Tensor Products of Ternary Semimodules over Ternary Semifields

Abstract

We introduce tensor products of ternary semimodules over ternary semifields and the n-fold tensor products for general case. We prove the universal mapping property of the n-fold tensor products. Moreover, we later provide a condition for preserving the flatness of ternary semimodules with tensor products of ternary semimodule homomorphisms.

1. Introduction

The subject of ternary algebra was started by D. H. Lehmer [1]. He studied some concepts of ternary systems called triplexes, which obtained the generalization of abelian groups. Afterward, Los [2] mentioned this algebraic structure studied by Banach and showed an example of a ternary semigroup, which is not a semigroup. In 1971, W. G. Lister [3] studied the abstract structure of a ternary ring, which is a ternary product of abelian groups. In 2003, R. Intarawong [4] studied and provided the universal mapping property of tensor products of modules over semifields. T. K. Dutta and S. Kar [5] studied ternary semiring and gave some properties of ternary semifields. In addition, H. J. M. Al-Thani [6] studied the flat semimodules, which is constructed by the exact sequences of semimodules and tensor products of semimodule homomorphisms.

The universal mapping property in the branch of modules over rings gives that there exists a unique group homomorphism from $M{\otimes }_{R}N$ to a module $A$_, which is composed with a bilinear map $M\times N\to M{\otimes }_{R}N$ is a bilinear map $M\times N\to A$. The structure of the ternary semimodules also leads us to derive a similar result.

The goal of this research is to investigate some properties of ternary semimodules over ternary semifields. We also study the universal mapping property of tensor products of ternary semimodules and investigate some types of ternary semimodules defined by tensor products of homomorphisms on ternary semimodules and exact sequences.

2. Preliminaries

The following familiar definitions and theorems from Refs. [7, 8, 9] regarding the notion of free abelian groups are needed to define tensor products of ternary semimodules over ternary semifields.

A ternary semifield $K$ is a system $\left(K,+,\cdot \right)$ with a binary operation ($+$) and a ternary operation $\left(\cdot \right)$ satisfying the following conditions for all $u,v,w,s,t\in K$.

1. $\left(K,+\right)$ is a commutative semigroup with identity $0$,

2. $\left(u\cdot v\cdot w\right)\cdot s\cdot t=u\cdot \left(v\cdot w\cdot s\right)\cdot t=u\cdot v\cdot \left(w\cdot s\cdot t\right)$,

3. $\left(u+v\right)\cdot w\cdot s=u\cdot w\cdot s+v\cdot w\cdot s$,

4. $u\cdot \left(v+w\right)\cdot s=u\cdot v\cdot s+u\cdot w\cdot s$, and.

5. $u\cdot v\cdot \left(w+s\right)=u\cdot v\cdot w+a\cdot v\cdot s$.

6. $u\cdot v\cdot w=v\cdot u\cdot w=w\cdot v\cdot u=u\cdot w\cdot v$,

7. $\exists 0\in K\forall u,v\in K,0+u=u$ and $0\cdot u\cdot v=u\cdot 0\cdot v=u\cdot v\cdot 0=0$, and.

8. $\forall u\in K\backslash \left\{0\right\}\exists v\in K\forall t\in K,u\cdot v\cdot t=v\cdot u\cdot t=t\cdot u\cdot v=t\cdot v\cdot u=t$.

(An element $v$ is called an inverse of $u$. In addition, the inverse of $u$ is unique and $u^{-1}$ denotes the inverse of $u$.)

For convenience, let $\mathit{uvw}$ denote $u\cdot v\cdot w$ for all $u,v,w\in K$.

Let $\left(R,+,\cdot \right)$ be a commutative ring. For a nonempty set $S$ of $R$, a subring $S$ of $R$ is said to be a positive cone if $S\cup \left(-S\right)=R$ and $S\cap \left(-S\right)=\left\{0\right\}$. $-S$ is called a negative cone of $S$.

It is clear that every negative cone of a cone $S$ of $R$ is a ternary semiring. For example, ${\mathbb{Z}}_0^{-}$ is a natural example of a ternary semiring. Moreover, ${\mathbb{Q}}_0^{-}$ and ${\mathbb{R}}_0^{-}$ become ternary semifields.

A group $\mathcal{F}$ is a free abelian group if $\mathcal{F}$ is an abelian group and for every nonzero element $g$ of $\mathcal{F}$, there exist unique nonzero integers ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ and unique distinct $x_1,x_2,\dots ,x_n$ in $X\subseteq \mathcal{F}$ such that $g={\alpha }_1{x}_1+{\alpha }_2{x}_2+\cdots +{\alpha }_n{x}_n$. We sometimes call $X$ a basis for $\mathcal{F}$.

For a nonempty set $X$, let

$$\mathcal{F}\mathcal{A}\left(X\right)=\left\{f:X\to \mathbb{Z}|\exists F\subseteq X\text{such that}|F|<\infty \text{and}f\left(x\right)=0\text{for}\mathrm{all}x\in X\backslash F\right\}.$$E1

Define $+$ on $\mathcal{F}\mathcal{A}\left(X\right)$ by for any $f,g\in \mathcal{F}\mathcal{A}\left(X\right)$,

$$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\text{for}\mathrm{all}x\in X.$$E2

Then, $\left(\mathcal{F}\mathcal{A}\left(X\right),+\right)$ is an abelian group. For any $x\in X$, define $f_x:X\to \mathbb{Z}$ by

$$f_x\left(y\right)=\left\{\begin{array}{ll}1,& \text{if}x=y,\\ 0,& \text{otherwise}.\end{array}\right.$$E3

Then, $f_x\in \mathcal{F}\mathcal{A}\left(X\right)$ for all $x\in X$. A group $\left(\mathcal{F}\mathcal{A}\left(X\right),+\right)$ is a free abelian group on $\left\{f_x|x\in X\right\}$ ($\subseteq \mathcal{F}\mathcal{A}\left(X\right)$). We sometimes say instead that $\mathcal{F}\mathcal{A}\left(X\right)$ is a free abelian group on $X$.

