Adela Vraciu

University of South Carolina

Free Resolutions and Representation Theory ICERM Workshop,August 3–7, 2020

Joint work with Andy Kustin and Rebecca R.G.

Problem:

k = field of char. zero, P = k[x,y, z,w], N,n ≥ 1 integers

We study the minimal free resolution of R over P , where

I = (xN , yN , zN , wN ) : (xn + yn + zn + wn)

R = P/I

This can then be used to build theP/(xn + yn + zn + wn)-resolution of

P/(xN , yN , zN , wN , xn + yn + zn + wn)

Problem:

k = field of char. zero, P = k[x,y, z,w], N,n ≥ 1 integers

We study the minimal free resolution of R over P , where

I = (xN , yN , zN , wN ) : (xn + yn + zn + wn)

R = P/I

This can then be used to build theP/(xn + yn + zn + wn)-resolution of

P/(xN , yN , zN , wN , xn + yn + zn + wn)

Notation

Let N = dn+ r, with 0 ≤ r ≤ n− 1.The answer will be given in terms of d and r instead of n and N .

Observation

I is a homogenous Gorenstein ideal of grade 4The resolution is self-dual and it has the form:

0→ F4At

→ F3Bt

→ F2B→ F1

A→ P → 0

F4 = P (−s− 4), where s=the socle degree of I,

F1 = P (−d1)⊕ · · · ⊕ P (−dk),where d1, . . . , dk = degrees of the generators of I

At : F4 = P (−s− 4)→ F3 and A : F1 → P preserve degrees; itfollows that

F3 = P (−s− 4 + d1)⊕ · · ·P (−s− 4 + dk)

Observation

I is a homogenous Gorenstein ideal of grade 4The resolution is self-dual and it has the form:

0→ F4At

→ F3Bt

→ F2B→ F1

A→ P → 0

F4 = P (−s− 4), where s=the socle degree of I,

F1 = P (−d1)⊕ · · · ⊕ P (−dk),where d1, . . . , dk = degrees of the generators of I

At : F4 = P (−s− 4)→ F3 and A : F1 → P preserve degrees; itfollows that

F3 = P (−s− 4 + d1)⊕ · · ·P (−s− 4 + dk)

Observation

I is a homogenous Gorenstein ideal of grade 4The resolution is self-dual and it has the form:

0→ F4At

→ F3Bt

→ F2B→ F1

A→ P → 0

F4 = P (−s− 4), where s=the socle degree of I,

F1 = P (−d1)⊕ · · · ⊕ P (−dk),where d1, . . . , dk = degrees of the generators of I

At : F4 = P (−s− 4)→ F3 and A : F1 → P preserve degrees; itfollows that

F3 = P (−s− 4 + d1)⊕ · · ·P (−s− 4 + dk)

Observation

I is a homogenous Gorenstein ideal of grade 4The resolution is self-dual and it has the form:

0→ F4At

→ F3Bt

→ F2B→ F1

A→ P → 0

F4 = P (−s− 4), where s=the socle degree of I,

F1 = P (−d1)⊕ · · · ⊕ P (−dk),where d1, . . . , dk = degrees of the generators of I

At : F4 = P (−s− 4)→ F3 and A : F1 → P preserve degrees; itfollows that

F3 = P (−s− 4 + d1)⊕ · · ·P (−s− 4 + dk)

Observation

I is a homogenous Gorenstein ideal of grade 4The resolution is self-dual and it has the form:

0→ F4At

→ F3Bt

→ F2B→ F1

A→ P → 0

F4 = P (−s− 4), where s=the socle degree of I,

F1 = P (−d1)⊕ · · · ⊕ P (−dk),where d1, . . . , dk = degrees of the generators of I

At : F4 = P (−s− 4)→ F3 and A : F1 → P preserve degrees; itfollows that

F3 = P (−s− 4 + d1)⊕ · · ·P (−s− 4 + dk)

If we know the socle degree of I and the degrees of thegenerators of I, then we know the graded free modules F1, F3, F4

in the resolution.

To find the graded free module F2: in addition to the above,we also need to know the Hilbert function of R = P/I.

