gorenstein

J. K. VERMA

Introduction

In these lectures, we discuss how Gorenstein rings evolved starting from a result of Gorenstein about colength of the conductor ideal of the cordinate ring of an irreducible affline plane curve. A modern proof of Gorenstein’s theorem will be presented. In so doing, we get introduced to some important notions in commutative algebra such as regular local rings, Cohen-Macaulay local rings, complete intersections, irreducible ideals, socle of a local ring, free and injective resolutions, Matlis duality and canonical modules. The exposition follows an excellent survey article by C. Huneke:

C. Huneke, Hyman Bass and Ubiquity: Gorenstein Rings, Algebra, K -theory, groups, and education (New York, 1997), 55–78, Contemp. Math., 243, Amer. Math. Soc., Providence, RI, 1999.

The foundations of modern theory of Gorenstein rings was laid by Hyman Bass in his classic paper:

H. Bass, On the ubiquity of Gorenstein rings, Math. Z.82(1963), 8-28. These rings are named after Daniel Gorenstein. His thesis on plane curves written under the guidance of Oscar Zariski was published in the Transac-tions of American Mathematical Society in 1952. The origins of the theory go back to F. S. Macaulay who studied inverse systems and Hilbert series of Gorenstein graded rings and Gr¨obner who investigated zero-dimensional Gorenstein rings and showed the role played by the socle of a local ring.

We will prove the characterization of Gorenstein graded rings in terms of their Hilbert series due to Macaulay and Stanley. This will provide us with a simple technique to recover a beutiful theorem of Watanabe which identifies the Gorenstein invariant subrings of finite subgroups of SL(n,C)

1. Plane Curves

Let kbe a field and f(x, y) be an equation of an irreducible affine plane curve. Then its coordinate ring is R = k[x, y]/(f). The ring R is a one dimensional Noetherian domain. Its integral closure R in its fraction field is a finitely generated R-algebra. Hence R corresponds to a smooth affine curve.

Definition 1.1. Theconductor ofR and R is the ideal

C=C(R, R) ={r∈R|rR⊆R}.

The conductorCis the largest ideal common between R and R.

Example 1.2. Consider the affine plane curve C defined by the equa-tion f(x, y) = X7 −Y3 = 0 over a field k. The coordinate ring k(C) =

k[X, Y]/(f)'k[t3_{, t}7_{].The integral closure of R in its fraction field k(t) is}

k[t] as tis integral overR and k[t] is integrally closed. Thek-bases ofR, R

and Care displayed below:

R 1 t t2 t3 t4 t5 t6 t7 t8 t9 t10 t11 t12 t13 . . .

R 1 t3 t6 t7 t9 t10 t12 t13 . . .

C t12 _t13 _{. . .}

We observe a striking fact from the above table: dimkR/R= dimkR/C= 6.

This was the main theorem of Gorenstein ! Apery in 1943 [3], Samuel in 1951 [14] and Gorenstein in 1952 [7] proved that for every prime idealQ of the coordinate ringR of an irreducible plane curve and conductor C,

dimk(RQ/RQ) = dimkRQ/CQ.

This property essentially characterizes one dimensional Gorenstein domains. Complete intersectons are more general than plane curves. LetI be an ideal of the polynomial ring S =k[X1, X2, . . . , Xn] over a field k.We say that I

is a complete intersection if it can be generated byhtI elements.

Theorem 1.3 (Rosenlicht [13]). Let p = (F1, F2, . . . , Fn−1) be a prime

ideal of height n−1in S =k[X1, X2, . . . , Xn].Let R=S/p.Then for every

prime ideal Qof R,

2. Nonsigular points and regular local rings

Letkbe an algebraically closed field. LetAn be then-dimensional affine

space over k. We recall the notions of smooth point of an affine variety

V ⊆_An_{and that of the tangent space at a pointp∈V.By an affine change of}

coordinates we may assume thatpis the origin. LetI(V) = (F1, F2, . . . , Fr)

be the ideal of V in S=k[X1, X2, . . . , Xn].Let` be a line passing through

the origin and a point q = (a1, a2, . . . , an). Then the intersection points of V and` are given by tq wheretis the solution of the equations:

F1(tq) =F2(tq) =· · ·=Fr(tq) = 0.

EachFi(qt) is a polynomial int.Hence it can be factored into linear factors.

The maximum power oftthat divides eachFi(qt) is called themultiplicity

of V ∩`atp.

Definition 2.1. The line`is calledtangent toV atpif the the multiplicity of`∩V atpexceeds one. Thetangent space TpV to V at p is the union

of all tangent lines atp toV.

Theorem 2.2. Letp= (p1, p2, . . . , pn)∈V where V is an irreducible affine

subvariety withI(V) = (F1, F2, . . . , Fr).The tangent spaceTp(V) is a linear

subvariety of An defined by n

X

j=1

∂Fi ∂Xj

(p)(Xj −pj) = 0

for i= 1,2, . . . , r.

Definition 2.3. The pointp is called asmooth point of V if dimTp(V) = dimpV.

If p is not smooth, we say it is singular. A Noetherian local ring (R,m) is called a regular local ringif dimm/m2 = dimR. A Noetherian ring R is called regular if Rp is regular for every prime ideal pof R.

Example 2.4. Polynomial and power series rings over any field are regular rings. The ring of integers is regular. The integral closure of a one dimen-sional Noetherian domain is regular. Any Dedekind domain is regular. Theorem 2.5. The point p of a d-dimensional affine variety V ⊂ An is

a smooth point if and only of the local ring Op,V is a regular local ring.

Regular local rings have many interesting properties. Their investigations by Serre and Auslander-Buchsbaum in the early 50’s introduced homological methods in commutative algebra.

Theorem 2.6 (Auslander-Buchsbaum [1], Serre [15]). The following are equivalent for a local ring (R,m) :

(1)R is a regular local ring.

(2) Any finite R-module M has a finite free resolution: 0−→Fk

fk

−→Fk−1

fk−1

−→ · · · −→F0−→M−→0

where Fi are free R -modules of finite rank.

