EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 4, 2020, 873-892
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Self-orthogonal Codes over Fq + uFq and Fq + uFq + u2Fq
Lucky Erap Galvez1,∗, Rowena Alma Betty1, Fidel Nemenzo1

1 Institute of Mathematics, University of the Philippines Diliman, Quezon City, Philippines

Abstract. In this paper, we establish a mass formula for Euclidean and Hermitian self-orthogonal
codes over the finite ring Fq + uFq, where Fq is the finite field of order q and u2 = 0. We also
establish a mass formula for Euclidean self-orthogonal codes over the finite ring Fq + uFq + u2Fq,
with u3 = 0 and characteristic of Fq is odd. These mass formulas are used to give a classification
of Euclidean and Hermitian self-orthogonal codes over F2 + uF2 and F3 + uF3 of small lengths.

2020 Mathematics Subject Classifications: 94B05

Key Words and Phrases: Codes over rings, self-orthogonal codes, mass formula

1. Introduction

Self-dual codes have rich mathematical theory and are of great interest to researchers
because many of the best known codes are self-dual. A fundamental problem in coding
theory is the classification of self-dual codes, that is, an enumeration of a complete set of
representatives for the equivalence classes of self-dual codes. In the past years, self-dual
codes over finite fields have been extensively studied and classified up to various lengths
(see [8, 11]). Since the discovery in 1994 [7] that certain non-linear binary codes can be
viewed as linear codes over the ring Z4, there has been much interest in the study of
self-dual codes over various finite rings.

A key problem is to establish an explicit formula for the number of distinct self-dual
codes of length n over a ring R, given by∑

C

|En|
|Aut(C)|

where C runs through the set of all inequivalent self-dual codes of length n over R, En
is the full group of transformations allowed in defining the equivalence for code C and
Aut(C) is the automorphism group. This is called the mass formula, and is an important
computational tool for the classification of such codes. Mass formulas for self-dual codes

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3838

Email addresses: legalvez@math.upd.edu.ph (L. E. Galvez), rabetty@math.upd.edu.ph (R. A. Betty),
fidel@math.upd.edu.ph (F. Nemenzo)

https://www.ejpam.com 873 c© 2020 EJPAM All rights reserved.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 874

over finite rings rings such as Z4 and Fq + uFq were given in [6], while [3] gave the mass
formula for self-dual codes over Fq + uFq + u2Fq.

In this paper, we focus on the more general mass formula for self-orthogonal codes,
which will include the mass formula for self-dual codes as a special case. Mass formulas
for self-orthogonal codes over Zp2 , where p is a prime, were given in [2], while the mass
formula for even codes over Z8, i.e., self-orthogonal codes whose codewords have Euclidean
weights divisible by 16, was computed in [1].

Codes over Fq + uFq and Fq + uFq + u2Fq have an invariant called type, denoted by
{k0, k1} and {k0, k1, k2}, respectively, where k0, k1 and k2 are nonnegative integers. The
type of a code is determined by its residue and torsion codes. To obtain the mass formula,
we will determine the number of self-orthogonal codes of length n over Fq + uFq with
given residue and torsion, and compute the number of self-orthogonal codes of length
n over Fq + uFq + u2Fq, for odd q, with given u2-Residue and torsion. We also give a
classification of Euclidean and Hermitian self-orthogonal codes over F2+uF2 and F3+uF3,
up to some short lengths.

2. Codes over Fq + uFq

We begin with some basic concepts about codes over rings. A linear code C of length
n over a ring R is a submodule of the Rn module. A generator matrix for C is a matrix
G ∈ Mk×n(R) whose rows generate the code. For a matrix G ∈ Mk×n(R), we denote by
RkG the code {aG | a ∈ Rk} of length n over R.

Let q be a power of a prime and Fq denote the finite field of q elements. Let R1 be the
commutative ring Fq[u]/(u2) = Fq + uFq, where u2 = 0. This finite chain ring is a local
ring with unique maximal ideal (u) and residue field Fq + uFq/(u) = Fq. Every code C of
length n over R1 is permutation-equivalent to a code with the following generator matrix,[

Ik0 A+ uB
0 uD

]
, (1)

where Ik0 is the k0 × k0 identity matrix, A,B ∈Mk0×(n−k0)(Fq) and D ∈Mk1×(n−k0)(Fq).
Such a code C is said to be of type {k0, k1}. The code C is said to be free if k1 = 0. The
type is the analog of the dimension of a code over a finite field. Such a code C contains
q2k0+k1 codewords.

Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) be elements of Rn1 . Define the Eu-
clidean inner product on Rn1 as 〈x, y〉E =

∑n
i=1 xiyi. Now, let z = a+ ub ∈ R1 and define

z = a−ub. The Hermitian inner product on Rn1 is defined as 〈x, y〉H =
∑n

i=1 xiyi. The set

C⊥ = {x ∈ Rn1 | 〈x, y〉 = 0 ∀y ∈ C}

is called the Euclidean or Hermitian dual of C, depending on which inner product 〈x, y〉
is used. The code C is said to be Euclidean or Hermitian self-orthogonal if C ⊆ C⊥. If
C = C⊥, then we say C is Euclidean or Hermitian self-dual.

