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

On the homology of semigroup rings

Porto 2008 Ralf Fr¨

• berg

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 tn+s ∈ Soc(k[[S]]/(ts)). Hence dimk Soc(k[[S]]/(ts)) = |T(S)| (the CM-type of k[[S]]). Proof n ∈ T(S) iff tn / ∈ k[[S]] and tnm ⊂ k[[S]] iff tn+s / ∈ tsk[[S]] and tn+sm ⊂ tsk[[S]] iff tn+sm = ¯ 0 and tn+s = 0 in k[[S]]/(ts) iff tn+s ∈Soc(k[[S]]/(ts).

2

Let S = n1, . . . , nk and let ci be the smallest positive integer such that cini ∈ n1, . . . , ˆ ni, . . . , nk, and suppose that cini =

j=i rijnj.

Theorem 1 (Herzog) Assume S = n1, n2, n3 and not symmetric. Then k[[S]] ≃ k[[X1, X2, X3]]/I where I = (Xc1

1 −Xr12 2

Xr13

3

, Xc2

2 −Xr21 1

Xr23

3

, Xc3

3 −Xr31 1

Xr32

2

) and rij < cj for all i, j.

3

Theorem 2 (Bresinsky) Let S = n1, n2, n3, n4 to be symmetric but not a complete intersec-

• tion. Then k[[S]] ≃ k[[X1, X2, X3, X4]]/I where

I = (Xc1

1 − Xr13 3

Xr14

4

, Xc2

2 − Xr21 1

Xr24

4

, Xc3

3 − Xr31 1

Xr32

2

, Xc4

4 − Xr42 2

Xr43

3

, Xr43

3

Xr21

1

− Xr32

2

Xr14

4

) and rij < cj for all i, j.

4

Theorem 3 Assume S = n1, n2, n3 and not symmetric.Then there is a minimal R = k[[X1, X2, X3]]-resolution of k[[S]]: 0 → R[−n2c2 − n3r13] ⊕ R[−n3c3 − n2r12] → R[−n1c1]⊕R[−n2c2]⊕R[−n3c3] → R → k[[S]] → 0. Proof It is easy to get a minimal k[[X2, X3]]- resolution of T = k[[S]]/(tn1) = k[[X2, X3]]/(Xr12

2

Xr13

3

, Xc2

2 , Xc3 3 ),

and then lift it. Corollary 1 Assume S = n1, n2, n3 and not

• symmetric. Then g(S) = max{n2r12 + n3c3 −

(n1 + n2 + n3), n2c2 + n3r13 − (n1 + n2 + n3)}. Proof Soc(T) = (Xr12−1

2

Xc3−1

3

, Xc2−1

2

Xr13−1

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 Hk[S] = 1 − z49 − z20 − z52 + z59 + z62 (1 − z7)(1 − z10)(1 − z13) . The conductor k[S] : k[t] is C = t33k[t]. The Hilbert series of k[S]/C, is Hk[S]/C = Hk[S] −

z33 1−z = 1+z7+z10+z13+z14+z17+z20+z21+

z23+z24+z26+z27+z28+z30+z31. Thus the length of k[S]/C is l(k[S]/C) = Hk[S]/C(1) = 15 and l(k[t]/k[S]) = 33 − 15 = 18.

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Theorem 4 Assume S = n1, n2, n3, n4 to be symmetric but not a complete intersection. Then there is a minimal R = k[[X1, X2, X3, X4]- reso- lution of k[[S]]: 0 → R[−n2c2 − n3c3 − n4r14] → R5 → R5 → R → k[[S]] → 0. Proof It is easy to get a k[[X2, X3, X4]]-resolution

• f T = k[[S]]/(tn1) = k[[X2, X3, X4]]/I where

I = (Xr13

3

Xr14

4

, Xc2

2 , Xc3 3 , Xc4 4 −Xr42 2

Xr43

3

, Xr32

2

Xr14

4

), and then lift it. Corollary 2 Assume S = n1, n2, n3, n4 to be symmetric but not a complete intersection. Then g(S) = n2c2+n3c3+n4r14−(n1+n2+n3+n4)}. Proof Soc(T) = (Xc2−1

2

Xc3−1

3

Xr14−1

4

).

7

Example Let S = 5, 7, 9, 11. We get g(S) =

• 13. In the same way one calculates that

Hk[S]−z14/(1−z) = 1+z5+z7+z9+z10+z11+z12 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 = n1, . . . , nk be a complete intersection semigroup, and let T = dn1, . . . , dnk, a1n1 + · · · + aknk, where ai > 1 and (d, a1n1 + · · · + aknk) = 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[x1, . . . , xk]/(f1, . . . , fk−1) has Hilbert series

k−1

• i=1

(1 − zri)

k

• i=1

(1 − zni) , where deg fi = ri, deg xi = ni. With the lemma above, this gives that the Frobenius number is g = ri − ni. Note how this generalizes the 2-generated case S = n1, n2, where k[S] = k[x1, x2]/(xn2

1 −xn1 2 ) and g(S) = n1n2 −n1 −n2.

