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A class of Congruencies on Semilattice

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A class of Congruencies on Semilattice
For any subsemilattice L of a semilattice S define, ๐œƒ๐ฟ = (๐‘ฅ, ๐‘ฆ) ∈ ๐‘†x๐‘†: ๐‘ฅห„๐‘Ž =
๐‘ฆห„๐‘Ž ๐‘“๐‘œ๐‘Ÿ ๐‘ ๐‘œ๐‘š๐‘’ ๐‘Ž ∈ ๐ฟ}. then ๐œƒ๐ฟ is congruence on ๐‘† ๐‘Ž๐‘›๐‘‘ ๐ฟ is contained in a single equivalence
class of ๐œƒ๐ฟ. ๐œ‹๐ฟ denotes the natural epimorphism of S on the quotient semilattice ๐‘†/๐œƒ๐ฟ .
If S is distributive semilattice and ๐น is a filter of ๐‘†, then ๐œƒ๐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œƒ๐ฟ is distributive for every subsemilattice ๐ฟ ๐‘œ๐‘“ ๐‘†.
Proof: (i) ๏ƒž (ii) Let S be distributive semilattice and L a subsemilattice of S. Let ๐‘ฅ, ๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘ง ∈
๐‘† such that ๐œ‹๐ฟ(๐‘ฅ) ∧ ๐œ‹๐ฟ(๐‘ฆ) ≤ ๐œ‹๐ฟ(๐‘ง) ⇒ ∃ ๐‘Ž ∈ ๐ฟ ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅห„๐‘ฆห„๐‘Ž = ๐‘ฅห„๐‘ฆห„๐‘งห„๐‘Ž
⇒ (๐‘ฅห„๐‘Ž) ห„(๐‘ฆ ห„๐‘Ž) ≤ ๐‘ง ⇒ ∃ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ ∈ ๐‘† ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅห„๐‘Ž ≤ ๐‘,
๐‘ฆ ห„๐‘Ž ≤ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ห„๐‘ = ๐‘ง.
Now, ๐œ‹๐ฟ(๐‘ฅห„๐‘Ž) ≤ ๐œ‹๐ฟ(๐‘), ๐œ‹๐ฟ(๐‘ฆห„๐‘Ž) ≤ ๐œ‹๐ฟ(๐‘) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐ฟ(๐‘) ห„๐œ‹๐ฟ(๐‘) = ๐œ‹๐ฟ(๐‘ง).
Therefore ๐‘†/๐œƒ๐ฟ is distributive for every subsemilattice ๐ฟ ๐‘œ๐‘“ ๐‘†.
(ii) ๏ƒž (i) Let ๐‘Ž, ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ ∈ ๐‘† ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘Žห„๐‘ ≤ ๐‘. Then choose ๐‘‘ ∈ ๐‘† such that ๐‘Ž, ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ ๏ƒŽ
(๐‘‘]๐‘Ž๐‘›๐‘‘ ๐‘๐‘ข๐‘ก ๐ฟ = (๐‘‘]. Since S⁄θL is distributive, there exists ๐‘’, ๐‘“ ∈ ๐‘† s.t ๐œ‹๐ฟ(๐‘Ž) ≤
๐œ‹๐ฟ(๐‘’), ๐œ‹๐ฟ(๐‘) ≤ ๐œ‹๐ฟ(๐‘“)๐‘Ž๐‘›๐‘‘ ๐œ‹๐ฟ(๐‘’)ห„ ๐œ‹๐ฟ(๐‘“) = ๐œ‹๐ฟ(๐‘). Again there exist ๐‘ฅ, ๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘ง ∈ ๐น such that
๐‘’ห„๐‘“ห„๐‘ฅ = ๐‘ห„๐‘ฅ, ๐‘’ห„๐‘Žห„๐‘ฆ = ๐‘Žห„๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘“ห„๐‘ห„๐‘ง = ๐‘ห„๐‘ง which implies that ๐‘’ห„๐‘“ห„๐‘ฅ = ๐‘, ๐‘’ห„๐‘Ž =
๐‘Ž ๐‘Ž๐‘›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œƒ๐ฟ, then ๐ผ = {๐‘ฅ ∈ ๐‘†: ๐œ‹๐ฟ(๐‘ฅ) ∈
๐ผ’} is an ideal of S. ๐ผ’ is a proper ideal of ๐‘†/๐œƒ๐ฟ if and only if ๐ผ ∩ ๐ฟ = ∅. Also, if ๐‘ƒ is a prime
ideal of ๐‘† such that ๐‘ƒ ∩ ๐ฟ = ∅, then ๐‘ƒ’ = { ๐œ‹๐ฟ(๐‘ฅ) ๏ƒŽ ๐‘†/๐œƒ๐ฟ: ๐‘ฅ ๏ƒŽ ๐‘ƒ} is a prime ideal of ๐‘†/๐œƒ๐ฟ.