Definition

H(n) := dimkPn =

(n+ 3

3

)For any graded P -module M =

⊕n

Mn,

HM (n) := dimk(Mn)

If we know the socle degree of I and the degrees of thegenerators of I, then we know the graded free modules F1, F3, F4

in the resolution.

To find the graded free module F2: in addition to the above,we also need to know the Hilbert function of R = P/I.

Definition

H(n) := dimkPn =

(n+ 3

3

)For any graded P -module M =

⊕n

Mn,

HM (n) := dimk(Mn)

If we know the socle degree of I and the degrees of thegenerators of I, then we know the graded free modules F1, F3, F4

in the resolution.

To find the graded free module F2: in addition to the above,we also need to know the Hilbert function of R = P/I.

Definition

H(n) := dimkPn =

(n+ 3

3

)For any graded P -module M =

⊕n

Mn,

HM (n) := dimk(Mn)

Once the Hilbert function HR and the free graded modulesF1, F3, F4 are known, we have

HF2 = HR −H +HF1 +HF3 −HF4

since the alternating sum of Hilbert functions in the resolution iszero.

Observation

Knowing the Hilbert function HF2 of a graded free module F2

allows us to determine the graded shifts of F2.

Proof: LetF2 = P (−δ1)b1 ⊕ · · · ⊕ P (−δl)bl

with δ1 < δ2 < · · · < δl, and b1, . . . , bl ≥ 1.

Once the Hilbert function HR and the free graded modulesF1, F3, F4 are known, we have

HF2 = HR −H +HF1 +HF3 −HF4

since the alternating sum of Hilbert functions in the resolution iszero.

Observation

Knowing the Hilbert function HF2 of a graded free module F2

allows us to determine the graded shifts of F2.

Proof: LetF2 = P (−δ1)b1 ⊕ · · · ⊕ P (−δl)bl

with δ1 < δ2 < · · · < δl, and b1, . . . , bl ≥ 1.

Once the Hilbert function HR and the free graded modulesF1, F3, F4 are known, we have

HF2 = HR −H +HF1 +HF3 −HF4

since the alternating sum of Hilbert functions in the resolution iszero.

Observation

Knowing the Hilbert function HF2 of a graded free module F2

allows us to determine the graded shifts of F2.

Proof: LetF2 = P (−δ1)b1 ⊕ · · · ⊕ P (−δl)bl

with δ1 < δ2 < · · · < δl, and b1, . . . , bl ≥ 1.

Plugging in values of n in the Hilbert function, we have

HF2(n) = b1H(n− δ1) + · · ·+ blH(n− δl) for all n ≥ 0

H(n− δi) =

{0 for n < δi1 for n = δi

, so we obtain:

δ1 = min{n |HF2(n) 6= 0}, b1 = HF2(δ1)

δ2 = min{n |HF2(n) > b1H(n− δ1)},

b2 = HF2(δ2)− b1H(δ2 − δ1)

ETC.

Plugging in values of n in the Hilbert function, we have

HF2(n) = b1H(n− δ1) + · · ·+ blH(n− δl) for all n ≥ 0

H(n− δi) =

{0 for n < δi1 for n = δi

, so we obtain:

δ1 = min{n |HF2(n) 6= 0}, b1 = HF2(δ1)

δ2 = min{n |HF2(n) > b1H(n− δ1)},

b2 = HF2(δ2)− b1H(δ2 − δ1)

ETC.

Plugging in values of n in the Hilbert function, we have

HF2(n) = b1H(n− δ1) + · · ·+ blH(n− δl) for all n ≥ 0

H(n− δi) =

{0 for n < δi1 for n = δi

, so we obtain:

δ1 = min{n |HF2(n) 6= 0}, b1 = HF2(δ1)

δ2 = min{n |HF2(n) > b1H(n− δ1)},

b2 = HF2(δ2)− b1H(δ2 − δ1)

ETC.

Summary

In order to find the graded Betti numbers in the resolution ofR = P/I over P , it suffices to know:

• The socle degree of I

• The degrees of the generators of I

• The Hilbert function of R

Observation

The socle degree of I is s = 4N − 4− n.

Proof: Recall I = (xN , yN , zN , wN ) : (xn + yn + zn + wn).There is an injective homorphism:

R =P

I↪→ P

(xN , yN , zN , wN)

given by multiplication by xn + yn + zn + wn.

