# A class of Congruencies on Semilattice

A class of Congruencies on Semilattice
For any subsemilattice L of a semilattice S define, ๐๐ฟ = (๐ฅ, ๐ฆ) ∈ ๐x๐: ๐ฅห๐ =
๐ฆห๐ ๐๐๐ ๐ ๐๐๐ ๐ ∈ ๐ฟ}. then ๐๐ฟ is congruence on ๐ ๐๐๐ ๐ฟ is contained in a single equivalence
class of ๐๐ฟ. ๐๐ฟ denotes the natural epimorphism of S on the quotient semilattice ๐/๐๐ฟ .
If S is distributive semilattice and ๐น is a filter of ๐, then ๐๐น is the smallest congruence on S
containing ๐น in a single equivalence class and that S⁄θL is distributive.
Theorem: 3.1. Let ๐ be any semilattice which is directed above. Then the following are
equivalent:
(i)
S is distributive.
(ii)
๐/๐๐ฟ is distributive for every subsemilattice ๐ฟ ๐๐ ๐.
Proof: (i) ๏ (ii) Let S be distributive semilattice and L a subsemilattice of S. Let ๐ฅ, ๐ฆ ๐๐๐ ๐ง ∈
๐ such that ๐๐ฟ(๐ฅ) ∧ ๐๐ฟ(๐ฆ) ≤ ๐๐ฟ(๐ง) ⇒ ∃ ๐ ∈ ๐ฟ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅห๐ฆห๐ = ๐ฅห๐ฆห๐งห๐
⇒ (๐ฅห๐) ห(๐ฆ ห๐) ≤ ๐ง ⇒ ∃ ๐ ๐๐๐ ๐ ∈ ๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅห๐ ≤ ๐,
๐ฆ ห๐ ≤ ๐ ๐๐๐ ๐ห๐ = ๐ง.
Now, ๐๐ฟ(๐ฅห๐) ≤ ๐๐ฟ(๐), ๐๐ฟ(๐ฆห๐) ≤ ๐๐ฟ(๐) ๐๐๐ ๐๐ฟ(๐) ห๐๐ฟ(๐) = ๐๐ฟ(๐ง).
Therefore ๐/๐๐ฟ is distributive for every subsemilattice ๐ฟ ๐๐ ๐.
(ii) ๏ (i) Let ๐, ๐ ๐๐๐ ๐ ∈ ๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ห๐ ≤ ๐. Then choose ๐ ∈ ๐ such that ๐, ๐ ๐๐๐ ๐ ๏
(๐]๐๐๐ ๐๐ข๐ก ๐ฟ = (๐]. Since S⁄θL is distributive, there exists ๐, ๐ ∈ ๐ s.t ๐๐ฟ(๐) ≤
๐๐ฟ(๐), ๐๐ฟ(๐) ≤ ๐๐ฟ(๐)๐๐๐ ๐๐ฟ(๐)ห ๐๐ฟ(๐) = ๐๐ฟ(๐). Again there exist ๐ฅ, ๐ฆ ๐๐๐ ๐ง ∈ ๐น such that
๐ห๐ห๐ฅ = ๐ห๐ฅ, ๐ห๐ห๐ฆ = ๐ห๐ฆ ๐๐๐ ๐ห๐ห๐ง = ๐ห๐ง which implies that ๐ห๐ห๐ฅ = ๐, ๐ห๐ =
๐ ๐๐๐ ๐ห๐ = ๐ . Therefore a ≤ e ห x, b ≤ x หf and (e ห x) ห (f ห x) =c. Hence S is distributive.
Theorem: 3.2. Let ๐ฟ be a sub semilattice of S. If ๐ผ’ is an ideal of ๐/๐๐ฟ, then ๐ผ = {๐ฅ ∈ ๐: ๐๐ฟ(๐ฅ) ∈
๐ผ’} is an ideal of S. ๐ผ’ is a proper ideal of ๐/๐๐ฟ if and only if ๐ผ ∩ ๐ฟ = ∅. Also, if ๐ is a prime
ideal of ๐ such that ๐ ∩ ๐ฟ = ∅, then ๐’ = { ๐๐ฟ(๐ฅ) ๏ ๐/๐๐ฟ: ๐ฅ ๏ ๐} is a prime ideal of ๐/๐๐ฟ.
๐
Proof: Let I’ be an ideal of๐๐ฟ, and ๐ผ = {๐ฅ ∈ ๐: ๐๐ฟ (๐ฅ) ๏ ๐ผ’}, let ๐ ∈ ๐ผ, ๐ ∈ ๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ ≤
๐. ๐โ๐๐ ๐ ∧ ๐ฅ ≤ ๐ ๐๐๐ ๐ ๐๐๐ ๐ฅ ∈ ๐ฟ ⇒ ๐๐ฟ(๐) = ๐๐ฟ(๐ฅ ∧ ๐) ≤ ๐๐ฟ(๐)
⇒ ๐๐ฟ(๐) ∈ ๐ผ ′ ๐๐๐ โ๐๐๐๐ ๐ ∈ ๐ผ.
