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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.

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    $\begingroup$ The two different definitions for $J$ are not equivalent, and from what is written it sounds a bit more like you mean for $J$ to be any monomial ideal. Could you clarify? $\endgroup$ Aug 2, 2019 at 16:44
  • $\begingroup$ @HughThomas the ideal $J$ is generated by homogeneous monomials, like, for example $J=\langle x_1x_2x_3,x_2x_3x_4,x_3x_4x_5,x_1x_2x_4,x_1x_3x_4\rangle$ in the ring $R[x_1,x_2,x_3,x_4,x_5]$ in which case $d=3$, where $d$ is defined in the question $\endgroup$
    – vidyarthi
    Aug 2, 2019 at 22:41
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    $\begingroup$ The way you have written the monomials in the question, they are of a particular form (eg, the first one starts with $x_1$, the second with $x_2$ and so on). If you just want to say "Let J be an ideal generated by a collection of degree $d$ monomials $m_1,\dots,m_r$", it would be clearer if you just said that. If you mean for the monomials to have some special form, I still don't know exactly what form you want. It is also confusing that you write "I think the value of $d$ for such a membership...", since $d$ is part of the given data. $\endgroup$ Aug 3, 2019 at 18:27
  • $\begingroup$ @HughThomas edited the post. As to your second comment, I meant if $J=\langle x_1x_2x_3,x_2x_3x_4,x_3x_4x_5,x_1x_2x_4,x_1x_3x_4\rangle$, then $(x_1x_2x_3x_4x_5)^2\in J^3$, that is, $(x_1x_2x_3x_4x_5)^{d-1}\in J^d$ for $d=3$ $\endgroup$
    – vidyarthi
    Aug 4, 2019 at 20:30
  • $\begingroup$ Are all the monomials square-free, or do you allow them to have powers? Would you allow $J=(xyz,x^2y,z^3)$? Also, do all the monomials have the same degree? Would you allow $J=(xyz,yzuv)$? $\endgroup$ Aug 5, 2019 at 13:07

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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.)

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    $\begingroup$ thanks! actually this question arose from the colorings of graphs and its equivalence to ideal membership for a certain ideal. The ideal I mention is a generalized for of the $\textit{cover ideal}$ which is found here, theorem 3. Such a $d$ must exist for a $\textit{cover ideal}$ for, otherwise, the coloring of graphs would no proper solution, which is absurd. What extra features on the ideal guarantee the existence of such a $d$? $\endgroup$
    – vidyarthi
    Aug 5, 2019 at 14:46
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    $\begingroup$ so the worst case onus of determining $d$ is the cardinaity of the ideal! Thanks $\endgroup$
    – vidyarthi
    Aug 5, 2019 at 16:04
  • $\begingroup$ The last inequality you give, shouldnt it be reversed and the minus sign be a plus sign? As the bonferroni inequality says $\sum_i |E_i|\ge|\cup_iE_i|$ and here $|\cup E_i|=n$ $\endgroup$
    – vidyarthi
    Aug 6, 2019 at 9:40
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    $\begingroup$ I'm using $|\cup_i E_i| \geq \sum_i |E_i| - \sum_{i<j} |E_i \cap E_j|$, sorry for any confusion. See math.stackexchange.com/questions/514246/… or sms.math.nus.edu.sg/smsmedley/Vol-19-2/…, or Stanley's EC1, chapter 2 (inclusion-exclusion), exercise 4. $\endgroup$ Aug 8, 2019 at 2:41

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