(3) The residue fieldR/mhas a finite free resolution as in (2).

Auslander and Buchsbaum [2] used the above characterization to prove that localization of a regular local ring is regular and regular local rings are UFDs.

Definition 2.7. The free resolution as in (2) is called aminimal free res-olution ofM if the entries of the matrices of the mapsfi fori= 0,1, . . . , k

belong to the maximal idealm.The common length of all minimal free reso-lutions ofM,denoted by projdim_R(M),is called theprojective dimension of M overR.

3. Cohen-Macaulay rings

We will see later that Gorenstein rings belong to a larger class of rings called Cohen-Macaulay rings. These rings arise in most applications of com-mutative algebra, specially in combinatorics.

Definition 3.1. Let R be a commutative ring and M be an R-module. A sequence elements x1, x2, . . . , xn ∈ R is called an M-sequence if for

each i = 1,2, . . . , n, xi is a nonzerodivisor on M/(x1, x2, . . . , xi−1)M and

(x1, x2, . . . , xn)M 6= M. If (R,m) is a local ring of dimension d, then a

system of elements x1, x2, . . . , xd ∈ m is called a system of parameters

for R if (x1, x2, . . . , xd) is an m-primary ideal, equivalently there is an N

such that mN ⊆ (x1, x2, . . . , xn). We say R is Cohen-Macaulay (CM) if

every system of parameters is anR-sequence.

Theorem-Definition 3.2(Rees). LetI be an ideal of a Noetherian ringR

andM be a finiteR-module withIM 6=M. Then all maximalM-sequences inI have length ngiven by

n= min{i|Exti_R(R/I, M)6= 0}.

The number n, denoted by grade(I, M),is called thegrade of I on M. If (R,m) is local then we put grade(m, M) = depthM.

Definition 3.3. A finite R-module over a local ring R is called Cohen-MacaulayR-module if depthM = dimM.

The depth of a module and its projective dimension are related in an important formula of Auslander and Buchsbaum:

Theorem 3.4 (Auslander-Buchsbaum Formula). Let (R,m) be a local ring and M be a finiteR-module with finite projective dimension. Then

projdimM+ depthM = depthR.

Northcott and Rees related irreducibility of ideals generated by systems of parameters, called parameter ideals, to Cohen-Macaulay property. This provided another impetus for the theory of Gorenstein rings.

Definition 3.5. An idealI is calledreducibleif there are idealsJ, K (R

and properly containingI so thatI =J∩K.An ideal that is not reducible is called irreducible.

In a Noetherian ring every ideal is a finite intersection of irreducible ideals. Definition 3.6. LetM be a module over a local ring (R,m).The socleof

M is defined as

Soc(M) ={x∈M |xm= 0}.

Proposition 3.7. Let M be an Artinian module over a local ring (R,m).

Then the extension SocM ⊂ M is an essential extension, i.e. if N is a nonzero submodule of M thenN ∩SocM 6= 0.

Proof. Let 0 6= x ∈ N. Then xmn is a descending sequence of submodules of N. Hence there is a least n such that xmn = 0. Hence xmn−1 6= 0 and

xmn−1 _{⊂SocM. Hence N∩SocM 6= 0.}

Proof. Let (0) be an irreducible ideal. If dim SocR ≥2 then there are one dimensional subspaces of SocRwhich intersect in (0).But (0) is irreducible. Hence dim SocR= 1.Conversely let dim SocR= 1.If J and K are nonzero ideals ofRthenJ∩SocR6= 0 andK∩SocR6= 0.Since SocRis 1-dimensional, SocR⊂J ∩K.Hence J∩K6= 0.Thus (0) is irreducible. Example 3.9. The ring k[X, Y]/(X2, Y2) has one dimensional socle gener-ated by XY.On the other hand the ring R =k[X, Y]/(X2, XY, Y2) has a two dimensional socle generated by X and Y. Moreover (X)R∩(Y)R= 0.

Hence (0) is reducible. In fact (X, Y)2= (X, Y2)∩(Y, X2).

Irreducibility is closely related to Gorenstein rings.

Theorem 3.10 (Northcott-Rees, [12]). (1) If a parameter ideal is irre-ducible in a local ringR then it is Cohen-Macaulay.

(2) Every parameter ideal in a regular local ring is irreducible. (3) [Cohen] A regular local ring is Cohen-Macaulay.

In fact if a local ring has an irreducible parameter ideal then it is Goren-stein. Therefore the theorem of Northcott-Rees proves that regular local rings are Gorenstein and Gorenstein local rings are Cohen-Macaulay.

4. Grothendieck’s definition of Gorenstein rings

Grothendieck defined Gorenstein rings in terms of canonical module. To prepare the background we need some vanishing theorems for the extension functor.

Theorem 4.1 (Ischebeck). Let (R,m) be a local ring and let M, N be finite R-modules. Then

Exti_R(M, N) = 0 for i <depthN−dimM,

Proposition 4.2. LetS be a regular local ring of dimensionnandR=S/I

be a Cohen-Macaulay local ring of dimensiondwhere I is anS-ideal. Then Exti_S(S/I, S)6= 0 if and only if i=n−d.

Proof. SinceSis regular local, projdimS/I <∞.By Auslander-Buchsbaum formula, projdimS/I + depthS/I = depthS = d. Hence projdimS/I =

n−d. Hence Exti_S(S/I, S) = 0 for i > n −d. By Ischebeck’s theorem, Exti_S(S/I, S) = 0 fori <depthS−dimS/I =n−d.

Definition 4.3(Grothendieck). LetSbe a regular local ring andR=S/I

for an R-ideal I of height h. Suppose R is Cohen-Macaulay. Then the canonical module of R is defined to be the module

ωR= ExthS(S/I, S).

The ringR is called Gorenstein ifωR'R.

Using the above definition of Gorenstein rings, let us prove a useful cri-terion for Gorenstein property in terms of free resoltions.

