TANGENT CONE OF NUMERICAL SEMIGROUP RINGS WITH SMALL

arXiv:0808.2162v1 [math.AC] 15 Aug 2008

TANGENT CONE OF NUMERICAL SEMIGROUP RINGS WITH SMALL EMBEDDING DIMENSION YIHUANG SHEN Abstract. In this paper, we study the tangent cone of numerical semigroup rings with small embedding dimension d. For d = 3, we give characterizations of the Buchsbaum and Cohen-Macaulay properties and for d = 4, we give a characterization of the Gorenstein property. In particular, when d = 4 and the tangent cone is Gorenstein, the initial form ideal of the defining ideal is 5-generated.

1. Introduction Throughout this paper we fix N = {1, 2, 3, · · · } and N0 = {0, 1, 2, · · · }. Recall that P a numerical semigroup G = hn1 , · · · , nd i generated by n1 , . . . , nd ∈ N is the n set { i=1 ai ni : ai ∈ N0 }. For simplicity, we always assume that G is minimally generated by these generators, with n1 < · · · < nd and gcd(n1 , . . . , nd ) = 1, unless stated otherwise. Let k be a field of characteristic 0, and t an indeterminate over k. Then as a subring of the power series ring V = k[[t]], R = k[[tn1 , . . . , tnd ]] is the numerical semigroup ring associated to G with m = (tn1 , . . . , tnd )R being the unique maximal ideal. R is the homomorphic image of the power series ring S = k[[x1 , . . . , xd ]] by mapping xi to tni . Let n be the unique maximal ideal of S, and I the kernel of this surjective map. I is a binomial ideal. We denote the kernel of the natural map grn (S) → grm (R) by I ∗ , and call it the initial form ideal of I. When the embedding dimension d = 3, Herzog [11] gave a complete characterization of the defining ideal I. In particular the minimal number of generators µ(I) ≤ 3. It is also proven by Robbiano and Valla [14] and Herzog [12] that the associated graded ring grm (R) is Cohen-Macaulay if and only if the initial form ideal I ∗ is generated by at most 3 elements. In this paper, we give another characterization for the Cohen-Macaulay property in terms of the index of nilpotency sQ (m) and the reduction number where Q = (tn1 )R is a principal reduction rQ (m), of m. Recall that sQ (m) = min s | ms+1 ⊆ Q . For d = 3, we also study the 0-th local cohomology module of the tangent cone. And we are able to characterize the k-Buchsbaum properties of grm (R) for k = 1, 2, in terms of the length of this local cohomology module. In particular, this answers the conjectures raised by Sapko [15]. When d = 4 and the numerical semigroup G is symmetric, Bresinsky [3] gave a complete description of the defining ideal I. In particular, it is well-known now that µ(I) ≤ 5. In this paper, we also study the initial form ideal of I. When the tangent cone grm (R) is Gorenstein, we show that I ∗ is also 5-generated. Meanwhile, Arslan 1991 Mathematics Subject Classification. Primary 13A30, Secondary 13P10, 13H10. 1

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[1] and Shibuta [16] showed that µ(I ∗ ) could be arbitrarily large when grm (R) is only Cohen-Macaulay. How about d = 5? In this situation, Bresinsky [4] proved that µ(I) ≤ 13, under the condition that n1 + n2 = n3 + n4 . Here in G = hn1 , n2 , · · · , n5 i, he didn’t assume the generators to be in increasing order. Furthermore, our computations suggest a positive answer to the following question. Question 1.1. If the symmetric numerical semigroup G has embedding dimension 5, is µ(I) ≤ 13 in general? If the associated graded ring grm (R) is Gorenstein, is µ(I ∗ ) ≤ 13 as well? The main technique in this paper is to play with the standard basis of the defining ideal I. The standard basis algorithm can generate a set of standard basis from a minimal generating set of I. The new generators are s-polynomials. Recall that for a polynomial ring A = k[X] with monomial ordering f , thenPan1 +wi > wi +ws−1−i −n1 . a n with aj ∈ N0 , then Thus (a + 1)n1 > ws−1−i . If we write ws−1−i = P P Pj j j a1 = 0, and ws−1−i > aj n1 . Consequently a + 1 > j aj , i.e., a ≥ j aj . Hence a ≥ ordG (ws−1−i ). Now ordm (z) ≥ a + ordG (wi ) ≥ ordG (ws−1−i ) + ordG (wi ) ≥ min-ordG (f + n1 ). Since the elements of the form tz with z > f generate the ideal C, this completes the proof. Example 2.3. In general, it is not true that ordG (f +n1 ) ≤ ordm (C). For example, if G = h11, 14, 21i, then G is symmetric and the Frobenius number f = 73. It is not difficult to see that ordG (f + n1 ) = 6 while ordm (C) = 5. Inspired by the proof of 2.2, we define a partial P order G on elements P in′ G. We ′ say x G x′ for x, x′ ∈ G if we can write x = a n and x = i i i i ai ni with P P ′ ′ ′ a = ord (x ), and a ≤ a ai , a′i ∈ N0 , such that ai = ordG (x), G i i for all i. i i Roughly speaking, x G x′ if and only if “x has a maximal decomposition with respect to G that is dominated by a maximal decomposition of x′ with respect to G”. Proposition 2.4. Let (R, m) be a Gorenstein numerical semigroup ring, where the associated graded ring grm (R) is Cohen-Macaulay. If for every x ∈ Ap(G, n1 ), x G (f + n1 ), then equality holds for (‡) in the introduction. In addition, g(tf +n1 +1 ) = g(tn1 ) = rQ (m) = ordm (C) = ordG (f + n1 ).

Proof. According to the proof of 2.2, for any element z ∈ G with z > f , z −Pan1 = wj ∈ Ap(G, P n1 ) with a = ordQ (z). Since wj G (f +n1 ), we can write′ wj = ai ni , f + n1 = bi ni , with ai , bi ∈ N0 ,P ai ≤ bi for 2 ≤ i ≤ d. Now write ai = bi − ai and one has ws−1−j = f + n1 − wi = a′i ni . Hence by the proof of 2.2, X ordG (z) ≥ ordG (wj ) + ordG (ws−1−j ) ≥ bi = ordG (f + n1 ). This implies that ordm (C) ≥ ordG (f +n1 ). By virtue of 2.1, we have ordG (f +n1 ) ≥ rQ (m) = g(tn1 ). On the other hand, Theorem 4.1 and Corollary 5.8 of [10] yield

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g(tn1 ) ≥ g(tf +n1 +1 ) ≥ ordm (C). Since

ordm (C) = min {ordG (z) | z ∈ G, z > f } ,

ordm (C) ≥ ordG (f + n1 ). Now the equalities in the assertion follow immediately. In particular, equality holds for (‡). Remark 2.5. If ordG (f + n1 ) = min-ordG (f + n1 ), i.e., the elasticity of f + n1 with respect to G is 1, then every decomposition of f + n1 is maximal. Hence for any x ∈ Ap(G, n1 ), x G (f + n1 ). This elasticity condition is satisfied when the embedding dimension equals 4, the tangent cone grm (R) is Gorenstein but R is not complete intersection, as we demonstrate in 5.16. Corollary 2.6. For numerical semigroup ring (R, m) with embedding dimension d ≤ 4, if the associated graded ring grm (R) is Gorenstein, then g(tf +n1 +1 ) = g(tn1 ) = r(m) = ordm (C) = ordG (f + n1 ).

In particular, equality holds for (‡). When d = 4, it follows from 2.4 and 5.1. The case when d = 3 will be an easy exercise. It is pointed out by Lance Bryant that the partial order condition on the Apéry set in 2.4 is equivalent to the following condition for multiplicity e = n1 : (†) ∀wi , wj ∈ Ap(G, n1 ), wi + wj = we−1 ⇒ ordm (wi ) + ordm (wj ) = ordm (we−1 ).

And he gives the following characterization

Theorem 2.7 ([6, 3.20]). Suppose G is a symmetric numerical semigroup, and grm (R) is Cohen-Macaulay. Then grm (R) is Gorenstein if and only if the condition (†) holds. In particular, Corollary 2.6 holds with no restriction on the embedding dimension. 3. Initial form ideals Following our previous notation, S = K[[x1 , . . . , xd ]] with maximal ideal n maps onto R. For each nonzero x ∈ S, let o = ordn (x) < ∞ be the n-adic order of x. We denote by x∗ the residue class of x in no /no+1 and call it the initial form of x. The initial form ideal I ∗ ⊂ grn (S) is generated by x∗ for all x ∈ I. For our numerical semigroup ring R, the radical of the initial ideal I ∗ is very simple. √ Lemma 3.1. I ∗ = (x∗2 , . . . , x∗d ) grn (S). i ∈ I. Since n1 < ni , the initial form Proof. For all i with 2 ≤ i ≤ d, fi = xni 1 − xn1√ ∗ ∗ ∗ ∗ ∗ n1 of fi is (xi ) ∈ I . Hence (x2 , . . . , xd ) ⊆ I ∗ . Since√ht(x∗2 , . . . , x∗d ) = ht I ∗ = ht I = d − 1 and (x∗2 , . . . , x∗d ) is a prime ideal in grn (S), I ∗ = (x∗2 , . . . , x∗d ).

Since grn (S) ∼ = k[x1 , . . . , xd ], by abuse of notation, in the sequel of this paper we simply write x∗i as xi for 1 ≤ i ≤ d, when there is no confusion. We need to go over some basic definitions and notations of standard basis. Given a fixed local monomial order ring >, for a nonzero f ∈ S, there is a unique way to write xα1 > xα2 > · · · > xαt . f = a1 xα1 + a2 xα2 + · · · + at xαt , LM(f ) := xα1 is called the initial monomial of f . The initial ideal LI(I) of I is generated by LM(f ) for f ∈ I. A subset F := {f1 , . . . , fs } ⊂ I is a standard basis

TANGENT CONE OF NUMERICAL SEMIGROUP RINGS

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of I if LI(I) = (LM(f1 ), . . . , LM(fs )). F is called minimal if LM(fi ) 6 | LM(fj ) for i 6= j. β A binomial f = xα − xP with α = P (α(1), . . . , α(d)), β = (β(1), . . . , β(d)) ∈ Nd0 is called weakly balanced if i α(i)ni = i β(i)ni . f is called balanced if it is weakly balanced, deg xα = xβ , and xα and xβ form a regular sequence. For the numerical semigroup ring R, the defining ideal I is generated by weakly balanced binomials. Applying the standard basis algorithm with suitable monomial ordering to this generating set, one is able to get a minimal standard basis {f1 , . . . , fs }. And I ∗ is minimally generated by the corresponding initial forms: I ∗ = (f1∗ , . . . , fs∗ ). Since each fi is also a weakly balanced binomial, fi∗ is either a monomial or a balanced binomial. In the latter case, roughly speaking, fi = fi∗ . In the rest of the paper, when we say g is a minimal generator of I ∗ , it is understood that g ∈ {f1∗ , . . . , fs∗ } when a minimal generating set of I and the monomial ordering is clear. From now on, we need to choose the monomial ordering > more carefully. A monomial ordering > is nice in variable xi if the following holds: deg xα < deg xβ , or (xα − xβ is balanced, β(i) > 0) =⇒ xα > xβ .

