GORENSTEIN ALGEBRAS
RALF FRÖBERG
1. Definition of Gorenstein
To begin with, we consider finite dimensional (thus Artinian) k-algebras A, k a field of
characteristic 0, so A = k ⊕ A1 ⊕ · · · ⊕ As = k[x1 , . . . , xn ]/I, I homogeneous.
Definition Soc(A) = {f ; f xi = 0 ∀i}.
Definition A is Gorenstein if Soc(A) is 1-dimensional, so if Soc(A) = As and As is
1-dimensional.
V, W k-spaces, B : V × W → k bilinear (so V ⊗ W → k linear). B induces linear maps
L : V → W ∗ , v 7→ (w 7→ B(v, w))
R : W → V ∗ , w 7→ (v 7→ B(v, w))
Definition B is called a perfect (non-singular, non-degenerate) pairing if L and R are
isomorphisms.
Proposition If V and W are finite dimensional with bases e1 , . . . , en and f1 , . . . , fn ,
then B is a perfect pairing if and only if ∀w, B(v, w) = 0 ∀v =⇒ w = 0 and if and only if
the matrix (B(ei , fj )) is non-singular.
Theorem A is Gorenstein if and only if At × As−t (multiplication) is a perfect pairing
for 0 ≤ t ≤ s.
Proof =⇒ : Let {mi } be all monomials of degree s − t and let deg(a) = t < s, and
suppose ami = 0 ∀i. Now let m0 be a monomial of degree s − t − 1. Then am0 xi = 0 ∀i,
so am0 ∈ Soc(A). But deg am0 = s − 1, so am0 = 0. Continue! Finally axi = 0 ∀i, so
a ∈ Soc(A), deg(a) = t, so a = 0.
⇐= : As = A∗0 , so dimk As = 1. Suppose a ∈ Soc(A), deg(A) = t < s. Then aAs−t = 0,
so not perfect pairing.
P
Corollary The Hilbert series of a Gorenstein ring is symmetric, if HA (t) = si=0 ai ti ,
then ai = as−i .
Remark There are other ways to define Gorenstein rings, e.g. (0) is an irreducible ideal,
or A is selfinjective. To define Gorenstein for rings of dimension > d, one can say: A is
Gorenstein if and only if A/x is Gorenstein for some (any) nonzerodivisor x. Complete
intersecions are Gorenstein, and Gorenstein are Cohen-Macaulay.
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RALF FRÖBERG
2. Perp ideals
Let A = k[x1 , . . . , xn ] and B = k[∂1 , . . . , ∂n ]. A is a B-module with ∂i xj = δi,j and
extended. (A is the injective envelope of k, the Macaulay’s inverse system.)
Theorem The pairing Bj × Aj → k is perfect.
Example

3!
0
Then 
0
0
j = 3, A =
k[x1 , x2 ]. Take bases x31 , x21 x2 , x1 x22 , x32 and ∂13 , ∂12 ∂2 , ∂1 ∂22 , ∂23 .
0 0 0
2! 0 0 
 is nonsingular.
0 2! 0 
0 0 3!
Definition For F ∈ As , F ⊥ = {∂ ∈ B; ∂F = 0} = AnnB (F ) (perp ideal).
Lemma If V1 ⊆ V and V × W → k is perfect, dim V1 = m, dim V = dim W = n, then
dim V1⊥ = n − m.
Theorem F ⊥ is an ideal and B/F ⊥ is Gorenstein (artinian).
Proof That F ⊥ is an ideal is clear. If deg(∂) > s, then ∂ ∈ F ⊥ , so B/F ⊥ is artinian.
hF i ⊆ As and dimhF i = 1, so dimhF i⊥ = dim As − 1 = dim Bs − 1, so dim(B/F ⊥ )s = 1.
(With hF i, I mean the vector space generated by F .) Suppose deg(∂) = t < s, ∂ ∈
/ F ⊥ , and
⊥
that ∂∂i ∈ F for all i. ∂F has degree s − t > 0, so at least one ∂i has ∂∂i F = ∂i (∂F ) 6= 0.
Example If F = xi11 · · · xinn is a monomial, then F ⊥ = (∂1i1 +1 , . . . , ∂nin +1 ).
Problem 1 For which F is F ⊥ a complete intersection?
Problem 2 For which F is F ⊥ generated in degree 2?
Problem 3 Let µ = (µ1 , . . . , µk ) be a partition of d, and F = lµ , where lµ = l1µ1 · · · lkµk ,
li generic linear forms. What is F ⊥ ?
a.s.o.
3. Inverse system of ideals
Definition Let I be a homogeneous ideal in B. Then I −1 = {f ∈ A; I ◦ f = 0} = {f ∈
A; gf = 0 ∀g ∈ I}.
Remarks I −1 is a graded B-submodule of A, not finitely generated in general, and not
an ideal in general. If I = (g1 , . . . , gk ), then I −1 = {f ∈ A; gi f = 0 ∀i}.
Notation ha1 , . . . , an i denotes the vector space generated by a1 , . . . , an .
Example If I = (∂12 ), then (I −1 )i = hx1 , x2 i, (I −1 )2 = {f = ax21 + bx1 x2 + cx22 ; ∂12 f =
0} = hx1 x2 , x22 i, (I −1 )3 = {f = ax31 + bx21 x2 + cx1 x22 + dx32 ; ∂12 f = 0} = hx1 x22 , x32 i a.s.o.
