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GF(2
ne
)
GF(2
e
)
with
n>
1
and let
Proposition 1.
Take any
v
∈
\
v
2
2
l
+1
0
(
v
0
+
v
1
)
2
l
+1
V
=
.
(6)
Then
n−
1
2
jl
B
n
(
V
)=
Tr
n
e
(
v
0
)
(
v
1
+
v
2
)
v
0
v
0
+
v
1
.
j
=2
If
n
is odd (respectively,
n
is even) then the total number of distinct zeros of
B
n
(
x
)
in
GF(2
ne
)
is equal to
2
(
n
−
1)
e
−
1
2
2
e
2
(
n
−
1)
e
−
2
e
1
). All zeros have
the form of (6) with
Tr
n
e
(
v
0
)=0
and occur with multiplicity
2
l
. Moreover,
polynomial
B
n
(
x
)
splits in
GF(2
ne
)
if and only if
e
=
l
or
n<
4
.
(respectively,
−
1
2
2
e
−
Proof.
Most of the proof is completed in [7, Lemma 1] and the remaining part
follows. In particular, it was shown that
i
j
=1
v
j
(
v
1
+
v
2
)
i−
1
2
jl
v
0
v
0
+
v
1
B
i
(
V
)=
(7)
j
=2
n
+1 and that
B
n
(
x
) splits in GF(2
nl
) and zeros of
B
n
(
x
)in
GF(2
ne
) are exactly the elements obtained by (6) using
w
0
∈
for 2
≤
i
≤
GF(2
nl
)
GF(2
l
)
\
with Tr
nl
l
GF(2
ne
).
It also follows from the proof of [7, Proposition 4] that polynomial
f
b
(
y
)=
y
2
l
+1
+
by
+
b
with
b
(
w
0
) = 0 that result in
V
∈
GF(2
ne
)
∗
has exactly 2
e
+1 zeros in GF(2
ne
) if and only if
b
−
1
has the form of (6) with Tr
ne
e
∈
GF(2
ne
) obtained by (6)
(
v
0
) = 0. Take any
V
∈
GF(2
l
)withTr
n
l
(
w
0
)=0.Then
f
V
−
1
(
y
) splits in GF(2
nl
)
and, by [6, Corollary 7.2], this is equivalent to
f
V
−
1
(
y
)having2
e
+ 1 zeros in
GF(2
ne
). Thus, there exists some
v
0
∈
GF(2
nl
)
using
w
0
∈
\
GF(2
e
)withTr
ne
e
GF(2
ne
)
\
(
v
0
)=0that
gives this
V
using (6).
Corollary 1.
If
n
is odd (respectively,
n
is even) then the total number of dis-
tinct zeros of
Z
n
(
x
)
in
GF(2
ne
)
is equal to
2
(
n
+1)
e
−
2
2
e
2
2
e
2
(
n
+1)
e
−
2
e
1
).
All zeros have the form of (6) and occur with multiplicity one. Moreover, poly-
nomial
Z
n
(
x
)
splits in
GF(2
ne
)
if and only if
e
=
l
or
n
=1
.
(respectively,
−
1
2
2
e
−
Proof.
Most of the proof is completed in [7, Corollary 1] and the remaining part
follows.
In particular, it was shown that
Z
n
(
x
) splits in GF(2
nl
) and zeros of
Z
n
(
x
)in
GF(2
ne
) are exactly the elements obtained by (6) using
w
0
∈
GF(2
nl
)
GF(2
l
)
\
that result in
V
GF(2
ne
). It also follows from the proof of [7, Propositions 3, 4]
that polynomial
f
b
(
y
)=
y
2
l
+1
+
by
+
b
with
b
∈
GF(2
ne
)
∗
has exactly one or 2
e
+1
zeros in GF(2
ne
) if and only if
b
−
1
has the form of (6). Take any
V
∈
GF(2
ne
)
∈
GF(2
l
). Then
f
V
−
1
(
y
) has exactly one
or 2
l
+ 1 zeros GF(2
nl
) and, by [6, Corollaries 7.2, 7.3], this is equivalent to
f
V
−
1
(
y
)havingoneor2
e
+ 1 zeros in GF(2
ne
) respectively. Thus, there exists
some
v
0
∈
GF(2
nl
)
obtained by (6) using
w
0
∈
\
GF(2
ne
)
GF(2
e
)thatgivesthis
V
using (6).
\
