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Prime ideals and theorem and problem

Definition: An ideal A of a lattice L is called a prime ideal of L if A is properly contained in L and whenever a ^ b ∈ A then a ∈ A or b ∈ B.

Theorem: Prove that a lattice L is a chain if and only if all ideal in are prime.

Proof: Let us suppose that L is chain and A is any proper ideal in L. We have to show that A is prime ideal.

Let a ^ b ∈ A, we will show that a ∈ A or b∈ A since L is chain. So A and B are must be comparable.

i.e., a ≤ b or b ≤ a

Suppose a ≤ b, then a ^ b = a

∴ a ^ b ∈ A ⇒ a ∈ A

Again, let b ≤ a then a ^ b = b

∴ a ^ b ∈ A ⇒ b ∈ A

Thus a ^ b ∈ A ⇒ a ∈ A or b ∈ A

Therefore A is prime ideal.

Conversely, suppose every proper ideal A of L is prime ideal of L. We have to show that L is chain. For this purpose, we need to show that, every element of L is comparable to each other.

Let a and b be any arbitrary elements of L and

A = {x ∈ L | x ≤ a ^ b}

Then A is principle ideal of L. A = (a ^ b) and it is proper ideal of L.

∴ A is prime ideal.

So a ^ b ∈ A, A is prime, thus a ∈ A or b ∈ A

⇒ a ≤ a ^ b or b ≤ a ^ b

But by property of join (^) we know

a ^ b ≤a and a ^ b ≤ b

∴ a ≤ a ^ b ≤a or b ≤ a ^ b ≤ b

⇒ a = a ^ b     or b = a ^ b

⇒ a ≤ b or b≤a

Thus ∀ a, b ∈ L, we have a ≤ b or b ≤ a

Hence L is a chain. (Proved)

Alternative Proof: Let L be a chain. Let A be any proper ideal of L. If

a ^ b ∈ A then as a, b are in a chain, they are comparable. Let a ≤ b. Then a ^ b  = a.

Conversely, let every ideal in A be prime. To show that L is a chain,   let a, b ∈ L be any elements.

Let A = {x ∈ L | x ≤ a ^ b} then A is easily seen to be an ideal of L. Thus A is a prime ideal.

Now a ^ b ∈ A, A is prime, thus a ∈ A or b ∈ A

⇒ a ≤ a ^ b or b ≤ a ^ b

⇒ a ^ b ≤ a ≤ a ^ b

Or a ^ b ≤ b ≤ a ^ b ⇒ a = a ^ b or b = a ^ b

⇒ a ≤ b or b ≤ a ⇒ L is chain. (Proved)

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