CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-07-31

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On a two-element chain, an incidence function with a zero diagonal value is not convolution-invertible

Statement refuted

Every incidence function on a finite poset is convolution-invertible.

Facts & Assumptions

Given: A two-element chain $0 < 1$ , a nonzero commutative ring $R$ , and the incidence function $f$ with $f (0, 0) = 0_{R}$ , $f (0, 1) = 0_{R}$ , and $f (1, 1) = 1_{R}$ .

[L1]

An incidence function is convolution-invertible exactly when every diagonal value is a unit (An incidence function is convolution-invertible if and only if every diagonal value is a unit).

[F1]

The convolution identity has $δ (0, 0) = 1_{R}$ (The delta and zeta incidence functions).

Counterexample

technique · direct

1.1

If $g$ were a convolution inverse, evaluation at $(0, 0)$ would give $(f * g) (0, 0) = f (0, 0) g (0, 0) = 0_{R}$ .

given

2.1

But an inverse equation requires $(f * g) (0, 0) = δ (0, 0) = 1_{R}$ , and $0_{R} \neq 1_{R}$ because the ring is nonzero.

step 1.1F1

3.1

Therefore $f$ is not invertible, in agreement with [L1] because its diagonal value $0_{R}$ is not a unit.

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

F. Gotti, Incidence Algebras, MIT 18.211 notes (standard reference, not scraped)