Free sets and free subsemimodules in a semimodule

Linear Algebra and its Applications 496 (2016) 527–548

Contents lists available at ScienceDirect

Linear Algebra and its Applications www.elsevier.com/locate/laa

Free sets and free subsemimodules in a semimodule ✩ Yi-Jia Tan 1 Department of mathematics, Fuzhou University, Fuzhou, 350116, China

a r t i c l e

i n f o

Article history: Received 5 December 2014 Accepted 3 February 2016 Available online 17 February 2016 Submitted by R. Brualdi MSC: 15A03 16Y60 Keywords: Semiring Semimodule Free set Free subsemimodule Rank of semimodule

a b s t r a c t In this paper, we investigate the free sets and the free subsemimodules in a semimodule over a commutative semiring S. First, we discuss some properties of the free sets and give a sufficient condition for a nonempty finite set to be free in a finitely generated free S-semimodule and obtain a relation between free set and linear independent set in an S-semimodule. Then we consider the free subsemimodules and prove that the rank of any free subsemimodule of a finitely generated S-semimodule M does not exceed that of M. Also, we give some equivalent descriptions for a commutative semiring S to have the property that all nonzero subsemimodules of any finitely generated free S-semimodule are free. Partial results obtained in the paper develop and generalize the corresponding results for modules over rings and linear spaces over fields. © 2016 Elsevier Inc. All rights reserved.

1. Introduction The study of semimodules over semirings has a long history. In 1966, Yusuf [22] introduced the concept of inverse semimodule over a semiring and obtained some analogues ✩

1

Supported by the Natural Science Foundation of Fujian Province (2012J01008), China. E-mail address: [email protected]. Tel.: +86 591 22865166.

http://dx.doi.org/10.1016/j.laa.2016.02.006 0024-3795/© 2016 Elsevier Inc. All rights reserved.

528

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

to theorems in module theory for inverse semimodules (note that an inverse semimodule M is a semimodule in which the monoid (M, +) is an inverse semigroup). Since then, a number of works on semimodule theory were published (see e.g. [1–3,5,6,15,20,21,23]). In 1999, Golan described semirings and semimodules over semirings in his work [7] comprehensively. Semimodules over semirings appear in many areas of mathematics and are useful in the area of theoretical computer science as well as in the cryptography (see e.g. [10]). Recently, Shu and Wang [13,14] discussed the cardinality of bases and dimensional formulas of semimodules over commutative semirings, and Tan [17] gave some equivalent conditions for a basis to be a free basis in a finitely generated free S-semimodule. Also, Tan [18] introduced the inner products on semimodules and obtained an equivalent description for a semifield S satisfying the property that every standard orthogonal set in a finitely generated semimodule M over S can be extended to a standard orthogonal basis for M. In the present work, we will investigate the free sets and the free subsemimodules in a semimodule over a commutative semiring S. In Section 3, we discuss some properties of the free sets and give a sufficient condition for a nonempty finite set to be free in a finitely generated free S-semimodule (see Theorem 3.3) and obtain a relation between free set and linear independent set in an S-semimodule (see Theorem 3.4). In Section 4, we consider the free subsemimodules of an S-semimodule and prove that the rank of any free subsemimodule of a finitely generated S-semimodule M does not exceed that of M (see Theorem 4.2). Also, we give some equivalent descriptions for a commutative semiring S to have the property that all nonzero subsemimodules of any finitely generated free S-semimodule are free (see Theorem 4.3). Partial results obtained in the paper develop and generalize the corresponding results for modules over rings and linear spaces over fields. 2. Definitions and preliminaries In this section, we will give some definitions and preliminary lemmas. For convenience, we use N to denote the set of all positive integers and use n to denote the set {1, 2, . . . , n} for n in N. Also, we use |S| to denote the cardinality for any set S. Definition 2.1. (See [7].) A semiring is an algebraic system (S, +, ·) in which (S, +) is an Abelian monoid with identity element 0 and (S, ·) is another monoid with identity element 1, connected by ring-like distributivity. Also, 0a = a0 = 0 for all a in S and 0 = 1. The elements 0 and 1 are called the zero element and the identity element of S, respectively. A semiring S is called commutative if (S, ·, 1) is commutative; S is called a zerosumfree semiring (see [7]) or an antiring (see [16,19]) if a + b = 0 implies a = b = 0 for all a, b in S.

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

529

An element a in a semiring S is called a left (resp. right) zero divisor if ab = 0 (resp. ba = 0) for some nonzero element b in S. Clearly 0 is a left and a right zero divisor. The semiring S is called entire if it possesses no zero divisors = 0 (left or right). An entire ring is called a domain (see [8]). An element a in a semiring S is called cancellable if a + b = a + c implies b = c for all b, c in S. The semiring S is called cancellative if every element in S is cancellable. An element a in S is called left (resp. right) multiplicatively cancellable if ab = ac (resp. ba = ca) implies b = c for all b, c in S; a is called multiplicatively cancellable if it is both left and right multiplicatively cancellable (see [7]). Semirings are quite abundant, for example, every ring with identity is a semiring which is cancellative; every Boolean algebra, the fuzzy algebra F = ([0, 1], ∨, ∧), any bounded distributive lattice is a commutative semiring which is zerosumfree. Also, the positive cone Z0 (resp. Q0 , R0 ) of the ring Z (resp. Q, R) is a commutative semiring which is zerosumfree and cancellative. In addition, the max–plus algebra (R ∪ {−∞}, max, +) is a commutative semiring which is zerosumfree (see [4,24]). A nonempty subset I of a semiring S is called a left (resp. right) ideal of S if a + b, sa ∈ I (resp. a + b, as ∈ I) for any a, b in I and s in S; I is called an ideal of S if it is a left and right ideal of S (see [7]). For any element a in S, the set of all  finite sums of the form ri asi with ri , si ∈ S, denoted by (a), is the smallest ideal of S containing a which is called the principal ideal generated by a (see [7]). A ring S is called a principal ideal ring if each ideal of S is a principal ideal. An element a in a semiring S is called additively invertible if there exists b ∈ S such that a + b = 0. The element b is called an additive inverse of a in S. If a has an additive inverse then such an inverse is unique. The additive inverse of a in S is denoted by −a. We use V (S) to denote the set of all additive invertible elements in S (see [7]). It is clear that V (S) is an ideal of the semiring S. Definition 2.2. (See [7].) Let S be a semiring. A left S-semimodule is a commutative monoid (M, +) with additive identity θ for which we have a function S × M → M, denoted by (λ, α) → λα and called scalar multiplication, which satisfies the following conditions for all λ, μ in S and α, β in M: (1) (2) (3) (4) (5)

(λμ)α = λ(μα); λ(α + β) = λα + λβ; (λ + μ)α = λα + μα; 1α = α; λθ = θ = 0α.

Right S-semimodules are defined analogously. In this paper, S-semimodules will always mean left S-semimodules. S-semimodules were studied in [9,13,14,23] under the name S-semilinear spaces.

