In this section, we introduce the concept of *f*-semiprime ideals in ordered semigroups and obtain that the *f*-prime radical of an ideal *I* is the least *f*-semiprime ideal containing *I*.

#### Definition 3.1

*Let* *S* *be an ordered semigroup*. *A subset* *A* *of* *S* *is said to be an* *fn*-*system of* *S* *if*
$\begin{array}{}{\displaystyle A=\bigcup _{i\in \mathrm{\Gamma}}{F}_{i}}\end{array}$
*where* {*F*_{i}|*i* ∈ *Γ*} ⊆ *fS*(*S*).

#### Definition 3.2

*An ideal* *P* *of an ordered semigroup* *S* *is said to be* *f*-*semiprime if its complement* *C*(*P*) *in* *S* *is an* *fn*-*system*.

Clearly, every *f*-system is an *fn*-system. Therefore, every *f*-prime ideal is an *f*-semiprime ideal.

#### Proposition 3.3

*Let* *S* *be an ordered semigroup and* *I* *a weakly semiprime ideal of* *S*. *Then* *I* *is an* *f*-*semiprime ideal of* *S*.

#### Proof

Since *I* is a weakly semiprime ideal, *C*(*I*) is a *n*-system. Thus *C*(*I*) is the union of some *m*-systems of *S*. Since every *m*-system is an *f*-system, *C*(*I*) is the union of some *f*-systems of *S*. Hence *C*(*I*) is an *fn*-system. Therefore, *I* is an *f*-semiprime ideal. □

#### Proposition 3.4

*Let* *P* *be an* *f*-*semiprime ideal of an ordered semigroup* *S* *and* *a* ∈ *S*. *If* *f*(*a*)*f*(*a*) ⊆ *P*, *then* *a* ∈ *P*.

#### Proof

Suppose that *a* ∈ *C*(*P*). Since *P* is an *f*-semiprime ideal, *C*(*P*) is an *fn*-system. Thus
$\begin{array}{}{\displaystyle C(P)=\bigcup _{i\in \mathrm{\Gamma}}{F}_{i},}\end{array}$
where {*F*_{i}| *i* ∈ *Γ*} ⊆ *fS*(*S*). Hence, *a* ∈ *F*_{i} for some *i* ∈ *Γ*. Therefore, *f*(*a*) ∩
$\begin{array}{}{\displaystyle {F}_{i}^{\ast}}\end{array}$
≠ ∅, where
$\begin{array}{}{\displaystyle {F}_{i}^{\ast}}\end{array}$
is the kernel of *F*_{i}. Let *x* ∈ *f*(*a*) ∩
$\begin{array}{}{\displaystyle {F}_{i}^{\ast}}\end{array}$
. Since
$\begin{array}{}{\displaystyle {F}_{i}^{\ast}}\end{array}$
is an *m*-system, (*xrx*] ∩
$\begin{array}{}{\displaystyle {F}_{i}^{\ast}}\end{array}$
≠ ∅ for some *r* ∈ *S*. Thus (*xrx*] ∩ *F*_{i} ≠ ∅ and so (*xrx*] ∩ *C*(*P*) ≠ ∅. Also *xrx* ∈ *f*(*a*)*f*(*a*) ⊆ *P*. Therefore, (*xrx*] ⊆ *P* which is a contradiction. Hence *a* ∈ *P*. □

#### Proposition 3.5

*Let* *S* *be an ordered semigroup and* *I* *be an ideal of* *S*. *Then* *r*_{f}(*I*) *is an* *f*-*semiprime ideal of* *S*.

#### Proof

By Theorem 2.12, we know that *r*_{f}(*I*) = ∩_{P∈Γ}*P* where *Γ* = {*P* ∈ *fPI*(*S*)| *I* ⊆ *P*}. Thus *C*(*r*_{f}(*I*)) = ⋃_{P∈Γ} *C*(*P*). Since every *C*(*P*) is an *f*-system, *C*(*r*_{f}(*I*)) is an *fn*-system. Therefore, *r*_{f}(*I*) is an *f*-semiprime ideal of *S*. □

#### Proposition 3.6

*Let* *S* *be an ordered semigroup and* *I* *be an* *f*-*semiprime ideal of* *S*. *Then* *r*_{f}(*I*) = *I*.

