Free Commutative Monoid is Commutative Monoid

Theorem
The free commutative monoid on a set $\left\{{X_j: j \in J}\right\}$ is a commutative monoid.

Proof
Let $\displaystyle M$ be the set of all mononomials on the indexed set $\displaystyle \{X_j:j\in J\}$.

We are required to show that the following properties hold:

First note that using the multiindex notation described in the definition of mononomials, for $r \in \N$, $m_i=\mathbf X^{k^i}\in M$, $i=1,\ldots,r$, the product of the $m_i$ is given by


 * $m_1 \circ \cdots \circ m_r = \mathbf X^{k^1 + \cdots + k^r}$.

Here the superscripts enumerate the multiindices, and do not indicate raising to a power. So to show the closure, associativity and commutativity of mononomials under $\circ$, it is sufficient to show the corresponding properties for multiindices under addition defined by $(k^1 + k^2)_j=k^1_j + k^2_j$.

In the following $k^1, k^2, k^3$ are multiindices, that is, families of non-negative integers indexed by $J$ such that only finitely many entries are non-zero.

Closure

If $\left({k^1 + k^2}\right)_j=k^1_j+k^2_j\neq 0$. Then at least one of $k^1_j$ and $k^2_j$ must be non-zero, and this can only be true for a finite number of entries. Furthermore, since $k^1_j,\ k^2_j\geq 0$ we have $k^1_j+k^2_j\geq 0$. Therefore $k^1 + k^2$ has finitely many non-zero entries, and these are all positive, and multiindices are closed under addition.

Assosiativity

Using associativity of integer addition, we have


 * $\displaystyle \left({\left({k^1 + k^2} \right) + k^3} \right)_j = (k^1_j + k^2_j) + k^3_j = k^1_j + (k^2_j + k^3_j) = \left( {k^1 + \left({k^2 + k^3} \right)} \right)_j$

So addition of multiindices is associative.

Commutativity

Using commutativity of integer addition, we have


 * $\displaystyle \left( { k^1 + k^2} \right)_j = k^1_j + k^2_j = k^2_j + k^1_j = \left({k^2 + k^1}\right)_j$

So addition of multiindices is commutative.

Identity

Let $e_M$ be the multiindex such that $\left({e_M}\right)_j = 0$ for all $j \in J$. Then


 * $\displaystyle \left({ e_M + k^1 }\right)_j = \left({e_M}\right)_j + k^1_j = k^1_j$

so $e_M$ is a neutral element for the set of mononomials.