The Eisenbud-Green-Harris Conjecture Ben Richert California - - PDF document

The Eisenbud-Green-Harris Conjecture

Ben Richert California Polytechnic State University October 3, 2004

Let 1 ≤ a1 ≤ · · · ≤ an be integers in N. Then we call L an {a1, . . . , an} lex-plus-powers ideal if:

• 1. L is a monomial ideal minimally generated by

xa1

1 , . . . , xan n , m1, . . . , ml, and

• 2. for each i = 1, . . . , l, if r ∈ Rdeg(mi) and r ≥ mi,

then r ∈ L. We say that such an ideal L is lex-plus-powers with respect to A = {a1, . . . , an}.

x1, x2, x3, x4, x5 x2

1, x1x2, x1x3, x1x4, x1x5, x2 2, x2x3, x2x4, x2x5, x2 3

x3x4, x3x5, x2

4, x4x5, x2 5

x3

1, x2 1x2, x2 1x3, x2 1x4, x2 1x5, x1x2 2, x1x2x3, x1x2x4, x1x2x5,

x1x2

3, x1x3x4, x1x3x5, x1x2 4, x1x4x5, x1x2 5, x3 2, x2 2x3, x2 2x4,

x2

2x5, x2x2 3, x2x3x4, x2x3x5, x2x2 4, x2x4x5, x2x2 5, x3 3, x2 3x4,

x2

3x5, x3x2 4, x3x4x5, x3x2 5, x3 4, x2 4x5, x4x2 5, x3 5

x4

1, x3 1x2, x3 1x3, x3 1x4, x3 1x5, x2 1x2 2, x2 1x2x3, x2 1x2x4, x2 1x2x5,

x2

1x2 3, x2 1x3x4, x2 1x3x5, x2 1x2 4, x2 1x4x5, x2 1x2 5, x1x3 2, x1x2 2x3,

x1x2

2x4, x1x2 2x5, x1x2x2 3, x1x2x3x4, x1x2x3x5, x1x2x2 4,

x1x2x4x5, x1x2x2

5, x1x3 3, x1x2 3x4, x1x2 3x5, x1x3x2 4,

x1x3x4x5, x1x3x2

5, x1x3 4, x1x2 4x5, x1x4x2 5, x1x3 5, x4 2, x3 2x3,

x3

2x4, x3 2x5, x2 2x2 3, x2 2x3x4, x2 2x3x5, x2 2x2 4, x2 2x4x5, x2 2x2 5,

x2x3

3, x2x2 3x4, x2x2 3x5, x2x3x2 4, x2x3x4x5, x2x3x2 5, x2x3 4,

x2x2

4x5, x2x4x2 5, x2x3 5, x4 3, x3 3x4, x3 3x5, x2 3x2 4, x2 3x4x5, x2 3x2 5,

x3x3

4, x3x2 4x5, x3x4x2 5, x3x3 5, x4 4, x3 4x5, x2 4x2 5, x4x3 5, x4 5

Defining lex-plus-powers ideals allows us to make the following conjectures. Let H be a Hilbert function, A be a sequence of degrees, and suppose that there exists an ideal at- taining H and minimally containing an A-regular sequence (that is, a regular sequence in the de- grees A).

• 1. There is a lex-plus-powers ideal with respect to

A which attains H. (Eisenbud, Green, Harris)

• 2. The lex-plus-powers ideal with respect to A

has the largest graded Betti numbers among all ideals attaining H and containing a regular sequence in degrees A. (The lex-plus-powers conjecture— Charalambous, Evans)

These conjectures mimic Macaulay’s theorem and the Bigatti-Hulett-Pardue theorem (as well as im- plying them). Given a Hilbert function H:

• There is a lex ideal attaining H.

(Macaulay’s theorem)

• The lex ideal attaining H has largest graded

Betti numbers among all ideal attaining H. (Bigatti-Hulett-Pardue theorem) (One needs to prove that the lex ideal attaining H contains the “latest” possible regular sequence, that is, if the lex ideal contains a regular sequence in degrees {a1, . . . , an}, then so does every ideal attaining H).

A slightly broader context:

• If Ω ⊂ Pn is a complete intersection of quadrics,

then any hypersurface of degree k that con- tains a subscheme Γ ⊂ Ω of degree strictly greater than 2n − 2n−k must contain Ω. [Gen- eralized Cayley-Bacharach conjecture]

• Suppose that I is an ideal containing a maxi-

mal regular sequence in degree 2 and H(R/I, d) =

ad

d

• + · · · +

a1

1

• is the d-th Macaulay representation of H(R/I, d).