530

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

Example 2.1. Let S be a semiring and S n = {(a1 , a2 , . . . , an )T | ai ∈ S, i ∈ n}, where (a1 , a2 , . . . , an )T is the transpose of (a1 , a2 , . . . , an ) and n ≥ 1. Define x + y = (x1 + y1 , x2 + y2 , . . . , xn + yn )T and λx = (λx1 , λx2 , . . . , λxn )T for x = (x1 , x2 , . . . , xn )T , y = (y1 , y2 , . . . , yn )T in S n and λ in S. Then S n is an S-semimodule. In particular, S 1 = S is an S-semimodule. A nonempty subset N of an S-semimodule M is called a subsemimodule of M if N is closed under addition and scalar multiplication. It is clear that if {Ni | i ∈ Ω} is a family  of subsemimodules of M then i∈Ω Ni is a subsemimodule of M. Let A be a nonempty subset of an S-semimodule M . Then the intersection of all subsemimodules of M containing A is a subsemimodule of M, called the subsemimodule generated by A and denoted by Span(A). It is easy to verify that k  Span(A) = { λi αi | λi ∈ S, αi ∈ S, i ∈ k, k ∈ N}. i=1

The expression

k 

λi αi is called a linear combination of the elements α1 , α2 , . . . , αk .

i=1

If A = {α1 , α2 , . . . , αm }, then we denote Span(A) by Span(α1 , α2 , . . . , αm ). If Span(A) = M, then A is called a generating set for M. An S-semimodule having a finite generating set is called finitely generated. The rank of an S-semimodule M, denoted by r(M), is the smallest n for which there exists a generating set for M having cardinality n. It is clear that the rank r(M) exists for any finitely generated S-semimodule M. Definition 2.3. Let M be an S-semimodule. A nonempty subset A in M is called linearly independent if α ∈ / Span(A\{α}) for any α in A. If A is not linearly independent then it is called linearly dependent. The set A is called free if each element in M can be expressed as a linear combination of elements in A in at most one way. It is clear that any free set is linearly independent. Definition 2.4. Let M be an S-semimodule. A linearly independent generating set for M is called a basis for M and a free generating set for M is called a free basis for M. An S-semimodule having a free basis is called a free S-semimodule. Clearly, any finitely generated S-semimodule has at least a basis and any free basis is a basis. Example 2.2. The S-semimodule S n in Example 2.1 is a finitely generated free S-semimodule and the set E = {e1 , e2 , . . . , en } is a free basis for S n , where e1 = (1, 0, . . . , 0)T ,

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

531

e2 = (0, 1, . . . , 0)T , . . . , en = (0, 0, . . . , 1)T . The basis E is called the standard basis for S n . It can be seen that r(S n ) = n. Let M and N be S-semimodules. A mapping ϕ : M → N is called an S-homomorphism (see [7]) if the following conditions are satisfied: (1) ϕ(α + β) = ϕ(α) + ϕ(β) for all α, β in M; (2) ϕ(λα) = λϕ(α) for all α in M and λ in S. An S-homomorphism ϕ is called surjective if ϕ is a surjective mapping, and in this case, we write M ∼ N . An S-homomorphism ϕ is called S-isomorphism if ϕ is a bijective mapping, and in this case, we write M ∼ = N. Let S be a semiring. We denote by Mm×n (S) the set of all m × n matrices over S. Especially, we put Mn (S) = Mn×n (S). For A ∈ Mm×n (S), we denote by aij or [A]ij the (i, j)-entry of A. The trace of a matrix A ∈ Mn (S), written tr(A), is defined as the sum n  of all the elements lying on the main diagonal of A, i.e., tr(A) = aii . i=1

For any A ∈ Mm×n (S), we denote by Ai∗ the i-th row of A (i ∈ m) and denote by A∗j the j-th column of A (j ∈ n). It is clear that (Ai∗ )T ∈ S n (i ∈ m) and A∗j ∈ S m (j ∈ n). For any {α1 , α2 , . . . , αm } ⊆ S n , the symbol (α1 , α2 , . . . , αm ) stands for the n × m matrix with elements α1 , α2 , . . . , αm as columns. For any A, B ∈ Mm×n (S) and C ∈ Mn×l (S), we define: A + B = (aij + bij )m×n ; AC = (

n 

aik ckj )m×l .

k=1

It is easy to verify that (Mn (S), +, ·) is a semiring and the identity element in the semiring Mn (S) is the identity matrix In . Let S be a commutative semiring and A ∈ Mn (S). We define the positive determinant |A|+ and the negative determinant |A|− of A as follows: |A| = +

n  

aiσ(i)

σ∈An i=1

and |A|− =



n 

aiσ(i) ,

σ∈Sn \An i=1

where Sn and An denote the symmetric group and the alternating group on n, respectively (see [7]). Clearly, if S is a commutative ring and A ∈ Mn (S) then |A| = |A|+ −|A|− . Note that for any A ∈ M1 (S), we define |A|+ = a11 and |A|− = 0.

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

532

A matrix A in Mn (S) is called nonsingular if a|A|+ + b|A|− = b|A|+ + a|A|− implies a = b for any a, b in S. Otherwise, A is called singular. Let A ∈ Mn (S). An r × r submatrix of A (r ≤ n) is an r × r matrix which is obtained from A by striking out n − r rows and n − r columns. We use Aij to denote the (n − 1) × (n − 1) submatrix of A obtained by striking out the i-th row and j-th column. For any n ∈ N, we define π(n) = + if n is even, and π(n) = − if n is odd. For a matrix A ∈ Mn (S), we define the positive adjoint adj + (A) and the negative adjoint adj − (A) of A as follows (see [11]): ⎛

⎞ ... ...⎟ ⎟ ⎟ ...⎠ ...

⎛

⎞ ... ...⎟ ⎟ ⎟. ...⎠ ...

|A11 |+ |A21 |− |A31 |+ ⎜ |A |− |A |+ |A |− ⎜ 12 22 32 adj + (A) = ⎜ ⎝ |A13 |+ |A23 |− |A33 |+ ... ... ... and |A11 |− |A21 |+ |A31 |− ⎜ |A |+ |A |− |A |+ ⎜ 12 22 32 adj − (A) = ⎜ ⎝ |A13 |− |A23 |+ |A33 |− ... ... ...

