Free Commutative Monoid is Commutative Monoid

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