Consider the polynomial ring $R[x\_1,x\_2,\ldots,x\_n]$, where $R$ be an algebraically closed field (preferably $\mathbb{C}$) and the ideal $J=\langle m\_1, m\_2,\ldots,m\_n\rangle$ generated by monomials . The monomials are homogeneous and each variable has maximum degree $1$. Let $|m\_i|$ denotes the number of variables in a monomial $m\_i$ 


Is there a way to determine whether the monomial power $\prod\_{i=1}^nx\_i^{d-1}$ belongs to the ideal $J^d$? I think the value of $d$ for such a membership depends on the minimum number of variables intersecting between any two monomials in the generating set of the ideal. Specifically, I think if $m\_i\cap m\_j=\phi$, then $d=2$. If $m\_i\cap m\_j\neq\phi$, then the value of $d$ is the minimum $d$ such that $|m\_i|+d(min|m\_i\cap m\_j|)=n$. Any hints? Thanks beforehand.



Condition for a monomial to belong to a particular ideal
Such a value $d$ does not necessarily exist. For $J=(xy,xz)$ we have $(xyz)^{d-1} \notin J^d$ for any $d$. It’s even worse (but maybe a little degenerate) if $J$ is principal.


A necessary and sufficient condition for existence of such a $d$ is that the monomials generating $J$ don't have a common factor. To be explicit:


**Proposition:** Let $m\_1,\dotsc,m\_k \in R = \Bbbk[x\_1,\dotsc,x\_n]$ be square-free monomials, $J=(m\_1,\dotsc,m\_k)$. (We don't assume $k=n$, or that the monomials have the same degree.) There exists a $d$ such that $(x\_1 \dotsm x\_n)^{d-1} \in J^d$ if and only if for each $i=1,\dotsc,n$, there is (at least) one of the monomials $m\_j$ such that $x\_i$ does not appear in $m\_j$.


*Proof:* As in the example above, if each $m\_j$ is divisible by the same $x\_i$ for some $i$, then every generator of $J^d$ is divisible by $x\_i^d$, so $(x\_1\dotsm x\_n)^{d-1} \notin J^d$. Conversely, if the condition is met, then in the product of the generators $m\_1 \dotsm m\_k$, each variable $x\_i$ appears at most $k-1$ times, so $m\_1 \dotsm m\_k$ divides $(x\_1 \dotsm x\_n)^{k-1}$, and $(x\_1\dotsm x\_n)^{k-1} \in J^k$, i.e., $d=k$ (the number of monomial generators) works. $\square$


For a square-free monomial $m$, let $\overline{m}$ be the set of variables that don't appear in $m$. So the condition above is that $\overline{m}\_1 \cup \dotsb \cup \overline{m}\_k = \{x\_1,\dotsc,x\_n\}$.


**Proposition:** With notation as above, $(x\_1 \dotsm x\_n)^{d-1} \in J^d$ if and only if there exist $j\_1,\dotsc,j\_d$ such that $\overline{m}\_{j\_1} \cup \dotsb \cup \overline{m}\_{j\_d} = \{x\_1,\dotsc,x\_n\}$.


Indeed in the product $m\_{j\_1} \dotsm m\_{j\_d}$ each variable is omitted from at least one factor, hence appears at most $d-1$ times. Conversely if $(x\_1 \dotsm x\_n)^{d-1}$ is divisible by a generator of $J^d$, say, $m\_{j\_1} \dotsm m\_{j\_d}$ (here there is no assumption of distinctness of factors), then each $x\_i$ must be omitted from at least one of the $m\_{j\_\ell}$'s.


So the minimum $d$ that works is the same as the smallest size of a cover of the set of variables by the $\overline{m}$'s. I'm not a combinatorialist but it seems to me that such $d$ cannot be determined only from the sizes of the $\overline{m}$s and pairwise intersections. It must involve triple intersections, etc. If you only want a bound then I suppose: the size of $\overline{m}\_1 \cup \dotsb \cup \overline{m}\_d$ is greater than or equal to the sum of the sizes of the $\overline{m}$'s minus the sum of the sizes of pairwise intersections (baby Bonferroni inequality). So if $d$ works, every $m$ has degree $e$, and each two $m$'s have at least $f$ factors in common, then the pairwise intersections of $\overline{m}$s have size at least $n-2e+f$, so
$$
 n \geq d(n-e) - \binom{d}{2}(n-2e+f),
$$
and you can solve this quadratic inequality for $d$ and see if that gives an at all useful bound. I hope it helps.


(Edit: The pairwise *unions* of $\overline{m}$s have size at most $n-f$, but I should have bounded the size of the pairwise *intersections*. Again, sorry any confusion.)