Then H(R/I, d + 1) ≤

ad

d + 1

• + · · · +

a1

1 + 1

• .
• Suppose that I is an ideal containing a max-

imal regular sequence in degree 2 and L is a {2, . . . , 2} lex-plus- powers ideal such that H(R/I, d) = H(R/L, d). Then H(R/I, d + 1) ≤ H(R/Ld, d + 1).

• [EGH] Suppose that I is an ideal containing a

regular sequence in degrees A, and L is a A lex-plus-powers ideal such that H(R/I, d) = H(R/L, d). Then H(R/I, d + 1) ≤ H(R/L≤d, d + 1).

• [Generator version] Suppose that I is an ideal

containing a regular sequence in degrees A, and L is an A lex-plus-powers ideal such that H(R/I) = H(R/L). Then βI

1,j ≤ βL 1,j for all j.

• [Socle version] Suppose that I is an ideal con-

taining a regular sequence in degrees A, and L is an A lex-plus-powers ideal such that H(R/I) = H(R/L). Then βI

n,j ≤ βL n,j for all j.

• [Socle version 2] Suppose that I is an ideal

containing a regular sequence in degrees A, and L is an A lex-plus-powers ideal such that H(R/I) = H(R/L). Then βI

n,ρ+n−1 ≤ βL n,ρ+n−1

(here ρ is the regularity of H(R/L)).

βL αL

0 αL 1 . . .

αL

n

1

• . . .
• .

. . . . . . . . . . . ρ − 2

• . . .
• ρ − 1
• . . .
• ρ
• . . .

= βI αI

0 αI 1 . . .

αI

n

1

• . . .
• .

. . . . . . . . . . . ρ − 2

• . . .
• ρ − 1
• . . .
• ρ
• . . .

=

The Generalized Cayley-Bacharach conjecture is known for:

• The case for which n ≤ 7. (Eisenbud, Green,

Harris)

• The generalized hypercube, that is, the case

for which Ω consists of the 2n common zeros defined by n quadratics, each of which is a product of linear forms. (Evans, Riehl)

The Eisenbud-Green-Harris conjecture is known for:

• ideals containing the powers of the variables

[Clements, Lindstrom],

• dimension 2,
• dimension ≤ 5 if there is a maximal regular

sequence in degree 2. It is also known that minimal resolutions for lex- plus-powers ideals can be computed by:

• removing elements divisible by xai

i

for i > 1 from the Eliahou-Kervaire resolution on L,

• then iteratively using colon ideals and the map-

ping cone to introduce terms corresponding to the pure powers. (Charalambous, Evans)

Special classes of counter examples: If EGH fails, then there is an ideal I with the fol- lowing properties:

• I contains an A-regular sequence, {f1, . . . , fn},
• H(R/I) = H(R/L) where L is an A lex-plus-

powers ideal,

• βI

n,ρ+n−1 > βL n,ρ+n−1 (where ρ is the regularity

• f H(R/I)),
• I≤ρ−1 = (f1, . . . , fn)≤ρ−1,
• L≤ρ−1 = (xa1

1 , . . . , xan n )≤ρ−1.

βL αL αL

1

αL

2

. . . αL

n−1

αL

n

1

• . . .
• 1
• . . .
• .

. . . . . . . . . . . . . . . . . ρ − 2

• . . .
• ρ − 1

= ? . . . ? < ρ ? ? . . . ? = βI αI αI

1

αI

2

. . . αI

n−1

αI

n

1

• . . .
• 1
• . . .
• .

. . . . . . . . . . . . . . . . . ρ − 2

• . . .
• ρ − 1

= ? . . . ? > ρ ? ? . . . ? =

If EGH fails, then there is an ideal I with the fol- lowing properties:

• I contains an A-regular sequence,
• H(R/I) = H(R/L) where L is an A lex-plus-

powers ideal,

• βI

1,ρ+1 > βL 1,ρ+1 (where ρ is the regularity of

H(R/I)),

• βI

1,j ≤ βL 1,j for all j ≤ ρ,

• R/I is level.