Let $K$ be a ternary semifield. A left K-ternary semimodule or left ternary semimodule over K is an additive abelian group $M$ together with a function from $K\times K\times M$ into $M$, defined by $\left(k_1,k_2,m\right)\mapsto k_1{k}_{2}m$ called ternary scalar multiplication, which satisfies the following conditions for all $m,m_1,m_2\in M$ and $k_1,k_2,k_3,k_4\in K$,

1. $k_1{k}_2\left(m_1+m_2\right)=k_1{k}_2{m}_1+k_1{k}_2{m}_2$,

2. $k_1\left(k_2+k_3\right)m=k_1{k}_{2}m+k_1{k}_{3}m$,

3. $\left(k_1+k_2\right)k_{3}m=k_1{k}_{3}m+k_2{k}_{3}m$,

4. $\left(k_1{k}_2{k}_3\right)k_{4}m=k_1\left(k_2{k}_3{k}_4\right)m=k_1{k}_2\left(k_3{k}_{4}m\right)$.

A right $K$-ternary semimodule or right-ternary semimodule over $K$ is defined similarly via a function $M\times K\times K$ into $M$ and satisfies the obvious analogs of (1)–(4).

For convenience, we simply write ${}_{K}M$ [$M_K$] as $M$ is a left [right] ternary semimodule over a ternary semifield $K$.

From now on, unless specified otherwise, “$K$-ternary semimodule ${}_{K}M$ [$M_K$]” means “left [right] $K$-ternary semimodule $M$ .” Moreover, “$K$-ternary semimodule” means “left $K$-ternary semimodule.”

Example 1.1 Let $F\left[0,1\right]=\left\{f|f:\left[0,1\right]\to {\mathbb{Q}}_0^{-}\right\}$ with the operations $+$ and $\cdot$, defined by.

$$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\text{and}\left(\alpha \cdot \beta \cdot f\right)\left(x\right)=\alpha \cdot \beta \cdot f\left(x\right)\text{for}\mathrm{all}x\in \left[0,1\right],$$E4

where $f,g\in F\left[0,1\right]$ and $\alpha ,\beta \in {\mathbb{Q}}_0^{-}$. Then, $F\left[0,1\right]$ is a left ${\mathbb{Q}}_0^{-}$-ternary semimodule.

Example 1.2 If $n\in \mathrm{\mathbb{N}}$ and $M_1,M_2,\dots ,M_n$ are ternary semimodules over a ternary semifield $K$; then, $M_1\times M_2\times \cdots \times M_n$ is a ternary semimodules over $K$ under usual componentwise addition and scalar multiplication.

Let $M$ be a left [right] ternary semimodule over a ternary semifield $K$. A left [right] ternary subsemimodule of $M$ is a subset of $M$, which is itself a left [right] ternary semimodule over $K$ with the addition and ternary scalar multiplication of $M$.

Let $K$ and $S$ be ternary semifields. An additive abelian group $M$ is a $\left(K,S\right)$-ternary bisemimodule if $M$ is a left $K$-ternary semimodule and also a right $S$-ternary semimodule, and $k_1{k}_2\left({\mathit{ms}}_1{s}_2\right)=\left(k_1{k}_{2}m\right)s_1{s}_2$ for all $k_1,k_2\in K,m\in M,$ and $s_1,s_2\in S$. We write ${}_K{M}_S$ for a $\left(K,S\right)$-ternary bisemimodule $M$.

3. Universal mapping properties

For given ternary semimodules $M$ and $N$, over a ternary semifield $K$, it is known from Example 1.2 that $M\times N$ is a ternary semimodule over $K$. To find another ternary semimodule over $K$, arising from $M$ and $N$, which is different from $M\times N$, the tensor product of $M$ and $N$ is the case. The notion of free abelian groups plays a major role in constructing the tensor product of ternary semimodules over $K$.

Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$. For each $m\in M$ and $n\in N$, a function $f_{\left(m,n\right)}:M\times N\to \mathbb{Z}$ is defined by

$$f_{\left(m,n\right)}\left(x,y\right)=\left\{\begin{array}{ll}1,& \text{if}\left(x,y\right)=\left(m,n\right),\\ 0,& \text{otherwise}\end{array}\right.$$E5

for all $x\in M_K$ and $y{\in }_{K}N$. Then, $f_{\left(m,n\right)}\in \mathcal{F}\mathcal{A}\left(M\times N\right)$ and $\mathcal{F}\mathcal{A}\left(M\times N\right)$ is a free abelian group on a basis $\left\{f_{\left(m,n\right)}|m\in M,n\in N\right\}$.

Definition 1.3 Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$, and let $\mathcal{F}$ be the free abelian group on $M\times N$, that is, $\mathcal{F}=\mathcal{F}\mathcal{A}\left(M\times N\right)$. Let $L$ be the subgroup of $\mathcal{F}$ generated by elements of the following forms:

1. $f_{\left(m+\tilde{m},n\right)}-f_{\left(m,n\right)}-f_{\left(\tilde{m},n\right)}$,

2. $f_{\left(m,n+\tilde{n}\right)}-f_{\left(m,n\right)}-f_{\left(m,\tilde{n}\right)}$,

3. $f_{\left(\mathit{m\alpha \beta },n\right)}-f_{\left(m,\mathit{\alpha \beta n}\right)}$

where $\alpha ,\beta \in K$, $m,\tilde{m}\in M$, and $n,\tilde{n}\in N$. We call $\mathcal{F}/L$ the tensor product of $M$ and $N$, and denoted by $M{\otimes }_{K}N$.

Note 1.4 If $M_K$ and ${}_{K}N$ are ternary semimodules over a ternary semifield $K$_. Then,

1. $\left(M{\otimes }_{K}N,+\right)$ is an abelian group.

2. $f_{\left(m,n\right)}\in \mathcal{F}$ for all $m\in M_K$ and $n{\in }_{K}N$.

Recall that $\mathcal{F}=\mathcal{F}\mathcal{A}\left(M\times N\right)$ and $\mathcal{F}/L=\left\{f+L|f\in \mathcal{F}\right\}$. For any $m\in M$ and $n\in N$, we write $m\otimes n$ for $f_{\left(m,n\right)}+L$. Moreover, for all $f+L\in \mathcal{F}/L$,

$$f+L=\sum _{\left(m,n\right)\in F\subseteq M\times N}{\alpha }_{\left(m,n\right)}{\beta }_{\left(m,n\right)}\left(m\otimes n\right),$$E6

where ${\alpha }_{\left(m,n\right)},{\beta }_{\left(m,n\right)}\in \mathbb{Z}$ and $\mid F\mid <\infty$, and we simply write $f+L$ as

$$\sum _{i=1}^p{\alpha }_{\left(m,n\right)}{\beta }_{\left(m,n\right)}\left(m_i\otimes n_i\right)\text{or}\sum _{i=1}^p{\alpha }_i{\beta }_i\left(m_i\otimes n_i\right)$$E7

where ${\alpha }_i={\alpha }_{\left(m_i,n_i\right)},{\beta }_i={\beta }_{\left(m_i,n_i\right)}\in \mathbb{Z}$, $m_i\in M$, and $n_i\in N$.