Let C be a code over R1. The code
{
v ∈ Fnq | ∃w ∈ Fnq , v + uw ∈ C

}
is called the

residue code of C and is denoted by res(C). The code
{
v ∈ Fnq |uv ∈ C

}
is called the



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 875

torsion code of C and is denoted by tor(C). If C has generator matrix (1), then res(C)
and tor(C) are [n, k0] and [n, k0 + k1] codes over Fq, with generator matrices

[
Ik0 A

]
and

[
Ik0 A
0 D

]
,

respectively. Clearly, res(C) ⊆ tor(C) and |C| = q2k0+k1 = |res(C)||tor(C)|.
The following lemma shows the relationship between the residue and torsion codes of

a self-orthogonal code over R1. The proof is given in [6].

Lemma 1. Let C be a (Euclidean or Hermitian) self-orthogonal code over R1. Then

(i) res(C) is self-orthogonal, i.e. res(C) ⊆ res(C)⊥;

(ii) tor(C) ⊆ res(C)⊥.

In particular, if C is (Euclidean or Hermitian) self-dual, tor(C) = res(C)⊥.

3. Codes over Fq + uFq with prescribed residue and torsion

Let C1 be a code of length n over Fq with dimension k0 and generator matrix[
Ik0 A

]
, (2)

and C2 a code of length n over Fq of dimension k0 + k1 and has generator matrix[
Ik0 A
0 D

]
, (3)

where A ∈Mk0×(n−k0)(Fq), and D ∈Mk1×(n−k0)(Fq) is of full row rank.

Lemma 2. If C is a code of length n over R1 with res(C) = C1 and tor(C) = C2, then
there exists a matrix N ∈Mk0×(n−k0)(Fq) such that the matrix[

Ik0 A+ uN
0 uD

]
(4)

is a generator matrix of C. Such matrix N is unique if C is a free code.

Proof. Since the residue and torsion codes of C are C1 and C2, respectively, then for
some M1 ∈Mk0(Fq) and M2 ∈Mk0×(n−k0)(Fq),

Rk0+k11

[
Ik0 + uM1 A+ uM2

0 uD

]
⊆ C.

By an elementary row operation,



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 876

C ⊇ Rk0+k11

[
Ik0 − uM1 0

0 Ik1

] [
Ik0 + uM1 A+ uM2

0 uD

]
= Rk0+k11

[
Ik0 A+ u(M2 −M1A)
0 uD

]
.

Taking N = M2 −M1A, we have

|C| ≥
∣∣∣∣Rk0+k11

[
Ik0 A+ uN
0 uD

]∣∣∣∣ = q2k0+k1 = |C1| |C2| = |C| .

Thus, C has a generator matrix (4).
Suppose C is a free code and there exist N1, N2 ∈Mk0×(n−k0)(Fq) such that

Rk01 [I A+ uN1] = Rk01 [I A+ uN2] .

Then A+ uN1 ≡ A+ uN2 (u2), which implies that N1 ≡ N2 (u). �

For the remainder of this section, assume that C1 ⊆ C2 ⊆ C⊥1 . Then

Ik0 +AAT ≡ 0 (u), (5)

DAT ≡ 0 (u). (6)

It follows from (5) that A is of full row rank.
Denote by Symk0(Fq) the set of k0×k0 symmetric matrices, Altk0(Fq) the set of k0×k0

alternating matrices, and Skewk0(Fq) the set of k0 × k0 skew-symmetric matrices over Fq.

Lemma 3. Let A ∈Mm×n(Fq) where rank A = m. We define the mappings

ΨA : Mm×n(Fq) −→ Mm(Fq)
N 7−→ ANT +NAT ,

and

ΦA : Mm×n(Fq) −→ Mm(Fq)
N 7−→ ANT −NAT .

Then

ΨA (Mm×n(Fq)) =

{
Symm(Fq), if q is odd

Altm(Fq), if q is even,

and the image of the map ΦA is Skewm(Fq).



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 877

Proof. The image of ΨA was shown in [2]. Since rank A = m, A(Mn×m(Fq)) = Mm(Fq).
Indeed,

ΦA(Mk0×(n−k0)(Fq)) =
{
ANT −NAT | N ∈Mk0×(n−k0)(Fq)

}
=

{
S − ST | S ∈Mk0(Fq)

}
= Skewk0(Fq).

�

Lemma 4. The number of free Euclidean self-orthogonal codes over R1 with residue code
C1 is

qk0(2n−3k0+ε)/2,

where ε = −1 if q is odd and ε = 1 if q is even.
The number of free Hermitian self-orthogonal codes over R1 with residue code C1 is

qk0(2n−3k0+1)/2.

Proof. If C is a free code with residue code C1, then by Lemma 2, C has generator
matrix

[
Ik0 A+ uN

]
, for some unique N ∈Mk0×(n−k0)(Fq). Observe that C is Euclidean

self-orthogonal if and only if

Ik0 +AAT + u(ANT +NAT ) ≡ 0 (u2).

Hence, the number of free Euclidean self-orthogonal codes C with residue code C1 is∣∣{N ∈Mk0×(n−k0)|Ik0 +AAT + u(ANT +NAT ) ≡ 0 (u2)}
∣∣ . (7)

By (5), we have ANT +NAT ≡ 0 (u).Therefore, (7) becomes

∣∣{N ∈Mk0×(n−k0)|AN
T +NAT ≡ 0 (u)

}∣∣ = |ker ΨA| =
∣∣Mk0×(n−k0)

∣∣
|Im ΨA|

.

Thus, we have

|ker ΨA| =

{
q

k0(2n−3k0−1)
2 , if q is odd

q
k0(2n−3k0+1)

2 , if q is even,

by Lemma 3.
Similarly, C is Hermitian self-orthogonal if and only if

Ik0 +AAT + u(ANT −NAT ) ≡ 0 (u2).

