CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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$x$ has no multiplicative inverse in $R ⟦ x ⟧$ although it is invertible in $K ((x))$ when $R = K$ is a field

Counterexample

Let $K$ be a field. The series $x$ is not a unit in $K ⟦ x ⟧$ , but its image is a unit in $K ((x))$ with inverse $x^{- 1}$ .

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

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

[F2]

Power series embed in Laurent series, and a nonzero Laurent series $h = x^{v_{x} (h)} u$ has inverse $x^{- v_{x} (h)} u^{- 1}$ ( $K ⟦ x ⟧$ embeds in $K ((x))$ as the nonnegative-order subring; every nonzero Laurent series is uniquely $x^{v_{x} (h)} u$ with $u \in K ⟦ x ⟧^{\times}$ and inverse $x^{- v_{x} (h)} u^{- 1}$ ).

Verification

technique · compare constant coefficient and Laurent order

1.1

The constant coefficient of $x$ is $0$ , which is not a unit in the field $K$ , so the unit criterion excludes a power-series inverse. Equivalently, every product $x g$ has constant coefficient $0$ .

givenF1

1.2

In $K ((x))$ , negative exponents are permitted and $x x^{- 1} = 1 = x^{- 1} x$ . Thus passing to Laurent series changes the answer.

givenF2

2.1

Steps 1.1 and 1.2 exhibit the claimed contrast.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 11 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)

xx has no multiplicative inverse in R⟦x⟧R[[x]] although it is invertible in K((x))K((x)) when R=KR=K is a field