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 $M$ be the set of all mononomials on the indexed set $\left\{{X_j: j \in J}\right\}$.

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:
 * $\left\langle{ k^1 + k^2 }\right\rangle_j := k^1_j + k^2_j$

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.

Proof of Closure
Let $\left\langle{ k^1 + k^2 }\right\rangle_j = k^1_j + k^2_j \neq 0$.

By definition of multiindex, 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 $k^1_j,\ k^2_j \ge 0$, we have $k^1_j + k^2_j \ge 0$.

Therefore $k^1 + k^2$ has finitely many non-zero entries, and these are all positive, and multiindices are closed under addition.

Proof of Associativity
Using associativity of integer addition, we have:


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

So addition of multiindices is associative.

Proof of Commutativity
Using commutativity of integer addition, we have


 * $\left\langle{ k^1 + k^2 }\right\rangle_j = k^1_j + k^2_j = k^2_j + k^1_j = \left\langle{ k^2 + k^1 }\right\rangle_j$

So addition of multiindices is commutative.

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


 * $\left\langle{ e_M + k^1 }\right\rangle_j = \left\langle{e_M}\right\rangle_j + k^1_j = \left\langle{k^1}\right\rangle_j$

so $e_M$ is an identity for the set of mononomials.