LECTURES ON MULTIPLICITY CONJECTURES

HEMA SRINIVASAN

1. Introduction

In these lectures, we will start from elementary commutative algebra to outline

the multiplicity conjectures and their proofs.

The multiplicity conjecture, as it has come to be known, was born when I visited

Purdue University in Fall of 1992. While there, I looked at a paper by Craig Huneke

and Matthew Miller in the Canadian Journal of Math where they prove that if I

is a perfect ideal of height h in a polynomial ring R with a pure resolution, then

the multiplicity of R/I is the product of the shifts in the resolution divided by

h!. The proof uses the equations satisfied by the shifts proved earlier in a paper

by Peskine and Szpiro and the residue theorem. I found a different (and better in

my mind) proof using simple linear algebra, by computing a determinant in two

different expansions, which yielded the result. Inspired by this proof, Craig Huneke

and I thought that there may be a bound for the multiplicity by the product of the

minimal and maximal shifts when it no longer has a pure resolution. For the sake

of completeness, I give the proof of the pure case is CorollaryCorPure5.13. We checked or

rather proved it in the codimension two Cohen Macaulay (CM) case. We quickly

realized that the lower bound does not hold if it is not Cohen Macaulay and that

the bounds itself were quite tight from examples worked out using the computer

program Macaualay.

Later, Jurgen Herzog and I set out to work on this and we proved it in many

different cases. These appear in the first paper on multiplicity conjecturesHS1[11]

which appeared in 1998. We proved it for quasi pure resolutions and also for all

stable monomial ideals and square free strongly stable ideals. Further, we also

proved it for complete intersections. For the non Cohen Macaulay case where the

lower bound does not necessarily hold, we could find no examples where the upper

bound does not hold. So, we conjectured that the upper bound holds in general.

We proved it whenever the resolution is q−linear and also when it is not pure but

almost pure; that is, the purity is disturbed in only one spot in the resolution. Since

then there have been several directions, ideas, special cases and attempts to settle

the conjecture. Further, as part of the proof, we observed that if the resolution can

be made to look quasi pure by canceling shifts that occur with opposite signs, it

remains true. We called something a ”virtual resolution” if it is a resolution which

satisfies the equational conditions for the shifts whether or not it is a resolution of1

an existing ideal. We call something numerically pure (or quasi pure or linear) if it

equals a virtual resolution which is pure (or quasi pure or linear) after cancellation.

This extends the proof to numerically quasi pure resolutions.

In 2003, Boij and SoderbergBS1[1] studied the problem and made stronger, natural,

conjectures which would imply our conjectured bounds. In fact, they conjectured

that every resolution can be written as the sum of positive rational multiples of pure

resolutions. They also conjectured that every set of positive integers satisfying the

equations come from a pure resolution. In 2008 these conjectures were proved by

Eisenbud, Schreyer and Weyman for CM modules in characteristic zero and extended

to non CM modules by Boij and Soderberg.

In these lectures, I will work over polynomial rings, as any extension to standard

graded algebras is immediate from the definitions. We will introduce the definitions

and results on graded modules, graded resolutions and Hilbert functions that are

needed to set up the problem on multiplicity and the conjectures. I have attempted

to make these self contained with some references. I have not included all the

partial results associated with the conjectures. The main thrust is the beginning of

the Huneke-Herzog-Srinivasan conjectures and how it led to the Boij and Soderberg

conjectures and the final solution of these conjectures.

For a complete bibliography of the articles on multiplicity conjectures, till 2008,

please refer toFS[8]. The important papers containing the proofs of the Boij Soderberg

conjectures and the multiplicity conjecturesES[6],

BS2[2] are posted on the Arxiv and are

available on the web.

2. Preliminaries - Graded Modules and Graded Resolutions.

In this section we will establish notation and some basic facts in commutative

algebra that we need.

Let k be a field, and R be a standard graded k-algebra; that is, R =⊕

i≥0Ri,

with R0 = k and R is generated as an algebra by R1. For instance, R can be the

ring of polynomials over a field k in n variables. For all practical purposes, we might

as well take R to be a polynomial ring.

Let I be a homogeneous ideal of R. A = R/I is then a standard graded k-algebra.

Through out these talks, all modules will be graded modules and all ideals will

be homogeneous ideals. R will always denote a standard graded k-algebra.

