Set of Monomials is Closed Under Multiplication

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


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

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

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

Their product is:


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

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.