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Home / Lattices and Boolean Algebras / Product of two lattices is a lattice.

# Product of two lattices is a lattice.

Proof: Let A and B be two lattices the we already proved that A × B = {(a, b) | a ∈ A, b ∈ B} is a poset under the relation ≤ defined by

(a1 , b1) ≤ (a2 , b2 )⇔ a1 ≤ a1 in A

b1 ≤ b2 in B

We show A × B forms a lattice.

Let (a1 , b1), (a2, b2) ∈ A × B be any elements.

Then a1, a2 ∈ A and b1, b2 ∈ B.

### Since A and B are lattices {a1, a2} and {b1, b2} sup and inf in A and B respectively.

Let a1 ˄ a2 = inf{a1, a2}, b1 ˄ b2 = inf{b1, b2}

then a1 ˄ a2 ≤ a1, a1 ˄ a2 ≤ a2

b1 ˄ b2 ≤ b1, b1 ˄ b2 ≤ b2

⟹ (a1 ˄ a2, b1 ˄ b2) ≤ (a1, b1)

(a1 ˄ a2, b1 ˄ b2) ≤ ( a2, b2)

⟹ (a1 ˄ a2, b1 ˄ b2) is a lower bound of {(a1, b1), ( a2, b2)}

Suppose (c, d) is any lower bound of {(a1, b1), ( a2, b2)}

Then (c, d) ≤ (a1, b1)

(c, d) ≤ ( a2, b2)

⟹ c ≤ a1, c ≤ a2, d ≤ b1, d ≤ b2

⟹ c is a lower bound of {a1, a2} in A.

D is a lower bound of {b1, b2} in B.

⟹ c ≤ a1 ˄ a2 = inf {a1, a2}

d ≤ b1 ˄ b2 = inf{b1, b2}

⟹ (c, d) ≤ (a1 ˄ a2, b1 ˄ b2)

i.e., (a1 ˄ a2, b1 ˄ b2) is g.l.b {(a1, b1), ( a2, b2)}

Similarly by duality we can say that

(a1 ˅a2, b1 ˅b2) is l.u.b {(a1, b1), ( a2, b2)}

Hence A × B is a lattice. (Proved)

## Prime ideals and theorem and problem

Definition: An ideal A of a lattice L is called a prime ideal of L ...