If M is a graded R module, then M =⊕

i≥0Mi and RiMj ⊂ Mi+j . For two

finitely generated graded R modules, M and N , a homomorphism ϕ : M → N is

called graded of degree d if ϕ(Mi) ⊂ Ni+d for all i.

Theorem 2.1. If M and N are finitely generated graded R modules and ϕ : M → N

is a graded homomorphism, then kerϕ and cokerϕ are both finitely generated graded

R modules.2

Define M(−d) to be the graded R module, whose ith graded piece is Mi−d. Thus,

as a set M(−d) is the same as M but the we will add d to the degree of every

element in M .

Example 2.2. Let R = k[x, y]. Consider the map ϕ : R2 → R given by ϕ(a, b) =

ax2+by3. Then ϕ is not homogeneous since deg(ϕ(f, 0)) = degf+2 and deg(ϕ(0, g)) =

deg(g) + 3. However, ϕ : R(−2)⊕

R(−3)→ R given by ϕ(f, g) = fx2 + gy3 is ho-

mogeneous of degree zero.

A free resolution F of an R module M is an exact complex of free R modules

· · ·Fnδn→· · · → Fi

δi→· · · → F1δ1→F0 →M → 0.

The resolution F is said to be of length n if Fi = 0, i > n. The projective dimension

of M over R is the minimal length of a resolution of M over R. If R is a polynomial

ring over a field k in n variables, then R is a regular ring and hence all modules M

have projective dimension less than or equal to n.

Lemma 2.3. Suppose M is a finitely generated graded R-module. Then M has a

graded free resolution over R.

Proof. Let f1, · · · , ft be generators in degrees d1, · · · , dt respectively. Thus,

M = Rf1

⊕Rf2

⊕· · ·⊕

Rfn.

Then ϕ0 :⊕t

i=0R(−di)→M given by ϕ0(ei) = fi is a graded surjective homomor-

phism. Thus, its kernel is a finitely generated graded R module. So, there exists a

graded map ϕ1 :⊕b2

i=0R(−d1j) → K → 0. Continuing this, we get a graded free

resolution of M . �

A map of free modules can be represented by a matrix with entries in R after

a choice of a basis for the free modules. If ψ :⊕t

i=1R(−ai) →⊕s

j=1R(−bj) is a

graded map of degree zero, then by choosing a basis for the free R modules, we

represent ψ by a t× s matrix of polynomials such that the entry in the i-th row and

j-th column is homogeneous of degree ai − bj.A free resolution F of M is called minimal if δi(Fi) ⊂ R1Fi−1; that is, after a

choice of basis, the non zero entries in all the matrices representing the δi’s are all

of degree at least 1. This is because, if in the resolution F, δi : Fi → Fi−1 is such

that δi(Fi) is not contained in R1Fi−1 for some i, then we can eliminate at least one

row and one column and decrease the ranks of Fi and Fi−1 by 1 giving a resolution

which has smaller rank for the free modules.

Example 2.4. Let ψ : Rt =⊕t

i=1R(−ai)→ Rs =⊕s

j=1R(−bj) be a map. Then ψ

can be represented by a t× s matrix (ψij)t×s such that ψij is homogenous of degree

ai − bj and if ai < bj, then ψij = 0. If one of the entries ψij is not in the ideal

generated by (x1, · · · , xn), then it is of degree zero and hence a unit and ai = bj.3

Without loss of generality, we may take it to be ψ11, so that a1 = b1 = a. Then the

matrix looks likea ψ12 · · · ψ1t

ψ21 .. . . . .....

......

...

ψs1 .. . . . ..

By performing row and column operations, which amounts to making a change of

basis for Rt and Rs, we can make the matrix have 1 on the top left and zeros in the

first row and column after that. Then the map ψ becomes

ψ = id⊕ ψ′ : R⊕

Rt−1 → R⊕

Rs−1

Thus, if ψ is a map in a free resolution of M , we can replace ψ,Rt, and Rs by ψ′,

Rt−1, and Rs−1 respectively, to get another free resolution which is ”smaller”. This

justifies the name that the minimal resolution is one in which all the entries of the

matrices are in the maximal ideal (x1, · · ·xn).