Being nice is really a mild condition. For instance, the modified negative degree reverse lexicographical ordering in the following is nice in x1 : xα > xβ :⇐⇒ deg xα < deg xβ , or (deg xα = deg xβ and ∃1 ≤ i ≤ n : α(1) = β(1), . . . , α(i − 1) = β(i − 1), α(i) < β(i)).

When d = 3, the normal negative degree reverse lexicographical ordering is also nice in x1 . An ideal J ⊂ T := k[x1 , . . . , xd ] is called almost balanced if it satisfies the following conditions: √ (a) J = (x2 , . . . , xd ); (b) there is a minimal standard basis {f1 , . . . , ft } of J, such that fi is either a monomial or a balanced binomial. The initial form ideal I ∗ is almost balanced. Lemma 3.2. Let > be a nice monomial ordering in x1 for T and J an almost T -ideal. Then T /J is Cohen-Macaulay if and only if x1 does not divide any of the monomial generators of J constructed above. Proof. T /J is Cohen-Macaulay if and only if x1 is a regular element. If in the previous definition that some fi = x1 xα , then xα 6∈ J, since {f1 , . . . , fs } is a minimal standard basis. Hence J is not a perfect ideal. On the other hand, suppose that x1 f ∈ J and 0 6= f 6∈ J. We may replace f by the normal form NF(f | {f1 , . . . , fs }) to assume that none of the monomials in f is divisible by any of LM(fi ). Notice that x1 LM(f ) = LM(x1 f ) is divisible by some LM(fi ). But LM(f ) is not divisible by LM(fi ), hence LM(fi ) is divisible by x1 . Since the ordering is nice in x1 , fi cannot be a balanced binomial, hence it is a monomial. Example 3.3. Let G be the numerical semigroup generated by 5, 6 and 13. Then the defining ideal I = (x22 x3 − x51 , x23 − x41 x2 , x1 x3 − x32 ). With the normal negative degree reverse lexicographical ordering, x22 x3 − x51 , x23 − x41 x2 , x1 x3 − x32 , x52 − x61 form a minimal standard basis. Thus the initial form ideal is I ∗ = (x52 , x22 x3 , x23 , x1 x3 ).

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Since x1 x3 is divisible by x1 , I ∗ is not a Cohen-Macaulay ideal. This property also follows from the fact that I ∗ is generated by more than 3 elements. For the rest of this paper, we fix αi = min {α ∈ N | αni ∈ hn1 , . . . , nbi , . . . , nd i}

for 1 ≤ i ≤ d. In the previous example, α1 = 5, α2 = 3 and α3 = 2. 4. When the embedding dimension d = 3 Throughout this section, we use negative degree reverse lexicographical ordering on grn (S) ∼ = k[x1 , x2 , x3 ]. Hence if f = xb2 − xa1 xc3 with a, b, c ∈ N, and b = a + c, then the initial monomial of f is xb2 . We need the following result on the structure of 3-generated numerical semigroups by Herzog. Theorem 4.1 ([11, 2.1, 3.8]). Let G be a numerical semigroup minimally generated by 3 elements. (a) If G is symmetric, then by a permutation (i, j, k) of (1, 2, 3), the defining ideal is αj αk rki rki i I = (xα i − xj , xk − xi xj ). And the Frobenius number of G is f = (αi − 1)ni + (αk − 1)nk − nj .

(b) If G is not symmetric, then

α31 α32 α3 α21 α23 α2 α12 α13 1 I = (xα 1 − x2 x3 , x2 − x1 x3 , x3 − x1 x2 ),

where all α’s are in N and αi = αji + αki for all permutation (i, j, k) of (1, 2, 3). In terms of the defining ideal given by 4.1, one can quickly give arithmetic conditions on when grm (R) will be Cohen-Macaulay. α31 α32 α3 α2 1 Corollary 4.2. (a) If I = (xα 1 − x2 , x3 − x1 x2 ), then grm (R) is a comα3 ∗ 2 plete intersection and I is generated by {xα 2 , x3 }. α1 α3 α2 α21 α23 (b) If I = (x1 − x3 , x2 − x1 x3 ), then grm (R) is Cohen-Macaulay if and only if α2 ≤ α21 + α23 . Whence, I ∗ is generated by α2 α2 α21 α23 3 {xα 3 , x2 or x2 − x1 x3 } .

α3 α1 α12 α13 2 (c) If I = (f1 := xα 2 − x3 , f2 := x1 − x2 x3 ) with α13 < α3 . Set f3 := α2 +α12 α1 α3 −α13 spoly(f1 , f2 ) = x2 − x1 x3 . Then grm (R) is Cohen-Macaulay if and only if α2 + α12 ≤ α1 + α3 − α13 . Whence, I ∗ is generated by α12 α13 ∗ 3 {xα 3 , x2 x3 , f3 }. α12 α13 α2 α21 α23 α3 α31 α32 1 (d) If I = (xα − x 1 2 x3 , x2 − x1 x3 , x3 − x1 x2 ), then grm (R) is Cohen-Macaulay if and only if α2 ≤ α21 + α23 .

Recall that for a one-dimensional graded ring A with a unique homogeneous max0 (A) = 0. 1-Buchsbaum imal ideal M, A is called a k-Buchsbaum ring if Mk · HM rings are simply called Buchsbaum. And 0-Buchsbaum rings are precisely the Cohen-Macaulay rings. In the following, denote the homogeneous maximal ideal of grn (S) by M. Since grm (R) = grn (S)/I ∗ , we write the image of f ∈ grn (S) in grm (R) as f¯.


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Sapko [15] investigated the tangent cone of ring R associated with the 3-generated numerical semigroup ring G, and made the following conjectures when the tangent cone is Buchsbaum. Conjecture 4.3 ([15, Conjecture 24]). If grm (R) is Buchsbaum, then for some k ≥ 1, 0 :grm (R) M = (x¯k3 ) grm (R). Conjecture 4.4 ([15, 25]). If grm (R) is Buchsbaum, then the initial form ideal I ∗ of I is generated by 4 elements. Conjecture 4.5 ([15, Conjecture 33]). grm (R) is Buchsbaum if and only if the 0 length ℓ(HM (grm (R))) ≤ 1. In this note, we want to give positive answers to the above conjectures when the tangent cone is Buchsbaum, and show similar results when the tangent cone is 2-Buchsbaum. Lemma 4.6. Suppose grm (R) is not Cohen-Macaulay. Then the 0-th local coho0 mology module HM (grm (R)) is principal, and generated by xγ3 for suitable γ ∈ N. Proof. The initial form ideal I ∗ is generated by forms of the following 4 types: 3 (a) xα 3 , γ2 (b) x2 or balanced xγ22 − xγ121 xγ323 , (c) xa1 xc3 , (d) xb2 xc3 . There is exactly one generator of the type (a). The same is true for generators of type (b). To see this, notice that if xγ22 − xγ121 xγ323 is balanced, then xγ22 is its initial monomial. There might be more than one generators of type (c) and (d). It follows from Corollary 3.2 that I ∗ is Cohen-Macaulay if and only if generators 0 of type (c) do not exist. If grm (R) is not Cohen-Macaulay, then HM (grm (R)) 6= 0. γ ¯ We claim that this local cohomology module is generated by x3 where γ = min {c | xa1 xc3 is a generator of I ∗ of type (c) for some a, c ∈ N} . √ 0 Since I ∗ = (x2 , x3 ), this x¯γ3 ∈ HM (grm (R)). On the other hand, I ∗ + (xγ3 ) is (not necessarily minimally) generated by xγ3 together with the remaining generators of I ∗ of type (b) and (d). This last ideal is Cohen-Macaulay by 3.2. 0 Corollary 4.7. grm (R) is Buchsbaum if and only if ℓ(HM (grm (R))) ≤ 1.

Proposition 4.8. If grm (R) is Buchsbaum, then the initial form ideal I ∗ is 4generated. Proof. When grm (R) is Buchsbaum and not Cohen-Macaulay, I ∗ has at least one 0 minimal generator of the form xγ11 xγ33 . x¯γ33 ∈ HM (grm (R)), hence in grn (S), γ3 γ1 γ3 ∗ Mx3 ⊆ I . Since x1 x3 is a minimal generator, one must have γ1 = 1. One also observe that x2 xγ33 , xγ33 +1 ∈ I ∗ . By the minimality of α3 , one has α3 = γ3 + 1. It is also clear now that the only minimal monomial generator that involves positive exponent in x1 is exactly x1 x3α3 −1 . Since x2 x3α3 −1 ∈ I ∗ , there exists a generator f = x2 xγ3 − xα 1 of the minimal standard basis with γ ≤ α3 − 1. However, since γ < α3 , this f should not be new basis generated from the standard basis algorithm. It should belong to one of the minimal binomial generators of I.

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Claim. If R is Gorenstein, then grm (R) is Cohen-Macaulay if and only if it is Buchsbaum. By 4.1, when R is Gorenstein, the defining ideal, by a permutation (i, j, k) of (1, 2, 3), is αj αk αki αki i I = (xα i − xj , xk − xi xj ).

By symmetry, we can always assume that i < j. Now one can characterize when the associated graded ring is Buchsbaum in terms of these α’s. By our discussions for x2 x3α3 −1 , the claim is set by checking the minimal binomial generators for the cases when (i, j, k) = (1, 2, 3) or (1, 3, 2). If (i, j, k) = (2, 3, 1), then α2 α1 α12 α13 3 I = (f1 := xα 3 − x2 , f2 := x1 − x2 x3 ).