Hence, by (5) and Lemma 3, the number of free Hermitian self-orthogonal codes C with
residue code C1 is∣∣{N ∈Mk0×(n−k0)|AN

T −NAT ≡ 0 (u)
}∣∣ = |ker ΦA| = q

k0(2n−3k0+1)
2 .



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 878

�

Define the sets

X =
{
C | C ⊆ Rn1 , type {k0, 0}, C ⊆ C⊥, res(C) = C1

}
and

X ′ =
{
C ′ | C ′ ⊆ Rn1 , C ′ ⊆ C ′⊥, res(C ′) = C1, tor(C ′) = C2

}
,

where self-orthogonality is either in the Euclidean or Hermitian sense.

Lemma 5. If C ′ ∈ X ′, then |{C ∈ X|C ⊆ C ′}| = qk0k1.

Proof. By Lemma 2, C ′ has a generator matrix (4). Consider the map

ψ : Mk0×k1(Fq) −→
{
C ∈ X | C ⊆ C ′

}
M 7−→ Rk01 [I A+ u(N +MD)] .

Clearly, ψ is well-defined. We will show that ψ is bijective. If M1,M2 ∈Mk0×k1(Fq) such
that ψ(M1) = ψ(M2), then

Rk01 [Ik0 A+ u(N +M1D)] = Rk01 [Ik0 A+ u(N +M2D)]

which means A + u(N + M1D) ≡ A + u(N + M2D) (u2). Therefore N + M1D ≡ N +
M2D (u). Since D is of full row rank, we have M1 ≡M2 (u). Hence, ψ is injective.

Suppose C ∈ X and C ⊆ C ′. By Lemma 2, C = Rk01 [Ik0 A+ uF ] , for some matrix
F . The inclusion C ⊆ C ′ implies that

A+ uF ≡ A+ u(N +MD) (u2)

for some matrix M . So F ≡ N + MD (u), which shows that ψ is surjective, and hence,
bijective. Therefore, ∣∣{C ∈ X|C ⊆ C ′}∣∣ = |Mk0×k1(Fq)| = qk0k1 .

�

Lemma 6. If C ∈ X, then there exists a unique code C ′ ∈ X ′ such that C ⊆ C ′.

Proof. Since C ∈ X, C has a generator matrix [I A+ uN ] for some unique matrix N ,
by Lemma 2. Let C ′0 be a code with generator matrix[

Ik0 A+ uN
0 uD

]
.

The code C ′0 satisfies res(C ′0) = C1 and tor(C ′0) = C2.
Clearly, C ⊆ C ′0. Since C ∈ X, (6) implies C ′0 is self-orthogonal and hence, C ′0 ∈ X ′.

Suppose C ⊆ C ′ for some C ′ ∈ X ′. Because C ′ has torsion code C2, by Lemma 2,
Rk11 [0 uD] ⊆ C ′ and so C ′0 ⊆ C ′. Note that |C ′0| = |C1| |C2| = q2k0+k1 = |C ′|. Hence,
C ′0 = C ′. �

Next, we count self-orthogonal codes C with given residue code and torsion code.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 879

Theorem 1. Let C1 and C2 be codes of length n over Fq where C1 ⊆ C2 ⊆ C⊥1 . If
dim C1 = k0 and dim C2 = k0 + k1, then

(i) the number of Euclidean self-orthogonal codes C of length n over Fq + uFq with
res(C) = C1 and tor(C) = C2 is

qk0(2n−3k0−2k1+ε)/2,

where ε = −1 if q is odd and ε = 1 if q is even, and

(ii) the number of Hermitian self-orthogonal codes C of length n over Fq + uFq with
res(C) = C1 and tor(C) = C2 is

qk0(2n−3k0−2k1+1)/2.

Proof. We may assume without loss of generality that C1 and C2 are codes with
generator matrices (2) and (3), respectively. Then we have to compute |X ′|. By Lemma 5
and Lemma 6, we have

qk0k1
∣∣X ′∣∣ =

∑
C′∈X′

∣∣{C ∈ X|C ⊆ C ′}∣∣
=

∑
C∈X

∣∣{C ′ ∈ X ′|C ⊆ C ′}∣∣
=

∑
C∈X

1

= |X| .

The results follow from Lemma 4. �

4. Mass formula for self-orthogonal codes over Fq + uFq

Let σq(n, k0) denote the number of distinct self-orthogonal codes over Fq of length n

and dimension k0 (see [9, 10]). We define the Gaussian coefficient

[
n
k

]
q

for k ≤ n as

[
n
k

]
q

=
(qn − 1)(qn − q) · · · (qn − qk−1)
(qk − 1)(qk − q) · · · (qk − qk−1)

,

which gives the number of subspaces of dimension k contained in an n-dimensional vector
space over Fq.

We now have the following mass formula for self-orthogonal codes over R1.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 880

Theorem 2. Let Mq(n, k0, k1)E and Mq(n, k0, k1)H denote the number of distinct Eu-
clidean and Hermitian self-orthogonal codes of length n over Fq + uFq of type {k0, k1},
respectively. We have

Mq(n, k0, k1)E = σq(n, k0)

[
n− 2k0
k1

]
q

qk0(2n−3k0−2k1+ε)/2

where ε = −1 if q is odd and ε = 1 if q is even, and

Mq(n, k0, k1)H = σq(n, k0)

[
n− 2k0
k1

]
q

qk0(2n−3k0−2k1+1)/2.