Theorem 4.4. Let (S,n) be a regular local ring of dimension n and let I

be a height h ideal of S. Let R=S/I be d-dimensional with a minimal free resolution

(1) 0−→Fk

fk

−→Fk−1

fk−1

−→ · · · −→F0−→R−→0.

Then R is Gorenstein if and only if k= projdim_SS/I =h andFk'S.

Proof. By Auslander-Buchsbaum formula depthR+projdim_SR= depthS=

n.Hence depthR=n−k= dimR if and only ifk=n−d=h.Now we find the canonical moduleωR= ExthS(R, S) from the minimal free resolution of R as an S-module. Put M∗= HomS(M, S) for an S-moduleM. Apply the

left exact contravariant functor Hom(−, S) to the free resolution (1) ofRto get the complex:

(2) 0−→F₀∗ f

∗

1

−→F₁∗ f

∗

2

−→ · · · −→F_k−∗ ₁ f

∗

k

−→F_k∗−→0.

Let R be Gorenstein. Then it is Cohen-Macaulay and ωR ' R. Thus the

homology of the complex (2) Exti_S(R, S) = for all i= 0,1, . . . , k−1. More-over Exth_S(R, S) =cokerf_k∗.Hence the above is a resolution ofωR.SinceωR

is cyclic, rankF_k∗ = rankFk= 1.

Conversely let k = h and rankFk = 1. Since I ⊂ annSR, ωR ' S/J

for some ideal I ⊆J. The complex (2) is acyclic and it gives a minimal free resolution ofωRas anS-module. Thus projdimSωR=n−d=n−depthωR.

Hence depthωR = dimR = d. Thus ωR is a maximal Cohen-Macaulay R

-module. Moreover, applying Hom(−, S) to (2) we get Exth_S(Exth_S(R, S), S)'R.

Thus J ⊆ annS/I =I. Hence I =J and ωR 'R. Therefore, R is

we show that if I is generated by a regular sequence in a regular local ring

S then S/I is Gorenstein. For this we recall the construction of the Koszul complex of a sequence of elements.

The Koszul Complex

Let R be a commutative ring and x1, x2, . . . , xr ∈ R. Define the Koszul

complexK(x) =K(x1, x2, . . . , xr) : Put K0(x) =R

K1(x) = a freeR-module with basis {e1, e2, . . . , er}. Kp(x) = a freeR-module of rank r_p

with basis

{ei1∧ei2∧. . .∧eip |1≤i1< i2 <· · ·< ip≤r}

Define the boundary mapd(ei) =xifori= 1,2, . . . , r.The general boundary

mapd:Kp(x)→Kp−1(x) is defined as

d(ei1 ∧ei2 ∧. . .∧eip) = p

X

j=1

(−1)j−1xijei1 ∧ei2 ∧ · · · ∧eˆij∧ · · · ∧eip,

where ˆx indicates that x is ommitted. It is easy to see that d2 = 0. Hence we have a complex

K(x) : 0−→Kr(x)−→Kr−1(x)−→ · · · −→K1(x)−→R−→0.

If M is an R-module, the Koszul complex K(x, M) of M with respect to the sequencex1, x2, . . . , xr is defined to be the complex K(x)⊗RM.

Theorem 4.5. Let R be a commutative ring and M be an R-module. (a) If x1, x2, . . . , xr is an M-sequence then Hi(x, M) = 0 for i > 0 and H0(x, M) =M/(x)M. In other words the augmented Koszul complex 0→Kr(x)⊗M →Kr−1(x)⊗M → · · · →K1(x)⊗M →M →M/(x)M →0

is exact.

(b)Suppose (R,m) is local andx1, x2, . . . , xr ∈m.If M is a finiteR-module

and H1(x, M) = 0 then x1, x2, . . . , xr is an M-sequence.

Theorem 4.6. Let S be a regular local ring. Let I be an ideal generated by anS-sequence x=x1, x2, . . . , xr.Then S/I is a Gorenstein ring.

Proof. The Koszul complex K(x) is a minimal free resolution of S/I with lengthrand the last moduleS.By Theorem 4.4,it follows thatS/I is

5. Examples of Gorenstein ideals

An idealI of a ringS is called aGorenstein idealofS ifS/I is Goren-stein. In this section, we will present examples of Gorenstein ideals in regular local rings and polynomial rings.

Proposition 5.1. Let S be a regular local ring and I be an ideal of S such thatS/I is Gorenstein.

(a) If ht I = 1 then I is principal.

(b)[Serre] If ht I = 2 then I is a complete intersection.

Proof. (a) Since S/I is Gorenstein, it is Cohen-Macaulay. Hence I is an unmixed ideal of height one. Since S is regular, it is a UFD. Height one prime ideals are principal in a UFD. It follows thatp-primary ideals where

p is a height one prime, are also principal. HenceI is principal.

(b) Now let htI = 2. Since projdim_SS/I = 2, a minimal free resolution of

S/I looks like:

0→S→Sm →S→S/I →0.

Localize the above sequence at S \ {0} to see that m = 2. Hence I is a complete intersection.

Example 5.2. [6, Example 3.2.11] Height three Gorenstein ideals need not be complete intersections. We describe a family of Gorenstein Artin local rings to construct such examples. Letkbe a field andS =k[X1, X2, . . . , Xn]

be the polynomial ring overk.LetSmdenote thek-vector space of

homoge-neous polynomials of degreem. Letf :Sm→k be a nonzero k-linear map.

For eachj= 0,1, . . . , m, define a k-subspace ofSj : Ij ={a∈Sj |f(a.Sm−j) = 0}.

Put Ij = Sj for j > m. The ideal I = ⊕∞_j=0Ij is a graded ideal of S.