∗ 3 f1∗ = xα 3 and we can assume that 0 ≤ α13 < α3 , hence α12 > 0. Now f2 = α12 α13 ∗ −x2 x3 and it is non-comparable with f1 . Apply the standard basis algorithm, 2 +α12 1 α3 −α13 we get f3 := spoly(f1 , f2 ) = xα − xα . Now a necessary and sufficient 2 1 x3 ∗ condition for I to be Cohen-Macaulay is α2 + α12 ≤ α1 + α3 − α13 . Hence, if I ∗ is 1 α3 −α13 Buchsbaum and not Cohen-Macaulay, then f3∗ = −xα . Notice that f3 must 1 x3 belong to the minimal standard basis. Now by our discussion above, α12 = α12 = 1 and α3 − α13 = α3 − 1. But if G is minimally generated by 3 elements, then α1 > 2, and this is a contradiction. Thus, the claim if proved. If follows immediately that if grm (R) is Buchsbaum, not Cohen-Macaulay, then R is not Gorenstein. Now we can write the defining ideal α12 α13 α2 α21 α23 α3 α31 α32 1 I = (f1 := xα 1 − x2 x3 , f2 := x2 − x1 x3 , f3 := x3 − x1 x2 ).

12 α13 3 Obviously, f1∗ = −xα and f3∗ = xα 2 x3 3 . We can assume that α13 , α23 < α3 . Now by 4.2(d) and our discussion in the beginning, if grm (R) is Buchsbaum and not Cohen-Macaulay, then α2 > α21 + α23 , α12 = α21 = 1 and α23 = α3 − 1. Now by Theorem 4.1, α13 = 1, α31 = α1 − 1 and α32 = α2 − 1. Obviously for 1 +1 α3 −2 2 +1 f4 := spoly(f1 , f2 ) = xα x3 − xα . The Buchsbaumness would require that 1 2 α2 + 1 ≤ (α1 + 1) + (α3 − 2). An easy calculation would show that the standard basis algorithm will stop at this step, and {f1 , · · · , f4 } form a minimal standard basis for I. Now

α2 +1 3 2 +1 1 +1 α3 −2 I ∗ = (x2 x3 , x1 x3α3 −1 , xα or xα − xα x3 ). 3 , x2 2 1

Next, we study the 2-Buchsbaumness of the tangent cone, and still assume that the numerical semigroup G is minimally 3-generated. We want to show the following: Proposition 4.9. grm (R) is 2-Buchsbaum if and only if the length 0 ℓ(HM (grm (R))) ≤ 2.

Proof. Notice that we assume that the monomial ordering is nice in x1 . Hence if a monomial involving x1 is the initial monomial of a minimal generator constructed as before, this generator as to be a monomial. Now if the length is at most 2, the tangent cone is trivially 2-Buchsbaum. On the other hand, we assume that grm (R) is 2-Buchsbaum. Without loss of generality, we may assume that grm (R) is not Cohen-Macaulay. Hence in I ∗ , there is a monomial


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minimal generator xa1 xc3 . Since x21 xc3 , x23 xc3 ∈ I ∗ , one necessarily have 1 ≤ a ≤ 2 and α3 − 2 ≤ c ≤ α3 − 1. We claim that there is exactly one such minimal generator with the initial monomial xa1 xc3 . It is easy to see that this could fail only when both x1 x3α3 −1 and x21 x3α3 −2 are minimal generators of I ∗ . Since they are minimal, there exist β1 , β2 ∈ N such that both xβ2 1 − x1 x3α3 −1 and xβ2 2 − x21 x3α3 −2 are both in I. Since n3 > n2 > n1 , we must have β1 > β2 . Hence xβ2 1 −β2 x1 = x3 , and G is at most 2-generated, which contradicts our assumption. ′ Similarly, x22 xc3 ∈ I ∗ . Hence either it is divisible by the initial monomial xb2 xc3 of a minimal generator of I ∗ with 1 ≤ b ≤ 2 and 1 ≤ c′ ≤ c, or α2 = 2. If α2 = 2, then 4.2 implies that I ∗ is Cohen-Macaulay, and there is no need for further discussion. If α2 > 2, then by an argument similar to that in the previous paragraph, there is ′ exactly one such minimal generator in I ∗ with initial monomial of the form xb2 xc3 with 1 ≤ b ≤ 2 and 1 ≤ c′ ≤ c. And this generator must be a monomial. 0 Now we prove that ℓ(HM (grm (R))) ≤ 2.

(a) Suppose that xa1 xc3 = x1 x3α3 −2 . Notice that x22 x3α3 −2 , x2 x3α3 −1 ∈ I ∗ . Each of them has to be divisible by some monomial minimal generator of I ∗ in ′ the form xb2 xc3 with 1 ≤ b ≤ 2, 1 ≤ c′ ≤ c. But there are at most one such generator. Hence this generator must divide the gcd(x22 x3α3 −2 , x2 x3α3 −1 ) = 0 x2 x3α3 −2 . In particular, x2 x3α3 −2 ∈ I ∗ . Hence the vector space HM (grm (R)) = α3 −2 α3 −1 α¯ 3 −1 (x3 ) grm (R) is generated by x3 . , x3 (b) The case xa1 xc3 = x21 x3α3 −2 can never happen. Notice that the image of x3α3 −2 generates the local cohomology module. Hence x1 x3 · x3α3 −2 ∈ I ∗ . γ We know that there are no two such minimal generators of the form xα 1 x3 in I ∗ . Since x21 x3α3 −2 is assumed to be a minimal generator, hence x21 x3α3 −2 has to divide x1 x3α3 −1 , which is impossible. ∗ 3 (c) Assume that xa1 xc3 = x1 x3α3 −1 . Notice that xα Hence 3 ∈ I . the local coho- α3 −1 or x3α3 −1 , x2 x3α3 −1 . mology module is generated as vector space by x3 (d) Assume that xa1 xc3 = x21 x3α3 −1 . Notice x1 x2 x3α3 −1 ∈ I ∗ by the 2-Buchsbaumness. But it can never be a minimal generator. Hence either x1 x3α3 −1 ∈ I ∗ or x2 x3α3 −2 ∈ I ∗ . Because x21 x3α3 −1 is a minimal generator, the first option cannot happen. Hence x2 x3α3 −1 ∈ I ∗ and the local cohomology module as a vector space is generated by x3α3 −1 , x1 x3α3 −1 .

Proposition 4.10. If grm (R) is 2-Buchsbaum, the initial form ideal I ∗ is 4generated. Proof. We may assume from the beginning that grm (R) is not Cohen-Macaulay. In particular, α2 ≥ 3. If grm (R) is 2-Buchsbaum, then R is not Gorenstein, by an argument similar to that used in the proof for 4.8. Hence we can assume that R is not Gorenstein as well. (a) Suppose that x1 x3α3 −2 is one minimal generator for I ∗ , then α2 α3 −2 α1 −1 α2 −1 2 1 3 I = (f1 := xα , f3 := xα x2 ) 1 − x2 x3 , f2 := x2 − x1 x3 3 − x1

by case (a) of the proof for 4.9 together with 3.2 of [11]. spoly(f1 , f3 ) and (f1 , f2 ) will not contribute to the standard basis. Since grm (R) is not Cohen-Macaulay, α2 > 1 + (α3 − 2).

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α3 − 2 is an exponent in f2 , thus α3 − 2 ≥ 1. If α3 = 3, then x3 generates the local cohomology module. Hence x22 x3 ∈ I ∗ . And there is a standard basis generator f = xβ2 x3 − xγ1 ∈ I with β = 1 or 2. But then 2 +1 1 +1 f4 := spoly(f1 , f2 ) = xα x3 − xα . Since n2 > n1 and α2 ≥ 3 > β, this 2 1 will imply that γ < α1 , which is a contradiction. 2 +1 1 +1 α3 −4 Hence α3 ≥ 4 and f4 := spoly(f2 , f1 ) = xα − xα x3 . By the 2 1 α2 +1 2-Buchsbaumness, x2 has to be the initial monomial. And the standard basis algorithm will stop at this step. (b) Suppose that x1 x3α3 −1 is one minimal generator for I ∗ . Then α12 α2 α3 −1 α1 −1 α2 −α12 1 3 I = (f1 := xα , f3 := xα x2 ), 1 − x2 x3 , f2 := x2 − x1 x3 3 − x1

with α12 = 1 or 2. The standard basis algorithm generates f4 := spoly(f2 , f1 ) = 2 +α12 1 +1 α3 −2 xα −xα x3 . If the tangent cone is 2-Buchsbaum, then α2 +α12 ≤ 2 1 α1 + α3 − 1. And then the algorithm stop at this step. (c) If x21 x3α3 −1 is one minimal generate for I ∗ , then by the proof for 4.9, α12 = 1, and the defining ideal is α2 α1 −2 α2 −1 2 α3 −1 1 3 I = (f1 := xα , f3 := xα x2 ). 1 − x2 x3 , f2 := x2 − x1 x3 3 − x1

Similar to the previous case, the standard basis algorithm will only con2 +1 1 +2 α3 −2 tribute an additional basis element f4 := spoly(f1 , f2 ) = xα −xα x3 . 2 1

In the rest of this section, we establish a new characterization on when the tangent cone grm (R) will be Cohen-Macaulay in the case d = 3. First of all, we give a remark to 2.7. Remark 4.11. For symmetric G with arbitrary embedding dimension, if the order values of the Apéry set elements are symmetric in the sense of (†), and sQ (m) = rQ (m), then grm (R) is Gorenstein, even without assuming it to be Cohen-Macaulay in the beginning. For the proof, see [6, 3.14]. We thank Lance Bryant for the helpful comments regarding 4.12. Theorem 4.12. If G = hn1 , n2 , n3 i is minimally 3-generated, and for the principal reduction Q = (tn1 )R of the maximal ideal m, the index of nilpotency sQ (m) equals the reduction number rQ (m), then the tangent cone grm (R) is Cohen-Macaulay. Proof. In the following, for x ∈ S, write x ¯ for its image in R = S/I. First we study the case when G is symmetric, i.e., the numerical semigroup ring R is a complete intersection. Now for we−1 = f + n1 , ordm (we−1 ) = sQ (m) by 2.1. Using 4.1 again, we have three cases. (i) (i, j, k) = (1, 2, 3), then the tangent cone is automatically a complete intersection. (ii) (i, j, k) = (1, 3, 2). Now the last Apéry set element can be written as we−1 = (α2 − 1)n2 + (α3 − 1)n3 .

