On the homology of semigroup rings Porto 2008 Ralf Fr oberg 1 - PDF document

On the homology of semigroup rings Porto 2008 Ralf Fr¨ oberg 1

Let S be a numerical semigroup, g ( S ) the Frobe- nius number, T ( S ) = { n ∈ Z ; n + s ∈ S if s > 0 } , k [[ S ]] (or k [ S ]) the semigroup ring. Lemma 1 For s ∈ S \ { 0 } we have n ∈ T ( S ) if and only if t n + s ∈ Soc( k [[ S ]] / ( t s )) . Hence dim k Soc( k [[ S ]] / ( t s )) = | T ( S ) | (the CM-type of k [[ S ]] ). Proof n ∈ T ( S ) iff t n ∈ k [[ S ]] and t n m ⊂ / k [[ S ]] iff t n + s / ∈ t s k [[ S ]] and t n + s m ⊂ t s k [[ S ]] 0 and t n + s � = 0 in k [[ S ]] / ( t s ) iff iff t n + s m = ¯ t n + s ∈ Soc( k [[ S ]] / ( t s ). 2

Let S = � n 1 , . . . , n k � and let c i be the smallest positive integer such that c i n i ∈ � n 1 , . . . , ˆ n i , . . . , n k � , and suppose that c i n i = � j � = i r ij n j . Theorem 1 (Herzog) Assume S = � n 1 , n 2 , n 3 � and not symmetric. Then k [[ S ]] ≃ k [[ X 1 , X 2 , X 3 ]] /I where I = ( X c 1 1 − X r 12 X r 13 , X c 2 2 − X r 21 X r 23 , X c 3 3 − X r 31 X r 32 ) 2 3 1 3 1 2 and r ij < c j for all i, j . 3

Theorem 2 (Bresinsky) Let S = � n 1 , n 2 , n 3 , n 4 � to be symmetric but not a complete intersec- tion. Then k [[ S ]] ≃ k [[ X 1 , X 2 , X 3 , X 4 ]] /I where I = ( X c 1 1 − X r 13 X r 14 , X c 2 2 − X r 21 X r 24 , 3 4 1 4 X c 3 3 − X r 31 X r 32 , X c 4 4 − X r 42 X r 43 , 1 2 2 3 X r 43 X r 21 − X r 32 X r 14 ) and r ij < c j for all i, j . 3 1 2 4 4

Theorem 3 Assume S = � n 1 , n 2 , n 3 � and not symmetric.Then there is a minimal R = k [[ X 1 , X 2 , X 3 ]] -resolution of k [[ S ]] : 0 → R [ − n 2 c 2 − n 3 r 13 ] ⊕ R [ − n 3 c 3 − n 2 r 12 ] → R [ − n 1 c 1 ] ⊕ R [ − n 2 c 2 ] ⊕ R [ − n 3 c 3 ] → R → k [[ S ]] → 0 . Proof It is easy to get a minimal k [[ X 2 , X 3 ]]- resolution of T = k [[ S ]] / ( t n 1 ) = k [[ X 2 , X 3 ]] / ( X r 12 X r 13 , X c 2 2 , X c 3 3 ) , 2 3 and then lift it. Corollary 1 Assume S = � n 1 , n 2 , n 3 � and not symmetric. Then g ( S ) = max { n 2 r 12 + n 3 c 3 − ( n 1 + n 2 + n 3 ) , n 2 c 2 + n 3 r 13 − ( n 1 + n 2 + n 3 ) } . Proof Soc( T ) = ( X r 12 − 1 X c 3 − 1 , X c 2 − 1 X r 13 − 1 ). 2 3 2 3 5

Example If S = � 7 , 10 , 13 � we get the resolu- tion 0 → R [ − 59] ⊕ R [ − 62] → R [ − 49] ⊕ R [ − 20] ⊕ R [ − 52] → R → k [[ S ]] → 0 . This gives g ( S ) = 62 − (7 + 10 + 13) = 32 and T ( S ) = { 29 , 32 } . The Hilbert series of k [ S ] is H k [ S ] = 1 − z 49 − z 20 − z 52 + z 59 + z 62 . (1 − z 7 )(1 − z 10 )(1 − z 13 ) The conductor k [ S ] : k [ t ] is C = t 33 k [ t ]. The Hilbert series of k [ S ] /C , is H k [ S ] /C = H k [ S ] − z 33 1 − z = 1+ z 7 + z 10 + z 13 + z 14 + z 17 + z 20 + z 21 + z 23 + z 24 + z 26 + z 27 + z 28 + z 30 + z 31 . Thus the length of k [ S ] /C is l ( k [ S ] /C ) = H k [ S ] /C (1) = 15 and l ( k [ t ] /k [ S ]) = 33 − 15 = 18. 6

