On the Buchsbaumness of the associated graded ring of semigroup - - PowerPoint PPT Presentation

Marco D’Anna

On the Buchsbaumness of the associated graded ring of semigroup rings

Porto - March 2008 Joint work with: M. Mezzasalma and V. Micale

• 1. Setup

g1 < g2 < · · · < gn ∈ N, GCD(g1, . . . gn) = 1 S = g1, . . . , gn := {n1g1 + · · · + nngn | ni ∈ N, i = 1, . . . , n} R = k[[S]] := k[[tg1, . . . , tgn]] (or R := k[tg1, . . . , tgn](tg1,...,tgn)) R is a one-dimensional, local domain, with maximal ideal

m = (tg1, . . . , tgn) and quotient field Q = k((t)).

If we denote by v : k((t)) − → Z ∪ ∞ the natural valuation, we get v(R) = {v(r) | r ∈ R \ {0}} = S. The associated graded ring with respect to m will be denoted by G(m) :=

• i≥0

mi/mi+1

M :=

• i≥1

mi/mi+1

• 2. Problem, definition and first remarks

When is G(m) a Buchsbaum ring?

• Definition. (St¨

uckrad-Vogel) G(m) is Buchsbaum if M·H0

M = 0.

• Remarks. • As H0

M = (∪k≥1(0 :G(m) Mk)),

G(m) is Buchsbaum ⇐ ⇒ M · (∪k≥1(0 :G(m) Mk)) = 0.

• G(m) is Buchsbaum ⇐

⇒ (0 :G(m) M) = (0 :G(m) Mk), ∀k ≥ 1.

• (0 :G(m) M) = (0) ⇐

⇒ (0 :G(m) Mk) = (0), ∀k ≥ 1.

• G(m) Cohen Macaulay (C-M) ⇐

⇒ (0 :G(m) M) = (0)

• G(m) C-M =

⇒ G(m) Buchsbaum

• 3. Some references

The property for G(m) to be Cohen-Macaulay has been largely studied, while, not much is known about the Buchsbaum property of G(m) (except for the case that G(m) is Cohen-Macaulay). General case (R Noetherian, local ring, of dimension d):

• Goto, Buchsbaum rings of maximal embedding dimension,

(1982)

• Goto, Noetherian local rings with Buchsbaum associated

graded rings, (1984) The semigroup ring case (with S 3-generated):

• Sapko, Associated graded rings of numerical semigroup rings,

(2001)

• 4. More remarks

Let x be any element s.t. v(x) = g1 and x its image in G(m). The reduction number r of m is the minimal natural number such that mr+1 = xmr.

• mr/mr+1 ≃ mr+i/mr+i+1, as R-module, ∀ i ≥ 1.
• M is generated by m/m2, hence

(0 :G(m) Mi) = (0 :G(m) mi/mi+1).

• (0 :G(m) M) ⊆ · · · ⊆ (0 :G(m) Mr) = · · · = (0 :G(m) Mk) = . . .

(∀k ≥ r)

• G(m) is Buchsbaum (not C-M) ⇐

⇒ 0 = (0 :G(m) M) = (0 :G(m) Mr)

• 5. Graded description of (0 :G(m) Mk)

Recall that: G(m) is Buchsbaum ⇐ ⇒ (0 :G(m) M) = (0 :G(m) Mr) (1) (0 :G(m) Mk) =

• h≥1

(mh+k+1 :R mk) ∩ mh

mh+1

• Remark. For every h, (mh+1 :R m) ∩ mh ⊆ (mh+r+1 :R mr) ∩ mh

This means that each direct summand of (0 :G(m) M) is con- tained in the corresponding direct summand of (0 :G(m) Mr). Hence: G(m) is Buchsbaum ⇐ ⇒ (mh+1 :R m)∩mh = (mh+r+1 :R mr)∩mh ∀ h ≥ 1.

Let R′ = B(m) = ∪n≥1(mn :Q mn) = (mr :Q mr) = x−rmr (the last one is an equality of R-modules). Hence: (mh+r+1 :R mr) ∩ mh = (mh+r+1 :Q mr) ∩ mh = xh+1R′ ∩ mh (2) (0 :G(m) Mr) =

• h≥1

xR′ ∩ x−hmh x−hmh+1 xh Proposition. (3) (0 :G(m) Mr) =

r−2

• h=1

xR′ ∩ x−hmh x−hmh+1 xh (All direct summands are zero for h ≥ r−1, since x−hmh+1 = xR′, ∀ h ≥ r − 1.) Hence: G(m) is Buchsbaum ⇐ ⇒ (mh+1 :Q m)∩mh = (mh+r+1 :Q mr)∩mh ∀ h = 1, . . . , r − 2.

