On free resolutions of some semigroup rings Joint work with - PDF document

On free resolutions of some semigroup rings Joint work with Valentina Barucci and Mesut S ¸ahin J. Pure Appl. Algebra 218 (2014)

Let S = � n 1 , . . . , n k � be a numerical semigroup, i.e., n i are positive integers with greatest com- mon divisor 1, and S = { � k i =1 a i n i | a i ≥ 0 } . Let PF ( S ) = { n ∈ Z : n + s ∈ S, s ∈ S, s > 0 }\ S . The elements in PF ( S ) are called the pseud- ofrobenius numbers of S . Since S is a numer- ical semigroup, N \ S is finite. 1

The largest integer g ( S ) / ∈ S belongs to PF ( S ) and is called the Frobenius number of S . If PF ( S ) = { g ( S ) } , S is called symmetric, since then, for each n ∈ Z , exactly one of n and g ( S ) − n lies in S . If PF ( S ) = { g ( S ) / 2 , g ( S ) } , S is called pseudosymmetric. 2

Let K be a field and K [ S ] = K [ t n 1 , . . . , t n k ] be the semigroup ring of S , then K [ S ] ≃ K [ x 1 , . . . , x k ] /I S where I S is the kernel of the surjection K [ x 1 , . . . , x k ] φ 0 − → K [ t ] , where x i �→ t n i . If deg( x i ) = n i , this map is ho- mogeneous of degree 0. We will in the sequel denote K [ x 1 , . . . , x k ] by A . 3

If you have a graded ring R = A/I ( I generated by homogeneous elements), one is interested in the A -resolution of R . Example If R = k [ t a , t b ], ( a, b ) = 1, then the → R has kernel ( x b 1 − x a map A − 2 ) and x b 1 − x a 2 0 − → A − → A − → R − → 0 is exact. This is the resolution. 4

Example If R = k [ t 4 , t 5 , t 6 ], the kernel of the map is ( x 1 x 3 − x 2 2 , x 3 1 − x 2 3 ) = ( f 1 , f 2 ). Now there is a relation between these generators and → A φ 1 → A 2 ( f 1 ,f 2 ) 0 − → 0 , − − → A − → R − where φ 1 = ( f 2 , − f 1 ) t is exact. The semigroup � 4 , 5 , 6 � is symmetric, and all symmetric 3-generated semigroups have a res- olution like this. A k -generated semigroup has a resolution of length k − 1. 5

For some numerical semigroup rings of small embedding dimension, namely those of em- bedding dimension 3, and symmetric or pseu- dosymmetric of embedding dimension 4, pre- sentations has been determined in the litera- ture. We extend these results to whole graded minimal resolutions explicitly. Then we use these resolutions to determine some invariants of the semigroups and certain interesting rela- tions among them. 6

For completeness we start with 3-generated not symmetric semigroups. We will use Her- zog’s result. Theorem 1 (Herzog) Let ( n 1 , n 2 , n 3 ) = 1 . Let α i , 1 ≤ i ≤ 3 be the smallest positive integer such that α i n i ∈ � n k , n l � , { i, k, l } = { 1 , 2 , 3 } , and let α i n i = α ik n k + α il n l . Then S = � n 1 , n 2 , n 3 � is 3 -generated not symmetric if and only if α ik > 0 for all i, k , α 21 + α 31 = α 1 , α 12 + α 32 = α 2 , α 13 + α 23 = α 3 . Then K [ S ] = K [ � n 1 , n 2 , n 3 � ] = K [ x 1 , x 2 , x 3 ] / ( f 1 , f 2 , f 3 ) where ( f 1 , f 2 , f 3 ) = ( x α 1 1 − x α 12 x α 13 , x α 2 2 − x α 21 x α 23 , x α 3 3 − 2 3 1 3 x α 31 x α 32 ) . 1 2 7