Theorem 4 Assume S = � n 1 , n 2 , n 3 , n 4 � to be symmetric but not a complete intersection. Then there is a minimal R = k [[ X 1 , X 2 , X 3 , X 4 ] - reso- lution of k [[ S ]] : 0 → R [ − n 2 c 2 − n 3 c 3 − n 4 r 14 ] → R 5 → R 5 → R → k [[ S ]] → 0 . Proof It is easy to get a k [[ X 2 , X 3 , X 4 ]]-resolution of T = k [[ S ]] / ( t n 1 ) = k [[ X 2 , X 3 , X 4 ]] /I where I = ( X r 13 X r 14 , X c 2 2 , X c 3 3 , X c 4 4 − X r 42 X r 43 , X r 32 X r 14 ), 3 4 2 3 2 4 and then lift it. Corollary 2 Assume S = � n 1 , n 2 , n 3 , n 4 � to be symmetric but not a complete intersection. Then g ( S ) = n 2 c 2 + n 3 c 3 + n 4 r 14 − ( n 1 + n 2 + n 3 + n 4 ) } . Proof Soc( T ) = ( X c 2 − 1 X c 3 − 1 X r 14 − 1 ). 2 3 4 7

Example Let S = � 5 , 7 , 9 , 11 � . We get g ( S ) = 13. In the same way one calculates that H k [ S ] − z 14 / (1 − z ) = 1+ z 5 + z 7 + z 9 + z 10 + z 11 + z 12 so the elements in S below the conductor are 0,5,7,9,10,11,12. 8

A semigroup S is symmetric if and only if its semigroup ring is Gorenstein. I will now dis- cuss when it is even a complete intersection. A Gorenstein ring of codimension ≤ 2 is a com- plete intersection, so for semigroup rings in at most three variables Gorenstein and complete intersection are the same thing. 9

There is a general procure to construct new complete intersections from old. Namely, let S = � n 1 , . . . , n k � be a complete intersection semigroup, and let T = � dn 1 , . . . , dn k , a 1 n 1 + · · · + a k n k � , where � a i > 1 and ( d, a 1 n 1 + · · · + a k n k ) = 1. Then T is a complete intersection semigroup (Watanabe). 10

For 3-generated semigroups we get all sym- metric semigroups in this way (Herzog, Watan- abe). For complete intersections it is easy to use the Hilbert series to determine the Frobe- nius number. A complete intersection k [ x 1 , . . . , x k ] / ( f 1 , . . . , f k − 1 ) has Hilbert series k − 1 (1 − z r i ) � i =1 , k (1 − z n i ) � i =1 where deg f i = r i , deg x i = n i . With the lemma above, this gives that the Frobenius number is g = � r i − � n i . Note how this generalizes the 2-generated case S = � n 1 , n 2 � , where k [ S ] = k [ x 1 , x 2 ] / ( x n 2 1 − x n 1 2 ) and g ( S ) = n 1 n 2 − n 1 − n 2 . 11

For a symmetric semigroup in three variables, S = � dn 1 , dn 2 , a 1 n 1 + a 2 n 2 � , we have k [ S ] = k [ x 1 , x 2 , x 3 ] / ( x n 2 1 − x n 1 2 , x a 1 1 x a 2 2 − x d 3 ) . Thus g ( S ) = dn 1 n 2 + d ( a 1 n 1 + a 2 n 2 ) − ( dn 1 + dn 2 + a 1 n 1 + a 2 n 2 ). In general we have, if S = � a, dS 1 � , where S 1 = � n 1 , . . . , n k � is a com- plete intersection, a = � a i n i , and ( a, d ) = 1, it follows easily that k � g ( S ) = dg ( S 1 ) + ( d − 1) a i n i . i =1 Proof If the relations in S 1 are of degrees r 1 , . . . , r k − 1 , the relations in S are of degrees dr 1 , . . . , dr k − 1 and d ( a 1 n 1 + · · · + a k n k ). Thus g ( S ) = � dr i + d ( a 1 n 1 + · · · + c k n k ) − � dn i − ( a 1 n 1 + · · · + a k n k ) = d ( � r i − � n i ) + ( d − k 1)( a 1 n 1 + · · · + a k n k ) = dg ( S 1 )+( d − 1) a i n i . � i =1 12

