# Condition for a monomial to belong to a particular ideal

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.

• 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? Aug 2, 2019 at 16:44
• @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 Aug 2, 2019 at 22:41
• 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. Aug 3, 2019 at 18:27
• @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 $d=3$ Aug 4, 2019 at 20:30
• 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 $J=(xyz,yzuv)$? Aug 5, 2019 at 13:07

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.)

• 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 $d$? Aug 5, 2019 at 14:46
• so the worst case onus of determining $d$ is the cardinaity of the ideal! Thanks Aug 5, 2019 at 16:04
• 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 $|\cup E_i|=n$ Aug 6, 2019 at 9:40
• 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. Aug 8, 2019 at 2:41