Denham gave a minimal graded A -resolution of K [ S ], where A = K [ x 1 , x 2 , x 3 ]. Theorem 2 (Denham) If S is a 3 -generated semigroup which is not symmetric.Then K [ S ] = K [ x 1 , x 2 , x 3 ] /I S = A/I S has a minimal graded A -resolution φ 2 φ 1 → A 2 → A 3 0 − → 0 , − − → A − where φ 1 = ( x α 1 1 − x α 12 x α 13 , x α 2 2 − x α 21 x α 23 , x α 3 3 − 2 3 1 3 x α 23 x α 32   3 2 x α 31 x α 32 x α 31 x α 13 ) = ( f 1 , f 2 , f 3 ) , and φ 2 =  .   1 2 1 3  x α 12 x α 21 2 1 8

Next we look at 4-generated symmetric but not complete intersection semigroups. We will use a theorem by Bresinsky. Theorem 3 (Bresinsky) S = � n 1 , n 2 , n 3 , n 4 � is 4 -generated symmetric not a complete inter- section if and only if there are integers α i , 1 ≤ i ≤ 4 , α ij , ij ∈ { 21 , 31 , 32 , 42 , 13 , 43 , 14 , 24 } , such that 0 < α ij < α i , for all i, j , α 1 = α 21 + α 31 , α 2 = α 32 + α 42 , α 3 = α 13 + α 43 , α 4 = α 14 + α 24 and n 1 = α 2 α 3 α 14 + α 32 α 13 α 24 , n 2 = α 3 α 4 α 21 + α 31 α 43 α 24 , n 3 = α 1 α 4 α 32 + α 14 α 42 α 31 , n 4 = α 1 α 2 α 43 + α 42 α 21 α 13 , ( n 1 , n 2 , n 3 , n 4 ) = 1 . Then K [ S ] = K [ x 1 , x 2 , x 3 , x 4 ] / ( f 1 , f 2 , f 3 , f 4 , f 5 ) where f 1 = x α 1 1 − x α 13 x α 14 , f 2 = x α 2 2 − x α 21 x α 24 , f 3 = 3 4 1 4 x α 3 3 − x α 31 x α 32 , f 4 = x α 4 4 − x α 42 x α 43 , f 5 = x α 43 x α 21 − 1 2 2 3 3 1 x α 32 x α 14 . 2 4 9

We now give the whole minimal A -resolution of K [ S ]. Theorem 4 In case S is 4 -generated symmet- ric, not a complete intersection, then the fol- lowing is a minimal resolution of K [ S ] : → A φ 3 φ 2 φ 1 → A 5 → A 5 0 − − − − → A − → 0 where φ 1 = ( f 1 , f 2 , f 3 , f 4 , f 5 ) x α 32 x α 43 x α 24   0 0 2 3 4     x α 31 x α 14 x α 43  0 0  1 4 3       x α 21 x α 14 x α 42   φ 2 = 0 0   1 4 2       x α 13 x α 21 x α 32   0 0   3 1 2       − x α 13 − x α 31 x α 42 x α 24 0 3 1 2 4 and φ 3 = ( − f 4 , − f 2 , − f 5 , f 3 , f 1 ) t . 10

Next we look at 4-generated pseudosymmet- ric semigroups. We will use a theorem by Komeda. Theorem 5 (Komeda) S = � n 1 , n 2 , n 3 , n 4 � is 4 -generated pseudosymmetric if and only if there are integers α i > 1 , 1 ≤ i ≤ 4 , and α 21 , 1 < α 21 < α 1 , such that n 1 = α 2 α 3 ( α 4 − 1) , n 2 = α 21 α 3 α 4 + ( α 1 − α 21 − 1)( α 3 − 1) + α 4 , n 3 = α 1 α 4 + ( α 1 − α 21 − 1)( α 2 − 1)( α 4 − 1) − α 4 + 1 , n 4 = α 1 α 2 ( α 3 − 1) + α 21 ( α 2 − 1) + α 2 , ( n 1 , n 2 , n 3 , n 4 ) = 1 . Then, K [ S ] = K [ x 1 , x 2 , x 3 , x 4 ] / ( f 1 , f 2 , f 3 , f 4 , f 5 ) 1 − x α 3 − 1 x α 4 − 1 where f 1 = x α 1 , f 2 = x α 2 2 − x α 21 x 4 , f 3 = 3 4 1 3 − x α 1 − α 21 − 1 4 − x 1 x α 2 − 1 x α 3 − 1 x α 3 x 2 , f 4 = x α 4 , f 5 = 1 2 3 x α 3 − 1 x α 21 +1 − x 2 x α 4 − 1 . 3 1 4 11