Therefore ๐ผ is an initial segment of S. Also, if ๐๐ฟ(๐) ๐๐๐ ๐๐ฟ(๐) ∈ ๐ผ’
⇒ ∃ ๐ ๏ ๐ ๐๐ข๐โ ๐กโ๐๐ก ๐๐ฟ(๐) ∈ ๐ผ’, ๐๐ฟ(๐) ≤ ๐๐ฟ(๐) ๐๐๐ ๐๐ฟ(๐) ≤ ๐๐ฟ(๐)
⇒ ∃๐ฅ ๐๐๐ ๐ฆ ∈ ๐ฟ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ห๐ห๐ฅ = ๐หx ๐๐๐ ๐ห๐ห๐ฆ = ๐ห๐ฆ
⇒ ๐ห๐ห๐ฅห๐ฆ = ๐ห๐ฅห๐ฆ ๐๐๐ ๐ห๐ห๐ฅห๐ฆ = ๐ห๐ฅห๐ฆ, and ⇒ ∃๐ ∈ ๐, ๐ ๐ข๐โ ๐กโ๐๐ก ๐ ≤ ๐, ๐ ≤ ๐ ๐๐๐
๐ห๐ห๐ฅห๐ฆ = ๐ห๐ฅห๐ฆ. ⇒ ๐๐ฟ(๐) ≤ ๐๐ฟ(๐) ๐๐๐ โ๐๐๐๐ ๐ ∈ ๐ผ. ∴ ๐ผ’ is an ideal.
If ๐ ∈ ๐ผ ∩ ๐ฟ, then for any ๐ฅ ∈ ๐, ๐๐ฟ (๐ฅ) = ๐๐ฟ( ๐ฅห ๐ ) , and this implies ๐/๐๐ฟ = ๐ผ’.
Conversely if๐/๐๐ฟ = ๐ผ’, then for any ๐ฅ ∈ ๐ฟ, ๐๐ฟ(๐ฅ) ๏ ๐ผ’ and therefore ๐ผ ∩ ๐ฟ ≠ ∅.
Also let P be prime ideal of S such that ๐ ∩ ๐ฟ = ∅, then ๐’ is proper ideal of ๐/๐๐ฟ and
Let ๐๐ฟ(๐) ∧ ๐๐ฟ(๐) ∈ ๐′ , ๐๐ฟ(๐ ∧ ๐) ∈ ๐′ . Then ๐ ∧ ๐ ∈ ๐ ๐๐๐ ๐ ๐๐๐๐ ๐ ๐๐  ๐๐๐๐๐ ๐ ∈ ๐ ๐๐ ๐ ∈
๐. Hence ๐๐ฟ(๐) ∈ ๐′ ๐๐ ๐๐ฟ(๐) ∈ ๐′ . Therefore ๐’ is prime
Theorem: 3.3. Let ๐ฟ1 ๐๐๐ ๐ฟ2 be two sub semilattices of S such that ๐ฟ1 ๏ ๐ฟ2 and also let
๐: S⁄θL1 ๏ฎ S⁄θL2 be the epimorphism defined by ๐ ๐๐ฟ1(๐ฅ) = ๐๐ฟ2(๐ฅ), If ๐ is injection, then
๐ ∩ ๐ฟ1 = ∅ implies ๐ ∩ ๐ฟ2 = ∅, for any prime ideal P of S,
Proof: If P is a prime ideal of S such that ๐ ∩ ๐ฟ1 = ∅, then by theorem 3.2 above, {πL1(x) ∈
S⁄θL1} is a prime ideal of S⁄θL1 and hence { πL2(x) ∈S⁄θL2: x∈P} is a prime ideal of S⁄θL2, so that
๐ ∩ ๐ฟ2 = ∅.
Theorem: 3.4. Let ๐น be a filter of S, then the following are equivalent:
(i)
๐น ๐๐  ๐ ๐๐๐๐๐ ๐๐๐๐ก๐๐.
(ii)
๐ผ = {๐๐น(๐ฅ): ๐ฅ ∈ ๐ − ๐น} ๐๐  ๐๐ ๐๐๐๐๐ ๐๐ ๐/๐๐น .
(iii)
๐/๐๐น โ๐๐  ๐ ๐ข๐๐๐๐ข๐ ๐๐๐ฅ๐๐๐๐ ๐๐๐๐๐.