๐‘†
Proof: Let I’ be an ideal of๐œƒ๐ฟ, and ๐ผ = {๐‘ฅ ∈ ๐‘†: ๐œ‹๐ฟ (๐‘ฅ) ๏ƒŽ ๐ผ’}, let ๐‘Ž ∈ ๐ผ, ๐‘ ∈ ๐‘† ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ ≤
๐‘Ž. ๐‘‡โ„Ž๐‘’๐‘› ๐‘ ∧ ๐‘ฅ ≤ ๐‘Ž ๐‘“๐‘œ๐‘Ÿ ๐‘ ๐‘œ๐‘š๐‘’ ๐‘ฅ ∈ ๐ฟ ⇒ ๐œ‹๐ฟ(๐‘) = ๐œ‹๐ฟ(๐‘ฅ ∧ ๐‘) ≤ ๐œ‹๐ฟ(๐‘Ž)
⇒ ๐œ‹๐ฟ(๐‘) ∈ ๐ผ ′ ๐‘Ž๐‘›๐‘‘ โ„Ž๐‘’๐‘›๐‘๐‘’ ๐‘ ∈ ๐ผ.
Therefore ๐ผ is an initial segment of S. Also, if ๐œ‹๐ฟ(๐‘Ž) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐ฟ(๐‘) ∈ ๐ผ’
⇒ ∃ ๐‘ ๏ƒŽ ๐‘† ๐‘†๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐œ‹๐ฟ(๐‘) ∈ ๐ผ’, ๐œ‹๐ฟ(๐‘Ž) ≤ ๐œ‹๐ฟ(๐‘) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐ฟ(๐‘) ≤ ๐œ‹๐ฟ(๐‘)
⇒ ∃๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∈ ๐ฟ ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘Žห„๐‘ห„๐‘ฅ = ๐‘Žห„x ๐‘Ž๐‘›๐‘‘ ๐‘ห„๐‘ห„๐‘ฆ = ๐‘ห„๐‘ฆ
⇒ ๐‘Žห„๐‘ห„๐‘ฅห„๐‘ฆ = ๐‘Žห„๐‘ฅห„๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘ห„๐‘ห„๐‘ฅห„๐‘ฆ = ๐‘ห„๐‘ฅห„๐‘ฆ, and ⇒ ∃๐‘‘ ∈ ๐‘†, ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘Ž ≤ ๐‘‘, ๐‘ ≤ ๐‘‘ ๐‘Ž๐‘›๐‘‘
๐‘‘ห„๐‘ห„๐‘ฅห„๐‘ฆ = ๐‘‘ห„๐‘ฅห„๐‘ฆ. ⇒ ๐œ‹๐ฟ(๐‘‘) ≤ ๐œ‹๐ฟ(๐‘) ๐‘Ž๐‘›๐‘‘ โ„Ž๐‘’๐‘›๐‘๐‘’ ๐‘‘ ∈ ๐ผ. ∴ ๐ผ’ is an ideal.