#### Proof

By Theorem 2.12, we have *I* ⊆ *r*_{f}(*I*). Let *x* ∉ *I*. Since *C*(*I*) is an *fn*-system,
$\begin{array}{}{\displaystyle C(I)=\bigcup _{i\in \mathrm{\Gamma}}{F}_{i}}\end{array}$
where {*F*_{i}| *i* ∈ *Γ*} ⊆ *fS*(*S*). Thus
$\begin{array}{}{\displaystyle x\in \bigcup _{i\in \mathrm{\Gamma}}{F}_{i}}\end{array}$
and so *x* ∈ *F*_{i} for some *i* ∈ *Γ*, but *F*_{i} ∩ *I* = ∅. Hence *F*_{i} is an *f*-system containing *x* but not containing any element of *I*. By Definition 2.11, *x* ∉ *r*_{f}(*I*). It follows that *r*_{f}(*I*) ⊆ *I*. Therefore *r*_{f}(*I*) = *I*. □

#### Definition 3.7

*Let* *S* *be an ordered semigroup and* *f* *a good mapping on* *S*. *Let* *a* ∈ *S* *and* *I* ∈ *I*(*S*). *The set* {*x* ∈ *S*
| *f*(*a*)*f*(*x*) ⊆ *I*}, *denoted by I* : *a*, *is called the left* *f*-*quotient of* *I* *by* *a*. *Moreover*, *for any ideal* *J* *of* *S*, *the left* *f*-*quotient of* *I* *by* *J* *is defined to be*
$\begin{array}{}{\displaystyle \bigcap _{a\in J}(I:a),}\end{array}$
*denoted by I* : *J*.

We note that *I* : *a* may be empty. See the following example.

#### Example 3.9

*We consider the ordered semigroup S of Example 2.3. If we define f*(*a*) = *f*(*b*) = *f*(*c*) = *S*, *then the mapping f is a good mapping*. *Let I* = {*a*}. *Then I* : *a*, *I* : *b and I* : *c are all empty*.

The following result can be easily obtained from the above definition.

#### Proposition 3.10

*Let S be an ordered semigroup and f a good mapping on S*. *If I*, *I*^{′}, *I*^{″}, *J*, *J*^{′}, *J*^{″} ∈ *I*(*S*) *and a* ∈ *S*, *then*

*I*^{′} ⊆ *I*^{″} ⇒ *I*^{′} : *a* ⊆ *I*^{″} : *a and I*^{′} : *J* ⊆ *I*^{″} : *J*;

*J*^{′} ⊆ *J*^{″} ⇒ *I* : *J*^{′} ⊇ *I* : *J*^{″};

(*I*^{′} ∩ *I*^{″}) : *a* = (*I*^{′} : *a*) ∩ (*I*^{″} : *a*) *and* (*I*^{′} ∩ *I*^{″}) : *J* = (*I*^{′} : *J*) ∩ (*I*^{″} : *J*).

#### Proposition 3.11

*Let S be an ordered semigroup and f a good mapping on S*. *If I* ∈ *I*(*S*) *and a* ∈ *S*, *then I* : *a is
either empty or an ideal containing I of S*.

#### Proof

Suppose that *I* : *a* ≠ ∅. Let *x* ∈ *I* : *a* and *r* ∈ *S*. Then *rx*, *xr* ∈ *f*(*x*). Thus *f*(*rx*) ⊆ *f*(*x*) and *f*(*xr*) ⊆ *f*(*x*). Also *f*(*a*)*f*(*x*) ⊆ *I*. Therefore, *f*(*a*)*f*(*rx*) ⊆ *I* and *f*(*a*)*f*(*xr*) ⊆ *I*. Let *z* ≤ *y* ∈ *I* : *a*. Then *z* ∈ *f*(*y*) and *f*(*a*)*f*(*y*) ⊆ *I*. Thus *f*(*z*) ⊆ *f*(*y*). Therefore, *f*(*a*)*f*(*z*) ⊆ *I*, which implies that *z* ∈ *I* : *a*. Hence, *I* : *a* is an ideal of *S*. Next we prove that *I* ⊆ *I* : *a*.

Let *b* ∈ *I* and *x* ∈ *I* : *a*. Then *b* ∈ *f*(*x*) ∪ *I*. Thus *f*(*b*) ⊆ *f*(*x*) ∪ *I*. Also *f*(*a*)*f*(*x*) ⊆ *I*. It follows that *f*(*a*)*f*(*b*) ⊆ *f*(*a*)(*f*(*x*) ∪ *I*) = *f*(*a*)*f*(*x*) ∪ *f*(*a*)*I* ⊆ *I*. Hence *b* ∈ *I* : *a* and so *I* ⊆ *I* : *a*. □

Let *S* be an ordered semigroup and *f* a good mapping on *S*. Denote the following condition by (*α*):

$$\begin{array}{}{\displaystyle (\mathrm{\forall}F\in fS(S))\text{\hspace{0.17em}}(\mathrm{\forall}I\in I(S))\text{\hspace{0.17em}}\text{\hspace{0.17em}}F\cap I\ne \mathrm{\varnothing}\Rightarrow {F}^{\ast}\cap I\ne \mathrm{\varnothing}.}\end{array}$$

If *f*(*a*) = *I*(*a*) for every *a* ∈ *S*, then *S* satisfies the condition (*α*). But this is not true for any good mapping *f*. See the following example.