The following property can be derived directly from Definition 1.3.

Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$. Then,

1. $\left(m+m^{\prime }\right)\otimes n=\left(m\otimes n\right)+\left(m^{\prime }\otimes n\right)$,

2. $m\otimes \left(n+n^{\prime }\right)=\left(m\otimes n\right)+\left(m\otimes n^{\prime }\right)$,

3. $\mathit{m\alpha \beta }\otimes n=m\otimes \mathit{\alpha \beta n}$,

4. $m\otimes 0=0\otimes n=0\otimes 0=0$,

for all $\alpha ,\beta \in K$, $m,m^{\prime }\in M$ and $n,n^{\prime }\in N$.

Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$, and $A$ is an additive abelian group. A middle linear map (over $K$) from $M\times N$ to $A$ is a function $\tau :M\times N\to A$ such that for all $m,\tilde{m}\in M$, $n,\tilde{n}\in N$, and $\alpha ,\beta \in K$,

1. $\tau \left(m+\tilde{m},n\right)=\tau \left(m,n\right)+\tau \left(\tilde{m},n\right)$,

2. $\tau \left(m,n+\tilde{n}\right)=\tau \left(m,n\right)+\tau \left(m,\tilde{n}\right)$, and

3. $\tau \left(\mathit{m\alpha \beta },n\right)=\tau \left(m,\mathit{\alpha \beta n}\right)$.

Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$. The map $\mu :M\times N\to M{\otimes }_{K}N$ defined by $\mu \left(m,n\right)=m\otimes n$ is called the canonical middle linear map. The function $\pi :\mathcal{F}\mathcal{A}\left(M\times N\right)\to \mathcal{F}\mathcal{A}\left(M\times N\right)/L$ defined by $\pi \left(x\right)=x+L$ for all $x\in \mathcal{F}\mathcal{A}\left(M\times N\right)$ is an epimorphism of groups, which is called the canonical projection.

Lemma 1.5 ([9], p. 43): Let $A$ and $B$ be additive abelian groups. If $f:A\to B$ is a group homomorphism, and $C$ is a subgroup of $kerf$; then, there is a unique group homomorphism $\widehat{f}:A/C\to B$ such that $\widehat{f}\left(a+C\right)=f\left(a\right)$ for all $a\in A$, $\mathrm{im}\widehat{f}=\mathrm{im}f$ and $ker\widehat{f}=kerf/C$. Moreover, $\widehat{f}$ is a group isomorphism if and only if $f$ is a group epimorphism, and $C=kerf$. In particular, $A/kerf\cong \mathrm{im}f$.

By making use of Lemma 1.5, we get the following theorem.

Theorem 1.6. Let $M_K$ and ${}_{K}N$ be ternary semimodules over a ternary semifield $K$ and $\mu :M\times N\to M{\otimes }_{K}N$ be the canonical middle linear map. For any additive abelian group $U$ over $K$, and any middle linear map $\psi :M\times N\to U$, there exists a unique group homomorphism $\tilde{\psi }:M{\otimes }_{K}N\to U$ such that $\psi =\tilde{\psi }\circ \mu$, that is, the following diagram commutes.

Proof: Let $\left(U,+\right)$ be an abelian group. By Theorem 1.1 ([9], p. 71) there exists a unique group homomorphism $\widehat{\psi }:\mathcal{F}\mathcal{A}\left(M\times N\right)\to U$ such that $\psi =\widehat{\psi }\circ \phi$.

Let $L$ be the subgroup defined in Definition 1.3. Then, $L$ is a subgroup of $ker\widehat{\psi }$ because $\psi$ is a middle linear map and $\psi =\widehat{\psi }\circ \phi$.

Let $\pi :\mathcal{F}\mathcal{A}\left(M\times N\right)\to \mathcal{F}\mathcal{A}\left(M\times N\right)/L$ be the canonical projection. Since $L$ is a subgroup of $ker\widehat{\psi }$, by Lemma 1.5, there exists a unique group homomorphism $\widehat{\psi }:\mathcal{F}\mathcal{A}\left(M\times N\right)/L\to U$ such that $\widehat{\psi }=\tilde{\psi }\circ \pi$. Now, we obtain a group homomorphism $\tilde{\psi }:M{\otimes }_{K}N\to U$. Next, we will consider the diagram.

For each $m\in M$ and $n\in N$, the canonical middle linear map $\mu :M\times N\to M{\otimes }_{K}N$ satisfies $\mu \left(m,n\right)=m\otimes n=\phi \left(m,n\right)+L=\pi \left(\phi \left(m,n\right)\right)=\pi \circ \phi \left(m,n\right).$ That is, $\mu =\pi \circ \phi$. Hence, $\tilde{\psi }\circ \mu =\tilde{\psi }\circ \left(\pi \circ \phi \right)=\left(\tilde{\psi }\circ \pi \right)\circ \phi =\widehat{\psi }\circ \phi =\psi$. Last, let $\rho :M{\otimes }_{K}N\to U$ be a group homomorphism such that $\psi =\rho \circ \mu$, and let $\theta =\rho \circ \pi$. Consider the following diagram.

We have that $\theta \circ \phi =\left(\rho \circ \pi \right)\circ \phi =\rho \circ \left(\pi \circ \phi \right)=\rho \circ \mu =\psi$. So, $\theta =\widehat{\psi }$ because of the uniqueness of $\widehat{\psi }$. Moreover, $\rho \circ \pi =\theta =\widehat{\psi }=\tilde{\psi }\circ \pi$. By the uniqueness of $\tilde{\psi }$, we obtain that $\rho =\tilde{\psi }$.