Proof. If C is a self-orthogonal code of length n over Fq +uFq of type {k0, k1}, then by
setting C1 = res(C) and C2 = tor(C), we see that C1 and C2 satisfies Lemma 1. There are

σq(n, k0) self-orthogonal codes C1 of length n over Fq. Given C1, there are

[
n− 2k0
k1

]
q

codes C2 such that C1 ⊆ C2 ⊆ C⊥1 . Then the result follows from Theorem 1. �

We have the following mass formula for self-dual codes over R1 as a direct consequence
of the previous theorem.

Corollary 1. The number of distinct Euclidean self-dual codes of length n over Fq + uFq
is given by ∑

0≤k0≤bn2 c

σq(n, k0)q
k0(k0+ε)/2, (8)

where ε = −1 if q is odd and ε = 1 if q is even, and the number of distinct Hermitian
self-dual codes of length n over Fq + uFq is given by∑

0≤k0≤bn2 c

σq(n, k0)q
k0(k0+1)/2. (9)

Proof. Note that the number of distinct Euclidean self-dual codes and the number of
distinct Hermitian self-dual codes of length n over Fq + uFq are given by∑

0≤k0≤bn2 c

Mq(n, k0, n− 2k0)E , and
∑

0≤k0≤bn2 c

Mq(n, k0, n− 2k0)H ,

respectively. �

In [6, Theorem 3], Gaborit establishes the mass formula for Hermitian self-dual codes
over Fq + uFq, but gives the formula for Euclidean self-dual codes instead. The formula
(9) corrects this.

Next, we establish another formula for the number of distinct Euclidean self-orthogonal
codes when the given torsion is self-orthogonal.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 881

Corollary 2. Suppose q is odd. The number of distinct Euclidean self-orthogonal codes
of length n over Fq + uFq of type {k0, k1} with self-orthogonal torsion is

M̃q(n, k0, k1)E =

[
k0 + k1
k0

]
q

σq(n, k0 + k1)q
k0(2n−3k0−2k1−1)/2.

Proof. Let C1 and C2 be self-orthogonal codes where dim C1 = k0, dim C2 = k0 + k1
and C1 ⊆ C2 . By Theorem 2, we have

M̃q(n, k0, k1)E q
−k0(2n−3k0−2k1−1)/2 =

∑
C2⊆C⊥2

|{C1 | C1 ⊆ C2}|

=

[
k0 + k1
k0

]
q

∣∣∣{C2 | C2 ⊆ C⊥2
}∣∣∣

=

[
k0 + k1
k0

]
q

σq(n, k0 + k1).

�

This corollary will be useful in our mass formula computations on later chapters.

5. Classification of self-orthogonal codes over Fq + uFq

Using Theorem 2, we classify Euclidean and Hermitian self-orthogonal codes over F2 +
uF2 and F3 + uF3, of given type for small lengths. Note that two codes over F2 + uF2

are equivalent if one can be obtained from the other by permuting the coordinates and (if
necessary) multiplying certain coordinates by 1 + u. On the other hand, two Euclidean
self-orthogonal codes over F3+uF3 are equivalent if one can be obtained from the other by
permuting the coordinates and (if necessary) multiplying certain coordinates by 2, and two
Hermitian self-orthogonal codes over F3 + uF3 are equivalent if one can be obtained from
the other by permuting the coordinates and (if necessary) multiplying certain coordinates
by r, where r ∈ {2, 1 + u, 1 + 2u, 2 + u, 2 + 2u}.

To illustrate, we classify Euclidean self-orthogonal codes over F3 +uF3 of length 4 and
type {2, 0}. Let C1 and C2 be inequivalent Euclidean self-orthogonal codes over F3 + uF3

of length 4 and type {2, 0} with generator matrices[
1 0 2 2
0 1 2 1

]
and

[
1 0 2 + 2u 2 + u
0 1 2 + u 1 + u

]
,

respectively. The order of their automorphism groups are 48 and 24, respectively. Hence,

2∑
j=1

|E4|
|Aut(Cj)|

=
24 · 4!

48
+

24 · 4!

24
= 8 + 16 = 24.

From Theorem 2,

M3(4, 2, 0)E = σ3(4, 2)

[
4− 2 · 2

0

]
3

32(8−6−0−1)/2 = 8 · 1 · 3 = 24.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 882

Therefore, there are two Euclidean self-orthogonal codes of length 4 and type {2, 0} over
F3 + uF3, up to equivalence.

We note that Euclidean self-orthogonal and Hermitian self-orthogonal codes coincide
over F2 + uF2, as well as in codes over F3 + uF3 of type {0, k1}.

Table 1 gives the number of inequivalent Euclidean self-orthogonal codes over F2 +
uF2 of lengths 2 up to 7, while Table 2 gives the number of inequivalent Euclidean and
Hermitian self-orthogonal codes over F3+uF3 of lengths 2 up to 6, for each type. Note that
the code of length 1 with generator matrix [u] is a Euclidean self-orthogonal and Hermitian
self-orthogonal code over Fq+uFq. Therefore, there is a self-orthogonal code for any length
n, since one can just form a direct sum of this length 1 code. Our classification of self-
orthogonal codes over F2+uF2 agrees with the enumeration in [5] for self-dual codes (codes
of type {k0, n− 2k0}) up to length n = 7. Generators and the order of the automorphism
group of each code in Table 1 and Table 2 may be requested by the interested reader
from the authors. All computer calculations in this paper were done with the help of
Magma[4].

