- Email: [email protected]
Bases of a free semimodule are small
Linear Algebra and its Applications 466 (2015) 38–40
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Bases of a free semimodule are small Yaroslav Shitov National Research University Higher School of Economics, 20 Myasnitskaya Ulitsa, Moscow 101000, Russia
a r t i c l e
i n f o
Article history: Received 9 May 2014 Accepted 5 October 2014 Available online xxxx Submitted by R. Brualdi
a b s t r a c t A generating set G of a left semimodule S over a semiring R is called a basis if no proper subset of G generates S. We prove that Rn has no basis of cardinality exceeding qn, where q is the largest cardinality of bases of R. © 2014 Elsevier Inc. All rights reserved.
MSC: 16Y60 13C15 Keywords: Semiring theory Linear algebra
A set R equipped with two binary operations + and · is called a semiring if the following conditions are satisfied: (i) (R, +) is a commutative monoid, (ii) (R, ·) is a monoid, (iii) multiplication distributes over addition from both sides, and (iv) the additive identity 0 satisfies 0x = x0 = 0, for any x ∈ R. We also assume that the multiplicative identity 1 ∈ R is different from 0. An element a ∈ R is called additively invertible (or simply invertible) if there is b ∈ R such that a + b = 0. A commutative monoid (S, +, 0) turns into a left R-semimodule if one defines a scalar multiplication R × S → S satisfying, for all λ, μ ∈ R and a, b ∈ S, the relations
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2014.10.002 0024-3795/© 2014 Elsevier Inc. All rights reserved.
Y. Shitov / Linear Algebra and its Applications 466 (2015) 38–40
39
λ(μa) = (λμ)a, (λ + μ)a = λa + μa, λ(a + b) = λa + λb, 1a = a and 0a = λ0 = 0. The concept of a semimodule provides a further generalization of modules and vector spaces. Note that R turns itself to a left R-semimodule if we define the scalar multiplication by (r, s) → rs. A subset G of a left R-semimodule S is said to generate an element s ∈ S, if there are scalars λg ∈ R only finitely many of which are non-zero satisfying s = g∈G λg g. A basis of S is an inclusion minimal subset which generates every element from S. A standard result of linear algebra states that, when R is a field, every basis of the free semimodule Rn has cardinality n. The following generalization of this result was suggested by Tan [2]. Conjecture 1. (See [2, Remark 4.4].) Assume that q < ∞ is the largest cardinality of a basis of a commutative semiring R (considered as a semimodule over itself). Then, the possible cardinalities of bases of Rn are n, n + 1, . . . , qn. This problem has been studied in a number of recent publications, see [1–3] and references therein. In particular, Tan proves this conjecture for zero-sum-free semirings in [2], and Shu and Wang [3] prove it in the special case when q = 1. Our note aims to prove this conjecture in full generality. In the rest of our note, we think of Rn as a left R-semimodule, that is, we define λ(r1 , . . . , rn ) as (λr1 , . . . , λrn ). By πj we denote a natural injection from R to Rn sending an element s to a tuple in which the jth coordinate is s and all other are zero. Now fix an arbitrary tuple σ ∈ Rn ; we denote by ρj (σ) the jth coordinate of σ. By invi (σ) we will denote the set of all λσ ∈ Rn such that, for any j = i, the element ρj (λσ) is invertible in R. Lemma 2. Let Rn be generated by a set G as a left R-semimodule. Then, πi (1) is generated by the set Γ = g∈G invi (g). Proof. There are scalars λg satisfying πi (1) = g∈G λg g; acting with ρj on this equation, we get 0 = g∈G ρj (λg g), for j = i. This shows that ρj (λg g) is invertible in R, so that λg g ∈ invi (g). 2 Our main result is the following. Theorem 3. Let G be a set which generates Rn as a left R-semimodule. If the largest cardinality of a basis of R1 is q < ∞, then there is a set H ⊂ G which generates Rn and satisfies |H| nq. Proof. For n = 1, the result is easy. We assume n > 1 and proceed by induction. From Lemma 2 it follows that, for every i, the set Γi = g∈G invi (g) generates πi (1). Since the elements of Γi have the form νg for some ν ∈ R and g ∈ G, we can assume without loss of generality that Γi ⊂ G. By inductive assumption, there is a set H ⊂ Γ2 ∪ · · · ∪ Γn which satisfies |H | (n − 1)q and generates the tuples
40
Y. Shitov / Linear Algebra and its Applications 466 (2015) 38–40
α2 = (a2 , 1, 0, . . . , 0), . . . , αn = (an , 0, . . . , 0, 1). The definition of Γi shows that a2 , . . . , an are invertible in R. Since G is a generating set for Rn , we have π1 (1) = g∈G λg g, for some λg ∈ R. Since g∈G ρt (λg g) = 0, for t ∈ {2, . . . , n}, the elements ρt (λg g) are invertible in R. In other words, there are μtg ∈ R satisfying ρt (λg g) + μtg = 0, for any t and g. Then, we have π1 (1) =
g∈G
λg g +
n
0 = ρt
μτ g ατ
+
τ =2
The definition of μtg shows that λg g +
n τ =2 g∈G
n
ρτ (λg g)ατ .