If ๐‘Ž ∈ ๐ผ ∩ ๐ฟ, then for any ๐‘ฅ ∈ ๐‘†, ๐œ‹๐ฟ (๐‘ฅ) = ๐œ‹๐ฟ( ๐‘ฅห„ ๐‘Ž ) , and this implies ๐‘†/๐œƒ๐ฟ = ๐ผ’.
Conversely if๐‘†/๐œƒ๐ฟ = ๐ผ’, then for any ๐‘ฅ ∈ ๐ฟ, ๐œ‹๐ฟ(๐‘ฅ) ๏ƒŽ ๐ผ’ and therefore ๐ผ ∩ ๐ฟ ≠ ∅.
Also let P be prime ideal of S such that ๐‘ƒ ∩ ๐ฟ = ∅, then ๐‘ƒ’ is proper ideal of ๐‘†/๐œƒ๐ฟ and
Let ๐œ‹๐ฟ(๐‘Ž) ∧ ๐œ‹๐ฟ(๐‘) ∈ ๐‘ƒ′ , ๐œ‹๐ฟ(๐‘Ž ∧ ๐‘) ∈ ๐‘ƒ′ . Then ๐‘Ž ∧ ๐‘ ∈ ๐‘ƒ ๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘–๐‘›๐‘๐‘’ ๐‘ƒ ๐‘–๐‘  ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘Ž ∈ ๐‘ƒ ๐‘œ๐‘Ÿ ๐‘ ∈
๐‘ƒ. Hence ๐œ‹๐ฟ(๐‘Ž) ∈ ๐‘ƒ′ ๐‘œ๐‘Ÿ ๐œ‹๐ฟ(๐‘) ∈ ๐‘ƒ′ . Therefore ๐‘ƒ’ is prime
Theorem: 3.3. Let ๐ฟ1 ๐‘Ž๐‘›๐‘‘ ๐ฟ2 be two sub semilattices of S such that ๐ฟ1 ๏ƒŒ ๐ฟ2 and also let
๐‘“: S⁄θL1 ๏‚ฎ S⁄θL2 be the epimorphism defined by ๐‘“ ๐œ‹๐ฟ1(๐‘ฅ) = ๐œ‹๐ฟ2(๐‘ฅ), If ๐‘“ is injection, then
๐‘ƒ ∩ ๐ฟ1 = ∅ implies ๐‘ƒ ∩ ๐ฟ2 = ∅, for any prime ideal P of S,
Proof: If P is a prime ideal of S such that ๐‘ƒ ∩ 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‘ƒ ∩ ๐ฟ2 = ∅.
Theorem: 3.4. Let ๐น be a filter of S, then the following are equivalent:
(i)
๐น ๐‘–๐‘  ๐‘Ž ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘“๐‘–๐‘™๐‘ก๐‘’๐‘Ÿ.
(ii)
๐ผ = {๐œ‹๐น(๐‘ฅ): ๐‘ฅ ∈ ๐‘† − ๐น} ๐‘–๐‘  ๐‘Ž๐‘› ๐‘–๐‘‘๐‘’๐‘Ž๐‘™ ๐‘œ๐‘“ ๐‘†/๐œƒ๐น .
(iii)
๐‘†/๐œƒ๐น โ„Ž๐‘Ž๐‘  ๐‘Ž ๐‘ข๐‘›๐‘–๐‘ž๐‘ข๐‘’ ๐‘š๐‘Ž๐‘ฅ๐‘–๐‘š๐‘Ž๐‘™ ๐‘–๐‘‘๐‘’๐‘Ž๐‘™.