Example 1.7

1. Let $u=\left[a_1,a_2,\dots ,a_m\right]\in {\mathbb{R}}^m$ and $v=\left[b_1,b_2,\dots ,b_n\right]\in {\mathbb{R}}^n$. Define a function $\psi :{\mathbb{R}}^m\times {\mathbb{R}}^n\to {\mathcal{M}}_{m\times n}\left(\mathbb{R}\right)$ given by

   $$\psi \left(u,v\right)=u^{t}v=\left[\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_m\end{array}\right]\left[b_1{b}_2\cdots b_n\right]=\left[\begin{array}{cccc}{a}_1{b}_1& a_1{b}_2& \cdots & a_1{b}_n\\ a_2{b}_1& a_2{b}_2& \cdots & a_2{b}_n\\ ⋮& ⋮& \ddots & ⋮\\ a_m{b}_1& a_m{b}_2& \cdots & a_m{b}_n\end{array}\right].$$E8

   We can see that $\psi$ is a middle linear map. By Theorem 1.6, there exists a unique group homomorphism $\tilde{\psi }:{\mathbb{R}}^m{\otimes }_{\mathbb{R}}{\mathbb{R}}^n\to {\mathcal{M}}_{m\times n}\left(\mathbb{R}\right)$ such that $\psi \left(u,v\right)=\tilde{\psi }\left(u\otimes v\right)=u^{t}v$.

2. Let $A={\left[a_{\mathit{ij}}\right]}_{n\times n}\in {\mathcal{M}}_n\left(\mathbb{R}\right)$ and $B={\left[b_{\mathit{ij}}\right]}_{n\times n}\in {\mathcal{M}}_n\left(\mathbb{R}\right)$. Define a function $\psi :{\mathcal{M}}_n\left(\mathbb{R}\right)\times {\mathcal{M}}_n\left(\mathbb{R}\right)\to {\mathcal{M}}_{n^2}\left(\mathbb{R}\right)$ given by

$$\psi \left(A,B\right)=\left[a_{\mathit{ij}}B\right]=\left[\begin{array}{cccc}{a}_{11}B& a_{12}B& \cdots & a_{1n}B\\ a_{21}B& a_{22}B& \cdots & a_{2n}B\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}B& a_{n2}B& \cdots & a_{\mathit{nn}}B\end{array}\right].$$E9

We can see that $\psi$ is a middle linear map. By Theorem 1.6, there exists a unique group homomorphism $\tilde{\psi }:{\mathcal{M}}_n\left(\mathbb{R}\right){\otimes }_{\mathbb{R}}{\mathcal{M}}_n\left(\mathbb{R}\right)\to {\mathcal{M}}_{n^2}\left(\mathbb{R}\right)$ such that $\psi \left(A,B\right)=\tilde{\psi }\left(A\otimes B\right)=\left[a_{\mathit{ij}}B\right]$.

Let $M_K$, ${}_{K}N$, and ${}_{K}U$ be ternary semimodules over a ternary semifield $K$. A bilinear map over $K$ from $M\times N$ to $U$ is a function $T:M\times N\to U$ such that for all $m,\tilde{m}\in M$, $n,\tilde{n}\in N$ and $\alpha ,\beta \in K$,

1. $T\left(m+\tilde{m},n\right)=T\left(m,n\right)+T\left(\tilde{m},n\right)$,

2. $T\left(m,n+\tilde{n}\right)=T\left(m,n\right)+T\left(m,\tilde{n}\right)$, and

3. $T\left(\mathit{m\alpha \beta },n\right)=\mathit{\alpha \beta T}\left(m,n\right)=T\left(m,\mathit{\alpha \beta n}\right)$.

The map $\mu :M\times N\to M{\otimes }_{K}N$ is given by $\mu \left(m,n\right)=m\otimes n$ and is called the canonical bilinear map (over a ternary semifield $K$).

By Theorem 1.6, the following result is derived.

Corollary 1.8 Let ${}_K{M}_K$, ${}_{K}N$, and ${}_{K}U$ be ternary semimodules over a ternary semifield $K$, and $\mu :M\times N\to M{\otimes }_{K}N$ the canonical bilinear map. For any bilinear map $\psi :M\times N\to U$, there exists a unique $K$-ternary semimodule homomorphism $\tilde{\psi }:M{\otimes }_{K}N\to U$ such that $\psi =\tilde{\psi }\circ \mu$, that is, the following diagram commutes.

Example 1.9 We will apply Corollary 1.8 to show that ${\mathbb{Q}}_0^{-}{\otimes }_{{\mathbb{Q}}_0^{-}}F\left[0,1\right]\cong F\left[0,1\right]$.

Let $\mu :{\mathbb{Q}}_0^{-}\times F\left[0,1\right]\to {\mathbb{Q}}_0^{-}{\otimes }_{{\mathbb{Q}}_0^{-}}F\left[0,1\right]$ be a canonical bilinear map. That is, $\mu \left(a,f\right)=a\otimes f$, for all $a\in {\mathbb{Q}}_0^{-}$ and $f\in F\left[0,1\right]$.

Define $\psi :{\mathbb{Q}}_0^{-}\times F\left[0,1\right]\to F\left[0,1\right]$ by $\psi \left(a,f\right)=\left(-1\right)\mathit{af}$, where $a\in {\mathbb{Q}}_0^{-}$ and $f\in F\left[0,1\right]$. We can show that $\psi$ is a bilinear map.

By the universal mapping property, there exists a unique ternary homomorphism $\tilde{\psi }:{\mathbb{Q}}_0^{-}{\otimes }_{{\mathbb{Q}}_0^{-}}F\left[0,1\right]\to F\left[0,1\right]$ such that $\psi =\tilde{\psi }\circ \mu$. That is, $\tilde{\psi }\left(a\otimes f\right)=\tilde{\psi }\left(\mu \left(a,f\right)\right)=\psi \left(a,f\right)=\left(-1\right)\mathit{af}$ for all $a\in {\mathbb{Q}}_0^{-}$ and $f\in F\left[0,1\right]$.

We will show that $\tilde{\psi }$ is an isomorphism. Let $f\in F\left[0,1\right]$. We have $-1\otimes f\in {\mathbb{Q}}_0^{-}{\otimes }_{{\mathbb{Q}}_0^{-}}F\left[0,1\right]$ and $\tilde{\psi }\left(-1\otimes f\right)=\left(-1\right)\left(-1\right)f=f$. So, $\tilde{\psi }$ is surjective.

Next, we will show that $ker\tilde{\psi }=\left\{0\right\}$. Let $a\otimes f\in {\mathbb{Q}}_0^{-}{\otimes }_{{\mathbb{Q}}_0^{-}}F\left[0,1\right]$. Consider $a\otimes f=\left(-1\right)\left(-1\right)a\otimes f=-1\otimes \left(-1\right)\mathit{af}$. Since $F\left[0,1\right]$ is a left ${\mathbb{Q}}_0^{-}$-ternary semimodule, $a\otimes f=-1\otimes g$ for some $g\in F\left[0,1\right]$.

