Y. Cao et al. / Discrete Mathematics 338 (2015) 922–937 925
(ii) R
n
= I
0
⊕ I
1
⊕ · · · ⊕ I
s
, where ε
i
(X) is the multiplicative identity of I
i
for each 0 ≤ i ≤ s. This direct sum is a ring
direct sum in that I
i
I
j
= {0} if i = j.
(iii) For any r + 1 ≤ i ≤ s, we have I
i
= I
i,0
⊕ I
i,1
⊕ · · · ⊕ I
i,l−1
, where ε
i,h
(X) is the multiplicative identity of I
i,h
for each
0 ≤ h ≤ l − 1. This direct sum is a ring direct sum in that I
i,h
I
i,j
= {0} if h = j.
(iv) For each 0 ≤ i ≤ r, the mapping ψ
i
: R[X ]/(m
i
(X)) → I
i
defined by
f (X) + (m
i
(X)) → ε
i
(X)f (X ) + (X
n
− 1) (∀f (X) ∈ R[X])
is a ring isomorphism. Hence I
i
is an extension Galois ring of R. In particular, I
i
is a free R-module of rank κ
i
.
(v) For each r + 1 ≤ i ≤ s and 0 ≤ h ≤ l − 1, the mapping ψ
i,h
: R[X ]/(m
i,h
(X)) → I
i,h
defined by
f (X) + (m
i,h
(X)) → ε
i,h
(X)f (X ) + (X
n
− 1) (∀f (X) ∈ R[X])
is a ring isomorphism. Hence I
i,h
is an extension Galois ring of R. In particular, I
i,h
is a free R-module of rank κ
i,h
.
From now on, for any a(X ) =
n−1
k=0
a
k
X
k
∈ R
n
with a
k
∈ R, we define
φ(a(X)) =
n−1
k=0
φ(a
k
)X
k
, µ(a(X)) =
n−1
k=0
a
k
X
−k
= X
n
a
1
X
=
n−1
k=0
a
k
X
n−k
.
Then we have the following results.
• φ is a ring automorphism of R
n
of multiplicative order l.
• µ is a ring automorphism of R
n
of multiplicative order 2. For any a(X ) ∈ R
n
, we will denote
a(X) = µ(a(X )) in some
cases.
• φµ = µφ.
• R
(p)
n
is precisely the subring of R
n
whose elements are fixed by φ; i.e., R
(p)
n
is the set of polynomials in R
n
with all
coefficients in Z
p
ϵ
.
In the rest of this paper, we denote
ϱ(Y ) = (Y − ω)(Y − ω
p
) · · · (Y − ω
p
l−1
).
Then ϱ(Y ) is a monic basic primitive polynomial in Z
p
ϵ
[Y ] of degree l as ord(ω) = p
l
− 1, R = Z
p
ϵ
[ω]
∼
=
Z
p
ϵ
[Y ]/(ϱ(Y )) and
every element c of R can be uniquely expressed as c
0
+ c
1
ω + · · · + c
l−1
ω
l−1
where c
0
, c
1
, . . . , c
l−1
∈ Z
p
ϵ
(cf. [15] Theo-
rem 14.8). Therefore, R is a free Z
p
ϵ
-module of rank l and {1, ω, . . . , ω
l−1
} is a Z
p
ϵ
-basis of the Galois ring R. In the following,
we gather some facts about the image of I
i
and I
i,h
under φ, and about the relationship between K
i
and I
i
.
Theorem 2.4. Using the notations above, the following hold.
(i) φ(ε
i
(X)) = ε
i
(X) for all 0 ≤ i ≤ s, and φ(ε
i,h
(X)) = ε
i,h+1(mod l)
(X) for all r + 1 ≤ i ≤ s and 0 ≤ h ≤ l − 1.
(ii) φ(I
i
) = I
i
for all 0 ≤ i ≤ s, and φ(I
i,h
) = I
i,h+1(mod l)
for all r + 1 ≤ i ≤ s and 0 ≤ h ≤ l − 1.
(iii) K
i
= I
i
∩ R
(p)
n
and hence I
i
is a K
i
-module for all 0 ≤ i ≤ s.