Table 1: The number of inequivalent self-orthogonal codes of lengths 2 ≤ n ≤ 7 over F2 + uF2

{n, k0, k1} Number of {n, k0, k1} Number of {n, k0, k1} Number of
Codes Codes Codes

{2, 1, 0} 1 {5, 2, 1} 2 {6, 0, 6} 1
{2, 0, 1} 2 {5, 0, 1} 5 {7, 1, 0} 12
{2, 0, 2} 1 {5, 0, 2} 10 {7, 1, 1} 54
{3, 1, 0} 2 {5, 0, 3} 10 {7, 1, 2} 100
{3, 1, 1} 1 {5, 0, 4} 5 {7, 1, 3} 73
{3, 0, 1} 3 {5, 0, 5} 1 {7, 1, 4} 24
{3, 0, 2} 3 {6, 1, 0} 9 {7, 1, 5} 3
{3, 0, 3} 1 {6, 1, 1} 29 {7, 2, 0} 43
{4, 1, 0} 4 {6, 1, 2} 36 {7, 2, 1} 74
{4, 1, 1} 5 {6, 1, 3} 16 {7, 2, 2} 40
{4, 1, 2} 2 {6, 1, 4} 3 {7, 2, 3} 5
{4, 2, 0} 2 {6, 2, 0} 19 {7, 3, 0} 22
{4, 0, 1} 4 {6, 2, 1} 18 {7, 3, 1} 5
{4, 0, 2} 6 {6, 2, 2} 5 {7, 0, 1} 7
{4, 0, 3} 4 {6, 3, 0} 4 {7, 0, 2} 23
{4, 0, 4} 1 {6, 0, 1} 6 {7, 0, 3} 43
{5, 1, 0} 6 {6, 0, 2} 16 {7, 0, 4} 43
{5, 1, 1} 13 {6, 0, 3} 22 {7, 0, 5} 23
{5, 1, 2} 10 {6, 0, 4} 16 {7, 0, 6} 7
{5, 1, 3} 2 {6, 0, 5} 6 {7, 0, 7} 1
{5, 2, 0} 6



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 883

Table 2: The number of inequivalent Euclidean and Hermitian self-orthogonal codes of lengths 2 ≤ n ≤ 6 over
F3 + uF3

{n, k0, k1} Number of Codes {n, k0, k1} Number of Codes
Euclidean Hermitian Euclidean Hermitian

{2, 1, 0} 0 0 {5, 2, 1} 2 1
{2, 0, 1} 2 2 {5, 0, 1} 5 5
{2, 0, 2} 1 1 {5, 0, 2} 12 12
{3, 1, 0} 2 1 {5, 0, 3} 12 12
{3, 1, 1} 1 1 {5, 0, 4} 5 5
{3, 0, 1} 3 3 {5, 0, 5} 1 1
{3, 0, 2} 3 3 {6, 1, 0} 12 5
{3, 0, 3} 1 1 {6, 1, 1} 57 27
{4, 1, 0} 4 2 {6, 1, 2} 64 34
{4, 1, 1} 6 4 {6, 1, 3} 20 13
{4, 1, 2} 1 1 {6, 1, 4} 2 2
{4, 2, 0} 2 1 {6, 2, 0} 22 8
{4, 0, 1} 4 4 {6, 2, 1} 18 9
{4, 0, 2} 7 7 {6, 2, 2} 4 3
{4, 0, 3} 4 4 {6, 3, 0} 0 0
{4, 0, 4} 1 1 {6, 0, 1} 6 6
{5, 1, 0} 6 3 {6, 0, 2} 20 20
{5, 1, 1} 19 11 {6, 0, 3} 31 31
{5, 1, 2} 10 7 {6, 0, 4} 20 20
{5, 1, 3} 1 1 {6, 0, 5} 6 6
{5, 2, 0} 4 2 {6, 0, 6} 1 1

6. Codes over Fq + uFq + u2Fq, where q is odd

For the rest of this paper, let R2 be the commutative ring Fq[u]/(u3) = Fq+uFq+u2Fq,
where u3 = 0 and q is odd. We will only consider Euclidean inner product.

A code C of length n over R2 is permutation-equivalent to a code with generator
matrix  Ik0 A0 B0 + uB1 + u2B2

0 uIk1 uD1 + u2D2

0 0 u2F2

 (10)

where F2 ∈Mk2×(n−k0−k1)(Fq) and A0, B0, B1, B2, D1, D2 are matrices of appropriate sizes
over Fq. We define the torsion codes of C as follows:

tor0(C) = {v ∈ Fnq | ∃w, z ∈ Fnq , v + uw + u2z ∈ C} and

tori(C) = {v ∈ Fnq | uiv ∈ C}, for i = 1, 2.

The code tor0(C) is also called the residue code of C. Observe that tor0(C) ⊆ tor1(C) ⊆
tor2(C). If C has generator matrix (10), then the residue code tor0(C) has dimension k0
and generator matrix [

Ik0 A0 B0

]
, (11)



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 884

tor1(C) has dimension k0 + k1 and generator matrix[
Ik0 A0 B0

0 Ik1 D1

]
(12)

and tor2(C) has dimension k0 + k1 + k2 and generator matrix Ik0 A0 B0

0 Ik1 D1

0 0 F2

 (13)

where F2 is of full row rank. The code C is of type {k0, k1, k2} and

|C| = |tor0(C)| |tor1(C)| |tor2(C)| = q3k0+2k1+k2 .

