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Ideals | Lattices and Boolean algebra

Definition: A non empty subset I of a lattice L is called an ideal of L if

(i) a, b ∊I ⟹a ˅ b ∊ I

(ii) a ∊ I, l ∊ L ⟹ a ˄ l ∊ I.

Example: Let {1, 2, 5, 10} be a lattice of factors of 10 under divisibility. Then {1}, {1, 2}, {1, 5}, {1, 2, 5, 10} are all the ideal of L.

Theorem: An ideal is a sublattice. Converse is not true.

Proof: Let I be an ideal of a lattice L. Let a, b ∊I then a ˅ b ∊ I

Again, a ∊ I, b ∊ I ⊆ L ⟹ a ˄ b ∊ I

Hence I is a sub lattice.

Again, Let L= {1, 2, 5, 10} and I = {5, 10} is a sublattice of L but is not an ideal as 5 ∊ I, 2 ∊ I but 2 ˄ 5 = 1 ∉ I. (Proved)

Theorem: Intersection of two ideals is an ideal.

Proof: Let A, B be two ideals of a lattice L.

Since A, B are nonempty, there exists some a ∊ A, b ∊ B.

Now a ∊ A, b ∊ B ⊆ L ⟹ a ˄ b ∊ A

Similarly a ˄ b ∊ B

Thus A ∩ B ≠ ?.

Let x, y ∊ A ∩ B be any elements

⟹ x, y ∊ A and x, y ∊ B

⟹ x ˅ y ∊ A and x ˅ y ∊ B as A b are ideals.

⟹ x ˅ y ∊ A ∩ B.

Again, if x ∊ A ∩ B and l ∊ L be any elements then x ∊ A, x ∊ B, l ∊ L

⟹ x ˄ l ∊ A and x ˄ l ∊ B

⟹ x ˄ l ∊ A ∩ B

Hence A ∩ B is an ideal. (Proved).

Theorem: Union of ideals is an ideal if and only if one of them contained in the other.

Proof: Suppose A and B are two ideals of a lattice L such that A ⊆ B or B ⊆ A.

If A ⊆ B, then A ∪ B ⊆ B

∴ A ∪ B is an ideal of L, similar case for B ⊆ A.

Conversely, let A and B be two ideals of a lattice L.

Such that A ∪ B is an ideal of L.

Suppose A ⊈ B and B ⊈ A

⟹ There exists x ∊ A such that x ∉ B there exists y ∊ B such that y ∉ A

⟹ x, y ∊ A ∪ B

⟹ x ˅ y ∊ A ∪ B

⟹ x ˅ y ∊ A and x ˅ y ∊B

If x ˅ y ∊ A, then for any y ∊B ⊆ L we get y ˄ (x ˅ y) ∊ A

⟹ y ∊ A

Which is contradiction.

Similarly x ˅ y ∊ B would lead us to the result that hence either A ⊆ B or B ⊆ A. (Proved).

Theorem: In any finite lattice, every ideal is principle.

Proof: Let, L be a finite lattice and A be any ideal of L. Then, since A is an ideal, so A is a sub lattice and this implies A is a lattice.

Further, L is finite implies A is finite.

Thus A has a greatest element. Let it be g.

Then for all x ∊ A, x ≤ g

Now, (g] = {x | x ≤ g} = A

Hence A is a principle ideal. (Proved)

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