Moreover, S/I is an Artinian graded ring. We show that R = S/I is a Gorenstein ring by proving that the socle of R is one dimensional k-vector space. Let j ∈ _Nand 0 < j < m. For a∈S, let ¯adenote the residue class of a in R. Let ¯a ∈ Rj \ {0}. Then a /∈ Ij. Hence there exists b ∈ Sm−j

with f(ab) 6= 0. Hence ¯a¯b 6= 0. Since ¯b belongs to the maximal ideal of R,

it follows that ¯a /∈ Soc(R). Hence SocR =Rm. But Im = kerf and hence

dimkIm = dimSm−1. Hence dimRm = 1.Therefore R is Gorenstein. In

defined on its basis by

f(XiXj) = 0, 1≤i < j ≤n, f(Xi2) = 1, i= 1,2, . . . , n.

For this map we obtain

I = (X₁2−X₂2, . . . , X₁2−X_n2, XiXj, 1≤i < j≤n).

The ideal I is a complete intersection if and only ifn≤2.

We now discuss a remarkable structure theorem for Gorenstein ideals of height three in regular local rings due to Buchsbaum and Eisenbud [8].

LetA = (aij) be a skew-symmetric square matrix with entries in a

com-mutative ring R. If A is 2n+ 1×2n+ 1 then detA = 0. If A is 2n×2n

then detA is a square of an element pf(A) ∈ R called the Pfaffian of A.

Pfaffians can be defined inductively just like determinants. Set pf(A) = 0 ifA has odd size. LetAij denote the submatrix of A obtained by deleting i-th andj-th rows and columns. Then for allj6=i.

pf(A) = 2n

X

j=1

(−1)i+j−1_σ(j−i)a

ijpf(Aij).

whereσ(j−i) is the sign of j−i.

Theorem 5.3 (Buchsbaum-Eisenbud). Let I be an ideal of height 3 in a regular local ringS.ThenR=S/I is Gorenstein if and only ifI is generated by 2n-order Pfaffians of a skew-symmetric2n+ 1×2n+ 1matrixA. In this case a minimal free resolution of R as an S-module has the form

0−→S−→S2n+1 −→A S2n+1−→S−→S/I−→0.

6. Injective modules and Matlis duality

In this section we describe zero-dimensional Gorenstein local rings in terms of injective modules. This motivates Bass’ theory of Gorenstein local rings which encompasses all the earlier approaches to Gorenstein rings. We will first recall Matlis duality from [6].

of two nonzero R-modules. Matlis [10] described the structure of injective modules over Noetherian rings.

Theorem 6.1 (Matlis). (1)Let R be a Noetherian ring. Then

(1) Every injective R-module can be written as as a direct sum of inde-compsable injective R-modules.

(2) Indecomposable R-modules, upto isomorphism are the modules E(R/P) where P is a prime ideal ofR.

Let (R,m) be a local ring and E = E(R/m). The Matlis dual of an

R-moduleM is the moduleM∨ = HomR(M, E).

Theorem 6.2(Matlis duality). Let(R,m)be a complete Noetherian local ring with residue field k and E =E(R/m).Set T = Hom(−, E). Let F(R) denote the category of finite R-modules and A(R) denote the category Ar-tinian R-modules. Let M ∈ F(R) and N ∈ A(R).Then

(1) T(R)'E andT(E)'R.

(2) T(M)∈ A(R) and T(N)∈ F(R).

(3)There are natural isomorphisms T(T(N))'N and T(T(M))'M.

(5) T establishes an anti-equivalence between A(R) andF(R).

(6) If N is an R-module of finite length then λ(N) =λ(T(N)).

(7) If R is zero-dimensional local ring then λ(E(R/m)) =λ(R).

Theorem 6.3. Let(R,m) be an Artinian local ring with residue fieldkand injective hull E =ER(k).Then the following are equivalent.

(1) R is an injective R-module. (2) R'E.

(3) (0) is an irreducible ideal of R.

(4) dimkSoc(R) = 1

(5) R is Gorenstein in the sense of Grothendieck.

Proof. (1) =⇒ (2) Let R be an injective R-module. Since R is local, it is indecomposable. The only indecomposable injective R-module is E. Hence

R'E.

(2) =⇒ (1) Clear.

(1) =⇒ (4) Let R be injective R-module. Since R is Artinian, it is an essential extension of V = Soc(R). Hence R is the injective hull of V. If dimV > 1, then V = U ⊕W for proper subspaces U and W of V. The

(4) =⇒ (1) Let dimV = 1.As R 'E(V), R is an injectiveR-module and therefore R'E.

(3)⇐⇒ (4) This is proved in 3.8

(4) ⇐⇒(5) Since R is Artinian, it is complete. By Cohen-Structure Theo-rem, there is a complete regular local ringSof dimensionnso thatS/I =R,

for a height nideal I of S. We need to show that ωR = ExtnS(R, S) 'R if

and only if dimkSocR= 1.The Gorenstein property ofS/I is equivalent to

the condition that the rank of the last module is one in a minimal resolution of S/I.

Let (S,n) be a local ring. If M is an S-module with a minimal free resolution

C.= 0→Sβn _→Sβn−1 _{→ · · · →S}β0 _→0

then to findβi,tensorC.withk=S/nto find TorSi (k, M).Since the entries

of the matrices in the resolution are inn,the maps in the complexC.⊗_Sk

are zero. Hence TorS_i(k, M) =kβi _{for alli.}

Coming back to our situation, the Koszul complexK(x) is a minimal free

S-resolution of kwheren= (x1, x2, . . . , xn) :

0→S →Sn→S(n2) → · · · →Sn→S →0.

Tensor with S/I to find Torn_S(, S/I). This is the kernel of the map fn : S/I →(S/I)n induced from the Koszul complex map. Hence

βn= TornS(k, S/I) = kerfn

= {¯s∈S/I |sx¯i= 0 for alli= 1,2, . . . , n}

= I :n

I = Soc(S/I)

ThusS/I is Gorenstein if and only if dimkSoc(S/I) = 1.

7. Bass’ definition of Gorenstein rings

Auslander, Buchsbaum and Serre investigated regular local rings in terms of minimal free resoultions. This approach revealed interesting properties of regular local rings such as they are unique factorization domains and their localizations at prime ideals are again regular.

Bass [4] characterized irreducibility of parameter ideals in Cohen-Macaulay local rings in terms of injective dimension.

