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Home / Lattices and Boolean Algebras / Show that the cardinal product of two self dual posets is self dual

# Show that the cardinal product of two self dual posets is self dual

Proof: Let A and B be the given self dual posets.

Let f: AA and g: BB be the isomorphisms.

Now we define h: A × BA × B, such that,

h((a, b)) = (f(a), g(b))

then h is well defined, one – one map as

(a1, b1) = (a2, b2)

⇔ a1 = a2, b1 = b2

f (a1) = f (a2), g (b1) = g (b2)

⇔ ( f (a1, g(b1)) = ( f (a2), g (b2))

⇔ h((a1, b1)) = h((a2, b2))

Again for any (x, y) ∈ A × B, since

h (f -1(x), g –1(y)) = (ff –1 (x), gg –1(y)) = (x, y)

and f -1, g -1 exist as f, g are one-one onto, we find h is onto.

Again (a1, b1) ≤ (a2, b2)

⇔ a1 ≤ a2, b1 ≤ b2

f (a1) ≥ f (a2), g (b1) ≥ g (b2)

⇔ ( f (a1, g(b1)) ≥ ( f (a2), g (b2))

h((a1, b1)) ≥ h((a2, b2))

Thus h is the required isomorphism and A × B ≅ A × B. (Proved)

## Prime ideals and theorem and problem

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