CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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A two-element idempotent monoid is an algebra for the free-monoid monad but is not free

Counterexample

Let $M = {1, e}$ with $1$ a two-sided identity and $e^{2} = e$ . This is an algebra for the free-monoid monad, but it is not a free algebra.

Facts & Assumptions

Given: The displayed multiplication on $M$ .

[L1]

A monoid is a set with associative multiplication and a two-sided identity (Semigroup and monoid).

[L2]

The algebras for the free-monoid monad are monoids, and its free algebras are free word monoids (The free-monoid monad has monoids as its Eilenberg–Moore algebras).

Verification

technique · direct

1.1

The multiplication table is $1 \cdot 1 = 1$ , $1 \cdot e = e \cdot 1 = e$ , and $e \cdot e = e$ . The element $1$ is an identity, and every triple product containing $e$ evaluates to $e$ , so multiplication is associative.

2.1

Hence $M$ is a monoid and therefore an Eilenberg–Moore algebra by [L2].

L2step 1.1

3.1

The free monoid on the empty set is a singleton. On a nonempty set, choosing a generator $x$ yields the distinct words $[], [x], [x, x], \dots$ , so the free monoid is infinite. Therefore the two-element algebra $M$ is not free.

L2step 2.1∎

Depends on

Used by

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