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œ‹๐น(๐‘ฅ) ∈
๐ผ ๐‘Ž๐‘›๐‘‘ ๐œ‹๐น(๐‘ฆ) ∈ ๐‘†/๐œƒ๐น ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐œ‹๐น(๐‘ฆ) ≤ ๐œ‹๐น(๐‘ฅ) ⇒ ∃ ๐‘Ž ∈ ๐น ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฆ ∧ ๐‘Ž = ๐‘ฅ ∧ ๐‘ฆ ∧
๐‘Ž ⇒ ๐‘ฆ ∧ ๐‘Ž ≤ ๐‘ฅ ⇒ ๐œ‹๐น(๐‘ฆ) = ๐œ‹๐น(๐‘ฆ ∧ ๐‘Ž) ≤ ๐œ‹๐น(๐‘ฅ) ⇒ ๐œ‹๐น(๐‘ฆ) ≤ ๐œ‹๐น(๐‘ฅ)
∴ I is initial segment.
Let ๐œ‹๐น(๐‘ฅ), ๐œ‹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„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅ ≤ ๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ≤ ๐‘ง.
⇒ ∃ ๐‘Ž ∈ ๐น ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅ ∧ ๐‘Ž = ๐‘ฅ ∧ ๐‘Ž ∧ ๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∧ ๐‘Ž = ๐‘ฆ ∧ ๐‘Ž ∧ ๐‘ง,
⇒ ๐‘ฅ ∧ ๐‘Ž ≤ ๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∧ ๐‘Ž ≤ ๐‘ง
⇒ ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘ฅ ∧ ๐‘Ž) ≤ ๐œ‹๐น(๐‘ง) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐น(๐‘ฆ) = ๐œ‹๐น(๐‘ฆ ∧ ๐‘Ž) ≤ ๐œ‹๐น(๐‘ง)
⇒ ๐œ‹๐น(๐‘ฅ) ≤ ๐œ‹๐น(๐‘ง) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐น(๐‘ฆ) ≤ ๐œ‹๐น(๐‘ง)
Therefore ๐ผ = {๐œ‹๐น(๐‘ฅ): ๐‘ฅ ∈ ๐‘† − ๐น} is an ideal of ๐‘†/๐œƒ๐น .
(ii)⇒ (iii) ๐น๐‘œ๐‘Ÿ ๐‘Ž๐‘›๐‘ฆ ๐‘Ž ∈ ๐น, ๐œ‹๐น(๐‘Ž) is the greatest element of ๐‘†/๐œƒ๐น. Hence, every ideal of ๐‘†/๐œƒ๐น is
contained in {๐œ‹๐น(๐‘ฅ): ๐‘ฅ ∈ ๐‘† − ๐‘ƒ}.Therefore ๐‘†/๐œƒ๐น has a unique maximal.
(iii)⇒ (i): Let G be the unique maximal ideal of ๐‘†/๐œƒ๐น and let ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ๐œ– ๐‘† − ๐น. Since the
principal ideals of ๐‘†/๐œƒ๐น generated by ๐œ‹๐น(๐‘ฅ) ๐‘Ž๐‘›๐‘‘ ๐œ‹๐น(๐‘ฆ) are proper, it follows that
๐œ‹๐น(๐‘ฅ)๐‘Ž๐‘›๐‘‘ ๐œ‹๐น(๐‘ฆ) ∈ ๐บ. Therefore there exists ๐œ‹๐น(๐‘ง) ∈ ๐บ ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐œ‹๐น(๐‘ฅ) ≤ ๐œ‹๐น(๐‘ง) and
๐œ‹๐น(๐‘ฆ) ≤ ๐œ‹๐น(๐‘ง), ๐‘ ๐‘œ ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅห„๐‘งห„๐‘Ž = ๐‘ฅห„๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ฆห„๐‘งห„๐‘Ž = ๐‘ฆห„๐‘Ž ๐‘“๐‘œ๐‘Ÿ ๐‘ ๐‘œ๐‘š๐‘’ ๐‘Ž๐œ–๐น.