We shall frequently write A + for adj + (A) and A − for adj − (A). It is clear that [A + ]ij = π(i+j) π(i+j−1) and [A − ]ij = |A|ji for all i, j in n. |A|ji The following lemmas are used. Lemma 2.1. Let S be a semiring and a, b ∈ S. Then a + b ∈ V (S) if and only if a, b ∈ V (S). 2 Lemma 2.2. (See [12].) Let S be a commutative semiring and A, B ∈ Mn (S). If AB = In , then BA = In . 2 Lemma 2.3. Let S be a commutative semiring and A ∈ Mm×n (S) and B ∈ Mn×m (S). Then tr(AB) = tr(BA). 2 Lemma 2.4. (See [11, Lemmas 3.2, 3.14 and 4.1].) If A ∈ Mn (S), then (1) |A|+ = |A|+ = (2)

n  k=1

n  j=1 n 

apj |Apj |π(p+j) , |A|− = aiq |Aiq |π(q+i) , |A|− =

i=1

apk |Aqk |π(q+k) =

n  k=1 +

n 

apj |Apj |π(p+j−1) (p ∈ n),

j=1 n 

aiq |Aiq |π(q+i−1) (q ∈ n).

i=1

apk |Aqk |π(q+k−1) (p, q ∈ n, p = q).

(3) [AA + ]pp = |A|+ = [A A]pp and [AA − ]pp = |A|− = [A − A]pp (p ∈ n). (4) [AA + ]pq = [AA − ]pq and [A + A]pq = [A − A]pq (p, q ∈ n, p = q). 2

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

533

Lemma 2.5. Let M be a finitely generated free S-semimodule. Then any generating set A for M with |A| = r(M) is a free basis for M. Proof. Since |A| = r(M), A is a minimal generating set for M and so A is a basis for M. By Theorem 3.4 in [17], A is a free basis for M. 2 3. Free sets in a semimodule In this section, we discuss some properties of free sets in a semimodule over a commutative semiring S and obtain a sufficient condition for a nonempty finite set to be free in a finitely generated free S-semimodule. We also give a relation between free set and linear independent set in an S-semimodule. The following result is obvious. Proposition 3.1. If A is a free set in an S-semimodule M and B is a nonempty subset of A, then B is free in M. 2 Proposition 3.2. Let M and M be S-semimodules and ϕ : M → M a surjective S-homomorphism. Then (1) If A is a finite free set in M , then there exists a free set A in M such that A = ϕ(A) and |A| = |A |, where ϕ(A) = {ϕ(α) | α ∈ A}. (2) If ϕ is an S-isomorphism and A is a nonempty finite set in M then A is a free set in M if and only if ϕ(A) is a free set in M . Especially, A is a free basis for M if and only if ϕ(A) is a free basis for M .  } be a free set in M . Since ϕ is a surjective Proof. (1). Let A = {α1 , α2 , . . . , αm S-homomorphism, α−1 (αi ) = φ for every αi in A , where α−1 (αi ) = {α ∈ M | ϕ(α) = αi }. Choose one of elements in α−1 (αi ), denote by αi , for each αi in A and put A = {α1 , α2 , . . . , αm }. It is clear that A = ϕ(A) and |A| = |A |. Suppose now that m 

λi αi =

i=1

m 

μi αi

i=1

for some λ1 , λ2 , . . . , λm , μ1 , μ2 , . . . , μm in S. Then m 

λi ϕ(αi ) =

i=1

m 

μi ϕ(αi ),

i=1

i.e., m  i=1

λi αi

=

m  i=1

μi αi .

534

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

 Therefore λi = μi for all i in m (because the set A = {α1 , α2 , . . . , αm } is free in M ). This means that the set A is free in M. (2). It is obvious. 2

Corollary 3.1. If M and M are finitely generated free S-semimodules then M ∼ = M if  and only if r(M) = r(M ). Proof. If M ∼ = M , then by Proposition 3.2(2), we have r(M) = r(M ). Conversely, if r(M) = r(M ), and A = {α1 , α2 , . . . , αn } and A = {α1 , α2 , . . . , αn } are free bases for M and M , respectively, where n = r(M) = r(M ), then, it is easy to verify that n n   the mapping ϕ : M → M , defined by ϕ(α) = λi αi for α = λi αi ∈ M, is an i=1

S-isomorphism and αi = ϕ(αi ) for all i in n. Thus M ∼ = M .

2

i=1

In the classical linear algebra, we know that if A = {α1 , α2 , . . . , αm } is a linear independent set (or free set) and the set {α1 , α2 , . . . , αm , β} is linearly dependent in a linear space over a field then β can be uniquely represented as a linear combination of the elements in A. In general, for a free set in an R-semimodule, we have Proposition 3.3. If A = {α1 , α2 , . . . , αm } is a free set and the set {α1 , α2 , . . . , αm , β} is linearly dependent in an S-semimodule M then β can be uniquely represented as a linear combination of the elements in A or the set (A \ {αi }) ∪ {β} is free in M for some i in m. Proof. Suppose that A = {α1 , α2 , . . . , αm } is a free set and the set {α1 , α2 , . . . , αm , β} is linearly dependent in M. Then either αi or β can be represented as a linear combination of the other elements in the set. If β can be represented as a linear combination of the elements in A, then β can be uniquely represented as a linear combination of the elements in A (because A is free in M). If αi can be represented as a linear combination of the elements in (A \ {αi }) ∪ {β} for some i in m, then αi = λβ +

m 

λj αj

(1)

j=1,j=i

for some λ, λ1 , . . . , λi−1 , λi+1 , . . . , λm in S. We say λ is a multiplicatively cancellable element in S. Indeed, if λx = λy for some x and y in S, then λxβ = λyβ.

(2)

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

Adding

m  j=1,j=i m 

m 

yλj αj +

535

xλj αj to both sides in Equality (2), we have

j=1,j=i m 

yλj αj + x(λβ +

j=1,j=i

m 

λj αj ) =

j=1,j=i

xλj αj + y(λβ +

j=1,j=i

m 

λj αj ).

j=1,j=i

By (1), we have m 

m 

yλj αj + xαi =

j=1,j=i

xλj αj + yαi .

j=1,j=i

This implies x = y (because the set A = {α1 , α2 , . . . , αm } is free in M). In the following we prove the set (A \ {αi }) ∪ {β} is free in M. Let m 

m 

xj αj + xi β =

j=1,j=i

yj αj + yi β

j=1,j=i

for some x1 , x2 , . . . , xm and y1 , y2 , . . . , ym in S. Then m  j=1,j=i

Adding

m 

m 

xi λj αj +

j=1,j=i

m 

xj λαj + xi λβ =

yj λαj + yi λβ.

j=1,j=i

yi λj αj to both sides in the above equality, we have

j=1,j=i m 

xi λj αj +

j=1,j=i

xi λj αj +

j=1,j=i

m 

yi λj αj +

j=1,j=i

m 

=

m 

m 

xj λαj + xi λβ =

j=1,j=i m 

yi λj αj +

j=1,j=i

yj λαj + yi λβ.

j=1,j=i

Then m 

(yi λj + xj λ)αj + xi (

j=1,j=i

=

m 

m 

λj αj + λβ) =

j=1,j=i

(xi λj + yj λ)αj + yi (

j=1,j=i

m 

λj αj + λβ).

j=1,j=i

By (1), we have m  j=1,j=i

(yi λj + xj λ)αj + xi αi =

m  j=1,j=i

(xi λj + yj λ)αj + yi αi .

