Expanded lectures on binomial ideals

Here is the gist of the Eisenbud–Sturmfels paper Binomial ideals, Duke Math. J. 84 (1996), 1–45. The main results are that the associated primes, the primary components, and the radical of a binomial ideal in a polynomial ring are binomial if the base ring is algebraically closed. Throughout, R = k[X1, . . . , Xn], where k is a field and X1, . . . , Xn are variables over k. A monomial is an element of the form X for some a ∈ Nn0 , and a term is an element of k times a monomial. The words “monomial” and “term” are often confused. In particular, a binomial is defined as the difference of two terms, so it should better be called a “biterm”, but this name is unlikely to stick. An ideal is binomial if it is generated by binomials. Here are some easy facts: (1) Every monomial is a binomial, hence every monomial ideal is a binomial ideal. (2) The sum of two binomial ideals is a binomial ideal. (3) The intersection of binomial ideals need not be binomial: (t−1)∩ (t−2), even over a field of characteristic 0. (4) Primary components of a binomial ideal need not be binomial: in R[t], the binomial ideal (t − 1) has exactly two primary components: (t− 1) and (t + t+ 1). (5) The radical of a binomial ideal need not be binomial: Let k = Z/2Z(t), R = k[X, Y ], I = (X + t, Y 2 + t + 1). Note that I is binomial (as t + 1 is in k), and √ I = (X+ t, X+Y +1), and this cannot be rewritten as a binomial ideal as there is only one generator of degree 1 and it is not binomial. Thus, we do need to make a further assumption, namely, from now on, all fields k are algebraically closed, and then the counterexamples to primary components and radicals do not occur. The ring is always R = k[X1, . . . , Xn], and t is always a variable over R. Comment: Can one repeat this for trinomial ideals (with obvious meanings)? The answer is that not really, because all ideals are trinomial – after adding variables and a change of variable. Namely, let f = a1 + a2 + · · · + am be a polynomial with m terms. Introduce new variables t3, . . . , tm. Then k[x1, . . . , xn]/(f) = k[x1, . . . , xn, t3, . . . , tm]/(a1+ a2 − t3,−t3 + a3 − t4,−t4 + a4 − t5, . . . ,−tm−2 + am−2 − tm−1,−tm−1 + am−1 − tm).