Recall that a map of complexes is a map Φ = ⊕ϕi : F =⊕

Fi → G =⊕

Gi such

that all the squares commute. An isomorphism of complexes is a map Φ which is

an isomorphism for each ϕi : Fi → Gi.

Theorem 2.5. The minimal homogeneous resolution of a graded module M is

unique up to isomorphism.

Theorem 2.6. Suppose M is a graded R module. Then there exist unique positive

integers βij such that

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)→M → 0

is the minimal resolution of M .

The length of the minimal resolution of M is the projective dimension of M ,

pd(M), the minimal length of a resoution of M by free (projective) R-modules.

3. Preliminaries: Hilbert Functions

Let M be a graded R module. Then M =⊕

Mi. The dimension of Mi as a vector

space over the field k is the value of the Hilbert function, denoted by H(M, i).

The Hilbert series of M is defined by

HM(x) =∑

H(M, i)xi.

This is a power series in x. We list a few important facts.

(1) H(M, i) is a polynomial for large i and this polynomial is of degree dim M−1

and is called the Hilbert polynomial PM(i), of M .4

(2) The multiplicity of M , denoted by e(M) is the leading coefficient of the

Hilbert polynomial times (dim M − 1)!.

(3) The Hilbert function and Hilbert series are additive on exact sequences.

Example 3.1. Let R = k[x1, · · · , xn].

dimkRi =

(n+ i− 1

i

)=

(n+ i− 1

n− 1

)=i(i+ 1) · · · (i+ n− 1)

(n− 1)!

is a polynomial of degree n− 1. The multiplicity e(R) = 1.

HR(x) =∑(

n+ i− 1

i

)xi = (1− x)−n.

Let M be a graded module of dimension d. Since H(M, i) = PM(i) is of degree

d-1 for large i,

(1− x)d∑

H(M, i)xi = Q(x) ∈ Z[x]

and Q(1) 6= 0.

Theorem 3.2. Let M be a graded R module with a graded resolution, F :

resMresM (1)

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)→M → 0

Let BM(x) =∑

i,j(−1)iβijxi be the Betti series of M .

Then, HM(x) = BM (x)(1−x)n where R = k[x1, · · ·xn].

Proof. Since the Hilbert Function is additive on sequences,

HM(x) =∑

(−1)i∑βijHR(−j)(x)

=∑

(−1)i∑βijx

jHR(x)

=∑

(−1)i∑βijx

j(1− x)n

= BM (x)(1−x)n .

�

betti1 Corollary 3.3. Suppose M is a finitely generated graded R module of dimension d.

Then (1− x)n−d divides BM(x) and if Q(x) = BM (x)(1−x)n−d , then Q(1) = e(M).

Proof. Since HM(t) = PM(t) is a polynomial of degree d − 1 for large enough t,

HM(x) = A(x) +∑

t>r PM(t)xt for some r.∑PM(t)xt(1 − x) =

∑(PM(t) −

PM(t − 1))xt =∑P ′M(t)xt, where P ′M(t) is the derivative with respect to t. Thus,∑

PM(t)xt(1− x)d−1 =∑

(d− 1)!P (1)xt. Hence, we get HM(t)(1− x)d = A(x)(1−x)d+e(M)

∑t>r x

t(1−x) = Q(x). Further, Q(1) = e(M) 6= 0. Combining this with

the above theorem, we see that Q(x) = HM(x)(1− x)d = BM (x)(1−x)n (1− x)d = BM (x)

(1−x)n−d

as desired. �5

CIExam Example 3.4. Suppose f1, · · · , fh is a regular sequence in R and I is the ideal

generated by f1, · · · fh. Then the multiplicity of R/I is∏h

i=1 deg(fi).

This is because we can consider the exact sequence,

0→ R/(f1, · · · , ft−1)(−degft)→ R/(f1, · · · , ft−1)→ R/(f1, · · · , ft), for t = 1, . . . , h−1.

Thus,

H(R/f1, · · · , ft) = H(R/f1, · · · , ft−1)xdegftH(R/f1 · · · , ft−1) = (1−xdegft)H(R/f1, · · · , ft−1).

