In this part, we obtain several characterizations of some elements in Lattice Modules under special conditions.

**Lemma 3.1:** *Let M be a C*— *lattice L*— *module. Let N*_{1}, *N*_{2} ∈ *M*. *Suppose B* ∈ *M satisfies the following properties*.(*) *If H* ∈ *M is compact with H* ≤ *B*, *then either H* ≤ *N*_{1} *or H* ≤ *N*_{2}.*Then either B* ≤ *N*_{1} *or B* ≤ *N*_{2}.

**Proof:** *Assume that **B* ≰ *N*_{1} and *B* ≰ *N*_{2}. Then since *B* is a join of compact elements, we can find compact elements *H*_{1} ≤ *B* and *H*_{2} ≤ *B* such that *H*_{1} ≰ *N*_{1} and *H*_{2} ≰ *N*_{2}. Since *H* = *H*_{1} ∨ *H*_{2} ≤ *B* is compact, then by hypothesis (*) we have *H* ≤ *N*_{1} or *H* ≤ *N*_{2}, a contradiction. Consequently, we have either *B* ≤ *N*_{1} or *B* ≤ *N*_{2}. □

**Theorem 3.2:** *Let L be a C*— *lattice, M be a C*— *lattice L*— *module and N be an element of M*. *Then the following statements are equivalent*.(1) *N is weakly prime in M*.(2) *For any a* ∈ *L such that a* ≰ (*N* :_{L} 1_{M}), *either* (*N* :_{M} a) = *N or* (*N*:_{M} a) = (0_{M}:_{M} a).(3) *For every a* ∈ *L*_{*} and every K ∈ *M*_{*}; 0_{M} ≠ *aK* ≤ *N implies either a* ≤ (*N* :_{L} 1_{M}) *or K* ≤ *N*.

**Proof:** *(1) ⇒ (2) Suppose (1) holds. Let **H* be a compact element of *M* such that *H* ≤ *B* = (*N* :_{M} a) and *a* ≰ (*N* :_{L} 1_{M}). Then *aH* ≤ *N*. We have two cases:Case 1: Let *aH* = 0_{M}. Then *H* ≤ (0_{M} :_{M} a).Case 2: Let *aH* ≠ 0_{M}. Since *aH* ≤ *N, a* ≰ (*N* :_{L} 1_{M}) and *N* is weakly prime, it follows that *H* ≤ *N*.Hence by Lemma 3.1, either (*N* :_{M} a) ≤ (0_{M} :_{M} a) or (*N* :_{M} a) ≤ *N*. Consequently, either (*N* :_{M} a) = (0_{M} :_{M} a) or (*N* :_{M} a) = *N*.(2) ⇒ (3) Suppose (2) holds. Let 0_{M} ≠ *aK* ≤ *N* and *a* ≰ (*N* :_{L} 1_{M}) for *a* ∈ *L*_{*} and *K* ∈ *M*_{*}. We will show that *K* ≤ *N*. Since *aK* ≤ *N*, it follows that *K* ≤ (*N* :_{M} a). If (*N* :_{M} a) = *N*, then *K* ≤ *N*. If (*N* :_{M} a) = (0_{M} :_{M} a), then *aK* = 0_{M}. This is a contradiction. Consequently, *K* ≤ *N*.(3) ⇒ (1) Suppose (3) holds. Let *aK* ≤ *N, a* ≰ (*N* :_{L} 1_{M}) and *K* ≰ *N* for some *a* ∈ *L* and *K* ∈ *M*. Choose *x*_{1} ∈ *L*_{*} and *Y*_{1} ∈ *M*_{*} such that *x*_{1} ≤ *a, x*_{1} ≰ (*N* :_{L} 1_{M}), *Y*_{1} ≤ *K* and *Y*_{1} ≰ *N*. Let *x*_{2} ≤ *a* and *Y*_{2} ≤ *K* be any two compact elements of *L, M*, respectively. Then by our assumption (3), we have (*x*_{2} ∨ *x*_{1})(*Y*_{2} ∨ *Y*_{1}) = 0_{M} and so *x*_{2}*Y*_{2} = 0_{M}. Therefore *aK* = 0_{M}. This shows that *N* is weakly prime in *M*. □

**Theorem 3.3:** *Let L be a C*— *lattice, M be a C*— *lattice L*— *module and N be an element of M*. *Then the following statements are equivalent*:(1) *N is almost prime in M*.(2) *For any a* ∈ *L such that a* ≰ (*N* :_{L} 1_{M}), *either* (*N* :_{M} a) = *N or* (*N* :_{M} a) = ((*N* :_{L} 1_{M})*N* :_{M} a).(3) *For every a* ∈ *L*_{*} and every K ∈ *M*_{*}, aK ≤ *N and aK* ≰ (*N* :_{L} 1_{M}) *N implies either a* ≤ (*N* :_{L} 1_{M}) *or K* ≤ *N*.

