TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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A formal power series is a unit exactly when its constant coefficient is a unit

Statement

Let $R$ be a commutative ring and $f = \sum_{n \geq 0} a_{n} x^{n} \in R ⟦ x ⟧$ . Then $f$ is a unit in $R ⟦ x ⟧$ if and only if $a_{0}$ is a unit in $R$ .

When $a_{0}$ is a unit, the inverse $g = \sum_{n \geq 0} b_{n} x^{n}$ is unique and is determined by

$b_{0} = a_{0}^{- 1}, b_{n} = - a_{0}^{- 1} \sum_{i = 1}^{n} a_{i} b_{n - i} (n \geq 1) .$

The criterion also holds in the zero ring.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

Product coefficients satisfy $[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g$ (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[F2]

Two formal series are equal if and only if all their coefficients are equal (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[F3]

An element $u$ is a unit exactly when there is $v$ with $u v = 1 = v u$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[F4]

Zero is a unit exactly in the zero ring (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

Proof

technique · solve the inverse equations recursively

1.1

If $f g = 1$ , constant-coefficient extraction gives $a_{0} [x^{0}] g = 1$ . Commutativity gives the reverse product too, so $a_{0}$ is a unit.

givenF1F3

1.2

Conversely suppose $a_{0}$ is a unit and define $b_{n}$ by the displayed recursion. The coefficient of $f g$ at $0$ is $a_{0} b_{0} = 1$ . For $n \geq 1$ it is $a_{0} b_{n} + \sum_{i = 1}^{n} a_{i} b_{n - i} = 0$ . Hence $f g = 1$ by extensionality, and commutativity gives $g f = 1$ .

givenF1F2

2.1

Any inverse must satisfy the same constant equation and then, successively, the same equation for each $b_{n}$ ; multiplication by $a_{0}^{- 1}$ makes every coefficient unique. In the zero ring, $0 = 1$ and the unit theorem makes the same recursion and equivalence valid.

step 1.1step 1.2givenF3F4

3.1

Steps 1.1 and 1.2 prove both directions, while step 2.1 proves uniqueness and the boundary case.

step 1.1step 1.2step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)