root-generated does not imply integral: counterexample at W strictly containing R

The statement is false (found 2026-08-08). The catalogue plan had this row in its implications table: root-generated ⟹ integral. The proof in the cell was: “b(x,x) = 2 ∈ R on a generating family forces 𝔰(b) ≤ R”. That proof assumes a root norm. Roots have no fixed norm. The statement fails on BilinMod(R, W) when W is larger than R.

Counterexample. Let R = ℤ, W = ℚ, L = ℤ, and b(x,y) = xy/3.

• Take α = 1. Then b(α,α) = 1/3, which is regular.
• The reflection s_α is −id. It is an isometry of L. So α is a root.
• Therefore ⟨Rt(L)⟩ = L, and L is root-generated.
• But 𝔰(b) = (1/3), which is not in ℤ. So L is not integral.

Every lattice of rank one is root-generated, because −id is always an isometry. This is enough to break any implication from root generation to a condition on values.

Where the statement is still true. On Lattices(R) = BilinMod(R, R), the form b has values in R. Integrality is then automatic. The implication is true but empty.

Why this was easy to miss. The plan states its property atoms for a general W. Readers check them at W = R. At W = R the statement looks correct.

Why this is in memory and not in the plan. A plan lists work to do. It does not list statements that are false. The row was deleted. Without this record, a later pass adds the implication again, because the b(x,x) = 2 proof looks obvious to a reader who assumes ADE roots.

The correct definition. Rt(L) = { α ∈ L : s_α ∈ O(L) }. A root is a vector whose reflection is an isometry. Norm has no part in it.

Related: a term of art belongs to its own field not to the nearest abstraction.