(3)

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

536

This implies xi = yi (because the set A is free in M). By Equality (3), we have m 

m 

xj αj + xi β =

j=1,j=i

yj αj + xi β,

j=1,j=i

and so m  j=1,j=i m 

Adding

m 

λxj αj + xi λβ =

λyj αj + xi λβ.

j=1,j=i

λj xi αj to both sides in the above equality, we have

j=1,j=i m 

λxj αj + xi (

j=1,j=i

m 

m 

λj αj + λβ) =

j=1,j=i

λyj αj + xi (

j=1,j=i

m 

λj αj + λβ).

j=1,j=i

By (1), we have m 

m 

λxj αj + xi αi =

j=1,j=i

λyj αj + xi αi .

j=1,j=i

This implies λxj = λyj for all j in m with j = i (because the set A is free in M) and so xj = yj for all j in m with j = i (because λ is multiplicatively cancellable in S). Consequently, the set (A \ {αi }) ∪ {β} is free in M. The proof is completed. 2 Theorem 3.1. Let B = {β1 , β2 , . . . , βm } be a free set in an S-semimodule M and let A = {α1 , α2 , . . . , αm } ⊆ M satisfy (α1 , α2 , . . . , αm ) = (β1 , β2 , . . . , βm )A for A ∈ Mm (S) m  (or αj = aij βi for all j in m). Then i=1

(1) If the set A is free in M then the matrix A is nonsingular in Mm (S). (2) If the semiring S is cancellative and A is nonsingular then A is free in M. Proof. Let (α1 , α2 , . . . , αm ) = (β1 , β2 , . . . , βm )A, i.e., αj =

m 

aij βi

(4)

i=1

for all j in m. (1). Suppose to a contrary that A is singular in Mm (S). Then |A|+ a + |A|− b = |A|+ b + |A|− a for some a, b in S with a = b.

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

537

If aaij = baij for all i, j in m, then, by Equalities (4), we have aαj =

m 

m 

aaij βi =

i=1

for all j in m, and so

m 

aαj =

j=1

baij βi = bαj

i=1

m 

bαj . This means that the set A is not free in M,

j=1

which is a contradiction. If aaij = baij for some i, j in m, then there exists an r ∈ {1, 2, . . . , m − 1} such that |B|+ a + |B|− b = |B|+ b + |B|− a for all (r + 1) × (r + 1) submatrices B of A and |D|+ a + |D|− b = |D|+ b + |D|− a for some r × r submatrix D of A. Without loss of generality, we assume ⎛

a11 ⎜a ⎜ 21 D=⎜ ⎝ ... ar1

a12 a22 ... ar2

... ... ... ...

⎞ a1r a2r ⎟ ⎟ ⎟. ... ⎠ arr

For any i ∈ m, we consider the matrix ⎛

A(i)

a11 ⎜a ⎜ 21 ⎜ = ⎜ ... ⎜ ⎝ ar1 ai1

a12 a22 ... ar2 ai2

... ... ... ... ...

a1r a2r ... arr air

⎞ a1,r+1 a2,r+1 ⎟ ⎟ ⎟ ... ⎟. ⎟ ar,r+1 ⎠ ai,r+1

If i ∈ {r + 1, . . . , m}, then A(i) is an (r + 1) × (r + 1) submatrix of A, and in this case |A(i) |+ a + |A(i) |− b = |A(i) |+ b + |A(i) |− a. Let ⎛

a11 ⎜a ⎜ 21 Aj = ⎜ ⎝ ... ar1 Then |A(i) |+ =

r+1 

. . . a1,j−1 a1,j+1 . . . a2,j−1 a2,j+1 ... ... ... . . . ar,j−1 ar,j+1

... ... ... ...

⎞ a1r a1,r+1 a2r a2,r+1 ⎟ ⎟ ⎟ , 1 ≤ j ≤ r + 1. ... ... ⎠ arr ar,r+1

aij |Aj |π(r+1+j) and |A(i) |− =

j=1

r+1 

aij |Aj |π(r+j) (by Lemma 2.4(1)),

j=1

and so r+1  j=1

aaij |Aj |π(r+1+j) +

r+1  j=1

baij |Aj |π(r+j) =

r+1  j=1

aaij |Aj |π(r+j) +

r+1  j=1

baij |Aj |π(r+1+j) .

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

538

If i ∈ {1, 2, . . . , r}, then, by Lemma 2.4(2), we have r+1 

aij |Aj |π(r+j+1) =

j=1

r+1 

aij |Aj |π(r+j) ,

j=1

and so r+1 

aaij |Aj |

π(r+j+1)

=

j=1

r+1 

aaij |Aj |π(r+j)

j=1

and r+1 

baij |Aj |π(r+j+1) =

j=1

r+1 

baij |Aj |π(r+j) .

j=1

Therefore r+1 

aaij |Aj |π(r+1+j) +

j=1

r+1 

baij |Aj |π(r+j) =

j=1

r+1 

aaij |Aj |π(r+j) +

j=1

r+1 

baij |Aj |π(r+1+j) .

j=1

Consequently, we have r+1 r+1   (a|Aj |π(r+1+j) + b|Aj |π(r+j) )aij = (a|Aj |π(r+j) + b|Aj |π(r+1+j) )aij j=1

j=1

for all i in m. Then r+1 r+1   (a|Aj |π(r+1+j) + b|Aj |π(r+j) )aij βi = (a|Aj |π(r+j) + b|Aj |π(r+1+j) )aij βi j=1

j=1

for all i in m, and so r+1 m r+1 m     (a|Aj |π(r+1+j) + b|Aj |π(r+j) ) aij βi = (a|Aj |π(r+j) + b|Aj |π(r+1+j) ) aij βi . j=1

i=1

j=1

i=1

By Equalities (4), we have r+1 

(a|Aj |π(r+1+j) + b|Aj |π(r+j) )αj =

j=1

r+1 

(a|Aj |π(r+j) + b|Aj |π(r+1+j) )αj .

j=1

Since D = Ar+1 , we have a|Ar+1 |+ + b|Ar+1 |− = a|Ar+1 |− + b|Ar+1 |+ ,

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

539

i.e., a|Ar+1 |π(r+1+r+1) + b|Ar+1 |π(r+r+1) = a|Ar+1 |π(r+r+1) + b|Ar+1 |π(r+1+r+1) . This means that the set {α1 , α2 , . . . , αr+1 } is not free in M, and so the set A = {α1 , α2 , . . . , αm } is not free in M (by Proposition 3.1), which is a contradiction. Therefore, A is nonsingular in Mm (S). m m   (2). Let xi αi = yi αi for some x1 , x2 , . . . , xm ; y1 , y2 , . . . , ym in S, i.e., (α1 , α2 , i=1

i=1

. . . , αm )x = (α1 , α2 , . . . , αm )y, where x = (x1 , x2 , . . . , xm )T , y = (y1 , y2 , . . . , ym )T . Then (β1 , β2 , . . . , βm )Ax = (β1 , β2 , . . . , βm )Ay, and so Ax = Ay (because the set B is free). In the following we prove the set A is free in M. To do this, we prove x = y. By Ax = Ay, we have (A + A)x = (A + A)y and (A − A)x = (A − A)y. From (A + A)x = (A + A)y, we have m 