The direct summands of (0 :G(m) M) and (0 :G(m) Mr) corre- sponding to h = r − 2, are equal: Lemma. (m2r−1 :Q mr) ∩ mr−2

mr−1

= (mr :Q m) ∩ mr−2

mr−1

• Corollary. (Goto) Let r be the reduction number of R. If r ≤ 3

then G(m) is Buchsbaum. Corollary. Let e = g1 be the multiplicity of R. If e ≤ 4 then G(m) is Buchsbaum.

• 6. A characterization

Propostion A. (i) G(m) is NOT C-M ⇐ ⇒ ∃ α ∈ R′ and h ∈ {1, . . . , r − 2} such that αxh+1 ∈ mh \ mh+1 (there is an element, αxh+1, in (0 :G(m) Mr)) (ii) Assume that G(m) is Buchsbaum not C-M; if α and h are as in (i), then αxh+2 ∈ mh+2 (any αxh+1 ∈ (0 :G(m) Mr) is also in (0 :G(m) M) ⊆ (0 :G(m) x)) More precisely: Theorem B. Let G(m) be not C-M. Then G(m) is Buchsbaum ⇐ ⇒ ∀ α ∈ R′ such that αxh+1 ∈ mh \ mh+1, for some h ∈ {1, · · · , r − 2}, then αxh+1m ⊆ mh+2.

• 7. Some well-known facts
• H ⊂ Z, H = ∅ is a relative ideal of a semigroup S if

H + S ⊆ H and H + s ⊆ S ( ∃ s ∈ S).

• If H and L are relative ideals of S, then H + L, nH and

H − L := {n ∈ Z | n + L ⊆ H} are also relative ideals of S.

• The ideal M = {s ∈ S | s = 0} is called the maximal ideal of S.
• The reduction number r of M is the minimal natural number

such that (r + 1)M = g1 + rM.

• If I and J are fractional ideals of R, then v(I) and v(J) are

relative ideals of S = v(R);

• if I and J are monomial ideals, then

v(I ∩ J) = v(I) ∩ v(J), v(In) = n · v(I) and v(I :Q J) = v(I) − v(J);

• if J ⊆ I, then lR(I/J) = |v(I) \ v(J)|.

The blow up of S = g1, g2, . . . , gn is the numerical semigroup S′ =

• n≥1

(nM−nM) = (rM−rM) = rM−rg1 = g1, g2−g1, . . . , gn−g1

• 8. Translation at semigroup level

v((mh+r+1 :Q mr) ∩ mh) = (((r + h + 1)M − rM) hM) = = ((S′ + (h + 1)g1) hM) v((mh+1 :Q m) ∩ mh) = ((h + 2)M − M) hM) G(m) C-M ⇐ ⇒ ((h+1)g1+S′) hM = (h+1)M ∀ h = 1, . . . , r−2 G(m) Buchs., not C-M ⇐ ⇒ ((S′ + (h + 1)g1) hM) \ (h + 1)M = = (((h + 2)M − M) hM) \ (h + 1)M = ∅ ⇐ ⇒ ∀ α ∈ S′ such that α + (h + 1)g1 ∈ hM \ (h + 1)M, for some h ∈ {1, · · · , r − 2}, then α + (h + 1)g1 + M ⊆ (h + 2)M.

• 9. The Apery set

Fix s ∈ S and set ωi := min{s ∈ S | s ≡ i (mod s)}. ω0 = 0 We call the Apery set S with respect of S, the set Aps(S) = {ω0, . . . , ωs−1} We will compare the Apery sets of S and S′, with respect to g1. We fix the following notations: Apg1(S) = {ω0, . . . , ωg1−1} Apg1(S′) = {ω′

0, . . . , ω′ g1−1}

• Definition. (Barucci-Fr¨
• berg) For each i = 0, 1, . . . , g1 − 1 let:

ai be the only integer such that ω′

i + aig1 = ωi;

bi = max{l | ωi ∈ lM}.

• Remark. b0 = a0 = 0 and 1 ≤ bi ≤ ai.
• Theorem. (Barucci-Fr¨
• berg) G(m) is C-M ⇐

⇒ ai = bi for each i = 0, 1, . . . , g1 − 1. Semigroup level: ai > bi ⇐ ⇒ ∃ s ≡ i (mod g1) such that s ∈ ((h + 1)g1 + S′) ∩ hM \ (h + 1)M. Idea of the proof of (⇒): (Recall that G(m) C-M ⇐ ⇒ ((h + 1)g1 + S′) hM = (h + 1)M) Assume ai > bi; let s′ = ω′

i + (ai − bi − 1)g1 ∈ S′.