This is obviously the unique representation of we−1 with respect to G. Hence the order values of the Apéry set element is symmetric in the sense of (†). If rQ (m) = sQ (m), then grm (R) is Gorenstein by 4.11. (iii) (i, j, k) = (2, 3, 1). Now the associated graded ring grm (R) is CohenMacaulay if and only if α2 + α12 ≤ α1 + α3 − α13 . We want to prove that the following conditions are equivalent:


11

(a) grm (R) is Cohen-Macaulay. (b) rQ (m) = sQ (m). (c) rQ (m) ≤ α1 + α3 − 2. Notice that the Frobenius number can be written as f = (α2 + α12 − 1)n2 + (α13 − 1)n3 − n1 , hence we−1 = (α2 + α12 − 1)n2 + (α13 − 1)n3 .

This gives the maximal representation of we−1 with respect to G, and ordm (we−1 ) = (α2 + α12 − 1) + (α13 − 1). That (a) implies (b) is clear. ¯ xα13 −1 ∈ mr . So xα2 +α12¯xα13 −1 ∈ 2 +α12 −1 If (b) holds, then for r = rQ (m), xα 2 3 2 3 ¯ ¯ α +α α −1 α α +α 1 ¯α3 −1 mr+1 = Qmr . x2 2 12 x3 13 = x2 12 x3 3 13 −1 = xα and x¯1 is a reg1 x3 ular element in the domain R. Hence x1α1 −1¯x3α3 −1 ∈ mr . We want to show that ordm (x1α1 −1¯x3α3 −1 ) = (α1 − 1) + (α3 − 1), hence (c) holds. It suffices to show that (α1 −1)n1 +(α3 −1)n3 is the unique representation of this element with respect to G. Suppose not, then (α1 − 1)n1 + (α3 − 1)n3 = an1 + bn2 + cn3 , with a, b, c ∈ N0 and b > 0. By the minimality of α1 and α3 , one must have a ≤ α1 − 1 and c ≤ α3 − 1. Now (α1 − 1 − a)n1 + (α3 − 1 − c)n3 = bn2 Since b > 0, b ≥ α2 by the minimality of α2 . Hence (α1 − 1 − a)n1 + (α3 − 1 − c)n3 = (b − α2 )n2 + α3 n3 , thus (α1 − 1 − a)n1 = (b − α2 )n2 + (c + 1)n3 ,

which is against the minimality of α1 . This shows that (b) implies (c). If (c) holds, then α2 + α12 + α13 − 2 = sQ (m) ≤ rQ (m) ≤ α1 + α3 − 2. It follows immediately that α2 + α12 ≤ α1 + α3 − α13 . Hence grQ (m) is Cohen-Macaulay and (a) holds. Now we consider the case when the group G is not symmetric. Recall that the defining ideal shall be α31 α32 α3 α21 α23 α2 α12 α13 1 I = (f1 := xα 1 − x2 x3 , f2 := x2 − x1 x3 , f3 := x3 − x1 x2 ).

Our aim is to show that if sQ (m) = rQ (m), then α2 ≤ α21 + α23 . We first show that the index of nilpotency is (1)

sQ (m) = max {α2 + α13 − 2, α3 + α12 − 2} .

Notice that α2 + α13 − 2 = α2 + α3 − 2 − α23, and α3 + α12 − 2 = α2 + α3 − 2 − α32. sQ (m) = max {ordm (w) | w ∈ Ap(G, e)}. Every w ∈ Ap(G, e) can be written as w = bn2 + cn3 for some b, c ∈ N0 . Obviously, b < α2 and c < α3 . And it cannot happen that both b ≥ α12 and c ≥ α13 . Hence for w with this representation, if b = α2 − 1 ≥ α12 , then c < α13 because of f1 . Similarly, if c = α3 − 1 ≥ α13 , then b < α12 because of f2 . On the other hand, it is not difficult to see that (α2 − 1)n2 + (α13 − 1)n3 and (α12 − 1)n2 + (α3 − 1)n3 are elements of Ap(G, e). For instance, suppose for contradiction that (α2 − 1)n2 + (α13 − 1)n3 = an1 + bn2 + cn3

12

YIHUANG SHEN

with a, b, c ∈ N0 and a > 0. By the minimality of α2 and α3 , it is clear that α2 − 1 ≥ b and α13 − 1 ≥ c. Now (α2 − 1 − b)n2 + (α13 − 1 − c)n3 = an1 . a > 0, hence a ≥ α1 . Now (α2 − 1 − b)n2 + (α13 − 1 − c)n3 = (a − α1 )n1 + α12 n2 + α13 n3 . This implies (α32 − 1 − b)n2 = (a − α1 )n1 + (c + 1)n3 . This contradicts the minimality of α2 . One can argue in a similar way for (α12 − 1)n2 + (α3 − 1)n3 . Now the formula (1) follows naturally. The case when sQ (m) = α2 + α13 − 2 is easy. Suppose the condition is satα13 −1 2 ¯ isfied, i.e., α2 + α13 − 2 = r = rQ (m). Then xα ∈ mr+1 = Qmr . No2 x3 ¯ α2 ¯ α13 −1 α21 ¯α3 −1 α21 −1 α3 −1 tice that x2 x3 = x1 x3 . Hence x1 x3 ∈ mr . Similar to the proof for (3) of Gorenstein case, one can show that this decomposition is unique, ¯xα3 −1 ) = α + α − 2 ≥ r = s = α + α − 2. Hence 21 −1 hence ordm (xα 21 3 2 13 1 3 α2 ≤ α21 + α3 − α13 = α21 + α23 . If sQ (m) > α2 +α13 −2 and rQ (m) = sQ (m), then δ := α23 −α32 > 0 and α2 +α13 − 13 −1 2 +δ¯ α13 −1 2 = r − δ. Now ordm (x2α2 −1¯xα ) ≥ r − δ, hence xα x3 ∈ mr+1 = Qmr . It 3 2 α21 −1 ¯δ α3 −1 r follows easily that g0 := x1 x2 x3 ∈ m . It is not true that this is the unique representation. However, suppose that associated graded ring grm (R) is not CohenMacaulay, then α2 = α12 +α32 > α21 +α23 . Hence δ = α23 −α32 < α12 −α21 < α12 . If we can modify the representation of g0 , then the smallest possible step to do it for the time being will be to use f3 . Now g takes a different representation 21 −α31 −1 ¯δ−α32 2α3 −1 g1 := xα x2 x3 . It is by the same reason that the only minimal 1 relation that can be possibly used to modify the representation of g1 is f3 . And either you go back to get g0 , or you continue to drop the exponents of x1 and x2 , but increase the exponent of x3 . This argument can continue, i.e., one can only use f3 for a multiple of times to modify the representation of g0 to get some new gi . Notice that α3 < α31 +α32 , hence ordm (gi ) is achieved when it is written as g0 . The rest of the proof is similar. ordm (g0 ) = δ + α21 + α3 − 2 ≥ r = s = α2 + α13 − 2 + δ. Hence α2 ≤ α21 + α3 − α13 = α21 + α23 . The previous theorem fails if the embedding dimension is 4. Example 4.13. Let G = h9, 10, 11, 23i. Then G is symmetric, and sQ (m) = rQ (m) = 4. But grm (R) is not Cohen-Macaulay. If the 1-dimensional local ring R is not associated to numerical semigroup, then theorem might still fail, even when it has embedding dimension 3. The prototype of the following example is due to Lance Bryant. Example 4.14. Let S = C[[a, b, c]] and R = S/I where I = (a3 + c5 + b6 , a2 b + ac3 + b6 ). Then R is a 1-dimensional reduced ring. The initial form ideal I ∗ = (b2 c5 + ac6 , abc5 , a2 c3 , a2 b, a3 ), hence C[a, b, c]/I ∗ is not Cohen-Macaulay. On the other hand, Q = (b−c)R is a principal reduction of the maximal ideal m = (a, b, c)R. And rQ (m) = sQ (m) = 6.


13

5. When the embedding dimension d = 4 In this section, we want to prove the following main result by studying the standard basis of the corresponding defining ideal. Theorem 5.1. Let the numerical semigroup G = hn1 , n2 , n3 , n4 i be minimally generated by 4 elements. If the associated graded ring grm (R) is Gorenstein, then for every x ∈ Ap(G, n1 ), x G (f + n1 ). The proof of the above theorem depends heavily on an important result by Bresinsky [3]: Theorem 5.2. Let G = hn1 , n2 , n3 , n4 i be a symmetric numerical semigroup, minimally generated by 4 elements. Then up to a permutation of generators, the defining ideal I can be classified into the following three cases. β1 β2 β3 β4 α2 α3 α4 1 (I) I = (xα 1 − x2 , x3 − x4 , x1 x2 − x3 x4 ). α1 α2 α3 α31 α32 α4 41 α42 α43 (II) I = (x1 − x2 , x3 − x1 x2 , x4 − xα 1 x2 x3 ) with αij ∈ N0 . α1 α13 α14 α2 α21 α24 α3 α4 α42 α43 α43 α21 31 α32 (III) I = {x1 −x3 x4 , x2 −x1 x4 , x3 −xα 1 x2 , x4 −x2 x3 , x3 x1 − α32 α14 x2 x4 } with 0 < αij < αj .

