33 \ {0}.
On the exact number of solutions of certain linearized equations
5
We find it interesting to present a connection of the factorization problem
above with a result of Bluher [1]. Now we want to find α, β ∈ Fpn \ {0} such
that
2
2
z q + az q + bz = (z q − αz) ◦ (z q − βz) = z q − (α + β q ) z q + αβz,
which means that
a = α + β q and b = αβ.
Then by substituting α = b/β in the first equality we get
b
+ βq
β
a=
that is, β is a solution of the equation
0 = xq+1 − ax + b ∈ Fpn [x]
which is studied in more detail in [1].
Now we note that, using Lemma 1 Trachtenberg obtained the following
result [9, Proposition 1].
Proposition 3 Let γ be a nonzero element of Fpn where p is prime and n is
odd. Then the equation
zp
4m
− (2γ)p
2m
zp
2m
+z =0
(5)
has exactly 1, pe , or p2e roots in Fpn , where e = gcd(m, n).
Here we remark that using Theorem 1 and Proposition 2 it possible to find the
exact number of roots of (5) depending on γ. So this gives an improvement.
We note that k = 2m in our notation.
Furthermore, using the above techniques it is possible to find the exact
number of solutions of the following linearized equations depending on the
coefficients of that equation.
3
2
z q + az q + bz q + cz = 0.
(6)
Similarly, the problem of finding the exact number of solutions of (6) can be
reduced to a “factorization” problem as above. Here we note that using Lemma
1 one can say that the number of solutions of (6) is in the set {1, q1 , q12 , q13 }.
But in many cases depending on the coefficients of the equations, the number
of solutions of (6) will not take all the values in the set {1, q1 , q12 , q13 }. For
example, we can give the following proposition. It is given in [9] and it is
noted in [9] that the proof of the proposition is suggested by L. Welch.
Proposition 4 Let γ be a nonzero element of Fpn where p is an odd prime
and n is odd. Then the equation
zp
e
6m
has exactly 1, p , or p
2e
− γp
3m
zp
4m
− γp
2m
zp
2m
+z =0
roots in Fpn , where e = gcd(m, n).
(7)
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Ferruh Özbudak, Zülfükar Saygı
In our notation by setting k = 2m, we see that the equation (7) is of the form
3
(k/2)
2
z q + bz q + z ∈ Fpn [z]. Note that, if the equation (7) has a
0 = z q + bp
nonzero root then it is shown in [9] that
3
z q + bp
(k/2)
2
z q + bz q + z = (z q − α1 z) ◦ (z q − α2 z) ◦ (z q − α3 z)
for some α1 , α2 , α3 ∈ Fpn . Using this observation it is proved in [9] that the
equation (7) can not have q13 solutions in Fpn . Furthermore, using similar techniques as in Theorem 1 it is possible to make further improvements depending
on the values of α1 , α2 and α3 .
Acknowledgments. The authors were partially supported by TÜBİTAK
under Grant No. TBAG–109T672. The work of Z. Saygı was also supported
by TÜBİTAK under Grant No. TBAG–109T344.
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