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


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

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

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


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

If $$k_j+k_j'\neq 0$$ then either $$k_j\neq 0$$ or $$k_j'\neq 0$$ (or both are nonzero). Therefore if $$k_j+k_j'\neq 0$$ for infinitely many $$j$$, then either $$m_1$$ or $$m_2$$ is not a mononomial.