Theorem 7.1 (Bass). Let (R,m) be a Noetherian local ring. Then the following conditions are equivalent:

(1) R is Cohen-Macaulay and every parameter ideal is irreducible. (2) injdim_RR <∞.

(3) injdim_RR= dimR.

Serre pointed out to Bass that rings of finite injective dimension and rings studied by Grothendieck via canonical modules are same at least in the geometric case. Bass in his classic paper [5] connected all the variations studied before and presented a cohesive theory of Gorenstein rigs. Here is his characterization:

Theorem 7.2 (Bass). The following conditions are equivalent for a Noe-therian local ring (R,m) :

(1) If R=S/I where I is an ideal of a regular local ring S,then ωR'R.

(2) injdim_R(R)<∞.

(3) injdim_R(R) = dimR.

(4) R is Cohen-Macaulay and some parameter ideal of R is irreducible. (5) R is Cohen-Macaulay and every parameter ideal of R is irreducible. (6) Put Fh=L

htp=hER(R/p).If

0→R→E0→E1 → · · · →Eh→ · · ·

is a minimal injective resolution ofR,thenFh 'Eh for allh= 0,1, . . . ,dimR.

Bass generalized the results of Gorenstein and Rosenlicht to one dimen-sional Gorenstein domains using his characterization of Gorenstein rings. Theorem 7.3. Let p be a prime ideal of a regular local ring S so that

R = S/p is a one dimensional Gorenstein local domain. Suppose that the integral closure T of R in its fraction field K is a finite R-module. Let C be the conductor. Then

λ(T /R) =λ(R/C).

Proof. Let dimS = n. Then the canonical module ωR = Extn−S 1(R, S).

Consider the exact sequence

where D=T /R.First we show that R/C'Ext1_R(D, R).For this, consider the injective resolution ofR :

0−→R−→K−→E−→0

where K is the fraction field of R and E = ER(R/m). This follows from

Theorem 7.2. Applying ( )∗= HomR(, R) we get the exact sequence:

0−→D∗−→T∗−→R∗−→Ext1_R(D, R)−→Ext1_R(T, R)−→0.

SinceTis Cohen-Macaulay of dimension one, Ext1_R(T, R) = 0.SinceDis tor-sion,D∗ = 0.Moreover R∗'R and T∗ 'C.Therefore R/C'Ext1_R(D, R).

Now we show that λ(Ext1_R(D, R)) = λ(D). Use the minimal injective resolution of R given above and apply the functor HomR(D, ) to get the

exact sequence:

0 → HomR(D, R)→HomR(D, K)→HomR(D, E)

→ Ext1_R(D, R)→Ext1_R(D, K)→Ext1_R(D, E)→0.

HomR(D, R) = HomR(D, K) = 0 as D is torsion. Since K is an injective R-module, ExtR(D, K) = 0. Thus Ext1R(D, R) ' HomR(D, E). By Matlis

duality, λ(HomR(D, E)) =λ(D).Thereforeλ(Ext1R(D, R) =λ(D).

8. Macaulay-Stanley Theorem on Gorenstein graded rings Definition 8.1. LetI be a heighthgraded ideal of the polynomial ringR=

k[x1, x2, . . . , xs] where k is a field. Suppose A = R/I is Cohen-Macaulay.

Then the canonical module ofA,denoted by ωA is the gradedA-module

Exth_R(R/I, R).We say that Ais Gorenstein ifωA'A(a) for some a∈Z.

The numberais called thea-invariant of A.

Definition 8.2. Let R=L∞

n=0Rn be a graded ring. Let M, N be graded R-modules. Define Hom_R(M, N)n to be the group

{f :M →N |f isR-linear and f(Mi)⊆Nn+i for all i∈Z}, and

Hom_R(M, N) =

∞

M

n=−∞

Hom_R(M, N)n.

Proof. It is clear that Hom_R(M, N)⊆HomR(M, N).For the other inclusion,

let us assume thatM is a finite free gradedR-module. Let e1, e2, . . . , en be

homogeneous generators of M having degrees di for i = 1,2, . . . , n. Let f :M →N be anR-linear map. Let

f(ei) =yiµ1 +yiµ2 +· · ·+yiµr

for some yiµj ∈ Nµj for i = 1,2, . . . , n and j = 1,2, . . . , r. Define the map fd(ei) =yi(d+di).Thenfd= 0 for all but finitely manyd.Moreoverf =

P

fd

and fd∈HomR(M, N)d.

IfM is not free then we take a finite graded presentation of M :

G→F →M →0

where G and F are finite free R-modules. Apply the left exact functor Hom_R(−, N) to get the commutative diagram with exact rows :

0 // Hom_R(M, N) //

f

Hom_R(F, N) //

g

Hom_R(G, N)

h

0 // HomR(M, N) // HomR(F, N) // HomR(G, N)

Since g andh are isomorphisms,f is so.

Lemma 8.4. Let R be a graded ring. Then for any d∈_Z,

Hom_R(R(d), R)'R(−d).

Proof. If f ∈ Hom_R((R(d), R)e then there is an x ∈ Re−d so that f = µx

whereµx is the multiplication map byx. Define ϕ: Hom_R((R(d), R)e→R(−d)e

by ϕ(µx) =x.It is easy to see thatϕis a graded isomorphism.

Theorem 8.5. Let R = k[x1, x2, . . . , xs] be a polynomial ring over a field k.Let degxi =ei for i= 1,2, . . . , swhere e1, e2, . . . , es are positive integers.

Let I be a graded ideal of height h so that A = R/I is Cohen-Macaulay. Then

H(ωA, t) = (−1)dimAt−(e1+e2+···+es)H(A,1/t).