Remark 5.3. Let T = k[x1 , · · · , xd ] be a polynomial ring of dimension d over the field k. Let m = (x1 , · · · , xd ) be the graded maximal ideal of T . If an T -ideal J is m-primary, monomial and Gorenstein, then it is generated by pure powers of the αd 1 form {xα 1 , . . . , xd }. The proof is an easy application of linkage theory. When the ideal is Gorenstein, one can use the generators that are pure powers to generate a link. The resulting quotient ideal is an almost complete intersection and monomial. Now use the same pure powers to generate a new link, and we shall get the original monomial ideal. But it will be easy to see that for an almost complete intersection monomial ideal, this link has to be generated by pure powers. See [13] for related discussion. As a generalization of the above idea, we can prove the following: ′

′

Proposition 5.4. Let ideal L ⊂ S = k[x, y, z] be generated by f = xa − y b z c ∈ (x, y, z) and a finite collection of monomials. If ht(L) = 3 and L is Gorenstein, then L is generated by up to 5 elements. Proof. In the following, fix the binomial f . We say an ideal K is almost monomial if we can write K = (f ) + K ′ where K ′ is an monomial ideal in k[x, y, z]. We fix a monomial ordering so that the initial monomial of f is xa . Depending on whether a ≥ b′ + c′ or not, this monomial ordering is either global or local. Applying the reduced standard basis algorithm to the existing generators of I, we may have a minimal standard basis {f1 , . . . , fs }. At most one of the fi is a binomial; the remaining are monomials. If f is part of this minimal standard basis, then we say K is strictly almost monomial. Following 5.3, we may always assume that L is strictly almost monomial. Since L has the maximal possible height, we can write L in a standard form L = (f, y b , z c )+L#, where L# is (minimally) generated by the remaining monomial generators of the standard basis. Let J = (f, y b , z c ), then \ J : L = J : L# = J : xα y β z γ . xα y β z γ ∈L#

14

YIHUANG SHEN

Notice that for every xα y β z γ ∈ S J : xα y β z γ =(f, y max(b−β,0) , z max(c−γ,0) , (2)

′

′

xmax(a−α,0) y max(b−b −β,0) , xmax(a−α,0) z max(c−c −γ,0) ).

To prove it, one can argue directly or use standard basis arguments which we give in the Appendix. Hence the quotient ideal J : L is an intersection of almost monomial ideals. We want to show that it is again almost monomial. This can be done by using induction. Suppose Ki = (f ) + Ki′ , i = 1, 2, are two almost monomial ideals, then (3)

K1 ∩ K2 = (f ) + (K1′ ∩ K2′ ).

To see this, observe that for every element g ∈ S, there is g ′ = NF(g| {f }) ∈ S. For every monomial m ∈ mono-Supp(g ′ ), m is not divisible by xa . Hence g ∈ K if and only if g ′ ∈ K ′ . Since L is Gorenstein, µ((J : L)/J) = 1. Hence the almost monomial ideal J : L = J + xα y β z γ for some monomial xα y β z γ ∈ S. And L = J : xα y β z γ is given by the formula (2). In particular, L is generated by up to 5 elements. Corollary 5.5 (Structure for Gorenstein strictly almost monomial ideals). Let ′ ′ L ⊂ k[x, y, z] be a strictly almost monomial ideal with f = xa − y b z c . Then L is Gorenstein and not a complete intersection if and only if L can written as (4)

′

′

L = (f, y b , z c , xa−α y b−b , xa−α z c−c ).

Proof. If L is almost monomial and Gorenstein, but not complete intersection, then we can write L in the format (2). Since we can assume that y b and z c belong to the minimal generators of L, one must have β = γ = 0 and L can be written in the given format (4). On the other hand, if L is written as in (4), then it is generated by the five submaximal Pfaffians of the anti-symmetric matrix   ′ 0 0 −y b 0 xa−α ′ ′  0 0 −z c−c y b−b 0    ′  b′  c−c α M = y z 0 x 0 .  ′ ′   0 −y b−b −xα 0 zc  a−α c′ −x 0 0 −z 0 Since ht(L) = 3 and L is Cohen-Macaulay, by the Buchsbaum-Eisenbud’s theorem [5, 3.4.1], L is Gorenstein. Remark 5.6. With the assumptions as in Corollary 5.5 and equip the polynomial ring k[x, y, z] with a monomial ordering, such that the leading monomial of f is xa . Let L be Gorenstein and strictly almost monomial with standard basis f, y b , z c , g4 , . . . , gs , where all the gi ’s are monomials. One observes the following from (4). (a) None of the gi ’s is divisible by yz. (b) If L is not a complete intersection, then it has the two additional monomial generators that are not pure powers. These two monomials share the same x component.


15

(c) None of the gi ’s is a pure power in x. In other words, If a pure power term in x is a minimal generator for a Gorenstein almost monomial ideal, then this ideal is generated by pure powers in x, y and z. These observations will play an essential role in the proof of 5.1. Remark 5.7. Return to our discussion of numerical semigroup √ ring R. When the embedding dimension d = 4, we know ht(I) = ht(I ∗ ) = 3. Since I ∗ = (x2 , x3 , x4 ), ht(I ∗ + (x1 )) = 4. Thus if we assume that I ∗ is Cohen-Macaulay, then x1 is regular modulo I ∗ . And I ∗ is Gorenstein if and only if I ∗ + (x1 ) is so. Furthermore, if f1 , · · · , fm is a minimal standard basis for I with respect to suitable local ordering, ∗ then the initial forms f1∗ , · · · , fm minimally generate I ∗ . For each i, if fi∗ is a ∗ monomial, since I is Cohen-Macaulay, fi∗ does not involve any x1 term by 3.2. β3 β2 β4 ∗ Let ˜ denote the image by modulo x1 . If ff i is a binomial, then fi = x3 − x2 x4 is balanced. On the other hand, it is easy to check that at most one fej can be binomial. Hence Ie∗ is almost monomial where x3 plays the same role as x in 5.5. It follows from the previous discussion that

Corollary 5.8. If grm (R) is Gorenstein, then µ(I ∗ ) ≤ 5. Now we are ready for the proof of 5.1. We proceed according to the three cases in 5.2. We always assume the monomial order for T = k[x1 , x2 , x3 , x4 ] ∼ = grn (S) is first nice in x1 , then nice in x2 . For instance, the usual negative degree reverse lexicographical ordering on k[x4 , x3 , x2 , x1 ] satisfies this requirement. In particular, α32 α34 3 3 if xα is balanced with α32 > 0, its initial monomial is xα 3 − x2 x4 3 . 5.1. Proof for Case I. Remark 5.9. The original statement of Theorem 4 in [3] contains an unnecessary condition 0 < βi < αi . This restriction is incorrect, as can be seen in 5.11. But the proof of it still holds without any further change. And the Lemma 5 before it only needs minimal modifications. It follows from this Lemma 5 and its corollary that (5)

f + n1 = (α2 − 1)n2 + (α3 + β3 − 1)n3 + (β4 − 1)n4 .

Proof for case I. If we insist that n1 < n2 < n3 < n4 , then by a permutation we need to consider the following three sub-cases: β1 β2 β3 β4 α2 α3 α4 1 (i) I = (f1 := xα 1 − x2 , f2 := x3 − x4 , f3 := x1 x2 − x3 x4 ); β2 β4 β1 β3 α4 α2 α3 1 (ii) I = (f1 := xα 1 − x3 , f2 := x2 − x4 , f3 := x1 x3 − x2 x4 ); β1 β4 α1 α4 α2 α3 (iii) I = (f1 := x1 − x4 , f2 := x2 − x3 , f3 := x1 x4 − xβ2 2 xβ3 3 ). Let’s prove according to these three cases. (i) Due to the generator f2 , we may assume that 0 ≤ β4 ≤ α4 − 1 in f3 . Hence β3 6= 0 by the choice of α4 . Similarly, we assume that 0 ≤ β2 ≤ α2 − 1, and hence β1 6= 0. Since spoly(f1 , f2 ) does not contribute to the standard β3 β4 α4 ∗ ∗ 2 basis, the initial forms f1∗ = −xα 2 , f2 = −x4 , f3 = −x3 x4 form part of the minimal basis of I ∗ . By 5.7 and 5.6, I˜∗ is almost monomial. If β4 6= 0, then I˜∗ is strictly almost monomial. Now by applying standard basis algorithm, we get f4 = 3 +β3 4 −β4 spoly(f2 , f3 ) = xα − xβ1 1 xβ2 2 xα . By the Cohen-Macaulay condition, 3 4 we must have α3 + β3 ≤ β1 + β2 + α4 − β4 . Since β1 6= 0, I˜∗ has one 3 +β3 minimal generator xα which is a pure power. This is impossible for 3

16

YIHUANG SHEN

strictly almost monomial ideals. Hence β4 = 0 and I ∗ is generated by pure powers. Notice that f1 , f2 and f3 form a standard basis for I, and β3 ≥ α3 . It follows from the formula (5) that (6)

f + n1 = (α2 − 1)n2 + (β3 − 1)n3 + (α4 − 1)n4 . The decomposition of f + n1 with respect to G might not be unique in general. But it is easy to see that the coefficient (α2 − 1) of n2 is fixed. Now if β3 = α3 , then the above representation is unique. Otherwise, β3 −1 ≥ α3 , and we have at least one extra representation f + n1 = (α2 − 1)n2 + (β3 − 1 − α3 )n3 + (2α4 − 1)n4 . Since n3 < n4 , we have α3 > α4 by the relation α3 n3 = α4 n4 . Hence the decomposition in (6) is the unique one such that gives us the maximal length of f + n1 in terms of n2 , n3 and n4 . Now ordG (f + n1 ) = (α2 − 1) + (β3 − 1) + (α4 − 1). Similarly, every element z2 n2 + z3 n3 + z4 n4 ∈ Ap(G, n1 ) can be written in such a way that 0 ≤ z2 ≤ α2 − 1, 0 ≤ z3 ≤ β3 − 1 and 0 ≤ z4 ≤ α4 − 1. ordG (z2 n2 + z3 n3 + z4 n4 ) = z1 + z2 + z3 . In particular, we have (z2 n2 + z3 n3 + z4 n4 ) G (f + n1 ). (ii) We may assume that 0 ≤ β3 ≤ α3 − 1 and 0 ≤ β4 ≤ α4 − 1, so that β1 6= 0 α4 ∗ 3 and β2 6= 0. We have f1∗ = −xα 3 and f2 = −x4 . By comparing β1 + β3 ∗ with β2 + β4 , we have three cases for f3 . (a) f3∗ = xβ1 1 xβ3 3 . Following 5.7, the associated graded ring grm (R) won’t be Cohen-Macaulay unless β1 = 0, which is impossible here. (b) f3∗ = −xβ2 2 xβ4 4 . Since I˜∗ is almost monomial, if I ∗ is Gorenstein, then β4 = 0. Now β2 < β1 + β3 . Thus considering the permutation, from (5) we have f + n1 = (β2 − 1)n2 + (α3 − 1)n3 + (α4 − 1)n4 . Similar to the discussion in case (i), this decomposition is the unique maximal one. The rest of the proof is also similar. (c) f3∗ = xβ1 1 xβ3 3 − xβ2 2 xβ4 4 . After modulo x1 , we are reduced to a case similar to (b). 3 ˜∗ (iii) f2∗ = −xα 3 is a pure power in x3 . Hence I is a complete intersection in pure powers. This implies that β3 = 0 and β2 ≤ β4 + β1 if we assume that 0 ≤ β4 ≤ α4 − 1 from the beginning. The remaining discussion is similar to that of case (i).