11

For a symmetric semigroup in three variables, S = dn1, dn2, a1n1 + a2n2, we have k[S] = k[x1, x2, x3]/(xn2

1 − xn1 2 , xa1 1 xa2 2 − xd 3).

Thus g(S) = dn1n2 + d(a1n1 + a2n2) − (dn1 + dn2 + a1n1 + a2n2). In general we have, if S = a, dS1, where S1 = n1, . . . , nk is a com- plete intersection, a = aini, and (a, d) = 1, it follows easily that g(S) = dg(S1) + (d − 1)

k

• i=1

aini. Proof If the relations in S1 are of degrees r1, . . . , rk−1, the relations in S are of degrees dr1, . . . , drk−1 and d(a1n1 + · · · + aknk). Thus g(S) = dri + d(a1n1 + · · · + cknk) − dni − (a1n1 + · · · + aknk) = d( ri − ni) + (d − 1)(a1n1+· · ·+aknk) = dg(S1)+(d−1)

k

• i=1

aini.

12

As an example, let n1, . . . , nk be pairwise rela- tively prime and let S be generated by {

k

• i=1

ni nj ; j = 1, . . . , k}. In this case we can show more, not only k[S] is a complete intersection, but also grm(k[S]) is (Barucci-Fr¨

• berg). One easily gets g(S) =

(k − 1)N −

k

• j=1

N/nj, where N =

k

• i=1

ni. 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 xa1

1 −

xa2

2 , xa3 3 − xa4 4 , xb1 1 xb2 2 − xb3 3 xb4 4 . This is the case

when S = n1, n2, n3, n4, d = (n1, n2), d′ = (n3, n4), dd′ = b1n1 +b2n2 = b3n3 +b4n4. Here g(S) = lcm(n1, n2) + lcm(n3, n4) + dd′ − (n1 + n2 + n3 + n4). 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 PM(z) = dimk TorA

i (k, M)zi.

It was for a long time an open question (Serre, Kaplansky, Shafarevich) if Pk(z) was always ra- tional. A first counterexample was given by

• Anick. There are in fact counterexamples even

for semigroup rings (Fr¨

• berg-Roos),

k[[t18, t24, t25, t26, t28, t30, t33]] is one. Definition A graded algebra R is Koszul if Pk(z) = 1/HR(−z).

15

Our example uses a classification of Roos-Sturmfels

• f monomial curves in P n (toric rings).

For graded rings R = k[x1, . . . , xn]/I the following are well known: I has a quadratic Gr¨

• bner 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¨

• bner ba-

sis in P 7.

16

Roos-Sturmfels uses a result by Laudal-Sletsj¨

• e:

Tork[M]

i

(k, k)λ = ˜ Hi−2(∆(λ), k), if M is a semi- 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[t2, t3], TorR

3 (k, k)8 ≃ H1((0, 8), k)

③ ③ ③ ③ ③

• ❆

❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✉ ✉ ✉ ✉ ✉ ✉

2 3 4 5 6

• ❅

❅ ❅ ❅ ❅ ❅

4 2 6 5 3

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If A is a commutative k-algebra, k a field, the module of derivations Derk(A) ⊆ Homk(A, A) is the set {ρ ∈ Homk(A, A) | ρ(ab) = aρ(b) + ρ(a)b}. Theorem 5 (Eriksson, Eriksen) Let S be a numerical semigroup and T(S) = {a1, . . . , ah}. The module of derivations of k[S] is the left k[S]-module generated by {t∂} ∪ {tai+1∂, i = 1, . . . , h} where ∂ =

∂ ∂t.

In particular the number of generators is |T(S)| + 1.

19

Theorem 6 (Fr¨

• berg-Micale) Let S be a nu-

merical semigroup and T(S) = {a1, . . . , ah}, and let A = k[S]. Then PDerk(A)(z) = 1 + hPk(z), so PDerk(A)(z) is rational if and only if Pk(z) is rational.

20

Theorem 7 Let A = k[S]. Then Pk(z) is ra- tional in the following cases:

• S is 3-generated.
• S is n-generated and symmetric, n ≤ 5.
• S is a complete intersection.
• S = a, a+l, . . . , a+ld, 2d ≥ a−1, gcd(a, l) =

1.

21

• m(S) ≤ 7, where m(S) is the multiplicity.
• S is of maximal embedding dimension.
• S is of maximal length of type ≥ 2 or of

almost maximal length.