Proof: (i)⇒(ii) Suppose that ๐น is a prime filter, then ๐ − ๐น is prime ideal of S. ๐ฟ๐๐ก ๐๐น(๐ฅ) ∈
๐ผ ๐๐๐ ๐๐น(๐ฆ) ∈ ๐/๐๐น ๐ ๐ข๐โ ๐กโ๐๐ก ๐๐น(๐ฆ) ≤ ๐๐น(๐ฅ) ⇒ ∃ ๐ ∈ ๐น ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฆ ∧ ๐ = ๐ฅ ∧ ๐ฆ ∧
๐ ⇒ ๐ฆ ∧ ๐ ≤ ๐ฅ ⇒ ๐๐น(๐ฆ) = ๐๐น(๐ฆ ∧ ๐) ≤ ๐๐น(๐ฅ) ⇒ ๐๐น(๐ฆ) ≤ ๐๐น(๐ฅ)
∴ I is initial segment.
Let ๐๐น(๐ฅ), ๐๐น(๐ฆ) ∈ ๐ผ, this implies ๐ฅ, ๐ฆ ∈ ๐ − ๐น which is prime ideal of S.
⇒ ∃ ๐ง ∈ ๐ − ๐น ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅ ≤ ๐ง ๐๐๐ ๐ฆ ≤ ๐ง.
⇒ ∃ ๐ ∈ ๐น ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅ ∧ ๐ = ๐ฅ ∧ ๐ ∧ ๐ง ๐๐๐ ๐ฆ ∧ ๐ = ๐ฆ ∧ ๐ ∧ ๐ง,
⇒ ๐ฅ ∧ ๐ ≤ ๐ง ๐๐๐ ๐ฆ ∧ ๐ ≤ ๐ง
⇒ ๐๐น(๐ฅ) = ๐๐น(๐ฅ ∧ ๐) ≤ ๐๐น(๐ง) ๐๐๐ ๐๐น(๐ฆ) = ๐๐น(๐ฆ ∧ ๐) ≤ ๐๐น(๐ง)
⇒ ๐๐น(๐ฅ) ≤ ๐๐น(๐ง) ๐๐๐ ๐๐น(๐ฆ) ≤ ๐๐น(๐ง)
Therefore ๐ผ = {๐๐น(๐ฅ): ๐ฅ ∈ ๐ − ๐น} is an ideal of ๐/๐๐น .
(ii)⇒ (iii) ๐น๐๐ ๐๐๐ฆ ๐ ∈ ๐น, ๐๐น(๐) is the greatest element of ๐/๐๐น. Hence, every ideal of ๐/๐๐น is
contained in {๐๐น(๐ฅ): ๐ฅ ∈ ๐ − ๐}.Therefore ๐/๐๐น has a unique maximal.
(iii)⇒ (i): Let G be the unique maximal ideal of ๐/๐๐น and let ๐ฅ ๐๐๐ ๐ฆ ๐ ๐ − ๐น. Since the
principal ideals of ๐/๐๐น generated by ๐๐น(๐ฅ) ๐๐๐ ๐๐น(๐ฆ) are proper, it follows that
๐๐น(๐ฅ)๐๐๐ ๐๐น(๐ฆ) ∈ ๐บ. Therefore there exists ๐๐น(๐ง) ∈ ๐บ ๐ ๐ข๐โ ๐กโ๐๐ก ๐๐น(๐ฅ) ≤ ๐๐น(๐ง) and
๐๐น(๐ฆ) ≤ ๐๐น(๐ง), ๐ ๐ ๐กโ๐๐ก ๐ฅห๐งห๐ = ๐ฅห๐ ๐๐๐ ๐ฆห๐งห๐ = ๐ฆห๐ ๐๐๐ ๐ ๐๐๐ ๐๐๐น.
๐โ๐๐๐๐๐๐๐ ๐กโ๐๐๐ ๐๐ฅ๐๐ ๐ก๐  ๐๐๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅ ≤ ๐, ๐ฆ ≤ ๐ ๐๐๐ ๐ห๐งห๐ = ๐ห๐,
This implies ๐ ∈ ๐ − ๐น, thus ๐ − ๐น is a prime ideal. Hence ๐น is prime filter.
Theorem: 3.5. Let F be a filter of S. Then ๐น is a maximal filter if and only if ๐/๐๐น is the two
element chain.
Proof: Suppose F is a maximal filter, then for any ๐ฅ ๐๐๐ ๐ฆ๐ ๐น, ๐๐น(๐ฅ) = ๐๐น(๐ฆ) and that for
any ๐ฅ ∈ ๐น ๐๐๐ ๐ฆ ∈ ๐, ๐๐น(๐ฅ) = ๐๐น(๐ฆ) implies ๐ฆ ∈ ๐น.