Then ω′

i + aig1 = s′ + (bi + 1)g1 ∈ biM \ (bi + 1)M;

for h = bi we get the thesis. In particular, tωi ∈ (0 :G(m) Mr).

• 10. An example

Let S = 13, 16, 23, 31, 41, 51, 56 (r = 5, R = k[[S]]) S′ = 3, 10 = {0, 3, 6, 9, 10, 12, 13, 15, 16, 18, − →} Ap13(S) = {0, 79, 41, 16, 56, 31, 32, 46, 47, 48, 23, 63, 51} Ap13(S′) = {0, 27, 15, 3, 30, 18, 6, 20, 21, 9, 10, 24, 12}. 41 ∈ M \ 2M = ⇒ b2 = max{l | ω2 ∈ lM} = 1 41 − 15 = 2 · 13 = ⇒ a2 = 2 = ⇒ t41 ∈ (0 :G(m) Mr). Hence G(m) is not C-M. We will see that G(m) is not even Buchsbaum.

• 11. More invariants

We define two more families of invariants for S. Recall that: M − g1 ⊆ 2M − 2g1 ⊆ · · · ⊆ rM − rg1

| |

|| M − M ⊆ 2M − 2M ⊆ · · · ⊆ rM − rM For each i = 0, 1, . . . , g1 − 1 set: ci := min{n | ω′

i ∈ nM − ng1}

di := min{n | ω′

i ∈ nM −Z nM}.

• Proposition. bi ≤ ai ≤ ci ≤ di. Moreover, bi < ai ⇐

⇒ ai < ci. In particular, if bi < ai and di = ai + 1, then ci = ai + 1.

Theorem C. G(m) is Buchsbaum = ⇒ ci − ai ≤ ai − bi, ∀ i. (Easy case: ai = bi + 1. Recall that tωi ∈ (0 :G(m) Mr). G(m) Buchsbaum = ⇒ tωi ∈ (0 :G(m) Mr) = (0 :G(m) M) ⊆ (0 :G(m) x) = ⇒ ωi+g1 ∈ (bi+2)M i.e. ω′

i+(a1+1)g1 ∈ (a1+1)M =

⇒ ci = ai+1.) In the example, 15 + 3 · 13 = 54 / ∈ 3M, 15 + 4 · 13 = 67 / ∈ 4M and 15 + 5 · 13 = 80 ∈ 5M = ⇒ c2 = min{n | 15 ∈ nM − ng1} = 5. Hence c2 − a2 > a2 − b2 and G(m) is not Buchsbaum. Note that: b2 = 1, a2 = 2, 41 + 13 = 15 + 3 · 13 / ∈ 3M hence t41 / ∈ (0 :G(m) M).

• Remark. R = k[[t5, t9, t22]] shows that the condition in the last

theorem is not sufficient.

Theorem D. Suppose di = ai + 1 for every i such that ai > bi. Then G(m) is Buchsbaum. (Let α ∈ S′ such that α + (h + 1)g1 ∈ hM \ (h + 1)M We need to show that α + (h + 1)g1 ∈ ((h + 2)M − M) Easy case: ai = bi + 1 and α = ω′

i

We have ω′

i + (bi + 1)g1 ∈ biM \ (bi + 1)M and

di = bi + 2 i.e. ω′

i ∈ ((bi + 2)M − (bi + 2)M)

= ⇒ β := α + (h + 1)g1 ∈ ((bi + 2)M − M) ) Remark. R = k[[t10, t17, t23, t82]] shows that the condition in the last theorem is not necessary.

• 12. A further example

S = 12, 23, 66, 99, 100, 110, 121, R = k[[S]]. S′ = 11, 12, 54 Ap12(S) = {0, 121, 110, 99, 100, 89, 66, 115, 92, 69, 46, 23}. Ap12(S′) = {0, 109, 98, 87, 76, 65, 54, 55, 44, 33, 22, 11}. 100 ∈ M \ 2M and 100 − 76 = 2 · 12, = ⇒ b4 = 1 and a4 = 2. (Hence G(m) is not C-M.) 76 + 2 · 12 = 100 / ∈ 2M = ⇒ 76 + 2M 2M, Moreover, 76 + 3M ⊆ 3M = ⇒ d4 = min{n | 76 ∈ nM −Z nM} = 3 By Theorem D, d4 = a4 + 1 = ⇒ G(m) is Buchsbaum. In fact 76 + 2 · 12 + M ⊆ 3M