βL αL αL

1

αL

2

. . . αL

n−1

αL

n

1 = = . . . = 1 ≥ ? . . . ? ≥ . . . . . . . . . . . . . . . . . . ρ − 2 ≥ ? . . . ? ≥ ρ − 1 ≥ ? . . . ? ≥ ρ < ? . . . ? = βI αI αI

1

αI

2

. . . αI

n−1

αI

n

1 = = . . . = 1 ≤ ? . . . ? . . . . . . . . . . . . . . . . . . ρ − 2 ≤ ? . . . ? ρ − 1 ≤ ? . . . ? ρ > ? . . . ? =

Using Macaulay’s inverse systems to control so- cles: Let S = k[y1, . . . , yn] be considered as an R-module where the action of xi on S is partial differentiation with respect to yi. There is a bijection between Artinian ideals I ⊆ R and finitely generated R-submodules I−1 of S where I−1 = {s ∈ S | m ◦ s = 0 for all m ∈ I}. For all d, H(I−1, d) = H(R/I, d). For all d, dim(SocR/I(d)) is equal to the number

• f minimal generators of I−1 in degree d.

Transfer the conjecture to S. Suppose that L is A lex-plus-powers, write L to denote L−1 and let MA(d) denote a monomial basis for the set {m ∈ Sd | yai

i

divides m for some i}. Then

• 1. Ld ∩ MA(d) = ∅,
• 2. if m ∈ Ld and m′ ∈ Sd − MA(d) such that

m′ < m, then m′ ∈ L. If N ⊆ Sd satisfies 1 and 2 above, we say that N is SLpp(?) with respect to A.

x4

1, x3 1x2, x3 1x3, x3 1x4, x3 1x5, x2 1x2 2, x2 1x2x3, x2 1x2x4, x2 1x2x5,

x2

1x2 3, x2 1x3x4, x2 1x3x5, x2 1x2 4, x2 1x4x5, x2 1x2 5, x1x3 2, x1x2 2x3,

x1x2

2x4, x1x2 2x5, x1x2x2 3, x1x2x3x4, x1x2x3x5, x1x2x2 4,

x1x2x4x5, x1x2x2

5, x1x3 3, x1x2 3x4, x1x2 3x5, x1x3x2 4,

x1x3x4x5, x1x3x2

5, x1x3 4, x1x2 4x5, x1x4x2 5, x1x3 5, x4 2, x3 2x3,

x3

2x4, x3 2x5, x2 2x2 3, x2 2x3x4, x2 2x3x5, x2 2x2 4, x2 2x4x5, x2 2x2 5,

x2x3

3, x2x2 3x4, x2x2 3x5, x2x3x2 4, x2x3x4x5, x2x3x2 5, x2x3 4,

x2x2

4x5, x2x4x2 5, x2x3 5, x4 3, x3 3x4, x3 3x5, x2 3x2 4, x2 3x4x5, x2 3x2 5,

x3x3

4, x3x2 4x5, x3x4x2 5, x3x3 5, x4 4, x3 4x5, x2 4x2 5, x4x3 5, x4 5

x3

1, x2 1x2, x2 1x3, x2 1x4, x2 1x5, x1x2 2, x1x2x3, x1x2x4, x1x2x5,

x1x2

3, x1x3x4, x1x3x5, x1x2 4, x1x4x5, x1x2 5, x3 2, x2 2x3, x2 2x4,

x2

2x5, x2x2 3, x2x3x4, x2x3x5, x2x2 4, x2x4x5, x2x2 5, x3 3, x2 3x4,

x2

3x5, x3x2 4, x3x4x5, x3x2 5, x3 4, x2 4x5, x4x2 5, x3 5

x2

1, x1x2, x1x3, x1x4, x1x5, x2 2, x2x3, x2x4, x2x5, x2 3

x3x4, x3x5, x2

4, x4x5, x2 5

x1, x2, x3, x4, x5

Another fact: suppose that I contains an A-regular sequence {f1, . . . , fn}, and write Mf(d) to denote a basis for the set {

q

• j−1

yα1,j

1

· · · yαn,j

n

∈ Sd |

q

• j=1

xα1,j

1

· · · xαn,j

n

∈ (f1, . . . , fn)d}. Then Id ∩ Mf(d) = ∅.

EGH is equivalent to the following: Suppose

• {f1, . . . , fn} is an A-regular sequence,
• I ⊆ S is such that I∩Mf(d) = ∅ and the largest

degree in which I has generators is d,

• L ⊆ S is SLpp with respect to A and the largest

degree in which L has generators is d,

• H(L, d) = H(I, d),

Then H(L≤d, d − 1) ≤ H(I, d − 1).

Hope: this approach will provide a non-combinatorial proof of EGH for monomial ideals (at least in di- mension 3). It does give a different perspective on how the degrees of the regular sequence affect the problem.