๐‘‡โ„Ž๐‘’๐‘Ÿ๐‘’๐‘“๐‘œ๐‘Ÿ๐‘’ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘’๐‘ฅ๐‘–๐‘ ๐‘ก๐‘  ๐‘๐œ–๐‘† ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅ ≤ ๐‘, ๐‘ฆ ≤ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ห„๐‘งห„๐‘Ž = ๐‘ห„๐‘Ž,
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œƒ๐น is the two
element chain.
Proof: Suppose F is a maximal filter, then for any ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ๐œ– ๐น, ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘ฆ) and that for
any ๐‘ฅ ∈ ๐น ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∈ ๐‘†, ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘ฆ) implies ๐‘ฆ ∈ ๐น.
Now, ๐‘™๐‘’๐‘ก ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∈ ๐‘† − ๐น, Since ๐น is maximal, there exist ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘๐œ– ๐น ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก
๐‘Žห„๐‘ฅ ≤ ๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘ห„๐‘ฆ ≤ ๐‘ฅ . ๐ป๐‘’๐‘›๐‘๐‘’ ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘Žห„๐‘ฅ) ≤ ๐œ‹๐น(๐‘ฆ) = ๐œ‹๐น(๐‘ห„๐‘ฆ) ≤ ๐œ‹๐น(๐‘ฅ).
๐‘†๐‘œ ๐‘กโ„Ž๐‘Ž๐‘ก ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘ฆ).
Conversely, suppose that ๐‘†/๐œƒ๐น is the two element chain. Suppose ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ ∈ ๐‘† − ๐น. Choose an
element ๐‘ง ∈ ๐น, then we have ๐œ‹๐น(๐‘ง) ≠ ๐œ‹๐น(๐‘ฅ) and Hence ๐œ‹๐น(๐‘ฅ) = ๐œ‹๐น(๐‘ฆ) so that there exists
๐‘Ž ∈ ๐น ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅห„๐‘Ž = ๐‘ฆห„๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘‡โ„Ž๐‘’๐‘Ÿ๐‘’๐‘“๐‘œ๐‘Ÿ๐‘’ 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 FโŸผθF 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‘Ž ∈
๐ผ, ๐‘ ๐œ–๐ฝ ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘ฅ ∧ ๐‘Ž = ๐‘ฆ ∧ ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ฅ ∧ ๐‘ = ๐‘ฆ ∧ ๐‘ ⇒ ∃๐‘ ∈ ๐‘† ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘Ž ≤ ๐‘, ๐‘ ≤ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ฅ ∧
๐‘ = ๐‘ฆ ∧ ๐‘. Now since ๐‘ ∈ ๐ผ ∩ ๐ฝ, ๐‘ค๐‘’ โ„Ž๐‘Ž๐‘ฃ๐‘’ (๐‘ฅ, ๐‘ฆ) ∈ ๐œƒ๐ผ ∩ ๐ฝ. Hence FโŸผθF is a lattice
homomorphism. ๐ด๐‘™๐‘ ๐‘œ ๐‘™๐‘’๐‘ก ๐ผ ๐‘Ž๐‘›๐‘‘ ๐ฝ ∈ ๐น0(๐‘†) ๐‘Ž๐‘›๐‘‘ ๐œƒ๐ผ ⊂ ๐œƒ๐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‘œ๐‘ค (๐‘ฅ, ๐‘ฆ) ∈ ๐œƒ๐ผ ⊂ ๐œƒ๐ฝ and since ๐‘ง ๐œ–๐ฝ, we have ๐‘ฅ ๐œ–๐ฝ and
therefore ๐ผ ⊂ ๐ฝ and hence FโŸผθF 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‘Ž, ๐‘}, with partial order given by 0 < ๐‘Ž < ๐‘›,
0 < ๐‘ < ๐‘›, for all ๐‘› ∈ ๐•ซ+ and ๐‘š < ๐‘› ๐‘คโ„Ž๐‘’๐‘› ๐‘’๐‘ฃ๐‘’๐‘Ÿ ๐‘š − ๐‘› is positive for all ๐‘š, 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).
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