**Proof:** *(1) ⇒ (2) Suppose (1) holds. Let **H* be a compact element of *M* such that *H* ≤ *B* = (*N* :_{M} a) and *a* ≰ (*N* :_{L} 1_{M}). Then *aH* ≤ *N*. We have two cases:Case 1: If *aH* ≤ (*N* :_{L} 1_{M})*N*, then *H* ≤ ((*N* :_{L} 1_{M})*N*:_{M} a).Case 2: If *aH* ≰ (*N* :_{L} 1_{M})*N*, since *aH* ≤ *N, a* ≰ (*N* :_{L} 1_{M}) and *N* is almost prime, it follows that *H* ≤ *N*.Hence by Lemma 3.1, we prove that either (*N* :_{M} a) ≤ ((*N* :_{L} 1_{M})*N* :_{M} a) or (*N* :_{M} a) ≤ *N*. One can see, as (*N* :_{L} 1_{M})*N* ≤ *N*, we get ((*N* :_{L} 1_{M})*N* :_{M} a) ≤ (*N*:_{M} a). Moreover, always *N* ≤ (*N* :_{M} a). Consequently, either (*N* :_{M} a) = ((*N* :_{L} 1_{M})*N*:_{M} a) or (*N* :_{M} a) = *N*.(2) ⇒ (3) Suppose (2) holds. Let *aK* ≤ *N* and *aK* ≰ (*N* :_{L} 1_{M})*N* for *a* ∈ *L*_{*} and *K* ∈ *M*_{*}. Assume that *a* ≰ (*N* :_{L} 1_{M}). We show that *K* ≤ *N*. Since *aK* ≤ *N*, it follows that *K* ≤ (*N* :_{M} a). If (*N* :_{M} a) = *N*, then *K* ≤ *N*. If (*N* :_{M} a) = ((*N* :_{L} 1_{M})*N* :_{M} a), then *K* ≤ ((*N* :_{L} 1_{M})*N*:_{M} a). So we have *aK* ≤ (*N* :_{L} 1_{M})*N*, a contradiction. Thus *K* ≤ *N*.(3) ⇒ (1) Suppose (3) holds. Let *aK* ≤ *N, aK* ≰ (*N* :_{L} 1_{M})*N* for some *a* ∈ *L* and *K* ∈ *M*. Assume that *a* ≰ (*N* :_{L} 1_{M}) and *K* ≰ *N*. Choose *x*_{1} ∈ *L*_{*} and *Y*_{1} ∈ *M*_{*} such that *x*_{1} ≤ *a, x*_{1} ≰ (*N* :_{L} 1_{M}), *Y*_{1} ≤ *K* and *Y*_{1} ≰ *N*. As *L* and *M* are *C*— lattices, there exist two compact elements of *x*_{2} ∈ *L* and *Y*_{2} ∈ *M* such that *x*_{2} ≤ *a* and *Y*_{2} ≤ *K*. Moreover, as *x*_{1}, *x*_{2} ∈ *L*_{*} and *Y*_{1}, *Y*_{2} ∈ *M*_{*}, we have *x*_{1} ∨ *x*_{2} ∈ *L*_{*} and *Y*_{1} ∨ *Y*_{2} ∈ *M*_{*}. Since *x*_{1} ≤ *a* and *x*_{2} ≤ *a*, we have *x*_{1} ∨ *x*_{2} ≤ *a*. Similarly, we have *Y*_{1} ∨ *Y*_{2} ≤ *K*. Thus (*x*_{2} ∨ *x*_{1})(*Y*_{2} ∨ *Y*_{1}) ≤ *aK* ≤ *N*. In addition, (*x*_{2} ∨ *x*_{1}*)(Y*_{2} ∨ *Y*_{1}) ≰ (*N* :_{L} 1_{M})*N*. Indeed, assume that (*x*_{2} ∨ *x*_{1})(*Y*_{2} ∨ *Y*_{1}) ≤ (*N* :_{L} 1_{M})*N*. Then we get *x*_{2}*Y*_{2} ≤ (*N* :_{L} 1_{M})*N*. Since *x*_{2}*Y*_{2} ≤ *aK*, we can write *aK* ≤ (*N* :_{L} 1_{M})*N*, for *x*_{2}*Y*_{2} ∈ *M*_{*}. But it is a contradiction.Consequently, as (*x*_{2} ∨ *x*_{1})(*Y*_{2} ∨ *Y*_{1}) ≤ *N* and (*x*_{2} ∨ *x*_{1})(*Y*_{2} ∨ *Y*_{1}) ≰ (*N* :_{L} 1_{M})*N*, by our assumption (3), we have (*x*_{2} ∨ *x*_{1}) ≤ (*N* :_{L} 1_{M}) or (*Y*_{2} ∨ *Y*_{1}) ≤ *N*. Then we get *x*_{1} ≤ (*N* :_{L} 1_{M}) or *Y*_{1} ≤ *N*, a contradiction. This shows that *N* is almost prime in *M*. □

