Set of Monomials is Closed Under Multiplication

Theorem
Let $M$ be the set of all mononomials on the set $\left\{ {X_j: j \in J}\right\}$, with multiplication $\circ$ defined by:


 * $\displaystyle \left({\prod_{j \mathop \in J} X_j^{k_j} }\right) \circ \left({\prod_{j\in J}X_j^{k_j' }}\right) = \left({\prod_{j \mathop \in J} X_j^{k_j + k_j'} }\right)$

Then $M$ is closed under $\circ$.

Proof
Let $\displaystyle m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two mononomials.

Their product is:


 * $\displaystyle m_1 \circ m_2 = \left({\prod_{j \mathop \in J} X_j^{k_j + k_j'} }\right)$

If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero).

Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a mononomial.