(iv) If 0 ≤ i ≤ r, then I
i
is a free K
i
-module of rank l with a basis given by {ε
i
(X), ωε
i
(X), . . . , ω
l−1
ε
i
(X)}.
(v) If r + 1 ≤ i ≤ s, then I
i
is a free K
i
-module of rank l with a basis given by {ε
i,0
(X), ε
i,1
(X), . . . , ε
i,l−1
(X)}.
Proof. (i) Because ε
i
(X) ∈ Z
p
ϵ
[X], φ(ε
i
(X)) = ε
i
(X). The remainder of (i) follows from the definitions of ε
i,h
(X) and φ.
(ii) It follows from (i) and the definitions of I
i
and I
i,h
.
(iii) Obviously, K
i
= ε
i
(X)R
(p)
n
⊆ I
i
∩ R
(p)
n
. Conversely, let a(X) ∈ I
i
∩ R
(p)
n
. Then a(X ) ∈ R
(p)
n
and a(X ) ∈ I
i
. Since
ε
i
(X) is the multiplicative identity of I
i
by Lemma 2.3(ii), we have a(X ) = ε
i
(X)a(X ) ∈ K
i
.
(iv) Let 0 ≤ i ≤ r. Then gcd(κ
i
, l) = 1. By Lemma 2.2, K
i
is a Galois ring of characteristic p
ϵ
and cardinality p
ϵκ
i
. Since
ω is a primitive (p
l
− 1)th root of unity, by ω ∈ R ⊆ R
n
and Lemmas 2.2 and 2.3 ε
i
(X)ω is a primitive (p
l
− 1)th root
of unity contained in I
i
. But K
i
is a subring of I
i
, and the minimal polynomial of ε
i
(X)ω over K
i
is g
i
(Y ) =
d−1
j=0
(Y −
(ε
i
(X)ω)
(p
κ
i
)
j
) where d = min{a ∈ Z
+
| (p
κ
i
)
a
≡ 1 (mod p
l
− 1)} =
l
gcd(κ
i
,l)
= l. From this we deduce that K
i
[ε
i
(X)ω]
∼
=
K
i
[Y ]/(g
i
(Y )) which is a Galois ring of characteristic p
ϵ
and cardinality |K
i
|
l
. Therefore, a K
i
-basis of K
i
[ε
i
(X)ω] is given
by {ε
i
(X), ε
i
(X)ω, . . . , ε
i
(X)ω
l−1
}. Obviously, we have K
i
[ε
i
(X)ω] ⊆ I
i
. By Lemma 2.3(iv) it follows that |I
i
| = p
ϵκ
i
l
=
|K
i
[ε
i
(X)ω]|. Hence I
i
= K
i
[ε
i
(X)ω].
(v) Let r + 1 ≤ i ≤ s and denote S
i
= {
l−1
h=0
φ
h
(α) | α ∈ I
i,0
}. First, we claim that K
i
= S
i
. In fact, for any α ∈ I
i,0
we have φ
h
(α) ∈ I
i,h
by (ii), which then implies
l−1
h=0
φ
h
(α) ∈
l−1
h=0
I
i,h
= I
i
by Lemma 2.3(iii), and so S
i
⊆ I
i
. Since the
multiplicative order of φ is l, we have φ(
l−1
h=0
φ
h
(α)) =
l−1
h=0
φ
h+1
(α) =
l−1
h=0
φ
h
(α), which implies
l−1
h=0
φ
h
(α) ∈ R
(p)
n
,
and so
l−1
h=0
φ
h
(α) ∈ I
i
∩ R
(p)
n
= K
i
by (iii). Hence S
i
⊆ K
i
. Moreover, by Lemma 2.2(iii) we have |K
i
| = |R
i
| = p
ϵκ
i
. Since
φ is a ring automorphism of R
n
, by Lemma 2.3(v) and κ
i,0
=
κ
i
l
we have |S
i
| = |R|
κ
i,0
= (p
ϵl
)
κ
i
l
= |K
i
|. As stated above, we
conclude that K
i
= S
i
.