Proof. As seen before, pd_RA=h.SinceωA= ExthR(R/I, R),to findωA,we

take a graded free resolution ofR/I: 0−→Fh

ϕh

−→Fh−1

ϕh−1

−→ · · · −→F1

ϕ1

For i = 0,1,2, . . . , h, let rank Fi = βi and Fi = Lβ_j=1i R(−gij) for some gij ∈Z. Put Hom(M, R) = M∗ for an R-module M. Apply Hom(−, R) to

the exact sequence 3 to get the exact sequence: 0−→F₀∗ ϕ

∗

1

−→F₁∗ ϕ

∗

2

−→ · · · −→F_h−∗ ₁ ϕ

∗

h

−→F_h∗ −→ωA−→0.

(4)

The resolutions of R/I and ωA help us in finding their Hilbert series. For

this we first find the Hilbert series of Fi andFi∗.

H(Fi, t) = βi

X

j=1

H(R(−gij), t)

= t

gi1 _+tgi2 _{+· · ·+t}giβi

Qs

i=1(1−tei)

H(A, t) =

h

X

i=0

(−1)iH(Fi, t)

=

h

X

i=0

(−1)itgi1 +tgi2 +· · ·+tgiβi

Qs

i=1(1−tei)

H(A,1/t) = (−1)

s_te1+e2+···+es

Qs

i=1(1−tei)

h X i=1 (−1)i βi X j=1

t−gij

To find the Hilbert series of ωA we use the resolution 4. Note that

F_i∗ = HomR( βi

M

j=1

R(−gij), R) = βi

M

j=1

HomR(R(−gij), R) = βi

M

j=1

R(gij).

H(ωA, t) = h

X

i=0

(−1)h−iH(F_i∗, t)

= (−1)

hPh

i=0(−1)i

Pβi j=1t

−gij

Qs

i=1(1−tei)

Comparing the expressions forH(A,1/t) and H(ωA, t) we get H(ωA, t) = (−1)dt−(e1+e2+···+es)H(A,1/t).

Theorem 8.6(Macaulay). LetR=L∞

n=0Rn be a Gorenstein graded ring

field. Then there exists ρ∈Zsuch that

H(R,1/t) = (−1)dtρH(R, t).

Proof. Since R is Gorenstein, there is an integer a such that ωA ' A(a).

Therefore by Theorem 8.5 we have

H(ωR, t) =t−aH(R, t) = (−1)dt−(e1+e2+···+es)H(R,1/t).

Hence

H(R,1/t) = (−1)dt(e1+e2+···+es)−a_{H(R, t).}

Corollary 8.7. Let R be d-dimensional standard graded Gorenstein ring over a field. Let

H(R, t) = h0+h1t+h2t

2_{+· · ·+h}

sts

(1−t)d

where hs6= 0. Thenhi=hs−i for alli= 0,1, . . . , s.

Proof. Exercise.

Example 8.8. LetR=k[X, Y]/(X3, XY, Y2) wherekis a field. Letx and

y denote the residue classes of X and Y in R respectively. Then R is an Artin ring. A basis of R as a k-vector space is given by {1, x, y, x2}. Thus

H(R, t) = 1 + 2t+t2.Hence the h-vector ofR is symmetric. However, R is not Gorenstein.

Stanley [17] proved a remarkable converse to Macaulay’s Theorem for graded Cohen-Macaulay domains. This has been used to provide simple proofs of many results about Gorenstein graded domains.

Theorem 8.9 (Stanley). Let R = L∞

n=0Rn be a d-dimensional graded

Cohen-Macaulay domain where R0=k is a field. Suppose for some integer

a the Hilbert series ofR satisfies the equation of rational functions:

H(R,1/t) = (−1)d_ta_{H(R, t).}

Then R is Gorenstein.

Proof. Let R = S/I where S = k[x1, x2, . . . , xs] and I is a homogeneous

ideal under the grading degxi =ei fori= 1,2, . . . , s. Since ωR is a graded

finite R-module, Rn = 0 for n ≤ N for some N. Hence We may shift the

grading on ωR so that [ωR]n = 0 for n < 0 and [ωR]0 6= 0. Therefore H(ωR, t) =H(R, t).However, since R is a Cohen-Macaulay domain, ωR is

9. Invariant subrings of finite groups

Letk[x] denote the polynomial ring in the indeterminates x1, x2, . . . , xn.

Let G < GL(n, k) be a finite subgroup. The ring of invariants of G is the ring

k[x]G ={f ∈k[x]|M(f) =f for all M ∈G}.

AnyM ∈GL(n, k) acts on the indeterminates linearly by the rule (x1, x2, . . . , xn)t7−→M(x1, x2, . . . , xn)t.

This action extends to allf ∈k[x] so thatf 7−→M(f) is an automorphism of k[x].

LetR be any ring,Ga finite group of automorphisms ofR such that|G| is invertible in R. Put S = RG, the ring of invariants of G acting on R.

Consider the map ρ:R−→S defined as:

ρ(r) = 1 |G|

X

g∈G g(r).

Then (i) ρ is S-linear (ii) ρ|_S = idS. Such a map ρ :R −→ S is called the

Reynolds operator of the pairS ⊂R.

Proposition 9.1. Let S be a subring of a ring R and ρ : R → S be a Reynolds operator. Then (i) IR∩S = I for all ideals I of S. (ii) If R is Noetherian then so is S.

Proof. (i) Let

n

P

i=1

airi =a∈ S wherea1, . . . , an ∈ I and r1, r2, . . . , rn∈ R.

Then a=ρ(a) =P

iρ(ri)ai ∈I.

(ii) Let I1 ⊂I2 ⊂. . . be an ascending chain of ideals in S.Then InR = In+1R=. . . for somenasR is Noetherian. Hence

In=InR∩S=In+1R∩S=In+1=. . . .

ThusS is Noetherian.

Theorem 9.2 (Hilbert’s finiteness theorem). Let G be a subgroup of

GL(n, k) acting linearly on k[x] =R. Put S =k[x]G.Suppose that there is a Reynolds operator ρ:R→S.Then S is a finitely generated k-algebra. Proof. LetM be the maximal ideal ofSgenerated by homogeneous elements of positive degree. Since R is Noetherian M R is finitely generated. Put

M = (f1, f2, . . . , fs) where f1, f2, . . . , fn are homogeneous. We claim that R = k[x]G = k[f1, f2, . . . , fs]. Let f ∈ R be homogeneous of degree d.