Let $-1\otimes g\in ker\tilde{\psi }$. Then, $\tilde{\psi }\left(-1\otimes g\right)=0$. Thus, $g=\left(-1\right)\left(-1\right)g=0$, that is, $a\otimes f=-1\otimes 0=0$. Hence, $ker\tilde{\psi }=\left\{0\right\}$, so $\tilde{\psi }$ is injective. Consequently, $\tilde{\psi }$ is an isomorphism.

By Theorem 1.6, the following results about the tensor products of two ternary semimodule homomorphisms are derived.

Proposition 1.10 Let $M_K$, $M_K^{\prime }$, ${}_{K}N$, and ${}_K{N}^{\prime }$ be ternary semimodules over a ternary semifield $K$. If $f:M\to M^{\prime }$ and $g:N\to N^{\prime }$ are right and left $K$-ternary semimodule homomorphisms, respectively; then, there exists a unique group homomorphism $h$ from $M{\otimes }_{K}N$ into $M^{\prime }{\otimes }_K{N}^{\prime }$ such that for all $m\in M$ and $n\in N$, $h\left(m\otimes n\right)=f\left(m\right)\otimes g\left(n\right)$.

The unique group homomorphism $h$ in Proposition 1.10 is denoted by $f\otimes g:M{\otimes }_{K}N\to M^{\prime }{\otimes }_K{N}^{\prime }$.

Lemma 1.11 Let $K$ and $S$ be ternary semifields, and ${}_S{M}_K$ and ${}_{K}N$ be ternary semimodules as indicated. Then, $M{\otimes }_{K}N$ is a left ternary semimodule over $S$ defined by $s_1{s}_2\left(m\otimes n\right)=s_1{s}_{2}m\otimes n$ for all $s_1$, $s_2\in S$, $m\in M$, and $n\in N$.

Theorem 1.12 Let $K$ and $S$ be ternary semifields, and ${}_S{M}_K$, ${}_S{M}_K^{\prime }$, ${}_{K}N$, and ${}_K{N}^{\prime }$ be ternary semimodules as indicated. If $f:M\to M^{\prime }$ is a right $K$-ternary semimodule homomorphism, and $g:N\to N^{\prime }$ is a left $K$-ternary semimodule homomorphism; then, $f\otimes g$ is a left $S$-ternary semimodule homomorphism.

Proof: Let $f:M\to M^{\prime }$ be an $\left(S,K\right)$-ternary bisemimodule homomorphism and $g:N\to N^{\prime }$ be a left $K$-ternary semimodule homomorphism. By Lemma 1.11, we obtain that $M{\otimes }_{K}N$ and $M^{\prime }{\otimes }_K{N}^{\prime }$ are left $S$-ternary semimodules. Moreover, there exists a unique group homomorphism $f\otimes g:M{\otimes }_{K}N\to M^{\prime }{\otimes }_K{N}^{\prime }$ such that $\left(f\otimes g\right)\left(m\otimes n\right)=f\left(m\right)\otimes g\left(n\right)$ for all $m\in M$ and $n\in N$. By Lemma 1.11, we obtain that $s_1{s}_1\left(\sum _{i=1}^p{m}_i\otimes n_i\right)=\sum _{i=1}^p{s}_1{s}_2{m}_i\otimes n_i$ for all $s_1,s_2\in S$ and $\sum _{i=1}^p{m}_i\otimes n_i\in M{\otimes }_{K}N$ then

$$\left(f\otimes g\right)\left(s_1{s}_1\left(\sum _{i=1}^p{m}_i\otimes n_i\right)\right)=s_1{s}_2\left(\left(f\otimes g\right)\left(\sum _{i=1}^p{m}_i\otimes n_i\right)\right).$$E10

Therefore, $f\otimes g$ is a left $S$-ternary semimodule homomorphism.

Proposition 1.13 Let $K$ be a ternary semifield, and $M_K$, $M_K^{\prime }$, ${M^{\prime }}_K^{\prime }$, ${}_{K}N$, ${}_K{N}^{\prime }$ and ${}_K{N^{\prime }}^{\prime }$ be ternary semimodules. If $f:M\to M^{\prime }$ and $f^{\prime }:M^{\prime }\to {M^{\prime }}^{\prime }$ are right $K$-ternary semimodule homomorphisms, $g:N\to N^{\prime }$ and $g^{\prime }:N^{\prime }\to {N^{\prime }}^{\prime }$ are left $K$-ternary semimodule homomorphisms. Then,

$$\left(f^{\prime }\otimes g^{\prime }\right)\circ \left(f\otimes g\right)=\left(f^{\prime }\circ f\right)\otimes \left(g^{\prime }\circ g\right):M{\otimes }_{K}N\to {M^{\prime }}^{\prime }{\otimes }_K{N^{\prime }}^{\prime }$$E11

is a group homomorphism. If $f$ and $g$ are right and left $K$-ternary semimodule isomorphisms, respectively; then, $f\otimes g$ is a group isomorphism and ${\left(f\otimes g\right)}^{-1}=f^{-1}\otimes g^{-1}$ is also a group isomorphism.

Proof: We can see that $\left(\tilde{f}\otimes \tilde{g}\right)\circ \left(f\otimes g\right)$ and $\left(\tilde{f}\circ f\right)\otimes \left(\tilde{g}\circ g\right)$ are $K$-ternary semimodule homomorphisms. It suffices to show that equality holds. Let $m\in M$ and $n\in N$. Then,

$$\begin{array}{ll}\left(\tilde{f}\otimes \tilde{g}\right)\circ \left(f\otimes g\right)\left(m\otimes n\right)& =\left(\tilde{f}\otimes g\right)\left(f\left(m\right)\otimes g\left(n\right)\right)\\ & =\left(\tilde{f}\circ f\right)\left(m\right)\otimes \left(\tilde{g}\circ g\right)\left(n\right)\\ & =\left(\tilde{f}\circ f\right)\otimes \left(\tilde{g}\circ g\right)\left(m\otimes n\right)\end{array}$$E12

Hence, $\left(f^{\prime }\otimes g^{\prime }\right)\circ \left(f\otimes g\right)=\left(f^{\prime }\circ f\right)\otimes \left(g^{\prime }\circ g\right)$. For the last statement, it is straightforward.