Theorem (I −1 )j = Ij⊥
Proof Ij × Ij⊥ → 0, so (I −1 )j ⊆ Ij⊥ . Let G ∈ Ij⊥ , then hG = 0 ∀h ∈ Ij . Claim
F G = 0 ∀F ∈ I. Clear if deg(F ) > j. Suppose then deg(F ) < j, and let f be an element,
GORENSTEIN ALGEBRAS
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deg(f ) = j − deg(F ). Then deg(f F ) = j and f F ∈ Ij , so f F G = 0, so F G is annihilated
by all f of degree j − deg(F ). Apolarity gives F G = 0.
Corollary dimk (I −1 )j = dimk (Bj /Ij ).
Proof (I −1 )j = Ij⊥ and dim Ij⊥ = dim Aj − dim Ij = dim Bj − dim Ij .
Corollary I −1 is finitely generated if and only if B/I is Artinian.
Proof I −1 is finitely generated if and only if (I −1 )j = 0, j >> 0 if and only if (R/J)j =
0, j >> 0.
Corollary If I is generated by monomials, then I −1 consists of those monomials which
do not belong to ”I”. (The reason for the quotation marks is that I is a monomial ideal
in B, not in A.)
Lemma If V1 , V2 ⊆ V and V × W → k is perfect, then (V1 + V2 )⊥ = V1⊥ ∩ V2⊥ and
(V1 ∩ V2 )⊥ = (V1 + V2 )⊥ = V1⊥ + V2⊥ .
4. Fat points
A point P ∈ Pn has a prime ideal p generated by n linear forms. A point is called fat of
order t if, for all F ∈ p all the partial derivatives of F of order ≤ t vanish at P , thus if and
only if F ∈ pt+1 , if and only if P is a singular point of V (F ) having multiplicity ≥ t + 1. A
∩ · · · ∩ pt+1
set of s fat points P1 , . . . , Ps of order t has ideal I = pt+1
s , the symbolic power
1
t+1
t+1
.
Let
P1 , . . . , Ps be points
of p1 ∩ · · · ∩ ps . We have (p1 ∩ · · · ∩ ps )
⊆ p1 ∩ · · · ∩ pt+1
s
of Pn and suppose that Pi = [pi0 : pi1 : · · · : pin ]. Let LP i = pi0 x0 + pi1 x1 + · · · + pin xn ∈
⊆ k[∂0 , . . . , ∂n ] we have (I −1 )j is the jth
k[x0 , . . . , xn ] = A . Then, if I = pt+1 ∩ · · · ∩ pt+1
s
j−t
j−t
graded piece of the ideal (LP1 , . . . , LPs ) for j ≥ t and (I −1 )j = Aj if j < t.
j−t
Corollary dimk (A/I)j = dimk (Lj−t
P1 , . . . , LPs )j .
Problem Are there any other classes of ideals I (besides monomial ideals and ideals of
fat points) where one can describe I −1 concretely?
Hard problem For which ideals is the symbolic powers equal to the real power?
5. Macaulay’s theorem
Theorem[Macaulay] B/I is Gorenstein with socle in degree s if and only if there exists
an F ∈ As such that I = {g ∈ B; gF = 0} = Ann(F ) if and only if I −1 is generated by F
as B-module.
Proof We have proved one half of the theorem. Suppose B/I is Gorenstein with socle
degree s. If g ∈ (B/I)t , t < s, we must have g(B/I)s−t ⊆ (B/I)t . Since (B/I)t is 1dimensional, we have that It has dimension dim Bt − 1 = dim At − 1, so It = hF i⊥ for some
F ∈ At . Let (hF i⊥ : Bs−t ) = {g ∈ Bt ; gBs−t ⊆ hF i⊥ }. Consider J = ⊕1i=t (hF i⊥ : Bi ) ⊕
(hF i⊥ ). It is not hard to see that J is an ideal (in fact the largest ideal which is equal to I in
degrees ≥ t). We claim that J = AnnB (F ). This is true in degrees ≥ s. Let G ∈ Jt , t < s,
so GBs−t ⊆ hf i⊥ . Now deg(GF ) = s − t. Let f ∈ Bs−t so deg(f G) = s, so f G ∈ hF i⊥ , so
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RALF FRÖBERG
(f G)F = f (GF ) = 0. Since Bs−t ×As−t → k is perfect, we have GF = 0, so J ⊆ AnnB (F ).
Now suppose G ∈ (AnnB (F ))t and f ∈ Bs−t . Then Gf ∈ hF i⊥ , so GBs−t ⊆ hF i⊥ , so
G ∈ J. What remains to show is that I = J. This is true in degrees ≥ s. Let f ∈ It , t < s.
We have f Bs−t ⊆ Is , so f ∈ J. The pairing Bt /It × Bs−t /Is−t → Bs /Is = k is perfect. Let
f ∈ Jt , so f Bs−t ⊆ Is . Then f¯h̄ = 0 for all h ∈ Bs−t , so f¯ = 0, so f ∈ It .
6. macaulay2
It is possible to make calculations in Macaulay2. Here is an example:
R = QQ[x, y, z]
g = matrix{{x ∗ y ∗ z}}
f = fromDual g
h = toDual (3, f )
f is a matrix of generators for the perp ideal of g, and Ann(h) = f . The number d (here
3) is needed. It should be so high so xd+1 = 0 for all variables in the Gorenstein ring. (So
in the example we could have d = 1.)