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 g 1 < g 2 < · · · < g n ∈ N , GCD ( g 1 , . . . g n ) = 1 S = � g 1 , . . . , g n � := { n 1 g 1 + · · · + n n g n | n i ∈ N , i = 1 , . . . , n } R = k [[ S ]] := k [[ t g 1 , . . . , t g n ]] (or R := k [ t g 1 , . . . , t g n ] ( t g 1 ,...,t gn ) ) R is a one-dimensional, local domain, with maximal ideal m = ( t g 1 , . . . , t g n ) and quotient field Q = k (( t )). → Z ∪ ∞ If we denote by v : k (( t )) − 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 m i / m i +1 m i / m i +1 � � G ( m ) := M := i ≥ 0 i ≥ 1

2. Problem, definition and first remarks When is G ( m ) a Buchsbaum ring? uckrad-Vogel) G ( m ) is Buchsbaum if M· H 0 Definition. (St¨ M = 0. Remarks. • As H 0 M = ( ∪ k ≥ 1 (0 : G ( m ) M k )), ⇒ M · ( ∪ k ≥ 1 (0 : G ( m ) M k )) = 0. G ( m ) is Buchsbaum ⇐ ⇒ (0 : G ( m ) M ) = (0 : G ( m ) M k ) , ∀ k ≥ 1. • G ( m ) is Buchsbaum ⇐ ⇒ (0 : G ( m ) M k ) = (0) , ∀ k ≥ 1. • (0 : G ( m ) M ) = (0) ⇐ • 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 ) = g 1 and x its image in G ( m ). The reduction number r of m is the minimal natural number such that m r +1 = x m r . • m r / m r +1 ≃ m r + i / m r + i +1 , as R -module, ∀ i ≥ 1. • M is generated by m / m 2 , hence (0 : G ( m ) M i ) = (0 : G ( m ) m i / m i +1 ) . • (0 : G ( m ) M ) ⊆ · · · ⊆ (0 : G ( m ) M r ) = · · · = (0 : G ( m ) M k ) = . . . ( ∀ k ≥ r ) • G ( m ) is Buchsbaum (not C-M) ⇐ ⇒ 0 � = (0 : G ( m ) M ) = (0 : G ( m ) M r )

5. Graded description of (0 : G ( m ) M k ) ⇒ (0 : G ( m ) M ) = (0 : G ( m ) M r ) Recall that: G ( m ) is Buchsbaum ⇐ ( m h + k +1 : R m k ) ∩ m h (0 : G ( m ) M k ) = � (1) m h +1 h ≥ 1 Remark. For every h , ( m h +1 : R m ) ∩ m h ⊆ ( m h + r +1 : R m r ) ∩ m h This means that each direct summand of (0 : G ( m ) M ) is con- tained in the corresponding direct summand of (0 : G ( m ) M r ). Hence: ⇒ ( m h +1 : R m ) ∩ m h = ( m h + r +1 : R m r ) ∩ m h G ( m ) is Buchsbaum ⇐ ∀ h ≥ 1.

Let R ′ = B ( m ) = ∪ n ≥ 1 ( m n : Q m n ) = ( m r : Q m r ) = x − r m r (the last one is an equality of R -modules). Hence: ( m h + r +1 : R m r ) ∩ m h = ( m h + r +1 : Q m r ) ∩ m h = x h +1 R ′ ∩ m h xR ′ ∩ x − h m h (0 : G ( m ) M r ) = x h � (2) x − h m h +1 h ≥ 1 Proposition. xR ′ ∩ x − h m h r − 2 (0 : G ( m ) M r ) = x h � (3) x − h m h +1 h =1 (All direct summands are zero for h ≥ r − 1, since x − h m h +1 = xR ′ , ∀ h ≥ r − 1.) Hence: ⇒ ( m h +1 : Q m ) ∩ m h = ( m h + r +1 : Q m r ) ∩ m h G ( m ) is Buchsbaum ⇐ ∀ h = 1 , . . . , r − 2.