This raises degrees by n, and sends the socle of I to the socle of(xN , yN , zN , wN ), which is (xyzw)N−1.

Observation

The socle degree of I is s = 4N − 4− n.

Proof: Recall I = (xN , yN , zN , wN ) : (xn + yn + zn + wn).There is an injective homorphism:

R =P

I↪→ P

(xN , yN , zN , wN)

given by multiplication by xn + yn + zn + wn.

This raises degrees by n, and sends the socle of I to the socle of(xN , yN , zN , wN ), which is (xyzw)N−1.

Observation

The socle degree of I is s = 4N − 4− n.

Proof: Recall I = (xN , yN , zN , wN ) : (xn + yn + zn + wn).There is an injective homorphism:

R =P

I↪→ P

(xN , yN , zN , wN)

given by multiplication by xn + yn + zn + wn.

This raises degrees by n, and sends the socle of I to the socle of(xN , yN , zN , wN ), which is (xyzw)N−1.

To find the generators of I and the Hilbert function of R: we use amulti-grading on the polynomial ring P by the Abelian group G:

Definition

G = Z× Zn × Zn × Zn × Zn

Let D ∈ Z and (r̄1, r̄2, r̄3, r̄4) ∈ Zn × Zn × Zn × Zn,

P(D,r̄1,r̄2,r̄3,r̄4) = the k-span of the monomials xρ1yρ2zρ3wρ4 suchthat:

• ρ1 + ρ2 + ρ3 + ρ4 = D, and

• the image of ρi in Zn is r̄i for each i = 1, . . . 4.

Note: P(D,r̄1,r̄2,r̄3,r̄4) = 0 unless D̄ = r̄1 + r̄2 + r̄3 + r̄4.

To find the generators of I and the Hilbert function of R: we use amulti-grading on the polynomial ring P by the Abelian group G:

Definition

G = Z× Zn × Zn × Zn × Zn

Let D ∈ Z and (r̄1, r̄2, r̄3, r̄4) ∈ Zn × Zn × Zn × Zn,

P(D,r̄1,r̄2,r̄3,r̄4) = the k-span of the monomials xρ1yρ2zρ3wρ4 suchthat:

• ρ1 + ρ2 + ρ3 + ρ4 = D, and

• the image of ρi in Zn is r̄i for each i = 1, . . . 4.

Note: P(D,r̄1,r̄2,r̄3,r̄4) = 0 unless D̄ = r̄1 + r̄2 + r̄3 + r̄4.

To find the generators of I and the Hilbert function of R: we use amulti-grading on the polynomial ring P by the Abelian group G:

Definition

G = Z× Zn × Zn × Zn × Zn

Let D ∈ Z and (r̄1, r̄2, r̄3, r̄4) ∈ Zn × Zn × Zn × Zn,

P(D,r̄1,r̄2,r̄3,r̄4) = the k-span of the monomials xρ1yρ2zρ3wρ4 suchthat:

• ρ1 + ρ2 + ρ3 + ρ4 = D, and

• the image of ρi in Zn is r̄i for each i = 1, . . . 4.

Note: P(D,r̄1,r̄2,r̄3,r̄4) = 0 unless D̄ = r̄1 + r̄2 + r̄3 + r̄4.

To find the generators of I and the Hilbert function of R: we use amulti-grading on the polynomial ring P by the Abelian group G:

Definition

G = Z× Zn × Zn × Zn × Zn

Let D ∈ Z and (r̄1, r̄2, r̄3, r̄4) ∈ Zn × Zn × Zn × Zn,

P(D,r̄1,r̄2,r̄3,r̄4) = the k-span of the monomials xρ1yρ2zρ3wρ4 suchthat:

• ρ1 + ρ2 + ρ3 + ρ4 = D, and

• the image of ρi in Zn is r̄i for each i = 1, . . . 4.

Note: P(D,r̄1,r̄2,r̄3,r̄4) = 0 unless D̄ = r̄1 + r̄2 + r̄3 + r̄4.

Observation

P is graded by G in the sense that

Pm1 · Pm2 ⊆ Pm1+m2 for all m1,m2 ∈ G.

