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Let $R=k[x_1,x_2,...,x_n]$ be the polynomial ring in $n$ indeterminates over a field $k$. An ideal (that can be) generated by monomials is called a monomial ideal. For the monomial ideal $M=(m_1,m_2,...,m_t)$, the radical of $M$ itself is monomial and can be written as, $Rad(M)=(\sigma(m_1),\sigma(m_2),...,\sigma(m_t))$ where $\sigma(x_1^{a_1}x_2^{a_2}...x_n^{a_n})$ is the product of indeterminates $x_i$ s.t. $a_i\geq 1$.

A binomial ideal in $R$ is generated by binomials. I was wondering if we have similar theorems for the case of binomial ideals where we can write down a generating set for the radical by just knowing a generating set of the ideal. Eisenbud and Sturmfels, in their monumental paper on binomial ideals, showed that the radical itself is binomial. I am especially interested in finding generators for radical of binomial ideals in the case where char$(k)=0$ (or even when $k=\mathbb{C}$) and what kind of binomials generate radical binomial ideals.

Becker, Grobe and Niermann discuss the case of zero dimensional binomial ideals. Ojeda and Sanchez prove some results for radicals of lattice (binomial) ideals. I have also seen some results in positive characteristic, but they are not relevant to my research.

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  • $\begingroup$ Did you check this out: front.math.ucdavis.edu/1009.2823? $\endgroup$
    – Hailong Dao
    Commented Dec 1, 2010 at 4:22
  • $\begingroup$ @Hailong: Thanks for the link. I hadn't seen this before, though curiously, the word "radical" does not appear even once in the article. I'll see if the primary decomposition methods are of any help. $\endgroup$
    – Timothy Wagner
    Commented Dec 1, 2010 at 6:32
  • $\begingroup$ Have you seen this: arxiv.org/abs/alg-geom/9401001? $\endgroup$
    – J.C. Ottem
    Commented Dec 1, 2010 at 9:24
  • $\begingroup$ @Tymothy: the radical is the intersection of all minimal primes, so in some sense you only need to know the minimal primes. $\endgroup$
    – Hailong Dao
    Commented Dec 1, 2010 at 13:59
  • $\begingroup$ @Hailong: Yes I understand that. But I am looking for a more concrete description of the ideals in terms of generators rather than as intersection of several prime ideals. $\endgroup$
    – Timothy Wagner
    Commented Dec 1, 2010 at 21:34

1 Answer 1

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The minimal primes (and sometimes also their intersection) can be computed relatively quickly (compared to primary decomposition) using Algorithm 4 of http://arxiv.org/pdf/0906.4873v3. I've looked at binomial ideals for some time and I doubt that there is an easy way to see the generators of the radical.

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  • $\begingroup$ @Thomas: Thanks. I had already looked over your paper earlier and also used your package "binomials" in Macaulay2. It has been extremely useful, though I was curious to know if there is any abstract description of radicals of binomial ideals. I am not optimistic about as general a result as in the case of monomial ideals, but I would definitely be interested in seeing some results under additional hypothesis (like the ones I mention in the last paragraph). $\endgroup$
    – Timothy Wagner
    Commented Dec 1, 2010 at 21:39
  • $\begingroup$ @Timothy, I think the most opaque part of this is why the intersection of the minimal primes comes out binomial in the first place. For me it has always been useful to think about the cellular case: An ideal is cellular if in the quotient every monomial is nilpotent or regular. Cellular decompositions into binomial ideals exist in polynomial rings over every field and in characteristic zero a radical cellular ideal is just a lattice ideal + variables. (Note that in characteristic zero lattice ideals themselves radical.) $\endgroup$
    – Thomas Kahle
    Commented Dec 2, 2010 at 7:47

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