**Lemma 3.4:** *Let L be a C*— *lattice and M be a multiplication C*— *lattice L*— *module. If N is an almost prime element of M, then* $\sqrt{((N{:}_{L}{1}_{M})N{:}_{L}{1}_{M})N=(N{:}_{L}{1}_{M})N}$.

**Proof:** *We first note that (**N* :_{L} 1_{M})^{2} ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}). Indeed, since *M* is a multiplication lattice module, we have (*N* :_{L} 1_{M})(*N* :_{L} 1_{M})*1*_{M} = (*N* :_{L} 1_{M})*N*, i.e., (*N* :_{L} 1_{M})^{2} ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}).Let *a* be a compact element in *L* and $a\phantom{\rule{thinmathspace}{0ex}}\le \sqrt{((N{:}_{L}{1}_{M})N{:}_{L}{1}_{M})}$.If *a* ≤ (*N* :_{L} 1_{M}), then we have *a*(*N* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M})^{2} ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}). Thus we obtain *aN* = *a*(*N* :_{L} 1_{M})*1*_{M} ≤ (*N* :_{L} 1_{M})*N* :_{L} 1_{M})*1*_{M} = (*N* :_{L} 1_{M})*N*.If *a* ≰ (*N* :_{L} 1_{M}), then we have either (*N* :_{M} a) = ((*N* :_{L} 1_{M})*N* :_{M} a) or (*N* :_{M} a) = *N* by Theorem 3.3(2).Case 1: Suppose that (*N* :_{M} a) = ((*N* :_{L} 1_{M})*N*:_{M} a). Since *N* ≤ (*N*:_{M} a), then we have *aN* ≤ *a*(*N* :_{M}a) = *a*((*N* :_{L} 1_{M})*N* :_{M} a) ≤ (*N* :_{L} 1_{M})*N*.Case 2: Suppose that (*N* :_{M} a) = *N*. Let *n* be the smallest positive integer such that *a*^{n} ≤ ((*N* :_{L} 1_{M})*N* :_{L}1_{M}). If *n* = 1, then we have *a1*_{M} ≤ (*N* :_{L} 1_{M})*N* ≤ *N*, a contradiction.So, we assume *n* ≥ 2. Then *a*^{n}1_{M} ≤ (*N* :_{L} 1_{M})*N* ≤ *N* with *a*^{k}1_{M} ≰ (*N* :_{L} 1_{M})*N* for every *k* ≤ *n*— 1. It follows that *a*^{n−1}1_{M} ≤ (*N* :_{M} a) = *N* and *a*^{n−1}1_{M} ≰ (*N* :_{L} 1_{M})*N*. If *n* = 2, we also get a contradiction. If *n* ≥ 3, we have *a*(*a*^{n—2}1_{M}) ≤ *N* and *a*(*a*^{n—2} 1_{M}) ≰ (*N* :_{L} 1_{M})(*N*. Thus, since *N* is an almost prime element, we obtain either *a* ≤ (*N* :_{L} 1_{M}) or *a*^{n—2}1_{M} ≤ *N*. Continuing this process, we conclude that *a* ≤ (*N* :_{L} 1_{M}), which is a contradiction. Therefore $\sqrt{(N{:}_{L}{1}_{M})N{:}_{L}{1}_{M})}N\le (N{:}_{L}{1}_{M})N$.For the second part, let *a* be a compact element in *L* and *a* ≤ (*N* :_{L} 1_{M}). Then we have *a*^{k} 1_{M} ≤ *a*1_{M} ≤ *N* for positive integer *k*, i.e., *a*^{k} ≤ (*N* :_{L} 1_{M}). Thus, we obtain *a*^{k+1} 1_{M} ≤ *a*^{k}*N* ≤ (*N* :_{L} 1_{M})*N*, i.e., *a*^{k+1} ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}). Consequently, $a\le \sqrt{((N{:}_{L}{1}_{M})N{:}_{L}{1}_{M})}$ □

