By Ralph Freese, Ralph McKenzie
Freese R., McKenzie R. Commutator thought for congruence modular types (CUP, 1987)(ISBN 0521348323)(O)(174s)
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Extra resources for Commutator theory for congruence modular varieties
Example text
Un ) = r(Fi(x)), Fi (r(x1 ), . . , r(xn )) /∆. ˆ then we have We can choose 0 = x, x /∆ as the identity element of Q; a + b = d(a, 0, b) for a, b ∈ Q. What remains is to show that for Fi , u, x as above with say π(xj ) = aj , uj for 1 ≤ j ≤ n, that we have π(Fi (x)) = Fi (a) + Ti (u), Fi (u) . This is equivalent to showing r(Fi(x)), Fi (x) /∆ = Fi (rx), Fi (x) /∆ + r(Fi (x)), Fi (rx) /∆. If we use that 0 = Fi (rx), Fi (rx) /∆ and a + b = d(a, 0, b), and that d(y, z, z) = y when y ζ z, then the computation checks out.
Xn−1 ) [θ, θ] t(y, y1, . . , yn−1 ) for any y ∈ A with x θ y. Applying this to the sixth variable of the above equation, we get mi+1 (mi (y, y, y, x), y, x, mi(y, y, y, x)) [θ, θ] mi+1 (mi (y, y, y, y), y, x, mi(x, x, x, x)) = mi+1 (y, y, x, x). Thus qi+1 (x, y, y) [θ, θ] mi+1 (y, y, x, x). Now suppose i is even and that (1) holds for i. Then qi+1 (x, y, y) [θ, θ] mi+1 (mi (y, y, x, x), x, y, mi(y, y, x, x)) as before. Now mi+1 (mi (y, y, x, x),x, x, mi (y, y, x, x)) = mi (y, y, x, x) = mi+1 (y, y, x, x) = mi+1 (mi (y, y, y, y), y, x, mi(x, x, x, x)).
4 showed that (8) holds if V is modular. If (8) holds let F = FV(x, y, z), and let α = Cg(x, y), β = Cg(y, z), and γ = Cg(x, z) in Con F. Since x, z is in the left side of (8) it is in the right. 4 yields terms q0 , . . 4(1)–(6). Hence V is modular. 4 shows that any Gumm difference term can serve as the p(x, y, z) of that theorem. One of the exercises proves the converse. Exercises 1. Prove for α, β ∈ Con A, k < ω that α ◦ β ⊆ β ◦ α ◦ [β]k . 2. ) 50 6. PERMUTABILITY 2. 2. to prove for all m, m ∈ ω α ◦ β ⊆ β ◦ [α]m ◦ [β]n ◦ α.