(1)
τ =2 g∈G
n τ =2
μτ g ατ ∈ π1 (R), for every g, so that
ρτ (λg g)ατ
=
ρt (λg g).
(2)
g∈G
From (1) and (2) it follows that π1 (1) is generated by the set G of all elements λg g + n τ =2 μτ g ατ , each of which belongs to π1 (R). By induction, there is a set H ⊂ G that generates π1 (R) and satisfies |H| q. Since a2 , . . . , an are invertible, the set H ∪ H generates Rn . It remains to note that every element of H is generated by H and a single element from G. 2 Conjecture 1 follows from Theorem 3, as results from [2] show. In fact, the construction of bases for Rn of cardinalities between n and qn is explicit in Theorem 4.1 of [2]. Further, no basis of Rn can have cardinality greater than qn, by Theorem 3. Finally, if R is commutative, any generating set of Rn contains at least n elements, by Theorem 3.3 from [2]. Acknowledgements I would like to thank the referees for their suggestions, which lead to a significant improvement in presentation of the result. References [1] A.M. Kanan, Z.Z. Petrovic, Note on cardinality of bases in semilinear spaces over zerosumfree semirings, Linear Algebra Appl. 439 (2013) 2795–2799. [2] Yi-Jia Tan, Bases in semimodules over commutative semirings, Linear Algebra Appl. 443 (2014) 139–152. [3] Qian-yu Shu, Xue-ping Wang, The cardinality of bases in semilinear spaces over commutative semirings, Linear Algebra Appl. 459 (2014) 83–100.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Bases of a free semimodule are small Yaroslav Shitov National Research University Higher School of Economics, 20 Myasnitskaya Ulitsa, Moscow 101000, Russia
a r t i c l e
i n f o
Article history: Received 9 May 2014 Accepted 5 October 2014 Available online xxxx Submitted by R. Brualdi
a b s t r a c t A generating set G of a left semimodule S over a semiring R is called a basis if no proper subset of G generates S. We prove that Rn has no basis of cardinality exceeding qn, where q is the largest cardinality of bases of R. © 2014 Elsevier Inc. All rights reserved.
MSC: 16Y60 13C15 Keywords: Semiring theory Linear algebra
A set R equipped with two binary operations + and · is called a semiring if the following conditions are satisfied: (i) (R, +) is a commutative monoid, (ii) (R, ·) is a monoid, (iii) multiplication distributes over addition from both sides, and (iv) the additive identity 0 satisfies 0x = x0 = 0, for any x ∈ R. We also assume that the multiplicative identity 1 ∈ R is different from 0. An element a ∈ R is called additively invertible (or simply invertible) if there is b ∈ R such that a + b = 0. A commutative monoid (S, +, 0) turns into a left R-semimodule if one defines a scalar multiplication R × S → S satisfying, for all λ, μ ∈ R and a, b ∈ S, the relations
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2014.10.002 0024-3795/© 2014 Elsevier Inc. All rights reserved.