Suppose C is self-orthogonal. Then

Ik0 +A0A
T
0 +B0B

T
0 + u(B0B

T
1 +B1B

T
0 )

+u2(B0B
T
2 +B1B

T
1 +B2B

T
0 ) ≡ 0 (u3)

u(A0 +B0D
T
1 ) + u2(B1D

T
1 +B0D

T
2 ) ≡ 0 (u3)

u2(B0F
T
2 ) ≡ 0 (u3)

u2(Ik1 +D1D
T
1 ) ≡ 0 (u3)

which give the following:

Ik0 +A0A
T
0 +B0B

T
0 ≡ 0 (u) (14)

B0B
T
1 +B1B

T
0 ≡ 0 (u) (15)

B0B
T
2 +B1B

T
1 +B2B

T
0 ≡ 0 (u) (16)

A0 +B0D
T
1 ≡ 0 (u) (17)

B1D
T
1 +B0D

T
2 ≡ 0 (u) (18)

F2B
T
0 ≡ 0 (u) (19)

Ik1 +D1D
T
1 ≡ 0 (u). (20)

From (14), tor0(C) is self-orthogonal and by (14), (17) and (20), we have

tor1(C) ⊆ tor1(C)⊥,

that is, tor1(C) is self-orthogonal. Moreover, by (14), (17) and (19) we have

tor0(C) ⊆ tor2(C)⊥.

We will introduce another type of residue for a code over R2.

Definition 1. Let C be a code over R2. The code over Fq + uFq obtained from C by
reduction modulo u2 is called the u2-Residue of C and will be denoted by Res(C).



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 885

It is easy to see that a generator matrix for Res(C) is[
Ik0 A0 B0 + uB1

0 uIk1 uD1

]
.

It is also clear that res(Res(C)) = tor0(C), tor(Res(C)) = tor1(C), and Res(C) is of type
{k0, k1}.

If C is self-orthogonal, by (14), (15) and (17) we have

Res(C) ⊆ Res(C)⊥,

that is, Res(C) is self-orthogonal of type {k0, k1}. Also, since

tor(Res(C)) = tor1(C) ⊆ tor2(C)

and
tor2(C) ⊆ tor0(C)⊥ = res(Res(C))⊥,

we have
tor(Res(C)) ⊆ tor2(C) ⊆ res(Res(C))⊥

which gives the following lemma.

Lemma 7. Let C be a self-orthogonal code over R2 of type {k0, k1, k2} and let C1 =Res(C)
and C2 =tor2(C). Then

(i) C1 ⊆ C⊥1 ,

(ii) tor(C1) ⊆ tor(C1)
⊥, and

(iii) tor(C1) ⊆ C2 ⊆ res(C1)
⊥, dim C2 = k0 + k1 + k2.

7. Codes over Fq + uFq + u2Fq with prescribed u2-Residue and torsion

For the rest of this chapter, we let C1 be a self-orthogonal code over R1 of type {k0, k1}
such that tor(C1) is self-orthogonal. We assume without loss of generality that C1 has
generator matrix

G1 =

[
Ik0 A0 B0 + uB1

0 uIk1 uD1

]
.

Since C1 is self-orthogonal, we have

Ik0 +A0A
T
0 +B0B

T
0 + u(A0A

T
1 +A1A

T
0 +B0B

T
1 +B1B

T
0 ) ≡ 0 (u2)

u(A0 +B0D
T
1 ) ≡ 0 (u2)

which are equivalent to (14), (15) and (17). Moreover, since tor(C1) is self-orthogonal, we
have

Ik0 +A0A
T
0 +B0B

T
0 ≡ 0 (u)



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 886

A0 +B0D
T
1 ≡ 0 (u)

Ik1 +D1D
T
1 ≡ 0 (u)

which are equivalent to (14), (17) and (19).
Now, notice that from (17), we have

A0 ≡ −B0D
T
1 (u).

By (14), we have

Ik0 +B0D
T
1D1B

T
0 +B0B

T
0 ≡ 0 (u)

Ik0 +B0

(
DT

1D1B + Ik0
)
BT

0 ≡ 0 (u)

which implies B0 is of full row rank.
We start by counting the number of self-orthogonal codes C of type {k0, k1, 0} such

that Res(C) = C1. Similar to what we did in the previous chapter, we first exhibit the
generator matrix of such code C.

Lemma 8. If C is a code over R2 of type {k0, k1, 0} and Res(C) = C1, then there exist
matrices N0 ∈Mk0×(n−k0−k1)(Fq) and N1 ∈Mk1×(n−k0−k1)(Fq) such that[

Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

]
(21)

is a generator matrix for C. The matrices N0 and N1 are unique.

Proof. If C is a code over R2 of type {k0, k1, 0} such that Res(C) = C1, then for some
matrices M1,M2,M3,M4 and M5 over Fq of appropriate sizes,

Rk0+k12

[
Ik0 + u2M1 A0 + u2M2 B0 + uB1 + u2M3

0 Ik1 + u2M4 D1 + u2M5

]
⊆ C.

Applying elementary row operations,[
Ik0 − u2M1 0

0 Ik1 − u2M4

] [
Ik0 + u2M1 A0 + u2M2 B0 + uB1 + u2M3

0 Ik1 + u2M4 D1 + u2M5

]
=

[
Ik0 A0 + u2(M2 −M1A0) B0 + uB1 + u2(M3 −M1B0)
0 Ik1 D1 + u2(M5 −M4D1)

]
and[

Ik0
−u2(M2 −M1A0)

0 Ik1

] [
Ik0

A0 + u2(M2 −M1A0) B0 + uB1 + u2(M3 −M1B0)
0 Ik1

D1 + u2(M5 −M4D1)

]

=

[
Ik0 A0 B0 + uB1 + u2(M3 −M1B0 −M2D1 +M1A0D1)
0 Ik1 D1 + u2(M5 −M4D1)

]
.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 887

Letting

N0 = M3 −M1B0 −M2D1 +M1A0D1

N1 = M5 −M4D1,

we have

Rk0+k12

[
Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

]
⊆ C.