Now, ๐๐๐ก ๐ฅ ๐๐๐ ๐ฆ ∈ ๐ − ๐น, Since ๐น is maximal, there exist ๐ ๐๐๐ ๐๐ ๐น ๐ ๐ข๐โ ๐กโ๐๐ก
๐ห๐ฅ ≤ ๐ฆ ๐๐๐ ๐ห๐ฆ ≤ ๐ฅ . ๐ป๐๐๐๐ ๐๐น(๐ฅ) = ๐๐น(๐ห๐ฅ) ≤ ๐๐น(๐ฆ) = ๐๐น(๐ห๐ฆ) ≤ ๐๐น(๐ฅ).
๐๐ ๐กโ๐๐ก ๐๐น(๐ฅ) = ๐๐น(๐ฆ).
Conversely, suppose that ๐/๐๐น is the two element chain. Suppose ๐ฅ ๐๐๐ ๐ฆ ∈ ๐ − ๐น. Choose an
element ๐ง ∈ ๐น, then we have ๐๐น(๐ง) ≠ ๐๐น(๐ฅ) and Hence ๐๐น(๐ฅ) = ๐๐น(๐ฆ) so that there exists
๐ ∈ ๐น ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅห๐ = ๐ฆห๐ ๐๐๐ ๐โ๐๐๐๐๐๐๐ ๐ฆ ∈ ๐น ∨ [๐ฅ). Hence, for any ๐ฅ ∉ ๐น, ๐น ∨ [๐ฅ) = ๐.
Thus ๐น is maximal.
Theorem: 3.6. The map FโผθF is an isomorphism from the lattice F0(S) of all non empty filters
of S into a permutable sublattice of the lattice C(S) of all congruencies on S.
Hence {θF: F ฯต F0(S)} is a distributive and permutable sublattice of C(S).
Proof: For any filters I and J of S. then θI∩J⊂ θI∩θJ and let (x, y) ∈ θI∩θJ, there exist ๐ ∈
๐ผ, ๐ ๐๐ฝ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ฅ ∧ ๐ = ๐ฆ ∧ ๐ ๐๐๐ ๐ฅ ∧ ๐ = ๐ฆ ∧ ๐ ⇒ ∃๐ ∈ ๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐ ≤ ๐, ๐ ≤ ๐ ๐๐๐ ๐ฅ ∧
๐ = ๐ฆ ∧ ๐. Now since ๐ ∈ ๐ผ ∩ ๐ฝ, ๐ค๐ โ๐๐ฃ๐ (๐ฅ, ๐ฆ) ∈ ๐๐ผ ∩ ๐ฝ. Hence FโผθF is a lattice
homomorphism. ๐ด๐๐ ๐ ๐๐๐ก ๐ผ ๐๐๐ ๐ฝ ∈ ๐น0(๐) ๐๐๐ ๐๐ผ ⊂ ๐๐ฝ. Let ๐ฅ ∈ ๐ผ, choose ๐ฆ ∈ ๐ฝ and then there
exist ๐ง ∈ ๐ such that ๐ฅ ≤ ๐ง and ๐ฆ ≤ ๐ง. ๐๐๐ค (๐ฅ, ๐ฆ) ∈ ๐๐ผ ⊂ ๐๐ฝ and since ๐ง ๐๐ฝ, we have ๐ฅ ๐๐ฝ and
therefore ๐ผ ⊂ ๐ฝ and hence FโผθF is a lattice isomorphism of F0(S) onto the sublattice {θF: F
∈F0(S)} of C(S).
Remark: The lattice C(S) of all congruencies on a semilattice S need not be distributive, even
when S is distributive. For example let ๐ = ๐ซ+ ∪ {0, ๐, ๐}, with partial order given by 0 &lt; ๐ &lt; ๐,
0 &lt; ๐ &lt; ๐, for all ๐ ∈ ๐ซ+ and ๐ &lt; ๐ ๐คโ๐๐ ๐๐ฃ๐๐ ๐ − ๐ is positive for all ๐, ๐ ∈ ๐ซ+. Then S
becomes distributive semilattice, with a∧b=glb{a, b} , ∀ a, b∈ S.
Let θ ={(a, b), (0,a), (0,b), (b,a), (a,0), (b,0) }∪ {(x,x):x∈S},
α = {(x, y)∈SxS:x∧a=y∧a}, β= {(x, y)∈SxS :x∧b=y∧b}.
Define θ∧α=θ∩α={(x,y)∈SxS:(x,y)∈θ and (x,y)∈α} and θ∨α=θοα={(x,y)∈SxS: ∃z∈S such
that=(x,z)∈α and (z,y)∈θ}.Then (θ∨α)∧β≠(θ∧β)∨(α∧β), Since (1, b)∈β∧(θ∨α) and (1,
b)∉(θ∧β)∨(α∧β).