(A + A)kj xj =

j=1

m  (A + A)kj yj j=1

for any k in m. Then, by Lemma 2.4(3)(4), we have |A|+ xk +



(A − A)kj xj = |A|+ yk +

j=k

 (A − A)kj yj . j=k

Likewise, by (A − A)x = (A − A)y and Lemma 2.4(3), we have |A|− xk +



(A − A)kj xj = |A|− yk +

j=k



(A − A)kj yj

j=k

for any k in m. Then |A|+ xk + |A|− yk +



(A − A)kj yj +

j=k

 (A − A)kj xj = j=k

  = |A|+ yk + |A|− xk + (A − A)kj xj + (A − A)kj yj j=k

j=k

and so |A|+ xk + |A|− yk = |A|+ yk + |A|− xk for any k ∈ m (because S is cancellative). Since A is nonsingular in Mm (S), we have xk = yk for all k in m, which means that the set A is free in M. The proof is completed. 2 Since E = {e1 , e2 , . . . , en } is a free set in the S-semimodule S n and (α1 , α2 , . . . , αn ) = (e1 , e2 , . . . , en )A for any {α1 , α2 , . . . , αn } ⊆ S n , where A = (α1 , α2 , . . . , αn ), by Theorem 3.1, we have Corollary 3.2. Let A = {α1 , α2 , . . . , αn } ⊆ S n and A = (α1 , α2 , . . . , αn ). Then

540

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

(1) If the set A is free in S n then the matrix A is nonsingular in Mn (S). (2) If the semiring S is cancellative and A is nonsingular then A is free in S n .

2

Since any ring with identity is a cancellative semiring, by Corollary 3.2, we have Corollary 3.3. Let S be a commutative ring and A = {α1 , α2 , . . . , αn } ⊆ S n and A = (α1 , α2 , . . . , αn ). Then (1) The set A is free in S n if and only if |A| is not a zero divisor (left or right). (2) If S is a field then the set A is linear independent if and only if |A| = 0. 2 In the classical linear algebra, it is well known that the cardinality of any linear independent set in a finitely generated linear space V over a field does not exceed the dimension of V. Similarly, we have Theorem 3.2. Let M be a finitely generated S-semimodule. Then the cardinality of any free set in M does not exceed the rank of M. Proof. Let r(M) = m and let B = {β1 , β2 , . . . , βm } be a generating set for M. Let ϕ be a mapping from S m to M defined by ϕ : (λ1 , λ2 , . . . , λm )T →

m 

λi βi

i=1

for all (λ1 , λ2 , . . . , λm )T in S m . It is easy to see that ϕ is a surjective S-homomorphism. Suppose to a contrary that there exists a free set A0 in M such that |A0 | > m. Then there exists a subset A of A0 such that |A | = m + 1. By Proposition 3.1, A is a free set in M and so there exists a free set A in S m such that |A| = |A | = m + 1 (by Proposition 3.2(1)). Let A = {α1 , α2 , . . . , αm+1 }, where αi = (ai1 , ai2 , . . . , aim )T ∈ S m , 1 ≤ i ≤ m + 1, and let γi = (ai1 , ai2 , . . . , aim , 0)T , 1 ≤ i ≤ m + 1, and C = {γ1 , γ2 , . . . , γm+1 } and C = (γ1 , γ2 , . . . , γm+1 ). Then C ⊆ S m+1 and C ∈ Mm+1 (S). Since A is a free set in S m , the set C is free in S m+1 . By Corollary 3.2(1), the matrix C is nonsingular. On the other hand, it is easy to see that |C|+ = |C|− = 0. Thus C is singular, which is a contradiction. Therefore the cardinality of any free set in M does not exceed the rank of M. The proof is completed. 2 In the following we give a sufficient condition for a finite set to be free in a finitely generated free S-semimodule. Theorem 3.3. Let S be a commutative and cancellative semiring and M a free S-semimodule with a free basis B = {β1 , β2 , . . . , βn }. Let A = {α1 , α2 , . . . , αm } ⊆ M (m ≤ n), and (α1 , α2 , . . . , αm ) = (β1 , β2 , . . . , βn )A, where A ∈ Mn×m (S). If A contains an m × m nonsingular submatrix then the set A is free in M.

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

541

Proof. Let C be an m × m nonsingular submatrix of A and Ck∗ = Aik ∗ for all k in m, where ik ∈ n with ik1 = ik2 for k1 = k2 , and let C = {C∗1 , C∗2 , . . . , C∗m }. Then the set C is free in S m (by Corollary 3.2(2)). m m   Suppose now that xi αi = yi αi for some x1 , x2 , . . . , xm ; y1 , y2 , . . . , ym in S, i=1

i=1

i.e., (α1 , α2 , . . . , αm )x = (α1 , α2 , . . . , αm )y, where x = (x1 , x2 , . . . , xm )T , y = (y1 , y2 , . . . , ym )T . Then (β1 , β2 , . . . , βn )Ax = (β1 , β2 , . . . , βn )Ay, and so Ax = Ay (because B is a free basis for M). Since C is an m × m submatrix of A, we have Cx = Cy, i.e., (C∗1 , C∗2 , . . . , C∗m )x = (C∗1 , C∗2 , . . . , C∗m )y, and so x = y (because the set C is free in S m ). This means that the set A is free in M. 2 Example 3.1. Consider the set S = {(a, b) | a ∈ Z, b ∈ Z0 }, where Z and Z0 denote the ring of the integers and the semiring of the nonnegative integers, respectively. Define operations of addition and multiplication on S by setting (a, b) + (c, d) = (a + c, b + d) and (a, b)(c, d) = (ac + ad + bc, bd) for all (a, b), (c, d) ∈ S. Then S can be easily verified to be a commutative and cancellative semiring with the zero element (0, 0) and the identity element (0, 1) (note that S is called the Dorroh extension of Z by Z0 (see [7])). Let ⎛