Y. Shitov / Linear Algebra and its Applications 466 (2015) 38–40
39
λ(μa) = (λμ)a, (λ + μ)a = λa + μa, λ(a + b) = λa + λb, 1a = a and 0a = λ0 = 0. The concept of a semimodule provides a further generalization of modules and vector spaces. Note that R turns itself to a left R-semimodule if we define the scalar multiplication by (r, s) → rs. A subset G of a left R-semimodule S is said to generate an element s ∈ S, if there are scalars λg ∈ R only finitely many of which are non-zero satisfying s = g∈G λg g. A basis of S is an inclusion minimal subset which generates every element from S. A standard result of linear algebra states that, when R is a field, every basis of the free semimodule Rn has cardinality n. The following generalization of this result was suggested by Tan [2]. Conjecture 1. (See [2, Remark 4.4].) Assume that q < ∞ is the largest cardinality of a basis of a commutative semiring R (considered as a semimodule over itself). Then, the possible cardinalities of bases of Rn are n, n + 1, . . . , qn. This problem has been studied in a number of recent publications, see [1–3] and references therein. In particular, Tan proves this conjecture for zero-sum-free semirings in [2], and Shu and Wang [3] prove it in the special case when q = 1. Our note aims to prove this conjecture in full generality. In the rest of our note, we think of Rn as a left R-semimodule, that is, we define λ(r1 , . . . , rn ) as (λr1 , . . . , λrn ). By πj we denote a natural injection from R to Rn sending an element s to a tuple in which the jth coordinate is s and all other are zero. Now fix an arbitrary tuple σ ∈ Rn ; we denote by ρj (σ) the jth coordinate of σ. By invi (σ) we will denote the set of all λσ ∈ Rn such that, for any j = i, the element ρj (λσ) is invertible in R. Lemma 2. Let Rn be generated by a set G as a left R-semimodule. Then, πi (1) is generated by the set Γ = g∈G invi (g). Proof. There are scalars λg satisfying πi (1) = g∈G λg g; acting with ρj on this equation, we get 0 = g∈G ρj (λg g), for j = i. This shows that ρj (λg g) is invertible in R, so that λg g ∈ invi (g). 2 Our main result is the following. Theorem 3. Let G be a set which generates Rn as a left R-semimodule. If the largest cardinality of a basis of R1 is q < ∞, then there is a set H ⊂ G which generates Rn and satisfies |H| nq. Proof. For n = 1, the result is easy. We assume n > 1 and proceed by induction. From Lemma 2 it follows that, for every i, the set Γi = g∈G invi (g) generates πi (1). Since the elements of Γi have the form νg for some ν ∈ R and g ∈ G, we can assume without loss of generality that Γi ⊂ G. By inductive assumption, there is a set H ⊂ Γ2 ∪ · · · ∪ Γn which satisfies |H | (n − 1)q and generates the tuples
40
Y. Shitov / Linear Algebra and its Applications 466 (2015) 38–40
α2 = (a2 , 1, 0, . . . , 0), . . . , αn = (an , 0, . . . , 0, 1). The definition of Γi shows that a2 , . . . , an are invertible in R. Since G is a generating set for Rn , we have π1 (1) = g∈G λg g, for some λg ∈ R. Since g∈G ρt (λg g) = 0, for t ∈ {2, . . . , n}, the elements ρt (λg g) are invertible in R. In other words, there are μtg ∈ R satisfying ρt (λg g) + μtg = 0, for any t and g. Then, we have π1 (1) =
g∈G
λg g +
n
0 = ρt
μτ g ατ
+
τ =2
The definition of μtg shows that λg g +
n τ =2 g∈G
n
ρτ (λg g)ατ .
(1)
τ =2 g∈G
n τ =2
μτ g ατ ∈ π1 (R), for every g, so that
ρτ (λg g)ατ
=
ρt (λg g).
(2)
g∈G
From (1) and (2) it follows that π1 (1) is generated by the set G of all elements λg g + n τ =2 μτ g ατ , each of which belongs to π1 (R). By induction, there is a set H ⊂ G that generates π1 (R) and satisfies |H| q. Since a2 , . . . , an are invertible, the set H ∪ H generates Rn . It remains to note that every element of H is generated by H and a single element from G. 2 Conjecture 1 follows from Theorem 3, as results from [2] show. In fact, the construction of bases for Rn of cardinalities between n and qn is explicit in Theorem 4.1 of [2]. Further, no basis of Rn can have cardinality greater than qn, by Theorem 3. Finally, if R is commutative, any generating set of Rn contains at least n elements, by Theorem 3.3 from [2]. Acknowledgements I would like to thank the referees for their suggestions, which lead to a significant improvement in presentation of the result. References [1] A.M. Kanan, Z.Z. Petrovic, Note on cardinality of bases in semilinear spaces over zerosumfree semirings, Linear Algebra Appl. 439 (2013) 2795–2799. [2] Yi-Jia Tan, Bases in semimodules over commutative semirings, Linear Algebra Appl. 443 (2014) 139–152. [3] Qian-yu Shu, Xue-ping Wang, The cardinality of bases in semilinear spaces over commutative semirings, Linear Algebra Appl. 459 (2014) 83–100.