Hence there exist g1, . . . , gs ∈ R such that f = g1f1 +· · ·+gsfs. Hence ρ(f) =f =ρ(g1)f1+· · ·+ρ(gs)fs.We may assume thatgi are homogeneous.

Then deg ρ(gi) =deg gi =deg f −deg fi < deg f. Since ρ(gi) are of smaller

degree thandeg f,by inductionρ(g1), . . . , ρ(gs)∈k[f1, f2, . . . fs].Hencef ∈

k[f1, f2, . . . , fs].

Corollary 9.3. Let G be a finite subgroup of GL(n, k) acting linearly on

k[x]. Suppose that (|G|,char k) = 1. Then k[x]G is a finitely generated k -algebra.

10. Molien’s Theorem

Theorem 10.1. Let G be a finite subgroup of GL(n,C) acting linearly on R =C[x]. Let C[x]Gi denote the vector space generated by all homogeneous

invariants of degrees i. Put H(C[x]G, λ) = ∞

P

i=0

dimC[x]Gi λi.Then

H(C[x]G, λ) = _|1 G|

X

M∈G

1 det(I−λM).

Proof. Putg:=|G|.The Reynolds operatorρ:R→RG is a C-linear map.

Hence ρ induces a linear map ρ|Ri = ρi : Ri → (R G₎

i where ( )i denotes

elements of degreei. Clearlyρ2_i =ρi.Hence 0 and 1 are the only eigenvalues

of ρi.Thus rank(ρi) =tr(ρi) = dim(RG)i.Therefore H(RG, λ) =

∞

X

i=0

dim(RG)iλi

=

∞

X

i=0

tr(ρi)λi

= 1

g

X

M∈G ∞

X

i=0

trM|_Riλi

= 1

g

X

M∈G ∞

X

i=0

trM|_Riλi

!

.

We now prove that

∞

X

i=0

trM|_Riλi = 1 det(I−λM).

SinceCis algebraically closed and eachM has finite order,M can be

C[x].Hence eigenvalues ofM|Riareλ α1

1 . . . λαnn whereα= (α1, α2, . . . , αn)∈ Nn and |α|=i. Thus

trM|_Ri = X

α1+···+αn=i λα1

1 λ

α1

2 . . . λαnn.

Hence

∞

X

i=0

trM|_Riλi =

n

Y

j=1 1 (1−λjλ)

=

n

Y

j=1

λ−_j1

(λ−_j1−λ)

= detM

−1

det(M−1_−λI)

= 1

det(I−λM). ThereforeH(RG, λ) = _|G|1 P

M∈G

1

det(I−λM).

Example 10.2. Consider the quaternion group Q8 acting on C[x, y]. The

Molien series is calculated as follows

M det(I−λM) M det(I−λM)

"

1 0 0 1

#

(1−λ)2

"

−1 0 0 −1

#

(1 +λ)2

"

i 0

0 −i

#

1 +λ2

"

−i 0 0 i

#

1 +λ2

"

0 1

−1 0

#

1 +λ2

"

0 −1

1 0

#

1 +λ2

"

0 i i 0

#

1 +λ2

"

0 −i

−i 0

#

Hence,

H(C[x, y]Q8, λ) = 1

8

1 (1−λ)2 +

1 (1 +λ)2 +

6 1 +λ2

= 1 +λ 6

(1−λ4₎2

= 1 + 2λ4+λ6+· · ·.

Hence there are 2 invariants of degree 4 and one of degree 6. The orbit of x

and y is{±x,±y,±ix,±iy}.Hence

ρ(x4) = 1 8

x4+x4+y4+y4+x4+x4+y4+y4 = 1 2(x

4_+y4_),

ρ(x2y2) = x2y2.

Let J denote the Jacobian. If f1, f2, . . . , fn ∈ C[x1, x2, . . . , xn] and M ∈ GL(n,C) then by chain rule:

J(M(f1), M(f2), . . . M(fn)) = detM J(f1, . . . , fn).

In the present situation,J(x4+y4, x2y2) = 8(x5y−xy5).Hence

α=x4+y4, β=x2y2, γ=x5y−xy5 ∈_C[x, y]Q8_.

Hence C[α, β, γ]∈C[x, y]Q8.It can be verified that γ2 =α2β−4β3.Hence

C[α, β, γ]' C[u, v, w]

(w2_−u2_{v+ 4v}3₎.

where degu= degv= 4 and degw= 6.Therefore the Hilbert series is

H(C[α, β, γ], λ) = 1−λ

12

(1−λ4₎2_(1−λ6₎ =

1 +λ6

(1−λ4₎2. ThereforeC[α, β, γ] =C[x, y]Q8.

11. Cohen-Macaulayness of rings of invariants

Theorem 11.1. Let G⊂GL(n, k) be a finite group acting linearly on k[x].

Suppose thatchar(k)and|G|are coprime. Then the ring of invariantsk[x]G is CM.

Proof. Put R = k[x] and S = k[x]G and let ρ : R → S be the Reynolds operator. Let I be an ideal ofS. Then we claim thatIR∩S =I.Suppose

a1, a2, . . . , an∈I andr1, r2, . . . , rn∈Rsuch thatS =a1r1+. . .+anrn∈S

0

.

Let k[a1, a2, . . . , an] be a graded Noether normalization of S. Then the

ideal (a1, a2, . . . , an)R is primary forR+.AsR is CM, a=a1, a2, . . . , an is

an R-regular sequence. We claim that ais also anS-regular sequence. Suppose that 1 ≤ i ≤ n and sai ∈ (a1, a2, . . . , ai−1)S. Then sai ∈

(a1, a2, . . . , ai−1)R. Hence s ∈ (a1, a2, . . . , ai−1)R∩S = (a1, a2, . . . , ai−1).