Corollary 5.10. In case I of 5.2, if grm (R) is Gorenstein, then it is a complete intersection. To illustrate the above proposition, we give several examples. Example 5.11. (i) If G = h8, 12, 14, 21i, then the defining ideal I = (x24 − 3 2 2 x3 , x3 − x1 x2 , x22 − x31 ) and the initial form ideal I ∗ = (x22 , x23 , x24 ). (ii) If G = h8, 10, 12, 15i, then I = (x24 − x32 , x23 − x31 , x22 − x1 x3 ) and I ∗ = (x24 , x23 , x22 − x1 x3 ).


17

(iii) If G = h30, 33, 44, 45i, then I = (x52 − x41 x4 , x33 − x42 , x24 − x31 ) and I ∗ = (x52 − x41 x4 , x33 , x24 ). 5.2. Proof for Case II. In this case, the defining ideal is the complete intersection ideals studied in [11]. In particular, by [11, 2.1] the Frobenius number (7)

f = (α2 − 1)n2 + (α3 − 1)n3 + (α4 − 1)n4 − n1 .

We prove in accordance with the permutation (i, j, k, l) of (1, 2, 3, 4). By symmetry, we can always conveniently assume that i < j, and consider the following 12 cases. (i) (i, j, h, k) = (1, 2, 3, 4). Then α2 α3 α31 α32 α4 α41 α42 α43 1 I = (f1 := xα 1 − x2 , f2 := x3 − x1 x2 , f3 := x4 − x1 x2 x3 ).

Since n1 < n2 < n3 < n4 , we have α2 < α1 , α3 < α31 + α32 and α4 < α41 + α42 + α43 . Now for every x ∈ Ap(G, n1 ), suppose x = a2 n2 + a3 n3 + a4 n4 with a2 + a3 + a4 = ordG (x). By the maximality of ordG (x), one must have ai < αi for i = 2, 3, 4. On the other hand, by (7), f + n1 = (α2 − 1)n2 + (α3 − 1)n3 + (α4 − 1)n4 . Now it follows immediately that x G (f + n1 ). (ii) (i, j, h, k) = (1, 2, 4, 3). Then α2 α4 α31 α42 α3 α31 α32 α34 1 I = (f1 := xα 1 − x2 , f2 := x4 − x1 x2 , f3 := x3 − x1 x2 x4 ).

One has α2 < α1 and α4 < α41 + α42 automatically. Now we can assume that in f3 , 0 ≤ α32 < α2 and 0 ≤ α34 < α3 . If α31 6= 0, then by the CohenMacaulay condition, α3 ≤ α31 + α32 + α34 . If α31 = 0, by our previous assumption on α32 and α34 , f3∗ is part of the minimal generators for I ∗ . But if I ∗ is Gorenstein, terms of the form xβ2 2 xβ4 4 with β2 , β4 > 0 will not be part of the minimal generators. Hence either α3 ≤ α32 + α34 or one of α32 and α34 is zero. The latter cannot happen, since it is against to the choice of α4 or α2 respectively. In short, if we assume that I ∗ is Gorenstein, then we should have α3 ≤ α31 + α32 + α34 . And the rest of the proof is similar to that of (i). (iii) (i, j, h, k) = (1, 3, 2, 4). Then α41 α42 α43 α4 α21 α23 α2 α3 1 I = (f1 := xα 1 − x3 , f2 := x2 − x1 x3 , f3 := x4 − x1 x2 x3 ).

We automatically have α3 < α1 and α4 < α41 + α42 + α43 . We can always assume that 0 ≤ α23 < α3 . Hence the initial form f2∗ is part of the minimal generators for I ∗ . The Cohen-Macaulay condition would force α2 ≤ α21 + α23 . The rest of the proof is similar to that of (i). (iv) (i, j, h, k) = (1, 3, 4, 2). Then α21 α23 α24 α2 α41 α43 α4 α3 1 I = (f1 := xα 1 − x3 , f2 := x4 − x1 x3 , f3 := x2 − x1 x3 x4 ).

We have α3 < α1 and α4 < α41 + α43 . In addition, we can assume that 0 ≤ α23 < α3 and 0 ≤ α24 < α4 . Now f3∗ is part of the minimal generators. 3 Notice that f1∗ = −xα 3 is also one such minimal generator. Hence the ∗ Gorenstein ideal I after modulo x1 has to be a complete intersection ideal generated by pure powers. This would force that α2 ≤ α21 + α23 + α24 . The rest of the proof is similar to that of (i).

18

YIHUANG SHEN

(v) (i, j, h, k) = (1, 4, 2, 3). Then α4 α2 α21 α24 α3 α31 α32 α34 1 I = (f1 := xα 1 − x4 , f2 := x2 − x1 x4 , f3 := x3 − x1 x2 x4 ).

We have α4 < α1 . Now we can assume that 0 ≤ α24 , α34 < α4 . If f2∗ is part of a system of minimal generators, then the Cohen-Macaulayness would require that α2 ≤ α21 + α24 . The rest of the proof is similar to that of (ii). If f2∗ is not part of the minimal generating set for I ∗ , then α32 = 0, 31 α34 21 α24 is divisible by xα α3 > α31 + α34 , α2 > α21 + α24 and xα 1 x4 . The 1 x4 problem will be reduced to the case (vi) with α23 > 0. (vi) (i, j, h, k) = (1, 4, 3, 2). Then α4 α3 α31 α34 α2 α21 α23 α24 1 I = (f1 := xα 1 − x4 , f2 := x3 − x1 x4 , f3 := x2 − x1 x3 x4 ).

Clearly α4 < α1 and we can assume that 0 ≤ α24 , α34 < α4 . Now if initial form f2∗ belongs to a system of minimal generators for I ∗ , then the CohenMacaulay condition implies α3 ≤ α31 + α34 . The rest of the proof is similar to that of (iv). If f2∗ does not belong to a system of minimal generators for I ∗ , then 31 α34 α23 = 0, α2 > α21 + α24 , α3 > α31 + α34 and xα is divisible by 1 x4 α21 α24 x1 x4 . Now the problem is reduced to the case (v) with α32 > 0. (vii) (i, j, h, k) = (2, 3, 1, 4). Then α3 α1 α12 α13 α4 α41 α42 α43 2 I = (f1 := xα 2 − x3 , f2 := x1 − x2 x3 , f3 := x4 − x1 x2 x3 ).

α12 α13 ∗ 3 4 Since n1 < n2 < n3 < n4 , f1∗ = −xα and f3∗ = xα 3 , f2 = −x2 x3 4 . We ∗ can always assume that 0 ≤ α13 < α3 . Those fi belong to a system of minimal generators for I ∗ . Notice that f1∗ is a pure power in x3 . Hence after modulo x1 , I ∗ is generated by pure powers in x2 , x3 and x4 . f2∗ does not involve x1 term, hence need to be a pure power. One cannot have α12 = 0, since this is against the choice of α13 and α3 . Hence α13 = 0 and α12 ≥ α2 . P P Now for every x ∈ Ap(G, n1 ), we can write x = i ai ni with i ai = ordG (x). Obviously a1 = 0, a2 < α12 , a3 < α3 and a4 < α4 . Meanwhile, by (7),

f + n1 = (α12 − 1)n2 + (α3 − 1)n3 + (α4 − 1)n4 . Clearly x G (f + n1 ). (viii) (i, j, h, k) = (2, 3, 4, 1). Then α3 α4 α42 α43 α1 α12 α13 α14 2 I = (f1 := xα 2 − x3 , f2 := x4 − x2 x3 , f3 := x1 − x2 x3 x4 ).

α4 α12 α13 α14 ∗ ∗ 3 Obviously f1∗ = −xα 3 , f2 = x4 and f3 = −x2 x3 x4 . In the f3 , using f1 and f2 , we can assume that 0 ≤ α13 < α3 and 0 ≤ α14 < α4 . The rest of the proof is similar to that of (vii). In particular, we have α13 = α14 = 0. (ix) (i, j, h, k) = (2, 4, 1, 3). Then α4 α1 α12 α14 α3 α31 α32 α34 2 I = (f1 := xα 2 − x4 , f2 := x1 − x2 x4 , f3 := x3 − x1 x2 x4 ).

α12 α14 ∗ 4 f1∗ = −xα 4 and f2 = −x2 x4 . Using f1 we can always assume that 12 α14 0 ≤ α14 , α34 < α4 . Using f2 , we can assume that xα does not divide 2 x4 α31 α32 α34 x1 x2 x4 . Notice that if grm (R) is Gorenstein, then after modulo x1 , I ∗ is an almost monomial ideal where xβ2 2 xβ4 4 is not part of the minimal


19

generators, for β2 , β4 > 0. Hence either f2∗ is not part of the minimal generators or it is a pure power. The first case happens if and only if α31 = 0, α12 ≥ α32 , α14 ≥ α34 and α3 > α32 + α34 . Then this problem can be reduced to the case (x) with α13 > 0. For the second situation, since α14 < α4 , f2∗ is a pure power if and only if α14 = 0 and α12 ≥ α2 . Now in addition, one may assume that 0 ≤ α32 ≤ α12 − 1. For f3 , if α3 ≤ α31 + α32 + α34 , the proof can continue as in (i). This condition must be satisfied when α31 6= 0 and grm (R) is Cohen-Macaulay. Otherwise, assume α31 = 0 and α3 > α32 + α34 . Then 32 α34 will be part of the minimal basis of I ∗ . α32 , α34 > 0 and f3∗ = −xα 2 x4 Whence grm (R) will not be Gorenstein. (x) (i, j, h, k) = (2, 4, 3, 1). Then α4 α3 α32 α34 α1 α12 α13 α14 2 I = (f1 := xα 2 − x4 , f2 := x3 − x2 x4 , f3 := x1 − x2 x3 x4 ).