The direct summands of (0 : G ( m ) M ) and (0 : G ( m ) M r ) corre- sponding to h = r − 2, are equal: Lemma. ( m 2 r − 1 : Q m r ) ∩ m r − 2 = ( m r : Q m ) ∩ m r − 2 m r − 1 m r − 1 Corollary. (Goto) Let r be the reduction number of R . If r ≤ 3 then G ( m ) is Buchsbaum. Corollary. Let e = g 1 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 αx h +1 ∈ m h \ m h +1 (there is an element, αx h +1 , in (0 : G ( m ) M r )) (ii) Assume that G ( m ) is Buchsbaum not C-M; if α and h are as in (i), then αx h +2 ∈ m h +2 (any αx h +1 ∈ (0 : G ( m ) M r ) is also in (0 : G ( m ) M ) ⊆ (0 : G ( m ) x )) More precisely: Theorem B. Let G ( m ) be not C-M. Then ⇒ ∀ α ∈ R ′ such that αx h +1 ∈ m h \ m h +1 , G ( m ) is Buchsbaum ⇐ for some h ∈ { 1 , · · · , r − 2 } , then αx h +1 m ⊆ m h +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 = g 1 + 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 ( I n ) = n · v ( I ) and v ( I : Q J ) = v ( I ) − v ( J ); • if J ⊆ I , then l R ( I/J ) = | v ( I ) \ v ( J ) | .

The blow up of S = � g 1 , g 2 , . . . , g n � is the numerical semigroup S ′ = � ( nM − nM ) = ( rM − rM ) = rM − rg 1 = � g 1 , g 2 − g 1 , . . . , g n − g 1 � n ≥ 1 8. Translation at semigroup level v (( m h + r +1 : Q m r ) ∩ m h ) = ((( r + h + 1) M − rM ) � hM ) = = (( S ′ + ( h + 1) g 1 ) � hM ) v (( m h +1 : Q m ) ∩ m h ) = (( h + 2) M − M ) � hM ) ⇒ (( h +1) g 1 + S ′ ) � hM = ( h +1) M G ( m ) C-M ⇐ ∀ h = 1 , . . . , r − 2 ⇒ (( S ′ + ( h + 1) g 1 ) � hM ) \ ( h + 1) M = G ( m ) Buchs., not C-M ⇐ = ((( h + 2) M − M ) � hM ) \ ( h + 1) M � = ∅ ⇒ ∀ α ∈ S ′ such that α + ( h + 1) g 1 ∈ hM \ ( h + 1) M , for some ⇐ h ∈ { 1 , · · · , r − 2 } , then α + ( h + 1) g 1 + M ⊆ ( h + 2) M .

9. The Apery set Fix s ∈ S and set ω i := min { s ∈ S | s ≡ i (mod s ) } . ω 0 = 0 We call the Apery set S with respect of S , the set Ap s ( S ) = { ω 0 , . . . , ω s − 1 } We will compare the Apery sets of S and S ′ , with respect to g 1 . We fix the following notations: Ap g 1 ( S ) = { ω 0 , . . . , ω g 1 − 1 } Ap g 1 ( S ′ ) = { ω ′ 0 , . . . , ω ′ g 1 − 1 }

Definition. (Barucci-Fr¨ oberg) For each i = 0 , 1 , . . . , g 1 − 1 let: a i be the only integer such that ω ′ i + a i g 1 = ω i ; b i = max { l | ω i ∈ lM } . Remark. b 0 = a 0 = 0 and 1 ≤ b i ≤ a i . Theorem. (Barucci-Fr¨ oberg) G ( m ) is C-M ⇐ ⇒ a i = b i for each i = 0 , 1 , . . . , g 1 − 1. Semigroup level: a i > b i ⇐ ⇒ ∃ s ≡ i (mod g 1 ) such that s ∈ (( h + 1) g 1 + S ′ ) ∩ hM \ ( h + 1) M . Idea of the proof of ( ⇒ ): ⇒ (( h + 1) g 1 + S ′ ) � hM = ( h + 1) M ) (Recall that G ( m ) C-M ⇐ s ′ = ω ′ i + ( a i − b i − 1) g 1 ∈ S ′ . Assume a i > b i ; let i + a i g 1 = s ′ + ( b i + 1) g 1 ∈ b i M \ ( b i + 1) M ; ω ′ Then for h = b i we get the thesis. In particular, t ω i ∈ (0 : G ( m ) M r ).