Then

H(R/f1, · · · , fh) =∏

(1− xdegfi)H(R) =

∏hi=1(1− xdegfi)

(1− x)n= Q(x)/(1− x)n−h,

for dim R/I = n− h. Now, Q(1) = e(R/I). So,

Q(x) =

∏hi=1(1− xdegfi)

(1− x)h=

h∏i=1

(1 + x+ · · ·+ xdegfi−1).

So, Q(1) =∏h

i=1 degfi.

4. Multiplicity Conjectures

Throughout this section, M is a finitely generated graded R-module with a free

resolution

F : 0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)→M → 0

Recall that M is Cohen Macaulay (CM) if s = codim M = n− dim M .resMM

If I is a homogeneous ideal of R, and M = R/I, then β0j = 0 for j 6= 0 and

β00 = 1. Thus the resolution looks like

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→R→ R/I → 0.

resI

Thus R/I is CM if s = h = height(I).

The multiplicity conjectures are as follows:

HSconj Conjecture 4.1. (Herzog, Huneke and SrinivasanHS1[11]) Let R/I have the minimal

resolutionresI4 as above and let e(R/I) be its multiplicity. Then,

• If R/I is Cohen Macaulay (CM) of codimension h, then

1

h!

h∏i=1

mi ≤ e(R/I) ≤ 1

h!

h∏i=1

mi.

6

• If R/I is not Cohen-Macaulay, then only the upper bound may hold; that is,

e(R/I) ≤ 1

h!

h∏i=1

mi.

The reason for restricting only to the upper bound in the non CM case is that

when R/I is not CM, it is easy to have the lower bound fail. Later on we will see

examples and reasons for this.

The conjectures as stated have a natural generalization to modules and the con-

jectures for the modules are stated inMN[14].

We will develop the notion of the conjecture, the proofs and the history in the

next section.

Now, one of the consequences of the conjectures is that if mi = mi, for 1 ≤ i ≤ h,

we get equality in the formulas of the conjectures. Such resolutions are called pure

resolutions.

Definition 4.2. M is said to have a pure resolution of type d0, d1, d2, . . . , ds if the

minimal resolution of M is

0→ Rbs(−ds)δs→· · · → Rbi(−di)

δi→· · · → Rb1j (−d1)δ1→Rb0j (−d0)→M → 0.

This is a theorem of Huneke and Miller which predates the conjecture.

Theorem 4.3.HM[13] Let M = R/I be Cohen Macaulay with a pure resolution given

by

0→ Rbh(−dh)δs→· · · → Rbi(−di)

δi→· · · → Rb1j (−d1)δ1→R→ R/I → 0.

Then the multiplicity of R/I is 1h!

∏hi=1 di.

While giving a different proof of this interesting formula, Huneke and Srinivasan

came up with conjectureHSconj4.1 in 1993 which was later improved upon and investigated

by Herzog, Srinivasan, Migliore, Nagel, Roemer, Boij and Soderberg, leading to the

final proofs by Eisenbud, Schreyer, Weyman, Boij and Soderberg in 2008.

Theorem 4.4. Suppose I = (f1, · · · , fh) is a complete intersection, that is, gener-

ated by a regular sequence f1, · · · , fh in R with the degree of fi being di. Then the

multiplicity conjecture holds for R/I.

Proof. If I = (f1, · · · , fh) is generated by a regular sequence, then from ExampleCIExam3.4, the multiplicity of R/I is

∏hi=1 di. The minimal resolution of R/I is given by

the Koszul complex associated to the regular sequence. By arranging the generators

fi in the ascending order of their degrees, we may take, d1 ≤ d2 ≤ · · · ≤ dh. Thus,

the resolution looks like

0→ R(−h∑t=1

dt)δh→· · · →

⊕1≤j1<···<ji≤h

R(−dj1−· · ·−dji)δi→· · · →

h⊕t=1

R(−dt)δ1→R→ R/I → 0.

7

Thus the minimal and maximal shifts are:

mi = d1 + · · ·+ di ≤ idi and mi = dh + dh−1 + · · · dh−i+1 ≥ idh−i+1.

So,h∏i=1

mi ≤h∏i=1

idi = h!h∏i=1

di = h!e(R/I) =h∏i=1

i(h− i+ 1) ≤h∏i=1

mi.

�

5. What do the shifts say?quasipure

In this section, we will look at the information that can be gleaned from the

shifts in the resolution of an R-module. The shifts in the resolution satisfy certain

equations.