We can assume that 0 ≤ α14 , α34 < α4 . (A) If α3 < α32 + α34 , we can also assume that 0 ≤ α13 < α3 . Now f1∗ , f2∗ 12 α13 α14 and f3∗ = −xα form part of a system of minimal generators 2 x3 x4 ∗ ∗ for I . f2 is a pure power in x3 , hence after modulo x1 , I ∗ is a complete intersection generated by pure powers as well. Thus α13 = α14 = 0 and the rest of the proof is similar to that of (i). (B) If α3 > α32 + α34 , like in (ix), either we have α34 = 0 and we have a contradiction, or α13 = 0 and we can reduce the problem to the (ix) with α31 > 0. (C) If α3 = α32 +α34 , then under the monomial ordering we specified early 3 this section, the initial monomial of f2 is xα 3 . Now we further assume ∗ ∗ ∗ that 0 ≤ α13 < α3 . Hence f1 ,f2 and f3 form part of the system of generators for I ∗ , which is an almost monomial ideal when modulo x1 . α12 α13 α13 α14 12 By the structure of such ideals, −f3∗ is xα 2 , or x2 x3 , or x3 x4 , with strictly positive exponents. 12 (a) If −f3∗ = xα 2 , we can follow the proof in (i). α12 α13 ∗ (b) If −f3 = x2 x3 , we apply standard basis algorithm to find the standard basis for I. During the process, we have spoly(f2 , f3 ) = α1 α3 −α13 34 x2α32 +α12 xα . It is part of a minimal standard basis 4 −x1 x3 for I. Hence according to the structure for Gorenstein almost monomial ideal, grm (R) is not Gorenstein. 13 α14 (c) If −f3∗ = xα 3 x4 , by the structure property again, one has α14 + α34 = α4 . Applying the standard basis algorithm, we 2 α13 1 α34 find the two extra generators f4 := xα − xα and 2 x3 1 x4 α2 +α32 α1 α3 −α13 f5 := x2 −x1 x3 . In particular, we have the necessary conditions α2 + α13 P ≤ α1 + α34 and α2 + α32 ≤Pα1 + α3 − α13 . Now for every x = ai ni ∈ Ap(G, n1 ) with ai = ordG (x), 2 α13 does a1 = 0, a2 < α2 + α32 and a4 < α4 . Furthermore, xα 2 x3 a2 a3 α13 α14 a3 a4 not divide x2 x3 and x3 x4 does not divide x3 x4 . From (7), one has f + n1 =(α2 + α32 − 1)n2 + (α13 − 1)n3 + (α4 − 1)n4

=(α2 − 1)n2 + (α13 + α3 − 1)n3 + (α14 − 1)n4 .

20

YIHUANG SHEN

These are two decompositions of f + n1 with respect to G and have the same length. It will be routine to check that x G (f + n1 ). (xi) (i, j, h, k) = (3, 4, 1, 2). Then α4 α1 α13 α14 α2 α21 α23 α24 3 I = (f1 := xα 3 − x4 , f2 := x1 − x3 x4 , f3 := x2 − x1 x3 x4 ).

α13 α14 ∗ 4 Obviously f1∗ = −xα 4 and f2 = −x3 x4 . We can assume from the 23 α24 13 α14 beginning that 0 ≤ α14 , α24 < α4 , and xα is not divisible by xα 3 x4 3 x4 . (A) If α14 = 0, then α13 ≥ α3 . We have two sub-cases. (a) If α21 6= 0 or α24 6= 0, then f1∗ , f2∗ and f3∗ form part of a minimal system of generators for I ∗ . Since f2∗ is a pure power in x3 , I ∗ has to be a complete intersection after modulo x1 . This can happen only when α2 ≤ α21 + α23 + α24 . The rest of the proof for this case is similar to that of (i). (b) If α21 = α24 = 0, then α3 ≤ α23 < α13 . One can use f3 to reduce the problem to the case (xii) with α12 6= 0. (B) If α14 6= 0, since α14 < α4 , one must have α13 > 0. Now consider the following two sub-cases. 2 (a) If α2 ≤ α21 + α23 + α24 , then the initial monomial of f3 is xα 2 . Now apply the standard basis algorithm. We get 3 +α13 1 α4 −α14 f4 := spoly(f1 , f2 ) = xα − xα . 3 1 x4

None of the two monomial terms is divisible by any of f1∗ , f2∗ or f3∗ . Hence f4 is one new generator for the standard basis. Since α1 > 0, by the Cohen-Macaulay condition, we need to 1 have α3 + α13 ≤ α1 + α4 − α14 . Now spoly(f2 , f4 ) = xα 1 f1 . Hence the standard basis algorithm stops at this step, and the standard basis has only 4 elements. In other words, the initial form ideal I ∗ is an almost complete intersection, and cannot be Gorenstein. One can also use the structure property for strictly almost monomial ideals, by noticing that it should not contain pure powers in x3 . 21 α23 α24 (b) If α2 > α21 + α23 + α24 , then f3∗ = −xα 1 x3 x4 . ∗ ∗ ∗ If f1 , f2 and f3 form part of the minimal system of generators, then by the Cohen-Macaulay condition, we must have α21 = 0. Since f2∗ is not a pure power, I ∗ need to be a strictly almost monomial ideal after modulo x1 . If f3∗ is a pure power in x3 , then I ∗ need to be a complete intersection. This is impossible, since α14 6= 0. If α24 6= 0, then α23 > 0. But in an almost monomial ideal of our kind, monomials in both x3 and x4 only show up once. Hence we have another contradiction. If f1∗ , f2∗ and f3∗ do not form part of the minimal system of 23 α24 generators, since by our assumption α24 < α4 and xα is not 3 x4 α13 α14 divisible by x3 x4 , this can only happen when f2∗ is divisible by f3∗ . In particular, α21 = 0. Again, we can use f3 to reduce the problem to the case (xii) with α12 6= 0. (xii) (i, j, h, k) = (3, 4, 2, 1). Then α4 α2 α23 α24 α1 α12 α13 α14 3 I = (f1 := xα 3 − x4 , f2 := x2 − x3 x4 , f3 := x1 − x2 x3 x4 ).


21

α23 α24 ∗ 4 This case is similar to that of (xi). f1∗ = −xα and 4 , f2 = −x3 x4 α12 α13 α14 ∗ 13 α14 f3 = −x2 x3 x4 . We can assume that 0 ≤ α14 , α24 < α4 and xα x 3 4 23 α24 is not divisible by xα 3 x4 . (A) Suppose f1∗ , f2∗ and f3∗ form part of a minimal system of generators for I ∗ . Since I ∗ is almost monomial after modulo x1 , we have the following cases to consider with all given exponents strictly positive. 12 (a) f3∗ = −xα with α12 ≥ α2 . Applying standard basis algorithm, 2 2 α4 −α24 3 +α23 we get f4 = spoly(f1 , f2 ) = xα − xα and it should 2 x4 3 be a new generator for the standard basis. In almost mono2 α4 −α24 mial ideal described in 5.7, monomial of the form xα 2 x4 ∗ is not a minimal generator. Hence if I is Gorenstein, we must have α3 + α23 ≤ α2 + α4 − α24 . Whence the initial monomial of f4 is −x3α3 +α23 . Now the algorithm stops at this step and I ∗ = (f1∗ , f2∗ , f3∗ , f4∗ ) is an almost complete intersection, hence not Gorenstein. 13 (b) f3∗ = −xα with α13 ≥ α3 . Since it is a pure power in x3 , 3 the initial form ideal I ∗ has to be a complete intersection after modulo x1 . However, f2∗ is also an irredundant generator. This gives a contradiction. ∗ ∗ 13 α14 (c) f3∗ = −xα 3 x4 . Since f3 is irredundant, f2 cannot have strictly positive exponents for both variables. But 0 ≤ α24 < α4 , hence α23 6= 0. Thus α24 = 0 and f2∗ is a pure power in x3 . As a result, I ∗ need to be an complete intersection after modulo x1 . This is impossible, because of f3∗ . 12 α13 (d) f3∗ = −xα 2 x3 . By the structure of almost monomial ideal discussed in 5.7, we must have α24 6= 0 and α13 = α23 . Apply the standard basis algorithm, we get two more expected generators 2 α4 −α24 3 +α23 f4 := spoly(f1 , f2 ) = xα − xα 2 x4 3

and 1 α24 2 +α12 − xα f5 := spoly(f2 , f3 ) = xα 1 x4 . 2

Meanwhile, we get two necessary conditions α3 + α23 = α2 + α4 − α24 and α2 + α12 ≤ α1 + α24 . By (7), we know f + n1 =(α2 + α12 − 1)n2 + (α13 − 1)n3 + (α4 − 1)n4

=(α12 − 1)n2 + (2α13 + α3 − 1)n3 + (α24 − 1)n4

The rest of the proof is similar to that of (c) in (x). (B) If f1∗ , f2∗ and f3∗ fail to be part of a system of minimal generators, then f3∗ must divide f2∗ . This means α12 = 0 as well. Now using f3 , the problem can be reduced to the case (xi) with α21 6= 0. 5.3. Proof for case III. Up to a permutation of variables, the defining ideal I has the following form: 3 xα 3

−

1 I = (xα 1 − α31 α32 4 x1 x2 , xα 4

α2 α21 α24 13 α14 xα 3 x4 , x2 − x1 x4 ,

α43 α21 42 α43 32 α14 − xα − xα 2 x3 , x3 x1 2 x4 ).

In this case, Bresinsky [3, Theorem 5] proved that α1 = α21 + α31 , α2 = α32 + α42 , α3 = α13 + α43 and α4 = α14 + α24 .

22

YIHUANG SHEN

To simplify the notation, we follow the convention in [2] and write it as f1 = (1, (3, 4)), f2 = (2, (1, 4)), f3 = (3, (1, 2)), f4 = (4, (2, 3)), f5 = ((1, 3), (2, 4)). Hence we have six cases to consider 1. f1 = (1, (3, 4)) (a) f2 = (2, (1, 4)), f3 = (3, (1, 2)), (b) f2 = (2, (1, 3)), f3 = (3, (2, 4)), 2. f1 = (1, (2, 3)) (a) f2 = (2, (3, 4)), f3 = (3, (1, 4)), (b) f2 = (2, (1, 4)), f3 = (3, (2, 4)), 3. f1 = (1, (2, 4)) (a) f2 = (2, (1, 3)), f3 = (3, (1, 4)), (b) f2 = (2, (3, 4)), f3 = (3, (1, 2)), We also need the following result.

f4 = (4, (2, 3)), f5 = ((1, 3), (2, 4)) f4 = (4, (1, 2)), f5 = ((1, 4), (2, 3)) f4 = (4, (1, 2)), f5 = ((2, 4), (1, 3)) f4 = (4, (1, 3)), f5 = ((1, 2), (4, 3)) f4 = (4, (2, 3)), f5 = ((1, 2), (3, 4)) f4 = (4, (1, 3)), f5 = ((2, 3), (1, 4))

1 Theorem 5.12 ([2, 2.10]). In case III of Theorem 5.2, write f1 = xα 1 − m1 , α4 α3 α2 f2 = x2 − m2 , f3 = x3 − m3 , f4 = x4 − m4 and f5 , where m1 , m2 , m3 and m4 are monomials. If α2 ≤ total degree of m2 and α3 ≤ total degree of m3 , then the tangent cone grm (R) is Cohen-Macaulay.

