Information Technology Reference
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where
f
1
(
z
)
,...,f
n
(
z
)
∈
IF [
z
]. Conversely, any polynomial column vector
c
(
z
)
∈
IF [
z
]
m
completely determines an associated element in
Λ
given by
|
Λ
:=
T
(
z
)
c
(
z
)
.
Pol(
T
(
z
)
c
(
z
))
c
(
z
)
z
−N−
1
−
σ
(
c
(
z
))
For 1
≤
i
≤
m
,let
S
i
(
Λ
)=
{
γ
∈
Λ
:
θ
(
γ
)
∈
[
β
i
]
}
and
IF [
z
]
m
:
c
(
z
)isan
N
th right
i
-annihilating polynomial
column vector of
T
Γ
i
(
T
)=
{
c
(
z
)
∈
}
.
Furthermore, we define a natural ordering, namely, both
S
i
(
Λ
)and
Γ
i
(
T
)are
ordered by the valuation of an element in them.
Theorem 2.
The mapping η restricted on S
i
(
Λ
)
is an order-preserving one-to-
one correspondence from S
i
(
Λ
)
to Γ
i
(
T
)
for
1
≤
i
≤
m,andσ restricted on
Γ
i
(
T
)
is its inverse.
Proof.
Denote
η
restricted on
S
i
(
Λ
)by
η
|
S
i
(
Λ
)
,
σ
restricted on
Γ
i
(
T
)by
σ
|
Γ
i
(
T
)
,
respectively. First we need to show that
η
|
S
i
(
Λ
)
is well-defined. For any vector
γ
∈
S
i
(
Λ
), we have
γ
=
T
(
z
)
η
(
γ
)
.
Pol(
T
(
z
)
η
(
γ
))
η
(
γ
)
z
−N−
1
−
Since
γ
∈
S
i
(
Λ
), we have
v
(
T
(
z
)
η
(
γ
)
−
Pol(
T
(
z
)
η
(
γ
)))
≤−
N
−
1+
v
(
η
(
γ
))
and so
η
(
γ
)
∈
Γ
i
(
T
). Similarly
σ
|
Γ
i
(
T
)
is well-defined. It is easy to check that
η
|
S
i
(
Λ
)
σ
|
Γ
i
(
T
)
=1
Γ
i
(
T
)
and
σ
|
Γ
i
(
T
)
η
|
S
i
(
Λ
)
=1
S
i
(
Λ
)
. For any two elements
γ
1
,γ
2
∈
S
i
(
Λ
)with
v
(
γ
1
)
≤
v
(
γ
2
), we have
v
(
η
(
γ
1
))
≤
v
(
η
(
γ
2
)) for all 1
≤
i
≤
m
.
Theorem 3.
Let ω
1
,
···
,ω
p
+
m
be a normal basis for Λ.Thenη
(
ω
p
+
i
)
,
1
≤
i
≤
m,isanNth right i-minimal polynomial column vector of T .
Proof.
By Theorem 2, it suces to show that
ω
p
+
i
is a minimal element in
S
i
(
Λ
)for1
≤
i
≤
m
. Suppose there exists a vector
γ
∈
S
i
(
Λ
) such that
v
(
γ
)
<
v
(
ω
p
+
i
)forsome
i
,1
≤
i
≤
m
.Inviewof
γ
∈
Λ
,wecanwrite
γ
=
f
1
(
z
)
ω
1
+
···
+
f
n
(
z
)
ω
n
with
f
j
(
z
)
∈
IF [
z
]for 1
≤
j
≤
n
.Since
θ
(
ω
1
)
,
···
,θ
(
ω
n
)are
linearly independent over IF and by convention deg(0) =
−∞
,wehave
v
(
γ
)=
max
{
v
(
ω
j
)+deg(
f
j
(
z
)) : 1
≤
j
≤
n
}
.Let
I
(
γ
)=
{
1
≤
j
≤
n
:
v
(
ω
j
)+
deg(
f
j
(
z
)) =
v
(
γ
)
and so
f
p
+
i
(
z
) = 0 because of
v
(
γ
)
<v
(
ω
p
+
i
). Let lc(
f
(
x
))
denote the leading coe
cient of a polynomial
f
(
x
). Thus it follows that
θ
(
γ
)=
j∈I
(
γ
)
}
lc(
f
j
(
z
))
θ
(
ω
j
)
/
∈
[
β
i
]
,
a contradiction with
γ
∈
S
i
(
Λ
).