The ideals (xN , yN , zN , wN ) and (xn + yn + zn + wn) arehomogeneous under the multi-grading by G.

deg(xN ) = (N, r, 0, 0, 0) deg(yN ) = (N, 0, r, 0, 0)

deg(zN ) = (N, 0, 0, r, 0), deg(wN ) = (N, 0, 0, 0, r)

deg(xn + yn + zn + wn) = (n, 0, 0, 0, 0)

Observation

P is graded by G in the sense that

Pm1 · Pm2 ⊆ Pm1+m2 for all m1,m2 ∈ G.

The ideals (xN , yN , zN , wN ) and (xn + yn + zn + wn) arehomogeneous under the multi-grading by G.

deg(xN ) = (N, r, 0, 0, 0) deg(yN ) = (N, 0, r, 0, 0)

deg(zN ) = (N, 0, 0, r, 0), deg(wN ) = (N, 0, 0, 0, r)

deg(xn + yn + zn + wn) = (n, 0, 0, 0, 0)

Observation

P is graded by G in the sense that

Pm1 · Pm2 ⊆ Pm1+m2 for all m1,m2 ∈ G.

The ideals (xN , yN , zN , wN ) and (xn + yn + zn + wn) arehomogeneous under the multi-grading by G.

deg(xN ) = (N, r, 0, 0, 0) deg(yN ) = (N, 0, r, 0, 0)

deg(zN ) = (N, 0, 0, r, 0), deg(wN ) = (N, 0, 0, 0, r)

deg(xn + yn + zn + wn) = (n, 0, 0, 0, 0)

• I = (xN , yN , zN , wN ) : (xn + yn + zn + wn) is homogeneousunder the multi-grading by G.

• the multi-grading is inherited by R = P/I.

Definition

If M is a k-module which is multi-graded by G,

HM (−) = the Hilbert function of M with respect to theG-grading on M i.e.

for each g ∈ G, HM (g) is the vector space dimension of thecomponent of M of degree g.

Definition

If M is a k-module which is multi-graded by G,

HM (−) = the Hilbert function of M with respect to theG-grading on M i.e.

for each g ∈ G, HM (g) is the vector space dimension of thecomponent of M of degree g.

We find

• the multi-degree of the socle of I

• the multi-degrees of the generators of I

• the multi-graded Hilbert function of R

This information allows us to find multi-graded Betti numbers ofthe P -resolution of R.

Recall the multiplication by xn + yn + zn + wn

R =P

I↪→ P

(xN , yN , zN , wN )

sends the socle of I to xN−1yN−1zN−1wN−1, which hasmulti-degree(4N − 4, r − 1, r − 1, r − 1, r − 1)

Therefore, the socle of I has multi-degree(4N − 4− n, r − 1, r − 1, r − 1, r − 1)

Recall the multiplication by xn + yn + zn + wn

R =P

I↪→ P

(xN , yN , zN , wN )

sends the socle of I to xN−1yN−1zN−1wN−1, which hasmulti-degree(4N − 4, r − 1, r − 1, r − 1, r − 1)

Therefore, the socle of I has multi-degree(4N − 4− n, r − 1, r − 1, r − 1, r − 1)

Notation

For a g ∈ P , g[n] ∈ P is obtained by replacing x, y, z, w in g byxn, yn, zn, wn.

Observation

Let D = nk + r1 + r2 + r3 + r4 with k ∈ Z,m = (D, r̄1, r̄2, r̄3, r̄4) ∈ Z× Zn × Zn × Zn × Zn.

The elements of Pm have the form

xr1yr2zr3wr4g[n] with g ∈ P of degree k

Notation

For a g ∈ P , g[n] ∈ P is obtained by replacing x, y, z, w in g byxn, yn, zn, wn.

Observation

Let D = nk + r1 + r2 + r3 + r4 with k ∈ Z,m = (D, r̄1, r̄2, r̄3, r̄4) ∈ Z× Zn × Zn × Zn × Zn.

The elements of Pm have the form

xr1yr2zr3wr4g[n] with g ∈ P of degree k

Notation

For a g ∈ P , g[n] ∈ P is obtained by replacing x, y, z, w in g byxn, yn, zn, wn.

Observation

Let D = nk + r1 + r2 + r3 + r4 with k ∈ Z,m = (D, r̄1, r̄2, r̄3, r̄4) ∈ Z× Zn × Zn × Zn × Zn.