Therefore,

|C| ≥
∣∣∣∣Rk0+k12

[
Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

]∣∣∣∣
= qk0+k1q2k0+k1

= q3k0+2k1

= |C|

and hence, (21) is a generator matrix for C.
Next, we show uniqueness of the matrices N0 and N1 over Fq. Suppose there exist

matrices N ′0 ∈Mk0×(n−k0−k1)(Fq) and N ′1 ∈Mk1×(n−k0−k1)(Fq) such that

Rk0+k12

[
Ik0 A0 B0 + uB1 + u2N ′0
0 uIk1 uD1 + u2N ′1

]
=

Rk0+k12

[
Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1.

]
.

This means that

B0 + uB1 + u2N ′0 ≡ B0 + uB1 + u2N0 (u3)

uD1 + u2N ′1 ≡ uD1 + u2N1 (u3)

which imply that

N ′0 ≡ N0 (u)

N ′1 ≡ N1 (u)

and hence, N0 and N1 are unique. �

This shows that the number of self-orthogonal codes C over R2 of type {k0, k1, 0} with
Res(C) = C1 is determined by the number of such matrices N0 and N1, which will be
given in the next lemma.

Lemma 9. The number of self-orthogonal codes C of type {k0, k1, 0} over R2 such that
Res(C) = C1 is

q(k0+k1)(n−k0−k1)−k0(k0+1)/2−k0k1 .



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 888

Proof. By Lemma 8, C has generator matrix[
Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

]
for some matrices N0, N1 over Fq. From this, C is self-orthogonal if and only if

Ik0 +AAT0 +BBT
0 + u(B0B

T
1 +B1B

T
0 )

+u2(B1B
T
1 +B0N

T
0 +N0B

T
0 ) ≡ 0 (u3) (22)

u(A0 +B0D
T
1 ) + u2(B1D

T
1 +B0N

T
1 ) ≡ 0 (u3). (23)

We want to count the number of such matricesN0 andN1 satisfying the above equivalences.
First, consider the map

ΦB0 : Mk0×(n−k0−k1)(Fq) −→ Mk0(Fq)
N0 7−→ B0N

T
0 +N0B

T
0

as defined in the previous chapter. By (14) and (15), (22) becomes

B1B
T
1 +B0N

T
0 +N0B

T
0 ≡ 0 (u).

Hence,

|{N0 ∈Mk0×(n−k0−k1)|N0 satisfies (22)}| = |{Φ−1B0
(−B1B

T
1 )}|

= |ker ΦB0 |

=

∣∣Mk0×(n−k0−k1)
∣∣∣∣Symk0(Fq)

∣∣
= qk0(n−k0−k1)−k0(k0+1)/2.

Define another map

β : Mk1×(n−k0−k1)(Fq) −→ Mk0×k1(Fq)
N1 7−→ B0N

T
1 .

This map is surjective because B0 is of full row rank. Therefore by (17),

|{N1 ∈Mk1×(n−k0−k1)|N1 satisfies (23)}| = |{β−1(−B1D
T
1 )}|

= |ker β|

=

∣∣Mk1×(n−k0−k1)(Fq)
∣∣

|Mk0×k1(Fq)|
= qk1(n−k0−k1)−(k0k1).

Finally, the number of self-orthogonal codes C of type {k0, k1, 0} over R2 such that
Res(C) = C1 is the number of such matrices N0 satisfying (22) multiplied to the number
of such matrices N1 satisfying (23) which is

qk0(n−k0−k1)−k0(k0+1)/2qk1(n−k0−k1)−k0k1 .



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 889

The result follows by simplifying the above expression. �

For the rest of this chapter, let C2 be a code over Fq with dimension k0 + k1 + k2 and
has a generator matrix

G2 =

 Ik0 A0 B0

0 Ik1 D1

0 0 F2


where F2 is of full row rank. We assume that tor(C1) ⊆ C2 ⊆ res(C1)

⊥. Hence,

Ik0 +A0A
T
0 +B0B

T
0 ≡ 0 (u)

A0 +B0D
T
1 ≡ 0 (u)

F2B
T
0 ≡ 0 (u)

which are equivalent to (14), (17) and (19), respectively.
Consider the following sets of codes over R2:

Y = {C | C is self-orthogonal of type {k0, k1, 0},Res(C) = C1} ;

Y ′ =
{
C ′ | C ′ is self-orthogonal,Res(C ′) = C1, tor2(C

′) = C2

}
.

Note that |Y | is already given in Lemma 9. Our next goal is to compute for |Y ′|. This
will be done in the same way as in the previous chapter.

Lemma 10. If C ∈ Y , then there exists a unique C ′ ∈ Y ′ such that C ⊆ C ′.

Proof. Since C ∈ Y , C has generator matrix (21) for some matrices N0 and N1.
Suppose C ⊆ C ′ for some C ′ ∈ Y ′ and there exists a code C ′′ with generator matrix Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

0 0 u2F2

 .
Clearly, C ⊆ C ′′. Note that C ′′ satisfies Res(C ′′) = C1 and tor2(C

′′) = C2. Using (19), we
conclude that C ′′ is self-orthogonal. Hence, C ′′ ∈ Y ′.