Thusa is anS-regular sequence. ThereforeS is Cohen-Macaulay. Since k[x]G is Cohen-macaulay, it is a finite free module over any graded Noether normalization S = k[a1, a2, . . . , an] of k[x]G. Therefore there are

invariantsb1, b2, . . . , br such that

k[x]G=

r

M

i=1

Sbi.

The polynomialsa1, a2, . . . , anare calledprimary invariantsandb1, b2, . . . , br

are called secondary invariants. The above decomposition of k[x]G is called aHironaka decomposition.

Theorem 11.2. Let d1, d2, . . . , dn be degrees of a collection of primary

in-variants of a finite matrix group G⊆GL(n,C).Then

(i) the number of secondary invariants is r=d1d2. . . dn/|G|and

(ii) the degrees (together with multiplicities) of the secondary invariants are the exponents of the generating function

H(C[x]G, t) Πni=1(1−tdi) =te1 +te2+. . .+ter. Proof. By Molien’s theorem

H(C[x]G, t) = 1

|G|

X

M∈G

1

det(I−M t) =

te1 _+te2 _{+. . .+t}er

Πn_i=1(1−tdi₎ .

Hence 1 |G|

X

M∈G

(1−t)n det(I −M t) =

te1_+te2 _{+. . .+t}er

Πn

i=1(1 +t+t2+. . .+tdi−1)

.

Put t = 1 on both sides. On the right hand side we get r/d1d2. . . dn. On

the left hand side the only nonzero term after putting t= 1 is the one for

M =I. Hence the result. (ii) Clear.

Example 11.3. Consider the matrix group G=        

1 0 0 0 1 0 0 0 1

  ,   

0 1 0

−1 0 0 0 0 −1

  ,   

−1 0 0 0 −1 0

0 0 1

  ,   

0 −1 0

1 0 0

0 0 −1

       

Gis a cyclic group of order 4. The ring of invariantsC[x1, x2, x2]G=Rhas the Hilbert series

H(R, t) = 1 4

1 (1−t)3 +

2

(1 +t)(1 +t2₎ +

1 (1 +t)2_(1−t)

= 1−t+t 2_+t3 (1 +t)2_{(1 +t}2_)(1−t)3 = 1 + 2t2+ 2t3+ 5t4+ 4t5+· · ·

The polynomialsθ1=x21+x22, θ2 =x23, θ3 =x41+x42are invariants. They are algebraically independent by Jacobian criterion. Hence these are primary invariants. The number of secondary invariants is degθ1degθ2degθ3/4 = 4. To find their degrees we find

H(C[x]G, t)(1−t2)2(1−t4) = 1−t+t

2_+t3

(1 +t)2_{(1 +t}2_)(1−t)3 ·(1−t

2₎2_(1−t4₎ = 1 + 2t3+t4.

Hence e1 = 1, e2 =e3 = 3 and e4 = 4.Apply Reynolds operator to get the secondary invariants:

a1= 1, a2 =x1x2x3, a3 =x21x3−x22x3, a4=x13x2−x1x32. It is easy to verify that a Hironaka decomposition ofC[x]G is given by

C[x]G=

4

M

i=1

C[θ1, θ2, θ3]ai.

Theorem 12.1 (Stanley). Let G ⊂ GL(n, k) be a finite group having r

pseudo-reflections.Let k be a field of characteristic coprime with |G| and

R=k[x1, x2, . . . , xn].Then RG is Gorenstein if and only if

X

g∈G

1

det(I−tg) =t

−rX

g∈G

detg

det(I−tg).

Proof. Let|G|=m.First we find the rational function H(RG,1/t).

H(RG,1/t) = 1

m

X

g∈G

1 det(I−t−1_g)

= 1

m

X

g∈G tn

det(tI−g)

= 1

m

X

g∈G

(−1)n_tn_detg

det(I−tg)

ThereforeRG is Gorenstein if and only if there is anr∈Zsuch that

X

g∈G

1

det(I−tg) = t

−rX

g∈G

detg

det(I −tg) (5)

It remains to show that the integer r in the equation 6 is the number of pseudo-reflections in G. For this purpose, expand both sides around t= 1.

Let P be the subset of pseudo-reflections in G. The order of the pole in 1/det(I −tg) at t = 1 is the multiplicity of 1 as an eigenvalue of g. The only summand with a pole of order n att = 1 on the left hand side of the equation 6 is the term for g = I. The terms with a pole of order n−1 at

t= 1 occur for g∈P.Ifg∈P then 1

det(I−tg) =

1

(1−t)n−1_(1−tdetg) = cn−1

(1−t)n−1 +

cn−2

(1−t)n−2 +· · ·

Multiply both sides of the above equation by (1−t)n−1 and putt= 1 to get

both sides of the equation 6 to get 1

(1−t)n +

1 (1−t)n−1

X

g∈P

1

(1−detg) + higher order terms in (1−t)

= (1−(1−t))−r





1 (1−t)n +

1 (1−t)n−1

X

g∈P

detg

(1−detg) +· · ·



.

Equating the coefficients of 1/(1−t)n−1 we get

X

g∈P

1

(1−detg) =

X

g∈P

detg

(1−detg)

r .

Hence r=|P|.

Theorem 12.2 (Watanabe). Let charkbe coprime to|G|.Suppose Ghas no pseudo-reflections. Then RG is Gorenstein if and only if G⊂SL(n, k).

Proof. If G ⊂ SL(n, k) then there is no pseudo-reflection in G. Hence the equation

X

g∈G

1

det(I−tg) = t

−rX

g∈G

detg

det(I −tg) (6)

is satisfied. Therefore RG is Gorenstein. Conversely suppose that RG is Gorenstein. Then (6) is satisfied with r = 0. Put t = 0 in (6) to get |G| = P

g∈Gdetg. Since the eigenvalues of g ∈ G are nth roots of unity,

equating the real parts in this equation, we have |G|= P|G|

j=1cosθj. Thus

cosθj =±1.Therefore detg= 1 for all g∈G. Hence G⊂SL(n, k).

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Department of Mathematics, Indian Institute of Technology Bombay