The elements of Pm have the form

xr1yr2zr3wr4g[n] with g ∈ P of degree k

Finding generators - First simplification

Finding generators for I reduces to finding generators for the ideals

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : (x+ y + z + w)

for each choice of (ε1, ε2, ε3, ε4) ∈ {0, 1}4.

Lemma

The elements of I = (xN , yN , zN , wN ) : (xn + yn + zn + wn)of degree m = (D, r̄1, r̄2, r̄3, r̄4) have the form

xr1yr2zr3wr4g[n]

with g ∈ (xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : (x+ y + z + w), where

εi =

{1, if ri < r,

0, otherwise,

Lemma

The elements of I = (xN , yN , zN , wN ) : (xn + yn + zn + wn)of degree m = (D, r̄1, r̄2, r̄3, r̄4) have the form

xr1yr2zr3wr4g[n]

with g ∈ (xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : (x+ y + z + w), where

εi =

{1, if ri < r,

0, otherwise,

Corollary

The ideal I is generated by is generated by all the elements of theform

(x1−ε1y1−ε2z1−ε3w1−ε4)rg[n]

with g ∈ (xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f ,

for all choices of (ε1, ε2, ε3, ε4) ∈ {0, 1}4.

Observation

The list of generators of I obtained above is redundant.

A minimal set of generators consists of the generators obtained as

above for4∑i=1

εi = 0, 2, 4, together with xN , yN , zN , wN .

Observation

The list of generators of I obtained above is redundant.

A minimal set of generators consists of the generators obtained as

above for4∑i=1

εi = 0, 2, 4, together with xN , yN , zN , wN .

Corollary

For m = (D, r1, r2, r3, r4) with D = nk +

4∑i=1

ri, we have

Hm(R) = Hk

(P

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f

)where f = x+ y + z + w.

The left-hand side is the multi-graded Hilbert function and theright-hand side is usual N-graded Hilbert function.

Corollary

For m = (D, r1, r2, r3, r4) with D = nk +

4∑i=1

ri, we have

Hm(R) = Hk

(P

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f

)where f = x+ y + z + w.

The left-hand side is the multi-graded Hilbert function and theright-hand side is usual N-graded Hilbert function.

The Hilbert function can be explicitely calculated, using thefollowing facts:

• the Hilbert function of the complete intersection

C :=P

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

is known (can be found from the Koszul complex resolution)

• the map Ck·f→ Ck+1 has maximal rank in each degree k, i.e.

it is either injective or surjective (the weak LefschetzProperty).

•

Ker(·f) =(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

The Hilbert function can be explicitely calculated, using thefollowing facts:

• the Hilbert function of the complete intersection

C :=P

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

is known (can be found from the Koszul complex resolution)

• the map Ck·f→ Ck+1 has maximal rank in each degree k, i.e.

it is either injective or surjective (the weak LefschetzProperty).

•

Ker(·f) =(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

The Hilbert function can be explicitely calculated, using thefollowing facts:

• the Hilbert function of the complete intersection

C :=P

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

is known (can be found from the Koszul complex resolution)

• the map Ck·f→ Ck+1 has maximal rank in each degree k, i.e.

it is either injective or surjective (the weak LefschetzProperty).

•

Ker(·f) =(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4) : f

(xd+ε1 , yd+ε2 , zd+ε3 , wd+ε4)

Comment:

The Weak Lefschetz Property is the reason why being able toreplace xn + yn + zn + wn by x+ y + z + w allows us to performthe calculations for the Hilbert function, as well as the degrees ofgenerators.

Second simplification

We reduce to a calculation in three variables.

It is enough to find generators of the ideals

Jε := (xd+ε1 , yd+ε2 , zd+ε3) : (x+ y + z)d+ε4 ⊆ k[x, y, z]

for all choices of ε = (ε1, ε2, ε3, ε4) ∈ {0, 1}4.