Next, notice that Rk22 [0 0 u2F2] ⊆ C ′. This, together with the fact that C ⊆ C ′, forces
C ′′ ⊆ C ′. But

|C ′′| = |C1||C2|
= q2k0+k1qk0+k1+k2

= q3k0+2k1+k2

= |C ′|

and therefore, C ′ = C ′′. �

Lemma 11. Let C ′ ∈ Y ′. Then | {C ∈ Y | C ⊆ C ′} | = q(k0+k1)k2.



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 890

Proof. Let C ′ ∈ Y ′ whose generator matrix is Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

0 0 u2F2

 .
Define the map Ψ : Mk0×k2(Fq)×Mk1×k2(Fq) −→ {C ∈ Y | C ⊆ C ′} as

Ψ
(
M ′,M ′′

)
= Rk0+k12

[
Ik0 A0 B0 + uB1 + u2(N0 +M ′F2)
0 uIk1 uD1 + u2(N1 +M ′′F2)

]
and claim that this map is bijective.

Indeed, Ψ is injective because F2 is of full row rank. Now, suppose C ∈ Y such that
C ⊆ C ′. Then by Lemma 8, C has generator matrix[

Ik0 A0 B0 + uB1 + u2F ′

0 uIk1 uD1 + u2F ′′

]
for some matrices F ′ and F ′′. Since C ⊆ C ′, there exist matrices M ′ and M ′′ such that[

Ik0 A0 B0 + uB1 + u2F ′

0 uIk1 uD1 + u2F ′′

]
≡

[
Ik0 0 M ′

0 Ik1 M ′′

] Ik0 A0 B0 + uB1 + u2N0

0 uIk1 uD1 + u2N1

0 0 u2F2

 (u3).

Then we have F ′ = N0 + M ′F2 and F2 = N1 + M ′′F2, so Ψ is surjective and hence,
bijective.

Therefore,

|
{
C ∈ Y | C ⊆ C ′

}
| = |Mk0×k2(Fq)×Mk1×k2(Fq)|

= qk0k2qk1k2

= q(k0+k1)k2 .

�

Given C1 and C2, we can now count the number of self-orthogonal codes over R2 having
u2-Residue C1 and torsion C2.

Theorem 3. Suppose C1 is a self-orthogonal code over Fq + uFq, where q is odd, of
type {k0, k1} such that tor(C1) is self-orthogonal and C2 is a code over Fq of dimension
k0 +k1 +k2 such that tor(C1) ⊆ C2 ⊆ res(C1)

⊥. Then the number of self-orthogonal codes
C ′ of length n over Fq + uFq + u2Fq such that Res(C ′) = C1 and tor2(C

′) = C2 is

qk0(2n−3k0−6k1−2k2−1)/2+k1(n−k1−k2).



L.E. Galvez, R.A. Betty, F. Nemenzo / Eur. J. Pure Appl. Math, 13 (4) (2020), 873-892 891

Proof. Without loss of generality, we assume that C1 has generator matrix G1 and C2

has generator matrix G2. Then we compute for |Y ′|. By Lemma 10 and Lemma 11, we
have

q(k0+k1)k2
∣∣Y ′∣∣ =

∑
C′∈Y ′

∣∣{C ∈ X|C ⊆ C ′}∣∣
=

∑
C∈Y

∣∣{C ′ ∈ X ′|C ⊆ C ′}∣∣
=

∑
C∈Y

1

= |Y | .

The results follow from Lemma 9. �

8. Mass formula for self-orthogonal codes over Fq + uFq + u2Fq, where q is
odd

We now have the following theorem.

Theorem 4. Suppose q is odd. The number of distinct self-orthogonal codes over Fq +
uFq + u2Fq of length n and type {k0, k1, k2}, denoted by MR2(n, k0, k1, k2) is[

n− 2k0 − k1
k2

]
q

[
k0 + k1
k0

]
q

σq(n, k0 + k1)q
k0(2n−3k0−4k1−k2−1)+k1(n−k1−k2).

Proof. If C is a self-orthogonal code of length n over Fq+uFq+u2Fq of type {k0, k1, k2},
then by setting C1 = Res(C) and C2 = tor2(C), we see that C1 and C2 satisfies (i)–(iii) of
Lemma 7. The number of self-orthogonal codes with given u2-Residue C1 and torsion C2 is
given in Theorem 3. The number of self-orthogonal codes C1 over Fq+uFq satisfying (i) and

(ii) is given in Corollary 2. The number of codes C2 satisfying (iii) is

[
n− 2k0 − k1

k2

]
q

.

The value of MR2(n, k0, k1, k2) is obtained by the product of these. �

We now have the following mass formula for self-dual codes over R2 as a direct conse-
quence of Theorem 4.

Corollary 3. Suppose q is odd. The number of distinct self-dual codes of even length n
over Fq + uFq + u2Fq is given by

n
2∑

k0=0

MR2(n, k0,
n

2
− k0,

n

2
− k0). (24)

Proof. By [3], we have k1 = k2 and n = 2(k0 + k1). The result follows from Theorem
4. �

The formula (24) agrees with [3, Theorem 1].



REFERENCES 892

Acknowledgements

The authors gratefully acknowledge the support of the University of the Philippines
Office of the Vice President for Academic Affairs for this project. The authors also thank
Dr. Markus Grassl for useful suggestions and the reviewers for their valuable comments.

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