4. The $n$-fold tensor products

Definition 1.14 Let $K$ be a ternary semifield, and let $M_1,M_2,\dots ,M_n$ and $N$ be appropriate $K$-ternary semimodules. An $n$-multilinear map or simply multilinear map (over $K$) from $M_1\times M_2\times \cdots \times M_n$ into $N$ is a function $f:M_1\times M_2\times \cdots \times M_n\to N$ such that for all $\alpha ,\beta \in K$, $m_i\in M_i$ and $m_j^{\prime }\in M_j$ for all $i,j=1,2,\dots n$,

1. $f\left(m_1,\dots ,m_{j-1},m_j+m_j^{\prime },m_{j+1},\dots ,m_n\right)=f\left(m_1,\dots ,m_{j-1},m_j,m_{j+1},\dots ,m_n\right)+f\left(m_1,\dots ,m_{j-1},m_j^{\prime },m_{j+1},\dots ,m_n\right)$, and

2. $f\left(m_1,\dots ,m_{j-1},\mathit{\alpha \beta }{m}_j,m_{j+1},\dots ,m_n\right)=\mathit{\alpha \beta f}\left(m_1,\dots ,m_{j-1},m_j,m_{j+1},\dots ,m_n\right)$

We can generalize the tensor products of two ternary semimodules over ternary semifields to tensor products of ternary semimodules $M_1,M_2,\dots ,M_n$ over the same ternary semifield $K$, which we simply call the n-fold tensor product and write this as $M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n$ by dropping all parentheses.

For example, let $n\ge 2$ and $M_1,M_2,\dots ,M_n$ be ternary semimodules over a ternary semifield $K$. Then, the function

$${\mu }_n:M_1\times M_2\times \cdots \times M_n\to M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n$$

defined by $\left(m_1,m_2,\dots ,m_n\right)\mapsto m_1\otimes m_2\otimes \cdots \otimes m_n$ is a multilinear map. We call the function ${\mu }_n$ a canonical multilinear map.

Theorem 1.15: (The universal mapping property of the n-fold tensor product) Let $n\ge 2$, and $M_1,M_2,\dots ,M_n$ and $N$ be appropriate ternary semimodule over the same ternary semifield $K$. For any multilinear map, $\psi :M_1\times M_2\times \cdots \times M_n\to N$, there exists a unique $K$-ternary semimodule homomorphism $\tilde{\psi }:M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\to N$ such that $\psi =\tilde{\psi }\circ {\mu }_n$, i.e., the following diagram commutes.

Proof: Let $\psi :M_1\times M_2\times \cdots \times M_n\times M_{n+1}\to N$ be a multilinear map. For each $m\in M_{n+1}$, let ${\psi }_m:M_1\times M_2\times \cdots \times M_n\to N$ be defined by

$${\psi }_m\left(m_1,m_2,\dots ,m_n\right)=\psi \left(m_1,m_2,\dots ,m_n,m\right)\text{for}\mathrm{all}{m}_i\in M_i,i=\mathrm{1,2,3},\dots ,n.$$E13

We can see that ${\psi }_m$ is a multilinear map. By the induction hypothesis, there exists a unique $K$-ternary semimodule homomorphism ${\tilde{\psi }}_m:M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\to N$ such that ${\psi }_m={\tilde{\psi }}_m\circ {\mu }_n$.

Let $\mathcal{B}:\left(M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\right)\times M_{n+1}\to N$ be defined by $\mathcal{B}\left(x,m\right)={\tilde{\psi }}_m\left(x\right)$ for all $x\times M_1{\otimes }_K{M}_2{\otimes }_K\circ {\otimes }_K{M}_n$ and $m\in M_{n+1}$. Since ${\tilde{\psi }}_m$ is unique, $\mathcal{B}$ is well-defined. Next, we show that $\mathcal{B}$ is a bilinear map.

Let $\alpha ,\beta \in K$, $x,y\in M_1{\otimes }_K{M}_2{\otimes }_K\circ {\otimes }_K{M}_n$ and $s,t\in M_{n+1}$. Then,

$$\mathcal{B}\left(x+y,s\right)={\tilde{\psi }}_s\left(x+y\right)={\tilde{\psi }}_s\left(x\right)+{\tilde{\psi }}_s\left(y\right)=\mathcal{B}\left(x,s\right)+\mathcal{B}\left(y,s\right),$$E14

and

$$\mathcal{B}\left(\mathit{\alpha \beta x},s\right)={\tilde{\psi }}_s\left(\mathit{\alpha \beta x}\right)=\mathit{\alpha \beta }{\tilde{\psi }}_s\left(x\right)=\mathit{\alpha \beta }\mathcal{B}\left(x,s\right).$$E15

Moreover, we have to show that $\mathcal{B}\left(x,s+t\right)=\mathcal{B}\left(x,s\right)+\mathcal{B}\left(x,t\right)$ and $\mathcal{B}\left(x,\mathit{\alpha \beta s}\right)=\mathit{\alpha \beta }\mathcal{B}\left(x,s\right)$. That is, ${\tilde{\psi }}_{s+t}={\tilde{\psi }}_s+{\tilde{\psi }}_t$ and ${\tilde{\psi }}_{\mathit{\alpha \beta s}}=\mathit{\alpha \beta }{\tilde{\psi }}_s$, respectively. Because ${\tilde{\psi }}_{s+t}$ is a unique $K$-ternary semimodule homomorphism such that ${\psi }_{s+t}={\tilde{\psi }}_{s+t}\circ {\mu }_n$_; therefore, it suffices to show that $\left({\tilde{\psi }}_s+{\tilde{\psi }}_t\right)\circ {\mu }_n={\psi }_{s+t}$. Let $m_i\in M_i$ for all $i=1,2,\dots ,n$. Then,

$$\begin{array}{ll}\left({\tilde{\psi }}_s+{\tilde{\psi }}_t\right)\circ {\mu }_n\left(m_1,m_2,\dots ,m_n\right)& =\left({\tilde{\psi }}_s\circ {\mu }_n\right)\left(m_1,m_2,\dots ,m_n\right)+\left({\tilde{\psi }}_t\circ {\mu }_n\right)\left(m_1,m_2,\dots ,m_n\right)\\ & ={\psi }_s\left(m_1,m_2,\dots ,m_n\right)+{\psi }_t\left(m_1,m_2,\dots ,m_n\right)\\ & =\psi \left(m_1,m_2,\dots ,m_n,s\right)+\psi \left(m_1,m_2,\dots ,m_n,t\right)\\ & =\psi \left(m_1,m_2,\dots ,m_n,s+t\right)\\ & ={\psi }_{s+t}\left(m_1,m_2,\dots ,m_n\right).\end{array}$$E16