Idea behind this:

Element of (xd1 , yd2 , zd3 , wd4) : (x+ y + z + w) ;

relation on xd1 , yd2 , zd3 , wd4, x+ y + z + w

By substituting w = −(x+ y + z), this leads to a relation onxd1 , yd2 , zd3 , (x+ y + z)d4 ;

element of (xd1 , yd2 , zd3) : (x+ y + z)d4

Idea behind this:

Element of (xd1 , yd2 , zd3 , wd4) : (x+ y + z + w) ;

relation on xd1 , yd2 , zd3 , wd4, x+ y + z + w

By substituting w = −(x+ y + z), this leads to a relation onxd1 , yd2 , zd3 , (x+ y + z)d4 ;

element of (xd1 , yd2 , zd3) : (x+ y + z)d4

For all d1, · · · , d4, there is an isomorphism

(xd1 , yd2 , zd3) : (x+ y + z)d4

(xd1 , yd2 , zd3)

Φ−→ (xd1 , yd2 , zd3 , wd) : (x+ y + z + w)

(xd1 , yd2 , zd3 , wd4)

given by multiplication by

Pd4 =wd4 − (−1)d4(x+ y + z)d4

x+ y + z + w,

which raises degrees by d4 − 1.

The ideals Jε are compressed Gorenstein ideals of grade 3, withsocle degree equal to

2d+ ε1 + ε2 + ε3 − ε4 − 3

Compressed: they have maximal Hilbert function amongGorenstein grade three ideals with the give socle degree.

This follows from the fact that the map

k[x, y, z]

(xd1 , yd2 , zd3)

·(x+y+z)d4−→ k[x, y, z]

(xd1 , yd2 , zd3)

has maximal rank in each degree (so the kernel is as small aspossible), i.e.

k[x, y, z]

(xd1 , yd2 , zd3)

has the Strong Lefschetz Property.

Compressed: they have maximal Hilbert function amongGorenstein grade three ideals with the give socle degree.This follows from the fact that the map

k[x, y, z]

(xd1 , yd2 , zd3)

·(x+y+z)d4−→ k[x, y, z]

(xd1 , yd2 , zd3)

has maximal rank in each degree (so the kernel is as small aspossible), i.e.

k[x, y, z]

(xd1 , yd2 , zd3)

has the Strong Lefschetz Property.

Facts about Compressed ideals

Resolution of compressed Gorenstein ideals have been studiedextensively (Boij, Iarrobino, Migliore-Miro-Roig-Nagel), in any nr.of variables.

Depending on the parity of the socle degree, exact betti numbersor bounds on the betti numbers are known.

Let J be a compressed grade 3 Gorenstein ideal of grade 3 withsocle degree s.

• If s is even, then J is minimally generated by s+ 3 elementsof degree (s/2) + 1, and all the relations are linear.

• If s is odd, then there are• (s+ 3)/2 generators of degree (s+ 1)/2• ν ≥ 0 additional generators in degree (s+ 1)/2 + 1

The number ν of additional generators is equal to the numberof linear relations in the generators of degree (s+ 1)/2

Let J be a compressed grade 3 Gorenstein ideal of grade 3 withsocle degree s.

• If s is even, then J is minimally generated by s+ 3 elementsof degree (s/2) + 1, and all the relations are linear.

• If s is odd, then there are• (s+ 3)/2 generators of degree (s+ 1)/2• ν ≥ 0 additional generators in degree (s+ 1)/2 + 1

The number ν of additional generators is equal to the numberof linear relations in the generators of degree (s+ 1)/2

The ideals Jε = (xd+ε1 , yd+ε2 , zd+ε3) : (x+ y + z)d+ε4 have

• odd socle degree for4∑i=1

εi = 0, 2, 4

• even socle degree for4∑i=1

εi = 1, 3

In order to handle the case of odd socle degree,

• we find the required number of explicit elements of therequired degree in Jε

• we prove that these elements minimally generate Jε byproving that there are no linear relations.

The result

Ignoring the G-grading, the minimal homogeneous resolution F ofR is F0 = P ,

F1 = P (−N)4 ⊕ P (−(2nd− 2n+ 4r))d

⊕ P (−(2nd− n+ 2r))6d ⊕ P (−2nd)d+1,

F2 = P (−(2nd− n+ 3r))8d+4 ⊕ P (−(2nd+ r))8d+4,

F3 = P (−(3nd− n+ 3r))4 ⊕ P (−(2nd+ n))d

⊕ P (−(2nd+ 2r))6d ⊕ P (−(2nd− n+ 4r))d+1

F4 = P (−(4nd− n+ 4r)).