We say a monomial ideal is one that is generated by monomials. Fix a field $k$ and consider the ring of polynomials $k[x_1,\ldots,x_n]$. Denote by $x^\alpha$ where $\alpha=(a_1,\ldots,a_n)\in\mathbb{Z}^n_{\geq0}$ to be the monomial $x^\alpha=x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\in k[x_1,\ldots,x_n]$.
I want to prove:
Let $A\subset\mathbb{Z}^n_{\geq0}$ and let $I=(x^\alpha:\alpha\in A)\subset k[x_1,\ldots,x_n]$ be a monomial ideal. Then a monomial $x^\beta$ belongs to $I$ if and only if $x^\beta$ is divisible by $x^\alpha$ for some $\alpha\in A$.
My thoughts: The fact that $x^\beta\in I$ if $x^\beta$ is divisible by some $x^\alpha$ is obvious. I'm struggling with the other direction. Suppose $x^\beta\in I$. Then we can write $$ x^\beta=\sum_{i=1}^s h_ix^{\alpha_i},\;\;h_i\in k[x_1,\ldots,x_n],\;\;\alpha_i\in A,\;\;s\in\mathbb{N}. $$ I need to show $x^\beta$ is divisible by some $x^\alpha$. My first thought is that, if we expand the RHS, then every term is divisible by some $x^{\alpha_i}$. I'd like to pick a minimal $x^{\alpha_i}$ (with respect to some monomial order, say lexicographic), and then $x^\beta$ should be divisible by that minimal $x^{\alpha_i}$. But I'm not sure if this last bit is true. I know that given some monomial order, it is true that $x^\alpha|x^\beta$ implies $\alpha\leq\beta$, but I don't know if the converse holds. How can I finish this proof?
EDIT: In fact, I know the converse can't be true, since $x_1>x_2$ but $x_2$ does not divide $x_1$. So how can we say $x^\beta$ is divisible by some $x^\alpha$ for $\alpha\in A$?