• S is a “monomial semigroup”.

22

Complete intersection semigroup rings were de- termined by Delorme. Maximal embedding dimension means edim(k[S] = m(S). Maximal length (almost maximal length) means l( ¯ R/R) = l(R/C)t(R) (l( ¯ R/R) = l(R/C)t(R) − 1, resp.), where t(R) = CM-type(R). A semigroup S is monomial if v(R) = S implies R ≃ k[S].

23

If U = R/(x), x a nonzerodivisor in m \ m2, then P R

k (z) = (1 + z)P U k (z).

We let U = k[[S]]/(tm(S)). If S is 3-generated, then U has embedding di- mension 2, and U is either a complete or a so called Golod ring. P k[[S]]

k

(z) = (1 + z)P U

k (z) =

(1 + z)/(1 − z)2 or (1 + z)3/(1 − 3z2 − 2z3). Gorenstein rings of codimension at most 4 has rational Poincar´ e series (Avramov-Kustin-Miller). If S = a, a+l, . . . , a+ld, 2d > a−1, gcd(a, l) = 1, then U = k[[x1, . . . , xd]]/I, I generated in de- gree 2 and constitute a Gr¨

• bner basis in De-
• grevlex. Then U is a so called Koszul algebra,

and P U

k (z) = 1/HU(−z) and so rational.

If 2d = a − 1, then U = k[[x1, . . . , xd]]/(I + m3), I as above, and P U

k (z) is rational (L¨

• fwall).

24

Kunz considers a classification of numerical

• semigroups. Fix an integer m ≥ 3 and denote

by Hm the set of all numerical semigroups H with m ∈ H. Using the Ap´ ery set of H with respect to m, he associates to Hm a polyhe- dral cone Pm ⊂ Rm−1 such that one has a bi- jection Hm → Pm ∩ Nm−1. Then the disjoint decomposition of Pm into open faces leads to a classification of the numerical semigroups. It turns out that the semigroups of multiplicity m belonging to a fixed open face of Pm have the same Betti numbers, in particular they have the same Hilbert series, Cohen-Macaulay type, embedding dimension. If m(S) ≤ 7 we use a classification of all possible U’s. There are about 70 different cases. In all of these we can use the methods above.

25

If S is of maximal embedding dimension, then U = k[[x1, . . . , xd]]/(m2), which is Koszul. If S has maximal length, then either k[[S]] is Gorenstein or of maximal embedding dimen- sion (Brown-Herzog). The rings of almost maximal length are classi- fied (Brown-Curtis). They are either Golod or

• f maximal embedding dimension.

26

Let C[[x1, . . . , xk]]/(f) be an analytically irre- ducible curve, i.e. the zero set of f, an irre- ducible power series. Then this curve can be parametrized as (tn1, tn2 + · · · , . . . , tnk + · · · ). The set of values is a semigroup which con- tains n1, . . . , nk, but is in general larger. If any curve with semigroup S is isomorphic to the semigroup ring C[[S]], then S is called a monomial semigroup. The monomial semi- groups were classified by Pfister-Steenbrink, but an error in their proof was corrected by (Micale. They are either of maximal embed- ding dimension, Koszul, or 2-generated.

27

Differential operators If R is a k-algebra, k a field, the ring of differ- ential operators D(R) on R is a subalgebra of Homk(R, R) which can be defined recursively as follows. D0(R) ≃ R consists of the multiplications θr with elements in R, θr(a) = ra. D1(R) = {θ; [θ, D0(R)] ⊆ D0(R)}. Thus D1(R)\ D0(R) are the derivations. [, ] is the commu- tator. In general Dn(R) = {θ; [θ, D0(R)] ⊆ Dn−1(R)} and D(R) = ∪n≥0Dn(R).

28

If R = k[x1, . . . , xn], then D(R) = k[x1, . . . , xn, ∂1, . . . , ∂n], the Weyl algebra. Nakai’s conjecture: D(R) is generated by D1(R) if and only if R is regular. Fact 1: If R = k[S], then D(R) ⊆ D(k[t, t−1]) = k[t, t−1, ∂]. Definition If θ = tm∂n, then deg(θ) = m − n. Then D(k[t, t−1]) is graded, and D(k[S]) inherits this grading.

29

Fact 2: If S = s1, . . . , sk, then gr(D(k[S])) is the subring of k[t, y] generated by {ts1, . . . , tsk}∪ {ys1, . . . , ysk}∪{ty}∪{tv(n)+nyv(n)}n∈Z\S, where v(n) = #{s ∈ S; s + n / ∈ S}. This gives that grD(k[S]) is Noetherian, so D(k[S]) is left and right Noetherian. A strong version of Nakai’s conjecture is true: D1(k[S])2 = D2(k[S]).

30

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