Let $k$ be a field and consider the polynomial algebra $k[x_1,\dots,x_n]$. Suppose $I$ is a monomial ideal (generated by monomials). Since $k[x_1,\dots,x_n]$ is Noetherian we can choose a minimal generating set $\{f_1,\dots,f_k\}$ of $I$ consisting of monomials. Suppose $f_1=gh$ with $g,h$ coprime and $\deg (g), \deg (h)\geq 1$. Then is it true that $g,h\notin I$?
If the statement [If $g\in I$ then $g\in (f_2,\dots,f_k)$] is true, then I can conclude using minimality, but I'm not sure that this statement holds. Any hints?
Best Answer
One can show that if $f_1, \ldots, f_n, g$ are monomials, then $g \in \langle f_1, \ldots, f_n \rangle$ if and only if some $f_k \mid g$.
This is theorem $1.1.8$ of Moore, Rogers, and Sather-Wagstaff's Monomial Ideals and their Decompositions, available here.
Now if $f_1 = gh$, then $g$ and $h$ must be monomials (this is intuitively clear, but it's also proven in Chapter $1$ of the same book). So if $g$ were in $\langle f_1, \ldots, f_n \rangle$, we would need some $f_k \mid g$. But then we would have $f_k \mid f_1$, so $f_1$ would be redundant, contradicting minimality.
I hope this helps ^_^