**Lemma 3.5:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. Then we have (*aN* :_{L} 1_{M}) = *a*(*N* :_{L} 1_{M}) *for every element a in L*.

**Proof:** *As **M* is a multiplication lattice module, then we have *a*(*N* :_{L} 1_{M})*1*_{M} = *aN* = (*aN* :_{L} 1_{M})*1*_{M} . By Theorem 5 in [14], we obtain *a*(*N* :_{L} 1_{M}) = (*aN* :_{L} 1_{M}). □

**Theorem 3.6:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice L—module. For 1_{M} ≠ *N* ∈ *M*, *the followings are equivalent*:(1) *N is prime*.(2) (*N* :_{L} 1_{M}) *is prime*.(3) *N* = *q*1_{M} *for some prime element q of L*.

**Proof:** *The proof can be easily seen with Corollary 3 in [14]. □*

**Theorem 3.7:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. For 1_{M} ≠ *N* ∈ *M*, *then the followings are equivalent*:(1) *N is weakly prime*.(2) (*N* :_{L} 1_{M}) *is weakly prime*.(3) *N* = *q*1_{M} for some weakly prime element qof L.

**Proof:** *(1) ⇒ (2): Suppose **N* is weakly prime and *a, b* ∈ *L* such that 0_{L} ≠ *ab* ≤ (*N* :_{L} 1_{M}). Then we have *ab*1_{M} ≤ *N*. Since *M* is faithful and 0_{L} ≠ *ab*, then we obtain 0_{M} ≠ *ab*1_{M}. Now, as *N* is weakly prime, then we get either *a* ≤ (*N* :_{L} 1_{M}) or *b*1_{M} ≤ *N* (and so *b* ≤ (*N* :_{L} 1_{M})). Hence (*N* :_{L} 1_{M}) is a weakly prime element in *L*.(2) ⇒ (1): Let (*N* :_{L} 1_{M}) be weakly prime in *L*. Let *r* ∈ *L* and *X* ∈ *M*, such that 0_{M} ≠ *rX* ≤ *N*. By Lemma 3. 5, we have *r*(*X* :_{L} 1_{M}) = (*rX* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}). Moreover *r*(*X* :_{L} 1_{M}) ≠ 0_{L} because otherwise, if *r*(*X* :_{L} 1_{M}) = 0_{L}, then *rX* = *r*(*X* :_{L} 1_{M})*1*_{M} = 0_{L}1_{M} = 0_{M}. As (*N* :_{L} 1_{M}) is weakly prime, then either *r* ≤ (*N* :_{L} 1_{M}) or (*X* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}). Since *M* is a multiplication lattice module, we obtain *r* ≤ (*N* :_{L} 1_{M}) or *X* = (*X* :_{L} 1_{M})(*1*_{M} ≤ (*N* :_{L} 1_{M})*1*_{M} = *N*. Thus, *N* is weakly prime in *M*.(2) ⇒ (3): Choose *q* = (*N* :_{L} 1_{M}).(3) ⇒ (2): Suppose that *N* = *q*1_{M} for some weakly prime element *q* of *L*. By Lemma 3.5, we have (*N* :_{L} 1_{M}) = (*q*1_{M} :_{L} 1_{M}) = *q*(1_{M} :_{L} 1_{M}) = *q*. Thus *q* = (*N* :_{L} 1_{M}) is a weakly prime element. □

**Theorem 3.8:** *Let L be a PG—lattice with* 1_{L}compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. For 1_{M} ≠ *N* ∈ *M*, *then the followings are equivalent*:(1) *N is almost prime*.(2) (*N* :_{L} 1_{M}) *is almost prime*.(3) *N* = *q*1_{M} for some almost prime element q of L.

