Set of Monomials is Closed Under Multiplication

Theorem

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

$\ds \paren {\prod_{j \mathop \in J} X_j^{k_j} } \circ \paren {\prod_{j \mathop \in J} X_j^{k_j'} } = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$

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

Proof

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

Their product is:

$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$

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 monomial.

$\blacksquare$