Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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An incidence function is convolution-invertible if and only if every diagonal value is a unit

Statement

Let PP be any locally finite poset, possibly infinite, let RR be a commutative ring, and let fI(P,R)f\in I(P,R). Then ff is invertible under convolution if and only if f(x,x)f(x,x) is a unit of RR for every xPx\in P.

Facts & Assumptions

Given: A locally finite poset PP, a commutative ring RR, and fI(P,R)f\in I(P,R).

[L1]

If every diagonal value of ff is a unit, the recursive interval formulas construct a two-sided convolution inverse (If every diagonal value of an incidence function is a unit, recursive interval formulas construct both a left and a right convolution inverse).

[F2]

The convolution identity satisfies δ(x,x)=1R\delta(x,x)=1_R (The delta and zeta incidence functions).

Proof

technique · direct
1.1

Suppose ff has a convolution inverse gg. Evaluating fg=δf*g=\delta at (x,x)(x,x) gives f(x,x)g(x,x)=1Rf(x,x)g(x,x)=1_R, and evaluating gf=δg*f=\delta gives g(x,x)f(x,x)=1Rg(x,x)f(x,x)=1_R. Thus f(x,x)f(x,x) is a unit for every xx.

F1F2
1.2

Conversely, if every f(x,x)f(x,x) is a unit, [L1] constructs a two-sided convolution inverse of ff.

L1
2.1

Steps 1.1 and 1.2 prove both directions of the criterion.

step 1.1step 1.2

Depends on

Used by

Dependency tree · next 3 levels

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