**Proof:** *(1) ⇒ (2): Suppose **N* is almost prime and *a, b* ∈ *L* such that *ab* ≤ (*N* :_{L} 1_{M}) and *ab* ≰ (*N* :_{L} 1_{M})^{2}. Then we have *ab*1_{M} ≤ *N* and *ab*1_{M} ≰ (*N* :_{L} 1_{M})*N*. Indeed, if *ab*1_{M} ≤ (*N* :_{L} 1_{M})*N*, by Lemma 3.5, *ab* ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}) = (*N* :_{L} 1_{M})(*N* :_{L} 1_{M}) = (*N* :_{L} 1_{M})^{2}, a contradiction. Now, *N* is almost prime implies that either *a* ≤ (*N* :_{L} 1_{M}) or *b*1_{M} ≤ *N* (and so *b* ≤ (*N* :_{L} 1_{M})). Hence (*N* :_{L} 1_{M}) is an almost prime element in *L*.(2) ⇒ (1): Let *r* ∈ *L* and *X* ∈ *M* such that *rX* ≤ *N* and *rX* ≰ (*N* :_{L} 1_{M})*N*. By Lemma 3.5, we have *r*(X :_{L} 1_{M}) = (*rX* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}). Moreover *r*(*X* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M})^{2}. Indeed, if *r*(*X* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M})^{2} = ((*N* :_{L} 1_{M})*N* :_{L} 1_{M}), then *rX* = *r*(*X* :_{L} 1_{M}(1_{M} ≤ ((*N* :_{L} 1_{M})*N* :_{L} 1_{M})1_{M} = (*N* :_{L}1_{M})*N*, a contradiction. As (*N* :_{L} 1_{M}) is almost prime, either *r* ≤ (*N* :_{L} 1_{M}) or (*X* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}). By Proposition 2.8, we have *X* = (*X* :_{L} 1_{M})1_{M} ≤ (*N* :_{L} 1_{M})1_{M} = *N*. Thus, we obtain *r* ≤ (*N* :_{L} 1_{M}) or *X* ≤ *N*, i.e., *N* is almost prime in *M*.(2) ⇒ (3): Choose *q* = (*N* :_{L} 1_{M}).(3) ⇒ (2): Suppose that *N* = *q*1_{M} for some almost prime element *q* of *L*. By Lemma 3.5, we have (*N* :_{L} 1_{M}) = (*q*1_{M} :_{L} 1_{M}) = *q*(1_{M} :_{L} 1_{M}) = *q*. Thus *q* = (*N* :_{L} 1_{M}) is an almost prime element. □

Now, we define a new multiplication over the multiplication lattice modules.

**Definition 3.9:** *If Mis a multiplication L—lattice module and N* = *a*1_{M}, *K* = *b*1_{M} are two elements of M, where a, b ∈ *L, the product of Nand Kis defined as NK* = (*a*1_{M})(*b*1_{M}) = *ab*1_{M}.

**Proposition 3.10:** *Let M be a multiplication L—lattice module and N* = *a*1_{M}, K = *b*1_{M} are two elements of M, where a, b ∈ *L*. *Then the product of N and K is independent of expression of Nand K*.

**Proof:** *Let **N* = *a*_{1}1_{M} = *a*_{2}1_{M} and *K* = *b*_{1}1_{M} = *b*_{2}1_{M} for *a*_{1}, *a*_{2}, *b*_{1}, *b*_{2} ∈ *L*. Then *NK* = (*a*_{1}*b*_{1})1_{M} = *a*_{1}(*b*_{1}1_{M}) = *a*_{1}(*b*_{2}1_{M}) = *b*_{2}(*a*_{1}1_{M}) = *b*_{2}(*a*_{2}1_{M}) = (*a*_{2}*b*_{2})1_{M}. □

With the help of the new defined multiplication, we obtain the following results.

**Theorem 3.11:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. Then N is an idempotent element in M if and only if N^{2} = *N*.