#### Example 3.12

*We consider the ordered semigroup S* = {*a*, *b*, *c*} *defined by multiplication and the order below*:

$$\begin{array}{}{\displaystyle \overline{)\begin{array}{cccc}\cdot & a& b& c\\ a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\\ b\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}b\\ c\text{\hspace{0.17em}}& \text{\hspace{0.17em}}a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}b\text{\hspace{0.17em}}& \text{\hspace{0.17em}}c\end{array}}}\end{array}$$

$$\begin{array}{}{\displaystyle \le :=\{(a,a),(a,b),(a,c),(b,b),(b,c),(c,c)\}.}\end{array}$$

*It is easy to check that S is an ordered semigroup*. *The ideals of S are the sets*:

$$\begin{array}{}{\displaystyle \{a\},\text{\hspace{0.17em}}\{a,b\}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}S.}\end{array}$$

*If we define f*(*a*) = {*a*}, *f*(*b*) = *f*(*c*) = *S*, *then it is easy to see that f is a good mapping*. *Let I* = {*a*, *b*} *and F* = {*b*, *c*}. *Then F is an f*-*system with kernel F*^{*} = {*c*} *and F* ∩ *I* ≠ ∅. *However*, *F*^{*} ∩ *I* = ∅.

#### Proposition 3.13

*Let S be an ordered semigroup and f a good mapping on S*. *If I*, *J* ∈ *I*(*S*), *then*

*I* ⊆ *J* ⇒ *r*_{f}(*I*) ⊆ *r*_{f}(*J*);

*r*_{f}(*r*_{f}(*I*)) = *r*_{f}(*I*);

*r*_{f}(*I* ∪ *J*) = *r*_{f}(*r*_{f}(*I*) ∪ *r*_{f}(*J*));

*r*_{f}(*I* ∩ *J*) = *r*_{f}(*I*) ∩ *r*_{f}(*J*), *if S satisfies the condition* (*α*).

#### Proof

(1) and (2) are obvious.

(3) From Theorem 2.12, we have *I*∪*J* ⊆ *r*_{f}(*I*)∪*r*_{f}(*J*). By condition (1), we have *r*_{f}(*I*∪*J*) ⊆ *r*_{f}(*r*_{f}(*I*)∪*r*_{f}(*J*)) and *r*_{f}(*I*)∪*r*_{f}(*J*) ⊆ *r*_{f}(*I*∪*J*). Combining conditions (1) and (2), we obtain *r*_{f}(*r*_{f}(*I*) ∪ *r*_{f}(*J*)) ⊆ *r*_{f}(*r*_{f}(*I*∪*J*)) = *r*_{f}(*I*∪*J*).

(4) It is obvious that *r*_{f}(*I* ∩ *J*) ⊆ *r*_{f}(*I*) ∩ *r*_{f}(*J*) from condition (1). Next we prove the other inclusion. Let *x* ∈ *r*_{f}(*I*) ∩ *r*_{f}(*J*) and *F* be an *f*-system containing *x*. Suppose that *a* ∈ *F* ∩ *I* and *b* ∈ *F* ∩ *J*. By assumption (*α*), there exist *a*^{*} ∈ *F*^{*} ∩ *I* and *b*^{*} ∈ *F*^{*} ∩ *J*. Since *F*^{*} is an *m*-system, (*a*^{*}*zb*^{*}] ∩ *F*^{*} ≠ ∅ for some *z* ∈ *S*. Thus (*a*^{*}*zb*^{*}] ∩ *F* ≠ ∅. Moreover, *a*^{*}*zb*^{*} ∈ *I* ∩ *J* and so (*a*^{*}*zb*^{*}] ⊆ *I* ∩ *J*. Hence *F* ∩ (*I* ∩ *J*) ≠ ∅. It follows that *x* ∈ *r*_{f}(*I* ∩ *J*). Therefore *r*_{f}(*I* ∩ *J*) = *r*_{f}(*I*) ∩ *r*_{f}(*J*). □

Combining Proposition 3.5, 3.6 and 3.13 (1), we have the following result.

#### Theorem 3.14

*Let I be an ideal of an ordered semigroup S. Then r*_{f}(*I*) *is the least f*-*semiprime ideal containing I*.

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