PSModule Theorem 5.1.PS[15],

HK[10] Let M be a graded R-module with a free resolution as

follows:

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)→M → 0.

Then

**** (2)n∑i=0

∑j>0

(−1)iβijjt =

−β00 if t = 0,

0 if 1 ≤ t ≤ h = n− d(−1)hh!e(M) if t = h

Proof. Since R is of dimension n and dim M = d, the codimension of M is h = n−d.

By Corollarybetti13.3, we see that BM(x) =

∑i,j(−1)iβijx

j = (1− x)hQ(x), where Q(x)

is a power series with Q(1) = e(M). By evaluating this equation at x = 1, we get∑i,j βij = 0. This implies

∑ni=0

∑j>0(−1)iβijj

0 = −β00. By differentiating this

h − 1 times and evaluating at x = 1, we see that the right hand side is zero since

it is a multiple of (1 − x)h. So,∑n

i=0

∑j>0(−1)iβijj(j − 1) · · · (j − t + 1) = 0, for

1 ≤ t < h. This means,∑n

i=0

∑j>0(−1)iβijj

t = 0, for 0 < t < h.

Finally, when we differentiate h times and set x = 1, we get a non zero number,

(−1)hh!Q(1). Thusn∑i=0

∑j>0

(−1)iβijjh = (−1)hh!e(M).

Remark 5.2. The above equations are valid whether or not the resolution is min-

imal. All it requires is that the resolution is graded. In fact, for any given graded

resolution it is true that there exists a positive integer h such that it is the smallest

positive integer t for which the alternating sum∑

i,j(−1)iβijjt is not zero. One can

take this to be the height or the codimension of the resolution. Further, in this way,8

for any resolution F, we can define e(F) In fact, we can speak of the codimension,

perfection, the property of being CM and multiplicity for a resolution without refer-

ence to the module that it resolves. If a resolution F has the same codimension as

it’s length we can call it perfect, for in that case its dual Hom(F, R) is exact and F

is the resolution of a CM module.

Definition 5.3. Suppose that R is a polynomial ring (or R can be any standard

graded algebra). Let

F : 0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)

be a finite exact complex of free graded R modules with graded maps of degree zero.

Let h be the first positive integer such that∑

i

∑j(−1)iβijj

h 6= 0. Then we define

the ”multiplicity” of the complex to be

e(F) =

∑i

∑j(−1)iβijj

h

h!.

The point of the above theoremPSModule5.1 is that this number depends only on the zero-th

homology of F.

Remark 5.4. Since the equationsn∑i=0

∑j>0

(−1)iβijjt =

{−β00 if t = 0,

0 if 1 ≤ t ≤ h = n− d

remain valid when we replace βij with βij + a for any positive integer a, we can

simply take the minimal degree of a generator of M to be 0 if desired.

PSIDEAL Theorem 5.5. Let I be a homogeneous ideal of height h in R and R/I have a

resolution over R given by

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→R→ R/I → 0

Thens∑i=1

∑j

(−1)iβijjt =

−1 if t = 0,

0 if 1 ≤ t ≤ h− 1,

(−1)hh!e(R/I) if t = h

Remark 5.6. These equations do not determine the modules in the sense that we

can have numbers that satisfy these equations without there being a module with

precisely these shifts in the resolution. For instance, when R/I is Cohen Macaulay

of codimension 2, we have by Hilbert Burch the structure of the resolution as the

ideal of n×n minors of an n× (n+ 1) matrix. This is true for perfect homogeneous

ideals I of height 2; a matrix can be found to generate n+1 homogeneous generators9

for the ideal I. This fact can be used to prove the multiplicity conjectureHSconj4.1 in

codimension 2. However, we can pick (n+1) numbers ai, and n numbers bi such

that∑ai =

∑bi but

∑b2i −

∑a2i is not bounded by (min ai)(min bi) from below

or (max ai)(max bi) from above. For instance, take n = 2, ai = 2, b1 = 1, b2 = 5.

However, it is easy to see that this will not be a Betti diagram, for b1 < ai for all i

and that is not possible.

We will use these equations to get a proof for the multiplicity conjecture in the

quasi pure case. A graded resolution is called quasi pure if mi ≥ mi−1 for all i.