⎛ ⎛ ⎛ ⎞ ⎞ ⎞ ⎞ (0, 1) (0, 0) (0, 0) (0, 0) ⎜ (0, 0) ⎟ ⎜ (0, 1) ⎟ ⎜ (0, 0) ⎟ ⎜ (0, 0) ⎟ ⎜ ⎜ ⎜ ⎜ ⎟ ⎟ ⎟ ⎟ β1 = ⎜ ⎟ , β2 = ⎜ ⎟ , β3 = ⎜ ⎟ , β4 = ⎜ ⎟, ⎝ (0, 0) ⎠ ⎝ (0, 0) ⎠ ⎝ (0, 1) ⎠ ⎝ (0, 0) ⎠ (0, 0) (0, 0) (0, 0) (0, 1) ⎛ ⎛ ⎛ ⎞ ⎞ ⎞ (−1, 1) (2, 0) (1, 0) ⎜ (0, 0) ⎟ ⎜ (−1, 0) ⎟ ⎜ (−1, 1) ⎟ ⎜ ⎜ ⎜ ⎟ ⎟ ⎟ α1 = ⎜ ⎟ , α2 = ⎜ ⎟ and α3 = ⎜ ⎟, ⎝ (1, 0) ⎠ ⎝ (0, 1) ⎠ ⎝ (1, 0) ⎠ (1, 2) (2, 0) (2, 1) and let B = {β1 , β2 , β3 , β4 } and A = {α1 , α2 , α3 }. Then B is a free basis for the free S-semimodule S 4 and A ⊆ S 4 . It is easy to verify that (α1 , α2 , α3 ) = (β1 , β2 , β3 , β4 )A,

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

542

⎛

⎞ (1, 0) (−1, 1) ⎟ ⎟ ⎟ ∈ M4×3 (R). (1, 0) ⎠ (2, 1) ⎞ ⎛ (−1, 1) (2, 0) (1, 0) ⎟ ⎜ Considering the submatrix C = ⎝ (0, 0) (−1, 0) (−1, 1) ⎠ of A, we have (1, 0) (0, 1) (1, 0)

(−1, 1) ⎜ (0, 0) ⎜ where A = ⎜ ⎝ (1, 0) (1, 2)

(2, 0) (−1, 0) (0, 1) (2, 0)

|C|+ = (−1, 1)(−1, 0)(1, 0) + (2, 0)(−1, 1)(1, 0) + (1, 0)(0, 0)(0, 1) = (0, 0) + (0, 0) + (0, 0) = (0, 0) and |C|− = (1, 0)(−1, 0)(1, 0) + (2, 0)(0, 0)(1, 0) + (−1, 1)(−1, 1)(0, 1) = (−1, 0) + (0, 0) + (−1, 1) = (−2, 1). If |C|+ (a, b) + |C|− (c, d) = |C|+ (c, d) + |C|− (a, b) for (a, b) and (c, d) in S, then (−2c − 2d + c, d) = (−2a − 2b + a, b). This implies a = c and b = d, i.e., (a, b) = (c, d), which means that the matrix C is nonsingular in M3 (S). By Theorem 3.3, the set A is free in S 4 . Open problem 3.1. For a given finitely generated free semimodule M over a commutative semiring S, find a necessary and sufficient condition for a nonempty finite subset in M to be free. At the end of this section, we give a relation between free set and linear independent set in an S-semimodule. Theorem 3.4. Let M be an S-semimodule and B = {β1 , β2 , . . . , βm } ⊆ M and U = {u1 , u2 , . . . , us } ⊆ S, and let U B = {ui βj | 1 ≤ i ≤ s, 1 ≤ j ≤ m}. Then (1) If B is a free set in M and U is a linear independent set in the S-semimodule S then the set U B is linear independent in M. (2) If B is a free basis for M and U is a basis for the S-semimodule S then U B is a basis for M. Proof. (1). Suppose to a contrary that the set U B is linear dependent in M. Then there exist i0 in s and j0 in m such that ui0 βj0 =

 i=i0 ,1≤j≤m

λij ui βj +

 j=j0

λi0 j ui0 βj ,

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

543

where λij ∈ S (1 ≤ i ≤ s, 1 ≤ j ≤ m), i.e., 

ui0 βj0 =

λij ui βj + (

i=i0 ,j=j0

and so ui0 =





λij0 ui )βj0 +

i=i0



λi0 j ui0 βj

j=j0

λij0 ui (because the set B is free in M). This contradicts the definition

i=i0

of U . Thus, the set U B is linear independent in M. (2). By (1), the set U B is linear independent in M. For any α ∈ M, we have α=

m 

λj βj

j=1

for some λj in S (because B is a free basis for M). Since U is a basis for the S-semimodule S, for any λj , we have 

λj =

λij ui

1≤i≤s

for some λij in S. Then α=



λij ui βj ,

1≤i≤s,1≤j≤m

i.e., U B is a generating set for M. Thus, U B is a basis for M.

2

4. Free subsemimodules of a semimodule In this section, we consider the free subsemimodules of a semimodule over a commutative semiring S and prove that the rank of any free subsemimodule N of a finitely generated free S-semimodule M does not exceed that of M. Also, we give some equivalent descriptions for a commutative semiring S to have the property that all nonzero subsemimodules of any finitely generated free S-semimodule are free. Theorem 4.1. Let M and M be S-semimodules and M ∼ M . If N  is a finitely generated free subsemimodule of M , then there exists a free subsemimodule N of M such that N ∼ = N . Proof. Let B  be a free basis for N  , and let ϕ : M → M be a surjective S-homomorphism. By Proposition 3.2(1), there exists a free set B in M such that |B| = |B |. Let N = Span(B). Then N is a finitely generated free S-semimodule and r(N ) = r(N  ). By Corollary 3.1, we have N ∼ = N . 2 Theorem 4.2. If M is a finitely generated S-semimodule and N is a free subsemimodule of M, then N is finitely generated and r(N ) ≤ r(M).

544

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

Proof. Let B be a free basis for N . Then B is a free set in M and so |B| ≤ r(M) (by Theorem 3.2). This means that N is finitely generated and r(N ) = |B| ≤ r(M). The proof is completed. 2 Since any field is a commutative semiring, the foregoing theorem can be specialized to the case in which S = F is a field. Then it reduces to the following well-known result of linear algebra. Corollary 4.1. If V is a finitely generated linear space over a field F and W is a subspace of V, then W is finitely generated and dim(W) ≤ dim(V). 2 Remark 4.1. In Theorem 4.2, the assumption that N is a free subsemimodule of M is necessary. For example, consider the semiring Z0 of the nonnegative integers. It is a commutative semiring. Clearly, (Z0 )2 is a finitely generated Z0 -semimodule and r((Z0 )2 ) = 2. 0 1 1 Let A = {α1 , α2 , α3 } ⊆ (Z ) , where α1 = , α2 = and α3 = , and 2 1 0 let N = Span(A). Then N is a subsemimodule of (Z0 )2 and r(N ) ≤ 3. If r(N ) ≤ 2, then N has a generating set with the cardinality two, say, {β2 , β2 }. In this case (α1 , α2 , α3 ) = (β1 , β2 )A and (β1 , β2 ) = (α1 , α2 , α3 )B, where A ∈ M2×3 (Z0 ) and B ∈ M3×2 (Z0 ), and so (α1 , α2 , α3 ) = (α1 , α2 , α3 )BA. Taking BA = D = (dij )3×3 , we have ⎞ ⎛



d d d 11 12 13 011 011 ⎜ ⎟ = ⎝ d21 d22 d23 ⎠ 210 210 d31 d32 d33

d21 + d31 d22 + d32 d23 + d33 . = 2d11 + d21 2d12 + d22 2d13 + d23 0 2

Then d21 + d31 = 0, d22 + d32 = 1, d23 + d33 = 1, 2d11 + d21 = 2, 2d12 + d22 = 1, 2d13 + d23 = 0. This implies d 11 = d