Similarly, ${\tilde{\psi }}_{\mathit{\alpha \beta s}}$ is a unique $K$-ternary semimodule homomorphism such that ${\psi }_{\mathit{\alpha \beta s}}={\tilde{\psi }}_{\mathit{\alpha \beta s}}\circ {\mu }_n$. So, we will show that $\mathit{\alpha \beta }{\tilde{\psi }}_s\circ {\mu }_n={\psi }_{\mathit{\alpha \beta s}}$. Let $m_i\in M_i$ for all $i=1,2,\dots ,n$. Then,

$$\begin{array}{ll}\left(\left(\mathit{\alpha \beta }{\tilde{\psi }}_s\right)\circ {\mu }_n\right)\left(m_1,m_2,\dots ,m_n\right)& =\mathit{\alpha \beta }{\tilde{\psi }}_s\circ {\mu }_n\left(m1,m2,\dots ,m_n\right)\\ & ={\mathit{\alpha \beta \psi }}_s\left(m_1,m_2,\dots ,m_n\right)\\ & =\mathit{\alpha \beta \psi }\left(m_1,m_2,\dots ,m_n,s\right)\\ & =\psi \left(m_1,m_2,\dots ,m_n,\mathit{\alpha \beta s}\right)\\ & ={\psi }_{\mathit{\alpha \beta s}}\left(m_1,m_2,\dots ,m_n\right).\end{array}$$E17

Hence, $\mathcal{B}$ is a bilinear map.

For each $m_i\in M_i$ for all $i=1,2,\dots ,n,n+1$, we consider the bilinear map $\mu :\left(M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\right)\times M_{n+1}\to \left(M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\right){\otimes }_K{M}_{n+1}$. We can see that

$$\begin{array}{ll}\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)& ={\mu }_n\left(m_1,m_2,\dots ,m_n\right)\otimes m_{n+1}\\ & =\left(m_1\otimes m_2\otimes \cdots \otimes m_n\right)\otimes m_{n+1}\\ & =m_1\otimes m_2\otimes \cdots \otimes m_n\otimes m_{n+1}\\ & ={\mu }_{n+1}\left(m_1,m_2,\dots ,m_n,m_{n+1}\right).\end{array}$$E18

Fix $i\in \left\{\mathrm{1,2,3},\dots ,n,n+1\right\}$ and let $\alpha ,\beta \in K$, $m_i,m_i^{\prime }\in M_i$ for all $i$. If $i=1,2,\dots ,n$. Then,

$$\begin{array}{ll}{\mu }_{n+1}\left(m_1,m_2,\dots ,m_i+m_i^{\prime },\dots ,m_n,m_{n+1}\right)& =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_i+m_i^{\prime },\dots ,m_n\right),m_{n+1}\right)\\ & =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_i,\dots ,m_n\right)+{\mu }_n\left(m_1,m_2,\dots ,m_i^{\prime },\dots ,m_n\right),m_{n+1}\right)\\ & =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_i,\dots ,m_n\right),m_{n+1}\right)+\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_i^{\prime },\dots ,m_n\right),m_{n+1}\right)\\ & ={\mu }_{n+1}\left(m_1,m_2,\dots ,m_i,\dots ,m_n,m_{n+1}\right)+{\mu }_{n+1}\left(m_1,m_2,\dots ,m_i^{\prime },\dots ,m_n,m_{n+1}\right),\end{array}$$E19

and

$$\begin{array}{ll}{\mu }_{n+1}\left(m_1,m_2,\dots ,\mathit{\alpha \beta }{m}_i,\dots ,m_n,m_{n+1}\right)& =\mu \left({\mu }_n\left(m_1,m_2,\dots ,\mathit{\alpha \beta }{m}_i,\dots ,m_n\right),m_{n+1}\right)\\ & =\mu \left({\mathit{\alpha \beta \mu }}_n\left(m_1,m_2,\dots ,m_i,\dots ,m_n\right),m_{n+1}\right)\\ & =\mathit{\alpha \beta \mu }\left({\mu }_n\left(m_1,m_2,\dots ,m_i,\dots ,m_n\right),m_{n+1}\right)\\ & ={\mathit{\alpha \beta \mu }}_{n+1}\left(m_1,m_2,\dots ,m_i,\dots ,m_n,m_{n+1}\right).\end{array}$$E20

If $i=n+1$. Then,

$$\begin{array}{ll}{\mu }_{n+1}\left(m_1,m_2,\dots ,m_n,m_{n+1}+m_{n+1}^{\prime }\right)& =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}+m_{n+1}^{\prime }\right)\\ & =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)+\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}^{\prime }\right)\\ & ={\mu }_n+1\left(m_1,m_2,\dots ,m_n,m_{n+1}\right)+{\mu }_n+1\left(m_1,m_2,\dots ,m_n,m_{n+1}^{\prime }\right),\end{array}$$E21

and

$$\begin{array}{ll}{\mu }_{n+1}\left(m_1,m_2,\dots ,m_n,\mathit{\alpha \beta }{m}_{n+1}\right)& =\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),\mathit{\alpha \beta }{m}_{n+1}\right)\\ & =\mathit{\alpha \beta \mu }\left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)\\ & ={\mathit{\alpha \beta \mu }}_{n+1}\left(m_1,m_2,\dots ,m_n,m_{n+1}\right).\end{array}$$E22

Hence, ${\mu }_{n+1}$ is an ($n+1$)-multilinear map.

Next, we consider the case $n+1$, where $\psi :M_1\times M_2\times \cdots \times M_{n+1}\to N$ is a multilinear map. Because $\mathcal{B}$ and $\mu$ are bilinear maps and by the induction hypothesis, so there exists a unique $K$-ternary module homomorphism $\tilde{\psi }:M1{\otimes }_{K}M2{\otimes }_K\cdots {\otimes }_K{M}_n{\otimes }_K{M}_{n+1}\to N$ such that $\mathcal{B}=\tilde{\psi }\circ \mu$.

