Integer-valued finite formal sums of words form unital convolution rings

Regard ${\mathbb{Z}}^{(X^{\ast })}$ as the free abelian group on the words. Define the product by bilinearly extending concatenation of basis words, equivalently by the displayed coefficient formula.

A finite word $w$ has only finitely many cuts $w=uv$, and finite supports leave only finitely many nonzero summands. Hence every coefficient sum is finite, and the product has finite support contained in the finite set of concatenations of support words.

For $\alpha ,\beta ,\gamma$, both coefficients of $(\alpha \beta )\gamma$ and $\alpha (\beta \gamma )$ are the sum of $\alpha (u)\beta (v)\gamma (r)$ over triples with $uvr=w$. The two bracketings are bijective reindexings of the same finite set, so [L3] proves associativity.

Splitting finite sums termwise proves $(\alpha +\beta )\gamma =\alpha \gamma +\beta \gamma$ and $\alpha (\beta +\gamma )=\alpha \beta +\alpha \gamma$. Together with the pointwise abelian-group structure over $\mathbb{Z}$, these are both distributive laws.

Let ${\delta }_{[]}$ be $1$ on the empty word and $0$ elsewhere. The only contributing cut with a nonzero ${\delta }_{[]}$ factor is $w=[]w$ or $w=w[]$, so ${\delta }_{[]}\alpha =\alpha =\alpha {\delta }_{[]}$. This verifies the multiplicative identity and all ring laws. If $X=\mathrm{\varnothing }$, then $X^{\ast }=\{[]\}$ and the coefficient at $[]$ identifies the ring with $\mathbb{Z}$.