We now give the whole minimal A -resolution of K [ S ]. Theorem 6 In case S is 4 -generated pseudosym- metric, then the following is a minimal resolu- tion of K [ S ] : φ 3 φ 2 φ 1 → A 2 → A 6 → A 5 0 − − − − → A − → 0 where φ 1 = ( f 1 , f 2 , f 3 , f 4 , f 5 ) φ 2 =  x α 3 − 1  0 0 0 x 2 x 4 3 x 1 x α 3 − 1 x α 1 − α 21 x α 4 − 1 0 f 3 0   3 1 4  x α 21 +1  x α 4 − 1 x 1 x α 2 − 1 − f 2 0 0   1 4 2  x α 21  0 0 0 x 2 x 3   1 − x α 1 − α 21 − 1 x α 2 − 1 − x 3 0 x 4 0 1 2 and φ 3 = � t � 0 0 x 4 − x 1 x 3 − x 2 . − x α 2 − 1 x α 3 − 1 x α 4 − 1 − x α 1 − 1 1 x α 3 − 1 x α 21 f 2 f 3 2 3 4 1 3 12

In all proofs we use the following theorem by Buchsbaum-Eisenbud adopted to our situation. Theorem 7 (Eisenbud-Buchsbaum) Let φ n − 1 → · · · φ 2 φ 1 φ n 0 − → F n − → F n − 1 − − → F 1 − → F 0 be a complex of free modules. Let rank( φ i ) be the size of the largest nonzero minor in the matrix describing φ i , and let I ( φ i ) be the ideal generated by the minors of rank ( φ i ) in φ i . Then the complex is exact if and only if for all i (a) rank( φ i +1 ) + rank( φ i ) = rank( F i ) and (b) I ( φ i ) contains an A -sequence of length i . In all theorems it is an easy, but sometimes tedious, task to check that we have complexes. 13

Applications We will use the following well known facts: If S is generated by k elements, and A = K [ x 1 , . . . , x k ], then the free minimal A -resolution of K [ S ] has length codim( K [ S ]) = k − 1 since K [ S ] is a 1-dimensional Cohen-Macaulay ring: → A β k − 1 φ k − 1 → A β k − 2 φ k − 2 → · · · φ 2 → A β 0 − 0 − − − − → K [ S ] − → 0 14

The alternating sum of the β i ’s, the Betti num- bers is zero. The Betti numbers of A/I are β i = dim K H i ( F ∗ ⊗ K ) = dim K Tor A i ( R, K ) . This gives us an alternative way to define the Betti numbers, since also Tor A i ( R, K ) = H i ( G ⊗ R ), where G is a minimal A -resolution of K (the Koszul complex). If R is Cohen-Macaulay, the highest nonzero Betti number is called the CM-type of R . 15

The ring is homogeneous if we set deg( x i ) = n i . If we concentrate F ∗ above to a certain degree d , we get an exact sequence of vector spaces β k − 1 ,j → ⊕ j A β 1 ,j 0 − → ( A/I ) d → ⊕ j A − → · · · − d − j − → A d − d − j where the β i,j are the graded Betti numbers of K [ S ]. 16

The alternating sum of the dimensions of these vector spaces is 0. Multiplying each dimension with z d and summing for d ≥ 0, we get k − 1 ( − 1) i β i,j z j ) . � � Hilb A/I ( z ) = Hilb A ( z )(1 + i =1 j If deg( x i ) = n i , then Hilb K [ x 1 ,...,x k ] ( z ) = 1 / � k i =1 (1 − z n i ) . Letting K S = 1 + � k − 1 j ( − 1) i β i,j z j , we observe that � i =1 K S ( z ) z s . � Hilb K [ S ] ( z ) = = � k i =1 (1 − z n i ) s ∈ S 17