22 = d33 = 1 and dij = 0 for all i, j in {1, 2, 3}, i.e., BA = I3 . A and B1 = (B, O2 ), where O1 ∈ M1×3 (Z0 ) and O2 ∈ M3×1 (Z0 ) Let A1 = O1

A = BA = I3 , and so A1 B1 = I3 (by are zero matrices. Then B1 A1 = (B, O2 ) O1



A AB 0 (B, O2 ) = Lemma 2.2). On the other hand, A1 B1 = . This is a contraO1 0 0 diction. Then r(N ) > 2, and so r(N ) = 3 > r((Z0 )2 ).

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

545

We say N is not a free subsemimodule of (Z0 )2 . Indeed, if N is free, then A = {α1 , α2 , α3 } is a free basis for N (by Lemma 2.5). On the other hand, we have 2α2 = α1 + 2α3 . This is a contradiction. In the classical module theory, it is well known that if R is a principal ideal domain and n ∈ N then any nonzero submodule of the free R-module Rn is free (e.g. see Theorem 3.7 in [8]). In the following we give some equivalent descriptions for a commutative semiring S to have the property that all nonzero subsemimodules of any finitely generated free S-semimodule are free. Theorem 4.3. For any commutative semiring S, the following statements are equivalent: (1) All nonzero subsemimodules of any finitely generated free S-semimodule are free. (2) All nonzero subsemimodules of the S-semimodule S 3 are free. (3) S is a principal ideal domain. Proof. (1) ⇒ (2). It is obvious. (2) ⇒ (3). Suppose that all nonzero subsemimodules of S 3 are free. We will prove S is a principal ideal domain. Firstly, we prove S is a cancellative semiring. ⎛ ⎞ a ⎜ ⎟ Let a, b, c ∈ S satisfy a + b = a + c. If a = 0 then b = c. If a = 0, we let α1 = ⎝ a ⎠ 0 ⎛ ⎞ 1 ⎜ ⎟ and α2 = ⎝ 0 ⎠, and let N = Span(α1 , α2 ). Then N is a nonzero subsemimodule of S 3 0 and r(N ) ≤ 2, and so N is a free S-semimodule. It is clear that r(N ) = 0. If r(N ) = 1, we let {γ} be a generating set for N . Then α1 = λ1 γ and α2 = λ2 γ for some λ1 , λ2 in S, and so λ2 α1 = λ1 α2 , i.e., λ2 a = λ1 and λ2 a = 0. Thus α1 = λ1 γ = (λ2 a)γ = θ. This implies a = 0, which contradicts the assumption a = 0. Therefore r(N ) = 2. By Lemma 2.5, the set {α1 , α2 } is a free basis for N . Since ⎛ ⎞ ⎛ ⎞ a b ⎜ ⎟ ⎜ ⎟ α1 + bα2 = ⎝ a ⎠ + ⎝ 0 ⎠ 0 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ a+b a+c a c ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ a ⎠ = ⎝ a ⎠ = ⎝a⎠+⎝0⎠ 0 0 0 0 = α1 + cα2 , we have b = c (because the set {α1 , α2 } is free). Thus, S is a cancellative semiring. Secondly, we prove S is a ring. To prove S is a ring, it is sufficient to prove the identity element 1 ∈ V (S).

546

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 1 0 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Let α1 = ⎝ 0 ⎠, α2 = ⎝ 1 ⎠, α3 = ⎝ 1 ⎠, α4 = ⎝ 0 ⎠ and N = Span(α1 , α2 , α3 , α4 ). 0 0 1 1 3 Then N is a nonzero subsemimodule of S , and so N is a free S-semimodule. By Theorem 4.2, r(N ) ≤ r(S 3 ) = 3. We say {α1 , α2 , α3 } is a free set in N . In fact, if λ1 α1 + λ2 α2 + λ3 α3 = μ1 α1 + μ2 α2 + μ3 α3 for some λ1 , λ2 , λ3 and μ1 , μ2 , μ3 in S, i.e., ⎞ ⎛ ⎞ ⎛ μ1 + μ2 λ1 + λ2 ⎟ ⎜ ⎟ ⎜ ⎝ λ2 + λ3 ⎠ = ⎝ μ2 + μ3 ⎠ , λ3 μ3 then λ1 + λ2 = μ1 + μ2 , λ2 + λ3 = μ2 + μ3 and λ3 = μ3 . Therefore λ2 + μ3 = μ2 + μ3 . This implies λ2 = μ2 (because S is a cancellative semiring). Then λ1 + μ2 = μ1 + μ2 , and so λ1 = μ1 (because S is a cancellative semiring). This shows that the set {α1 , α2 , α3 } is free in N . By Theorem 3.2, we have 3 ≤ r(N ). Then r(N ) = 3, and so |A| = 3 for any free basis A for N . Let {γ1 , γ2 , γ3 } be a free basis for N . Then (α1 , α2 , α3 , α4 ) = (γ1 , γ2 , γ3 )A and (γ1 , γ2 , γ3 ) = (α1 , α2 , α3 , α4 )B, where A ∈ M3×4 (S) and B ∈ M4×3 (S), and so (α1 , α2 , α3 , α4 ) = (α1 , α2 , α3 , α4 )BA and (γ1 , γ2 , γ3 ) = (γ1 , γ2 , γ3 )AB. By (α1 , α2 , α3 , α4 ) = (α1 , α2 , α3 , α4 )BA, we have ⎞ ⎞ ⎛ 1100 1100 ⎟ ⎜ ⎟ ⎜ ⎝ 0 1 1 0 ⎠ = ⎝ 0 1 1 0 ⎠ BA. 0011 0011 ⎛

Taking BA = D = (dij )4×4 , we have ⎛

⎞

⎛

⎞

⎛

d11 1100 1100 ⎜ ⎟ ⎜ d21 ⎜ ⎟ ⎜ ⎝0 1 1 0⎠ = ⎝0 1 1 0⎠⎜ ⎝ d31 0011 0011 d41

d12 d22 d32 d42

d13 d23 d33 d43

⎞ d14 d24 ⎟ ⎟ ⎟, d34 ⎠ d44

i.e., ⎛

⎞ ⎛ ⎞ 1100 d11 + d21 d12 + d22 d13 + d23 d14 + d24 ⎜ ⎟ ⎜ ⎟ ⎝ 0 1 1 0 ⎠ = ⎝ d21 + d31 d22 + d32 d23 + d33 d24 + d34 ⎠ . 0011 d31 + d41 d32 + d42 d33 + d43 d34 + d44 Then d11 + d21 = 1, d22 + d32 = 1, d23 + d33 = 1, d34 + d44 = 1 and d21 + d31 = 0, d13 + d23 = 0, d32 + d42 = 0, d24 + d34 = 0.