As an example, let n 1 , . . . , n k be pairwise rela- tively prime and let S be generated by k n i � i =1 { ; j = 1 , . . . , k } . n j In this case we can show more, not only k [ S ] is a complete intersection, but also gr m ( k [ S ]) is (Barucci-Fr¨ oberg). One easily gets g ( S ) = k k ( k − 1) N − N/n j , where N = n i . � � j =1 i =1 Example If S = � 2 · 3 · 5 , 2 · 3 · 7 , 2 · 5 · 7 , 3 · 5 · 7 � = � 30 , 42 , 70 , 105 � , then g ( S ) = 3 · 210 − (105 + 70 + 42 + 30) = 383. 13

For semigroups generated by four elements, there is also another kind of complete inter- section semigroups. These have relations x a 1 1 − x a 2 2 , x a 3 3 − x a 4 4 , x b 1 1 x b 2 2 − x b 3 3 x b 4 4 . This is the case when S = � n 1 , n 2 , n 3 , n 4 � , d = ( n 1 , n 2 ), d ′ = ( n 3 , n 4 ), dd ′ = b 1 n 1 + b 2 n 2 = b 3 n 3 + b 4 n 4 . Here g ( S ) = lcm( n 1 , n 2 ) + lcm( n 3 , n 4 ) + dd ′ − ( n 1 + n 2 + n 3 + n 4 ). As an example, if S = � 14 , 21 , 15 , 20 � , then g ( S ) = 42+60+35 − (14+21+15+20) = 67. 14

The Poincar´ e series For a local ring ( A, m, k ) (or a graded k -algebra), M an A -module, let P M ( z ) = � dim k Tor A i ( k, M ) z i . It was for a long time an open question (Serre, Kaplansky, Shafarevich) if P k ( z ) was always ra- tional. A first counterexample was given by Anick. There are in fact counterexamples even for semigroup rings (Fr¨ oberg-Roos), k [[ t 18 , t 24 , t 25 , t 26 , t 28 , t 30 , t 33 ]] is one. Definition A graded algebra R is Koszul if P k ( z ) = 1 /H R ( − z ). 15

Our example uses a classification of Roos-Sturmfels of monomial curves in P n (toric rings). For graded rings R = k [ x 1 , . . . , x n ] /I the following are well known: I has a quadratic Gr¨ obner basis implies that R is Koszul implies that I is generated by quadrics. It was for some time open if the converse of these implications were true in the toric case. With an extensive computer search (more than 70000 cases) they found counterexamples to both statements. There is a ring with quadratic I which is not Koszul in P 5 , and there is a Koszul algebra without quadratic Gr¨ obner ba- sis in P 7 . 16

Roos-Sturmfels uses a result by Laudal-Sletsj¨ oe: Tor k [ M ] ( k, k ) λ = ˜ H i − 2 (∆( λ ) , k ), if M is a semi- i group and ∆( λ ) is the poset (0 , λ ) considered as a simplicial complex. Then it is not so hard to see that k [ M ] is Koszul if and only if all intervals are Cohen- Macaulay (i.e. have CM Stanley-Reisner rings). 17

Example: R = k [ t 2 , t 3 ], Tor R 3 ( k, k ) 8 ≃ H 1 ((0 , 8) , k ) 6 ③ ❆ ❆ ❆ ❆ ❆ 4 5 ❆ ③ ③ ❆ � ❆ � � ❆ � ❆ � ❆ 2 3 � ❆ ③ ✉ ③ 6 3 ✉ ✉ � � � � � 4 � ✉ ❅ ❅ ❅ ❅ ❅ 2 5 ❅ ✉ ✉ 18

If A is a commutative k -algebra, k a field, the module of derivations Der k ( A ) ⊆ Hom k ( A, A ) is the set { ρ ∈ Hom k ( A, A ) | ρ ( ab ) = aρ ( b ) + ρ ( a ) b } . Theorem 5 (Eriksson, Eriksen) Let S be a numerical semigroup and T ( S ) = { a 1 , . . . , a h } . The module of derivations of k [ S ] is the left k [ S ] -module generated by { t∂ } ∪ { t a i +1 ∂, i = 1 , . . . , h } ∂ where ∂ = ∂t . In particular the number of generators is | T ( S ) | + 1 . 19