On the associated graded ring of a semigroup ring (original) (raw)
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Lance Bryant noticed in his thesis [3], that there was a flaw in our paper [2]. It can be fixed by adding a condition, called the BF condition in [3]. We discuss some equivalent conditions, and show that they are fulfilled for some classes of rings, in particular for our motivating example of semigroup rings. Furthermore we discuss the connection to a similar result, stated in more generality, by Cortadella-Zarzuela in [4]. Finally we use our result to conclude when a semigroup ring in embedding dimension at most three has an associated graded which is a complete intersection. 2000 Mathematics Subject Classification: 13A30 If x ∈ R is an element of smallest positive value, i.e. v(x) = e, then xR is a minimal reduction of the maximal ideal, i.e. m n+1 = xm n , for n >> 0. Conversely each minimal reduction of the maximal ideal is a principal ideal generated by an element x of value e. The smallest integer n such that m n+1 = xm n is called the reduction number and we denote it by r. Observe that, if v(x) = e, then Ap e (S) = S \(e+S) = v(R)\v(xR), therefore w j / ∈ v(xR), for j = 0,. .. , e − 1. Consider the m-adic filtration m ⊃ m 2 ⊃ m 3 ⊃. .. . If a ∈ R, we set ord(a) := max{i | a ∈ m i }. If s ∈ S, we consider the semigroup filtration v(m) ⊃ v(m 2) ⊃. .. and set vord(s) := max{i | s ∈ v(m i)}. If a ∈ m i , then v(a) ∈ v(m i) and so ord(a) ≤ vord(v(a)). According to [3], we say that the m-adic filtration is essentially divisible with respect to the minimal reduction xR if, whenever u ∈ v(xR), then there is an a ∈ xR with v(a) = u and ord(a) = vord(u). The m-adic filtration is essentially divisible if there exists a minimal reduction xR such that it is essentially divisible with respect to xR. We fix for all the paper the following notation. Set, for j = 0,. .. , e − 1, b j = max{i|w j ∈ v(m i)}, and let c j = max{i|w j ∈ v(m i + xR)}. Note that the numbers b j 's do not depend on the minimal reduction xR, on the contrary the c j 's depend on xR. Lemma 1.1 If I and J are ideals of R, then v(I +J) = v(I)∪v(J) is equivalent to v(I ∩ J) = v(I) ∩ v(J).