**Proof:** *⇒ : Since **N* is idempotent, then we have *N* = (*N* :_{L} 1_{M})*N*. As *M* is a multiplication lattice module, then we get *N*^{2} = *NN* = (*N* :_{L} 1_{M})1_{M} (*N* :_{L} 1_{M})1_{M} = (*N* :_{L} 1_{M})^{2}1_{M}. By Proposition 2.8 and Lemma 3.5, we obtain *N* = (*N* :_{L} 1_{M})*N* = ((*N* :_{L} 1_{M})*N* :_{L} 1_{M})1_{M} = (*N* :_{L} 1_{M})(*N* :_{L} 1_{M})1_{M} = (*N* :_{L} 1_{M})^{2}1_{M} . Thus we have *N*^{2} = (*N* :_{L} 1_{M})^{2}1_{M} = *N*.⇐: Suppose that *N*^{2} = *N*. Following the same steps in the first part of the proof, we obtain *N* = *N*^{2} = (*N* :_{L} 1_{M})^{2}1_{M} = (*N* :_{L} 1_{M})*N*, i.e., *N* = (*N* :_{L} 1_{M})*N*. Consequently, *N* is idempotent in *M*. □

**Theorem 3.12:** *Let L be a PG—lattice with* 1_{L}compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. Then N < 1_{M} is prime in M if and only if whenever X and Y are elements of M such that XY ≤ *N, either X* ≤ *Nor Y* ≤ *N*.

**Proof:** *⇒: Assume that **N* is prime in *M*. By Theorem 3.6, we get (*N* :_{L} 1_{M}) is prime in *L*. Suppose that *X* and *Y* are elements of *M* such that *XY* ≤ *N*, but *X* ≰ *N* and *Y* ≰ *N*. By Proposition 2.8, we have *X* = (*X* :_{L} 1_{M})1_{M} and *Y* = (*Y* :_{L} 1_{M})1_{M} and so *XY* = (*X* :_{L} 1_{M})(*Y* :_{L} 1_{M})1_{M} . Since *M* is a multiplication lattice module, then we have (*X* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M}) and (*Y* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M}). Indeed, if (*X* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}) and (*Y* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}), then we have (*X* :_{L} 1_{M})1_{M} ≤ (*N* :_{L} 1_{M})1_{M} and (*Y* :_{L} 1_{M})1_{M} ≤ (*N* :_{L} 1_{M})1_{M}. So, by Proposition 2.8, *X* ≤ *N* and *Y* ≤ *N*, a contradiction. Hence (*X* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M}) and (*Y* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M}). Thus, since (*N* :_{L} 1_{M}) is prime, we obtain (*X* :_{L} 1_{M})(*Y* :_{L} 1_{M}) ≰ (*N* :_{L} 1_{M}). Moreover, we have *XY* = (*X* :_{L} 1_{M})(*Y* :_{L} 1_{M})1_{M} ≤ *N*, i.e., (*X* :_{L} 1_{M})(*Y* :_{L} 1_{M}) ≤ (*N* :_{L} 1_{M}), a contradiction. Therefore, either *X* ≤ *N* or *Y* ≤ *N*.⇐: We assume that if *XY* ≤ *N*, then *X* ≤ *N* or *Y* ≤ *N*. To prove that *N* is prime in *M*, it is enough, by Theorem 3.6, to prove that (*N* :_{L} 1_{M}) is prime in *L*. Let *r*_{1}, *r*_{2} ∈ *L* such that *r*_{1}*r*_{2} ≤ (*N* :_{L} 1_{M}). Let *X* = *r*_{1}1_{M} and *Y* = *r*_{2}1_{M} . Then *XY* = *r*_{1}*r*_{2}1_{M} ≤ *N*. By assumption, either *r*_{1}1_{M} = *X* ≤ *N* or *r*_{2}1_{M} = *Y* ≤ *N* and so, either *r*_{1} ≤ (*N* :_{L} 1_{M}) or *r*_{2} ≤ (*N* :_{L} 1_{M}). Hence (*N* :_{L} 1_{M}) is prime in *L*. Consequently, *N* is prime in *M*. □

The proof of the next Theorem can be shown to be similar to the previous proof with using Proposition 2.8 and Theorem 3.7.

**Theorem 3.13:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. Then N < 1_{M} is weakly prime in M if and only if whenever X and Y are elements of M such that 0_{M} ≠ *XY* ≤ *N, either X* ≤ *N or Y* ≤ *N*.

Finally, the proof of the following Theorem is obtained, as in the case of Theorem 3.12, by using the proof of Proposition 2.8, Lemma 3.5 and Theorem 3.8.

**Theorem 3.14:** *Let L be a PG—lattice with* 1_{L} compact and M be a faithful multiplication PG—lattice module with 1_{M} compact. Then N < 1_{M} is almost prime in M if and only if whenever X and Y are elements of M such that XY ≤ *N and XY* ≰ (*N* :_{L} 1_{M})*N, either X* ≤ *N or Y* ≤ *N*.

## Comments (0)