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

547

This implies d21 , d23 , d32 , d34 ∈ V (S) and d11 = 1 + (−d21 ), d22 = 1 + (−d32 ), d33 = 1 + (−d23 ), d44 = 1 + (−d34 ). Then Tr(BA) =

4 

dii = 1 + 1 + 1 + 1 + (−d21 ) + (−d32 ) + (−d23 ) + (−d34 ).

i=1

On the other hand, by (γ1 , γ2 , γ3 ) = (γ1 , γ2 , γ3 )AB, we have AB = I3 (because the set {γ1 , γ2 , γ3 } is free). Then Tr(AB) = 1 + 1 + 1. By Lemma 2.3, we have 1 + 1 + 1 = 1 + 1 + 1 + 1 + (−d21 ) + (−d32 ) + (−d23 ) + (−d34 ). Since R is cancellative, we have 1 = d21 + d32 + d23 + d34 ∈ V (S). Therefore, for any a ∈ S, a = a · 1 ∈ V (S). This shows that S is a ring. Thirdly, we prove each ideal in S is a principal ideal. Let I be any ideal of S. If I = {0}, then I = (0) and in this case I is a principal ideal. If I = {0}, we let N = {λe1 | λ ∈ I}, where e1 = (1, 0, 0)T . Then N is a subsemimodule of S 3 , and so N is a free S-semimodule. Since N ⊆ Span(e1 ) and r(Span(e1 )) = 1, we have r(N ) ≤ 1 (by Theorem 4.2). Since N = {0}, we have r(N ) = 1, and so N = Span(γ) for some nonzero element γ in S 3 . Then there exists an element a in I such that γ = ae1 . In the following we prove I = (a). It is clear that (a) ⊆ I since a ∈ I. On the other hand, for any b ∈ I, we have be1 ∈ N = Span(γ), and so be1 = cγ for some c in S. Then be1 = cγ = cae1 , and so b = ca, i.e., b ∈ (a). Therefore I = (a) which is a principal ideal of S. Finally, we prove S is a domain. Let a ∈ S and a = 0. Then ae1 = θ and so Span(ae1 ) is a nonzero subsemimodule of S 3 , where e1 = (1, 0, 0)T . Therefore Span(ae1 ) is a free S-semimodule and r(Span(ae1 )) = 1. By Lemma 2.5, {ae1 } is a free basis for the S-semimodule Span(ae1 ). If ab = 0 for some b in S, then b(ae1 ) = 0(ae1 ). This implies b = 0 (because {ae1 } is free), which shows that a is not a left zero divisor. Since S is commutative, a is not a right zero divisor. This means that S possesses no zero divisors = 0 (left or right), and so S is a domain. Consequently, S is a principal ideal domain. (3) ⇒ (1). Similar to that of Theorem 3.7 in [8]. The proof is completed. 2

548

Y.-J. Tan / Linear Algebra and its Applications 496 (2016) 527–548

Acknowledgements The author would like to thank the referees for a number of constructive comments and valuable suggestions. References [1] S.K. Bhambri, Extensions of semimodules and injective semimodules, Southeast Asian Bull. Math. 34 (2010) 25–41. [2] W. Briec, C. Horvath, On the separation of convex sets in some idempotent semimodules, Linear Algebra Appl. 435 (2011) 1542–1548. [3] G. Cohen, S. Gaubert, J.P. Quadrat, Duality and separation theorems in idempotent semimodules, Linear Algebra Appl. 379 (2004) 395–422. [4] R.A. Cuninghame-Green, Minimax Algebra, Lecture Notes in Economics and Mathematical Systems, vol. 166, Springer-Verlag, Berlin, 1979. [5] R. Ebrahimi Atani, S. Ebrahimi Atani, On subsemimodules of semimodules, Bul. Acad. Ştiinţe Repub. Mold. Mat. 63 (2) (2010) 20–30. [6] R. Ebrahimi Atani, Prime subsemimodules of semimodules, Int. J. Algebra 4 (26) (2010) 1299–1306. [7] J.S. Golan, Semirings and Their Applications, Kluwer Academic Publishers, 1999. [8] N. Jacobson, Basic Algebra 1, W.H. Freeman and Company, New York, 1985. [9] A.M. Kanan, Z.Z. Petrović, Note on cardinality of bases in semilinear spaces over zerosumfree semirings, Linear Algebra Appl. 439 (2013) 2795–2799. [10] G. Maze, C. Monico, J. Rosenthal, Public key cryptography based on semigroup actions, Adv. Math. Commun. 1 (2007) 489–507. [11] P.L. Poplin, R.E. Hartwig, Determinantal identities over commutative semirings, Linear Algebra Appl. 387 (2004) 99–132. [12] C. Reutenauer, H. Straubing, Inversion of matrices over a commutative semiring, J. Algebra 88 (1984) 350–360. [13] Q.Y. Shu, X.P. Wang, The cardinality of bases in semilinear spaces over commutative semirings, Linear Algebra Appl. 459 (2014) 83–100. [14] Q.Y. Shu, X.P. Wang, Dimensional formulas of semilinear subspaces over semirings, Linear Multilinear Algebra 62 (2014) 1475–1490. [15] M. Takahashi, Structures of semimodules, Kobe J. Math. 4 (1987) 79–101. [16] Y.J. Tan, On nilpotency of matrices over antirings, Linear Algebra Appl. 433 (2010) 1541–1554. [17] Y.J. Tan, Bases in semimodules over commutative semirings, Linear Algebra Appl. 443 (2014) 139–152. [18] Y.J. Tan, Inner products on semimodules over a commutative semiring, Linear Algebra Appl. 460 (2014) 151–173. [19] E.M. Vechtomov, Two general structure theorems on submodules, in: Abelian Groups and Modules, vol. 15, Tomsk State University, Tomsk, 2000, pp. 17–23 (in Russian). [20] H.X. Wang, A note on endomorphism semirings of semimodules, Kobe J. Math. 5 (1988) 155–160. [21] H.X. Wang, M. Takahashi, On epimorphisms of semimodules, Kobe J. Math. 6 (1989) 297–298. [22] S.M. Yusuf, Inverse semimodules, J. Natur. Sci. Math. 6 (1966) 111–117. [23] S. Zhao, X.P. Wang, Invertible matrices and semilinear spaces over commutative semirings, Inform. Sci. 180 (2010) 5115–5124. [24] U. Zimmermann, Linear and Combinatorial Optimization in Ordered Algebraic Structures, Annals of Discrete Mathematics, vol. 10, North-Holland, 1981.