TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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A zero-constant formal series has a compositional inverse exactly when its linear coefficient is a unit

Statement

Let $R$ be a commutative ring and $f \in x R ⟦ x ⟧$ . There is a unique $g \in x R ⟦ x ⟧$ such that

$f \circ g = x = g \circ f$

if and only if $[x] f$ is a unit in $R$ . In the zero ring the assertion holds with the unique zero series; when $R$ is nonzero, a zero linear coefficient cannot satisfy the criterion.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

If $g$ and $h$ both have zero constant coefficient then $(f \circ g) \circ h = f \circ (g \circ h)$ ; also $f \circ x = f$ and $x \circ f = f$ (Substitution by a zero-constant series is a ring homomorphism, and composition is associative when both inner series have zero constant coefficient).

Proof

technique · recursive coefficient construction

1.1

If $f \circ g = x$ , then the coefficient of $x$ is $[x] f [x] g = 1$ . Thus $[x] f$ is a unit. The same equation also determines $[x] g$ as its inverse.

given

1.2

Conversely write $f = a_{1} x + a_{2} x^{2} + \dots$ with $a_{1}$ a unit. Choose $b_{1} = a_{1}^{- 1}$ . After $b_{1}, \dots, b_{n - 1}$ have been chosen, the coefficient of $x^{n}$ in $f \circ (b_{1} x + \dots + b_{n} x^{n})$ is $a_{1} b_{n} + c_{n}$ , where $c_{n}$ depends only on the earlier $b_{j}$ . Set $b_{n} = - a_{1}^{- 1} c_{n}$ . The resulting $g$ has $f \circ g = x$ , and the same equations show that it is the unique left inverse.

given

2.1

Apply the construction to $g$ : its linear coefficient $a_{1}^{- 1}$ is a unit, so there is $h$ with $g \circ h = x$ . Associativity gives $f = f \circ x = f \circ (g \circ h) = (f \circ g) \circ h = x \circ h = h$ . Thus $g \circ f = x$ as well, and any two-sided inverse is the already unique solution of $f \circ g = x$ .

step 1.2givenF1

3.1

In the zero ring, $x = 0$ and the sole series is its own inverse. In a nonzero ring, $0$ is not a unit, so a zero linear coefficient fails necessity. Together with steps 1.1-2.1 this proves the equivalence and uniqueness.

step 1.1step 1.2step 2.1given∎

Depends on

Used by

Dependency tree · two levels

8 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

Benjamin Sambale, An Invitation to Formal Power Series (standard reference, not scraped)
Herbert S. Wilf, generatingfunctionology (standard reference, not scraped)