Their conditions are equivalent to saying: case 1(a): α2 ≤ α21 + α24 ; case 1(b): α2 ≤ α21 + α23 , α3 ≤ α32 + α34 ; case 2(b): α2 ≤ α21 + α24 , α3 ≤ α32 + α34 ; case 3(a): α2 ≤ α21 + α23 , α3 ≤ α31 + α34 . α23 α24 2 No similar result is reproduced for 2(a) and 3(b), since f2 = xα is an 2 − x3 x4 element of the generator set in both cases, and thus α2 > α23 + α24 . It follows from 3.2 that if grm (R) is Cohen-Macaulay, then I ∗ does not contain any monomial minimal generator that is divisible by x1 . Corollary 5.13. Assume the setting of 5.2, if the defining ideal I is not a complete intersection and the associated graded ring grm (R) is Cohen-Macaulay, then the following inequalities hold: case 1(a): α2 ≤ α21 + α24 , α3 ≤ α31 + α32 , α21 + α43 ≥ α32 + α14 ; case 1(b): α2 ≤ α21 + α23 , α4 ≤ α41 + α42 , α21 + α34 ≥ α42 + α13 ; case 2(a): α3 ≤ α31 + α34 , α4 ≤ α41 + α42 , α41 + α23 ≥ α12 + α34 ; case 2(b): α2 ≤ α21 + α24 , α4 ≤ α41 + α43 , α41 + α32 ≥ α24 + α13 ; case 3(a): α2 ≤ α21 + α23 , α3 ≤ α31 + α34 , α31 + α42 ≥ α23 + α14 ; case 3(b): α3 ≤ α31 + α32 , α4 ≤ α41 + α43 , α41 + α34 ≥ α31 + α24 . Some of the inequalities are redundant, and hold automatically even without assuming the Cohen-Macaulayness of grm (R). They are kept here for completeness. But as an interesting result, we can see that Corollary 5.14. For cases 1(a) and 3(a), the conditions in [2, 2.10] are necessary and sufficient for grm (R) to be Cohen-Macaulay. Due to 2.5, our proof for 5.1 will be done by showing the following. Proposition 5.15. Assume the notations in 5.12. Then grm (R) is Gorenstein if and only the following conditions hold:


23

(a) α2 ≤ total degree of m2 , (b) f3 = (3, (2, 4)) and α3 = total degree of m3 . In particular, grm (R) could be Gorenstein only for case 1(b) and case 2(b). Proof. Sufficiency. Condition (2) only holds for cases 1(b) or 2(b). According to [2, 2.10], grm (R) is Cohen-Macaulay and the initial form ideal I ∗ ⊂ S = k[x1 , x2 , x3 , x4 ] of I is generated by the initial form of each generators. In case 1(b) we have α21 α34 α4 α32 α34 α3 α21 α23 α2 42 α13 ∗ 2 13 α14 − xα or xα I ∗ = (xα 2 x3 ) ). 2 , x3 − x2 x4 , x4 , (x1 x4 3 x4 , x2 − x1 x3

We want to point out that α21 + α34 ≥ α42 + α13 . To see this inequality, just note that by the assumptions we have α2 = α32 + α42 ≤ α21 + α23

and

α3 = α13 + α23 = α32 + α34 .

Now it follows that α21 α34 α42 α13 21 α34 42 α13 ∗ 42 α13 (xα − xα − xα 1 x4 2 x3 ) = (x1 x4 2 x3 ) or − x2 x3 .

Hence after modulo x1 , I ∗ becomes

α42 α13 α4 α32 α34 α3 α2 13 α14 Ie∗ = (xα 3 x4 , x2 , x3 − x2 x4 , x4 , x2 x3 ).

It follows from 5.5 that I ∗ is a Gorenstein ideal. 41 α32 13 α24 As for case 2(b), we have f5 = xα − xα 1 x2 3 x4 . Hence α41 n1 + α32 n2 = α13 n3 + α24 n4 . But α41 n1 + α32 n2 < α41 n2 + α32 n2 and α13 n3 + α24 n4 > α13 n2 + 13 α24 α24 n2 . This implies that α41 + α32 > α13 + α24 , and f5∗ = −xα 3 x4 . The rest of the discussion is similar to that of the previous case. Necessity. Since 0 < αij < αj , the five generators fi form part of a minimal standard basis for I, and their initial forms fi∗ form part of the minimal basis for the leading ideal I ∗ . By the Cohen-Macaulay assumption, after modulo x1 , none ∗ of the leading forms will go to zero. Hence ff i form part of the minimal basis for Ie∗ ⊂ k[x2 , x3 , x4 ], where ˜ denotes the image k[x1 , x2 , x3 , x4 ]/(x1 ) ∼ = k[x2 , x3 , x4 ]. By the Gorenstein assumption, this Ie∗ is again Gorenstein, and not a complete intersection. Hence by 5.7 it is strictly almost monomial and 5-generated. Now the assertion follows from 5.6 and 5.13. Corollary 5.16. Assume that the numerical semigroup G is 4-generated, and the defining ideal I is not a complete intersection. If the associated graded ring grm (R) is Gorenstein, then the elasticity of the special element f + n1 is 1 and ordG (f + n1 ) = α2 + α4 + α13 − 3. Proof. By 5.15, we only need to consider the two cases 1(b) and 2(b). We can write f + n1 = z1 n1 + z2 n2 + z3 n3 + z4 n4 . Since tf +n1 6∈ (tn1 ), we must have z1 = 0, and hence z2 ≤ α2 − 1 and z4 ≤ α4 − 1. Suppose that we can also write f + n1 = z2′ n2 + z3′ n3 + z4′ n4 . By symmetry, there are three cases to consider. (a) z2′ > z2 , z3′ ≤ z3 and z4′ ≤ z4 . (b) z3′ > z3 , z2′ ≤ z2 and z4′ ≤ z4 . (c) z4′ > z4 , z2′ ≤ z2 and z3′ ≤ z3 . By the choice of α2 , α3 and α4 , the first and third case cannot happen. Because otherwise, for instance in the first case, we would have (z2′ − z2 )n2 = (z3 − z3′ )n3 + (z4 − z4′ )n4 . Since z2′ − z2 < α2 , this cannot happen.

24

YIHUANG SHEN

By the same reason, in the second case, one must have z2′ = z2 − kα32 , z3′ = z3 +kα z4′ = z4 −kα34 for some k ∈ N0 . Since α3 = α32 +α34 , we can conclude P3 and P that zi = zi′ , i.e., the elasticity of f + n1 is 1. Now by the proof of [3, Theorem 5], one knows that f + n1 = (α2 − 1)n2 + (α13 − 1)n3 + (α4 − 1)n4 . It is clear then (α2 − 1) + (α13 − 1) + (α4 − 1) is one (and hence the) length of f + n1 . Corollary 5.17. In case III of 5.2, if grm (R) is Gorenstein, then the initial form ideal I ∗ is 5-generated. Remark 5.18. The converse of 5.17 is not true. For instance in case 1(a) of 5.14, when grm (R) is Cohen-Macaulay, I ∗ is 5-generated. But it can never be Gorenstein by 5.15. Acknowledgement The author wishes to thank Bernd Ulrich for the helpful suggestions throughout the preparation of this paper. Also, the author wishes to thank William Heinzer for the warm encouragement and helpful comments. He wants to acknowledge the support provided by Singular [9] and GAP [7]. In particular, he thanks Lance Bryant for the many helpful conversations and for the Singular library that facilitates the computations of initial form ideals. Appendix Here we use the standard basis method to give a strict proof of formula (2). Without loss of generality, we can assume that α < a, β < b − b′ and γ < c − c′ . Consider the following monomials and binomials in k[t, x, y, z] with lexicographical monomial ordering. ′

f1 =txa − ty b z c

′

f3 =tz c

′

′

f5 =ty b +β z c +γ − xa y β z γ

f7 =xα y β z c

f9 =xa y β z c−c

′

f2 =ty b f4 =txα y β z γ − xα y β z γ

f6 =xα y b z γ

′

f8 =xa y b−b z γ ′

f10 =xα+a y β z γ − xα y β+b z γ+c

′

The following table shows that every s-polynomial spoly(fi , fj ) is either 0 or can be reduced by some fk . By symmetry, we only need to consider for i < j. f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f1 f2 f3 f5 f5 , f8 f2 f3 f2 f3 0 f2 0 f6 f8 0 0 0 0 f2 f7 f9 0 0 0 0 f3 f3 f10 f6 f7 f8 f9 f5 f4 f8 f9 f10 , f8 f10 , f9 f10 f5 0 0 0 f6 f6 0 0 f7 f7 0 f6 f8 f7 f9 f10


25

Since f5 = spoly(f1 , f4 ), f6 = spoly(f2 , f4 ), f7 = spoly(f3 , f4 ), f8 = spoly(f2 , f5 ), f9 = spoly(f3 , f5 ) and f10 = spoly(f4 , f5 ), we know immediately that the set {fi : 1 ≤ i ≤ 10} forms a standard basis for I = (f1 , f2 , f3 , f4 ). Hence by [8, 1.8.10], the intersection ideal is ′

′

(xa − y b z c , y b , z c ) ∩ (xα y β z γ ) = (xα y β z γ · y b−β , xα y β z γ · z c−γ , ′

′

′

′

xα y β z γ · xa−α y b−b −β , xα y β z γ · xa−α z c−c −γ , xα y β z γ · (xa − y b z c ))

and by [8, 1.8.12], the quotient ideal is ′

′

′

′

′

′

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