Consider the following diagram.

where $\theta$ is a function from $\left(M_1\times M_2\times \cdots \times M_n\right)\times M_{n+1}$ into $\left(M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\right)\times M_{n+1}$ defined by $\theta \left(\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)=\left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)$. To show $\psi =\tilde{\psi }\circ {\mu }_{n+1}$, let $m_i\in M_i$ for all $i=\mathrm{1,2,3},\dots ,n+1$. Then,

$$\begin{array}{ll}\tilde{\psi }\circ {\mu }_{n+1}\left(m_1,m_2,\dots ,m_n,m_{n+1}\right)& =\tilde{\psi }\left(\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)\right)\\ & =\mathcal{B}\left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),m_{n+1}\right)\\ & ={\tilde{\psi }}_{m_{n+1}}\left({\mu }_n\left(m_1,m_2,\dots ,m_n\right)\right)\\ & ={\psi }_{m_{n+1}}\left(m_1,m_2,\dots ,m_n\right)\\ & =\psi \left(m_1,m_2,\dots ,m_n,m_{n+1}\right).\end{array}$$E23

Thus, $\psi =\tilde{\psi }\circ {\mu }_{n+1}$.

Finally, we show that $\tilde{\psi }$ is unique. Suppose that $\kappa :M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n{\otimes }_K{M}_{n+1}\to N$ is a $K$-ternary semimodule homomorphism such that $\psi =\kappa \circ {\mu }_{n+1}$. For each $s\in M_{n+1}$, let ${\kappa }_s:M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n\to N$ be defined by ${\kappa }_s\left(x\right)=\kappa \left(\mu \left(x,s\right)\right)$ for all $x\in M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n$. Then, ${\kappa }_s$ is a $K$-ternary semimodule homomorphism. Let $m_i\in M_i$ for all $i=\mathrm{1,2,3},\dots ,n$. Then,

$$\begin{array}{ll}\left({\kappa }_s\circ {\mu }_n\right)\left(m_1,m_2,\dots ,m_n\right)& =\kappa \left(\mu \left({\mu }_n\left(m_1,m_2,\dots ,m_n\right),s\right)\right)\\ & =\kappa \left({\mu }_{n+1}\left(m_1,m_2,\dots ,m_n,s\right)\right)\\ & =\psi \left(m_1,m_2,\dots ,m_n,s\right)\\ & ={\psi }_s\left(m_1,m_2,\dots ,m_n\right).\end{array}$$E24

Thus, ${\kappa }_s\circ {\mu }_n={\psi }_s$. By the uniqueness of ${\tilde{\psi }}_s$, we obtain that ${\kappa }_s={\tilde{\psi }}_s$. Now, we show that $\mathcal{B}=\kappa \circ \mu$. Let $x\in M_1{\otimes }_K{M}_2{\otimes }_K\cdots {\otimes }_K{M}_n$ and $s\in M_{n+1}$. Then, $\left(\kappa \circ \mu \right)\left(x,s\right)=\kappa \left(\mu \left(x,s\right)\right)={\kappa }_s\left(x\right)={\tilde{\psi }}_s\left(x\right)=\mathcal{B}\left(x,s\right)=\tilde{\psi }\circ \mu \left(x,s\right)$. By the uniqueness of $\tilde{\psi }$, we obtain that $\kappa =\tilde{\psi }$. Hence, $\tilde{\psi }$ is unique.

5. Exact sequences

For the last section, we investigate short exact sequences of ternary semimodules and flat ternary semimodules. By the way, we follow the definitions of short exact sequences and flat semimodules from Refs. [6, 10] but we change the semimodules to the ternary semimodules.

Theorem 1.16 Let $K$ be a ternary semifield, $M$ be a left $K$-ternary semimodule and $\left\{V_i|i\in I\right\}$ be a family of right $K$-ternary semimodules. If each $V_i$ for all $i\in I$ is $M$-flat; then, the direct sum $V_i$ is $M$-flat.

Proof: Let ${\mathrm{id}}_M$ be the identity function on a $K$-ternary semimodule $M$. Consider the following diagram

where ${\pi }_i:{\oplus }_{i\in I}\left(V_i{\otimes }_{K}M\right)\to V_i{\otimes }_{K}M$ given by ${\pi }_i\left(\left\{v_i\otimes m\right\}\right)=v_i\otimes m$ for all $i\in I$ and ${\iota }_i:V_i{\otimes }_{K}M\to {\oplus }_{i\in I}\left(V_i{\otimes }_{K}M\right)$ given by ${\iota }_i\left(v_i\otimes m\right)=\left\{v_j\otimes m\right\}$ where

$$v_j\otimes m=\left\{\begin{array}{ccc}{v}_i\otimes m& \text{if}& i=j,\\ 0& \text{if}& i\ne j\end{array}\right.$$E25

for all $i,j\in I$. There exists a group isomorphism $\psi :\left({\oplus }_{i\in I}{V}_i\right){\otimes }_{K}M\to {\oplus }_{i\in I}\left(V_i{\otimes }_{K}M\right)$ such that $\psi \left(\left\{v_i\right\}\otimes m\right)=\left\{v_i\otimes m\right\}$.

Assume that $V_i$ is $M$-flat for each $i\in I$. Then,

$$0=\left({\mathrm{id}}_{V_i}\otimes {\mathrm{id}}_N\right)\left(v_i\otimes n\right)={\mathrm{id}}_{V_i}\left(v_i\right)\otimes {\mathrm{id}}_N\left(n\right)=v_i\otimes m$$E26

for some $m\in M$. By diagram (*), we have that

$$0={\psi }^{-1}\left({\iota }_i\left(v_i\otimes m\right)\right)={\psi }^{-1}\left(\left\{v_i\otimes m\right\}\right)=\left\{v_i\right\}\otimes m.$$E27

So, $\left({\mathrm{id}}_{{\oplus }_{i\in I}{V}_i}\otimes {\mathrm{id}}_N\right)\left(\left\{v_i\right\}\otimes n\right)={\mathrm{id}}_{{\oplus }_{i\in I}{V}_i}\left(\left\{v_i\right\}\right)\otimes {\mathrm{id}}_N\left(n\right)=\left\{v_i\right\}\otimes m=0$.

That is, $ker\left({\mathrm{id}}_{{\oplus }_{i\in I}{V}_i}\otimes {\mathrm{id}}_N\right)=0$. Hence, ${\oplus }_{i\in I}{V}_i$ is $M$-flat.

Acknowledgments

We thank the referees for their suggestions and comments on the manuscript.