Let $I \subseteq \mathbb{K}[x_1, \dots, x_n]$ be an ideal of a polynomial ring over a field $\mathbb{K}$.

$I$ is called cellular if every variable $x_i$, with $i=1, \dots, n$, is either a nonzerodivisor modulo $I$ or is nilpotent modulo $I$.

$I$ is called primary if whenever $fg \in I$ then or $f \in I$ or $g$ is nilpotent modulo $I$.

Of course, if $I$ is primary $\Rightarrow$ $I$ is cellular.

My question is: consider a binomial ideal $I \subset \mathbb{K}[x_1, \dots, x_n]$, generated by binomials of the form $f_k=\prod_{i \in I_k} x_i - \prod_{j \in J_k} x_j$, for some $I_k,J_k \subset \{1, \dots, n\}$ with $I_k \cap J_k = \emptyset$ and both no empty.

What can I say about $I$ cellular $\Rightarrow$ $I$ primary? Under which hypothesis a cellular binomial ideal is primary?


  • $\begingroup$ The following two papers might be relevant: arxiv.org/abs/alg-geom/9401001 and arxiv.org/abs/0803.3846 $\endgroup$ Oct 27, 2018 at 17:12
  • $\begingroup$ Thanks @AviSteiner. I knew them, but I don't find any help to solve my question. $\endgroup$
    – Ella Smith
    Oct 29, 2018 at 10:03
  • 1
    $\begingroup$ @EllaSmith - Is the assumption that the ambient ring of $I$ is a polynomial ring over a field? (I am having trouble making sense of the phrase "every variable" in the definition of a cellular ideal.) $\endgroup$ Nov 3, 2018 at 1:19
  • $\begingroup$ @benblumsmith I edited my question in order to answer to your comment. $\endgroup$
    – Ella Smith
    Nov 5, 2018 at 10:56
  • $\begingroup$ Thank you @EllaSmith, this is what I needed to know. $\endgroup$ Nov 5, 2018 at 16:00

1 Answer 1


Consider the ideal $I=\left\langle x_{1}^{3} x_{3}-x_{1}^{3}, x_{1}^{4}, x_{1}^{2} x_{2} x_{4}-x_{1}^{2} x_{2}, x_{2}^{2}, x_{4}^{3}-1\right\rangle \subseteq \mathbb{k}\left[x_{1}, x_{2}, x_{3}, x_{4}\right]$ in Example 2.9 of this link. This is cellular but Macaulay2 says that it is not primary.


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