Definition 5.7. For any set of integers a1, a2, · · · , at, V (a1, a2, · · · , at) denotes the

Vandermonde determinant of the t × t matrix whose (i, j)-th entry is aj−1i . Thus

V (a1 · · · , at) =∏

1≤i<j≤t(aj − ai). To simplify notation, we will sometimes write

V (A) to mean the Vandermonde determinant V (a1, a2, · · · , at) for A = (a1 · · · , at) ∈Zt.

Remark 5.8. • The Vandermonde determininant V (a1, · · · , at) ≥ 0 if a1 ≤a2 · · · at.• If a resolution is quasi pure, with mi and mi as minimal and maximal shifts,

then V (j0, j1, · · · , jt) ≥ 0, whenever mi ≤ ji ≤ mi. In fact, as we will see

later in the proof of the quasi pure case, quasi pure is defined precisely to

make this happen.

We take βij = 0 if j < mi or j > mi.

The next lemma shows how multiplicity is a weighted sum.

multiplicity Lemma 5.9. Let R/I be Cohen Macaulay of codimension h. Then

(1)

e(R/I) =

∑(j1,···jh)∈Zh

∏hi=1 ji[

∏βijiV (j1, · · · jh)]∑

(j1,···jh)∈Zh

∏hi=1 βijiV (j1, · · · jh)

.

(2) Further,

e(R/I) =

∣∣∣∣∣∣∣∣∣

∑j β1jj · · ·

∑j βhjj∑

j β1jj2 · · ·

∑j βhjj

2

... · · · ...∑j β1jj

h · · ·∑

j βhjjh

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

∑j β1j · · ·

∑j βhj∑

j β1jj · · ·∑

j βhjj... · · · ...∑

j β1jjh−1 · · ·

∑j βhjj

h−1

∣∣∣∣∣∣∣∣∣

.

10

Proof. Let I be a perfect homogeneous of height h and R/I have the minimal res-

olution consisting of shifts βij. Let e = e(R/I). Consider the determinant Q given

by

Q =

∣∣∣∣∣∣∣∣∣

∑j β1jj · · ·

∑j βhjj∑

j β1jj2 · · ·

∑j βhjj

2

... · · · ...∑j β1jj

h · · ·∑

j βhjjh

∣∣∣∣∣∣∣∣∣ .By performing the row operation of replacing the first column by the alternating

sum of the columns and using the equations satisfied by the shifts in theoremPSIDEAL5.5,

we see that the determinant Q equals,

Q =

∣∣∣∣∣∣∣∣∣0 · · ·

∑j βhjj

0 · · ·∑

j βhjj2

... · · · ...

(−1)h+1h!e · · ·∑

j βhjjh

∣∣∣∣∣∣∣∣∣ = h!e

∣∣∣∣∣∣∣∣∣

∑j β2jj · · ·

∑j βhjj∑

j β2jj2 · · ·

∑j βhjj

2

... · · · ...∑j β2jj

h−1 · · ·∑

j βhjjh−1

∣∣∣∣∣∣∣∣∣ = h!eQ1.

Now the determinant Q1 is precisely∣∣∣∣∣∣∣∣∣

∑j β2jj · · ·

∑j βhjj∑

j β2jj2 · · ·

∑j βhjj

2

... · · · ...∑j β2jj

h−1 · · ·∑

j βhjjh−1

∣∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣

∑j β1j · · ·

∑j βhj∑

j β2jj · · ·∑

j βhjj... · · · ...∑

j β2jjh−1 · · ·

∑j βhjj

h−1

∣∣∣∣∣∣∣∣∣ = Q1,

as is computed by performing the same column operation to the second matrix by

replacing it with the alternating sum of columns and using the equations satisfied

by the shifts.

Thus we get the multiplicity e as the quotient of the two determinants,

h!e =Q

Q1

as desired in (3).

However, by computing the determinant Q directly, using multilinearity we see

that

Q =∑

mi≤ji≤mi

∣∣∣∣∣∣∣∣∣β1j1j1 · · · βhjhjhβ1j1j

21 · · · βhjhj

2h

... · · · ...

β1j1jh1 · · · βhjhj

hh

∣∣∣∣∣∣∣∣∣ =∑

mi≤ji≤mi

h∏i=1

jiβijiV (j1, · · · , jh).

