Let $R=k[x_1,\ldots,x_n]$ for some field $k$. Let $I\subset R$ be a monomial ideal. That is, $I$ is generated by monomials in $R$. I want to show that the monomials in $I$ form a $k$-vector space basis for $I$. It is clear to me that the monomials are $k$-linearly independent. I am stuck on showing that every element of $I$ can be written as a $k$-linear combination of the monomials in $I$. More specifically, I want to prove:
Let $f\in R$. Then $f\in I$ if and only if every monomial of $f$ belongs to $I$.
My work so far: the backward direction is obvious. So let $f\in I$. For any polynomial $p\in R$, we know $p$ is a unique $k$-linear combination of monomials. Define the support of $p$: $$ \text{supp}(p)=\{u\in R:u\text{ is a monomial in $p$}\}. $$ We need to show $\text{supp}(f)\subset I$. Since $I$ is a monomial ideal, there are monomials $u_i\in I$ and polynomials $h_i\in R$ such that $f=\sum_{i=1}^sh_iu_i$. Here is the part I don't quite understand:
It follows that $\text{supp}(f)\subset\bigcup_{i=1}^s\text{supp}(h_iu_i)$.
From here it is straightforward: each $\text{supp}(h_iu_i)\subset I$ since each $u_i\in I$ and $I$ absorbs multiplication. But how do we know $\text{supp}(f)\subset\bigcup_{i=1}^s\text{supp}(h_iu_i)$? What is a rigorous way to see this?
EDIT: Is it because two polynomials are equal if and only if their coefficients on distinct monomials (after combining like-terms) are equal? Then every monomial on the LHS of $f=\sum_{i=1}^sh_iu_i$ must appear on the RHS as well?