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 , − →} Ap 13 ( S ) = { 0 , 79 , 41 , 16 , 56 , 31 , 32 , 46 , 47 , 48 , 23 , 63 , 51 } Ap 13 ( S ′ ) = { 0 , 27 , 15 , 3 , 30 , 18 , 6 , 20 , 21 , 9 , 10 , 24 , 12 } . 41 ∈ M \ 2 M = ⇒ b 2 = max { l | ω 2 ∈ lM } = 1 ⇒ t 41 ∈ (0 : G ( m ) M r ). 41 − 15 = 2 · 13 = ⇒ a 2 = 2 = 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 − g 1 ⊆ 2 M − 2 g 1 ⊆ · · · ⊆ rM − rg 1 � | � | || M − M ⊆ 2 M − 2 M ⊆ · · · ⊆ rM − rM For each i = 0 , 1 , . . . , g 1 − 1 set: c i := min { n | ω ′ i ∈ nM − ng 1 } d i := min { n | ω ′ i ∈ nM − Z nM } . Proposition. b i ≤ a i ≤ c i ≤ d i . Moreover, b i < a i ⇐ ⇒ a i < c i . In particular, if b i < a i and d i = a i + 1, then c i = a i + 1.

Theorem C. G ( m ) is Buchsbaum = ⇒ c i − a i ≤ a i − b i , ∀ i . (Easy case: a i = b i + 1. Recall that t ω i ∈ (0 : G ( m ) M r ). G ( m ) Buchsbaum = ⇒ t ω i ∈ (0 : G ( m ) M r ) = (0 : G ( m ) M ) ⊆ (0 : G ( m ) x ) = ⇒ ω i + g 1 ∈ ( b i +2) M i.e. ω ′ i +( a 1 +1) g 1 ∈ ( a 1 +1) M = ⇒ c i = a i +1.) In the example, 15 + 3 · 13 = 54 / ∈ 3 M , 15 + 4 · 13 = 67 / ∈ 4 M and 15 + 5 · 13 = 80 ∈ 5 M = ⇒ c 2 = min { n | 15 ∈ nM − ng 1 } = 5. Hence c 2 − a 2 > a 2 − b 2 and G ( m ) is not Buchsbaum. Note that: b 2 = 1, a 2 = 2, 41 + 13 = 15 + 3 · 13 / ∈ 3 M hence t 41 / ∈ (0 : G ( m ) M ). Remark. R = k [[ t 5 , t 9 , t 22 ]] shows that the condition in the last theorem is not sufficient.

Theorem D. Suppose d i = a i + 1 for every i such that a i > b i . Then G ( m ) is Buchsbaum. (Let α ∈ S ′ such that α + ( h + 1) g 1 ∈ hM \ ( h + 1) M We need to show that α + ( h + 1) g 1 ∈ (( h + 2) M − M ) Easy case: a i = b i + 1 and α = ω ′ i We have ω ′ i + ( b i + 1) g 1 ∈ b i M \ ( b i + 1) M and ω ′ d i = b i + 2 i.e. i ∈ (( b i + 2) M − ( b i + 2) M ) = ⇒ β := α + ( h + 1) g 1 ∈ (( b i + 2) M − M ) ) R = k [[ t 10 , t 17 , t 23 , t 82 ]] shows that the condition in Remark. the last theorem is not necessary.