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Home / Lattices and Boolean Algebras / The set of all ideals of a lattice L forms a lattice under ⊆ relation.

The set of all ideals of a lattice L forms a lattice under ⊆ relation.

Or, Suppose L is a lattice, What do you mean by I(L), prove that (I(L), ⊆) is a lattice.
Proof: Let I(L) is a set of all ideals of a lattice L.
Then clearly, L ∊ I(L)
∴ I(L) is non empty.
Now, let us first, show that (I(L), ⊆) is a poset, clearly ⊆ is reflexive in I(L) , because A ⊆ A, ∀ A.
Next, ∀ A, B ∊ I(L)
A ⊆ B and B ⊆ C ⟹ A = B
∴ ⊆ is antisymetric in I(L)
Last of all, ∀ A, B , C∊ I(L)
A ⊆ B and B ⊆ C ⟹ A ⊆ C
∴ ⊆ is transitive in I(L)
Hence I(L) is a poset.
Now we have to show that I(L) is a lattice.
∀ A, B ∊ I(L), let us consider A ˄ B = A ∩ B
We know, intersection of two ideals is always an ideal.
∴ A ˄ B I(L)
Let us consider ∀ A, B ∊ I(L)
X ={x ∊ L | x ≤ a ˅ b, a ∊ A, b ∊ B}
Clearly ∀ a ∊ A
a ≤ a ˅ b, ∀ b ∊ B
∴ a ∊ X [by definition] ∴ A ⊆ X and so X is non empty.
Similarly we can show, B ⊆ X
∴ A ∪ B ⊆ X
Let us show that X is an ideal of L
Let ∀ p, q ∊ X
p ∊ X ⟹ there exists a1 ∊ A, b1 ∊ B such that p ≤ a1 ˅ b1
q ∊ X ⟹ there exists a2 ∊ A, b2 ∊ B such that q ≤ a2 ˅ b2
Now, p ˅ q ≤ (a1 ˅ b1) ˅ (a2 ˅ b2)
= (a1 ˅ a2) ˅ (b1 ˅ b2)
Since a1, a2 ∊ A and A is an ideal, so
a1 ˅ a2 ∊ A, similarly (b1 ˅ b2)∊ B
∴ p ˅ q ∊ X
Now, ∀ p ∊ X and ∀ l ∊ L
p ∊ X ⟹ there exists a1 ∊ A and there exists b1 ∊ B such that a1 ˅ b1 ——–(i)
Again, clearly p ˄ l ≤ p ————–(ii) [by property of meet] From (i) and (ii) we get,
p ˄ l ≤ a1 ˅ b1; for some a1 ∊ A, b1 ∊ B
∴ p ˄ l ∊ X
Hence X is an ideal of L
∴ X ∊ I(L)
If C is any ideal of L such that
A ⊆ C and B ⊆ C
Then ∀ a ∊ A and ∀ b ∊ B, we get,
a ∊ C and b ∊ C
⟹ X ⊆ C
Hence X is the smallest ideal of L containing A ∪ B
Therefore, X = sup {A, B} = A ˅ B
∴ A ˅ B ∊ I(L), ∀ A ∊ I(L) and ∀ B ∊ I(L)
Therefore, I(L) is a lattice which is called ideal lattice. (Proved)

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