Now the determinant Q1 becomes,

Q1 =

∣∣∣∣∣∣∣∣∣

∑j β2jj · · ·

∑j βhjj∑

j β2jj2 · · ·

∑j βhjj

2

... · · · ...∑j β2jj

h−1 · · ·∑

j βhjjh−1

∣∣∣∣∣∣∣∣∣11

=

∣∣∣∣∣∣∣∣∣

∑j β1j · · ·

∑j βhj∑

j β2jj · · ·∑

j βhjj... · · · ...∑

j β2jjh−1 · · ·

∑j βhjj

h−1

∣∣∣∣∣∣∣∣∣=

∑mi≤ji≤mi

∣∣∣∣∣∣∣∣∣β1j1 · · · βhjhβ1j1j1 · · · βhjhjh

... · · · ...

β1j1jh−11 · · · βhjhj

h−1h

∣∣∣∣∣∣∣∣∣ =∑

mi≤ji≤mi

h∏i=1

βijiV (j1, · · · , jh).

Thus, the multiplicity is,

h!e(R/I) =Q

Q1

=

∑(j1,···jh)∈Zh

∏hi=1 ji[

∏βijiV (j1, · · · jh)]∑

(j1,···jh)∈Zh

∏hi=1 βijiV (j1, · · · jh)

.

Remark 5.10. To make the weighted sum clear, we set for every A = (a1, · · · , ah) ∈Zh, σA =

∏i ai. We also let βij = 0, if j < mi or j > mi. Let β(A) =

∏hi=1 βiai

.

Then,

e(R/I) =

∑A∈Zh(σA

h!)β(A)V (A)∑

A∈Zh β(A)V (A).

The sum is a finite sum because β(A) = 0 for all but finitely many A ∈ Zh.

There is a similar expression for the multiplicity of a module too. The proof is

along the same lines, except we consider the (h+ 1)× (h+ 1) matrix

Q =

∣∣∣∣∣∣∣∣∣∣∣

∑i β0j · · ·

∑j βhj∑

j β0jj · · ·∑

j βhjj∑j βijj

2 · · ·∑

j βhjj2

... · · · ...∑j β1jj

h · · ·∑

j βhjjh

∣∣∣∣∣∣∣∣∣∣∣and compute its determinant using the equations in theorem

PSModule5.1 and then directly

using multilinearity. Comparing the two, gives the expression for the multiplicity.

We state it as a lemma below.

Lemma 5.11. Let M be a Cohen Macaulay R-module with a resolutionresM1 of length

h = codim M . Then the multiplicity e(M) is given by

e(M) =

∑hi=0

∑mi≤ji≤mi

∏βijiV (j0, · · · , jh)∑h

i=1

∑mi≤ji≤mi

∏βijiV (j1, · · · , jh)

=

∑A∈Zh σAβ(A)V (A)∑A∈Zh β(A)V (A)

,

where for A = (a1, · · · , an), σA =∑m0

a=m0β0a

∏(ai − a) and β(A) =

∏βiai

.12

Now we use these formulas to prove the multiplicity conjecture for the quasi pure

case.

Theorem 5.12. Let I be a perfect ideal of height h with a quasi pure minimal graded

resolution. Then the multiplicity conjecture holds for R/I.

Proof. By theoremmultiplicity5.9, we see that

e(R/I) =

∑(j1,··· ,jh)∈Zh

∏hi=1 ji[

∏βijiV (j1, · · · , jh)]∑

(j1,···jh)∈Zh

∏hi=1 βijiV (j1, · · · , jh)

.

So,

e(R/I) =

∑mi≤ji≤mi

∏hi=1 ji[

∏βijiV (j1, · · · , jh)]∑

mi≤ji≤mi

∏hi=1 βijiV (j1, · · · , jh)

.

Since the resolution of R/I is quasi pure, we see that V (j1, j2, · · · , jh) ≥ 0 in

this sum. Thus, we can estimate it by the maximum and minimum value of∏

i jiwhich are

∏imi and

∏imi respectively. Thus,

∏imi ≤ h!e ≤

∏imi. This gives

the conjectured bounds on the multiplicity. �

If the resolution is pure, mi = mi = di for all i and βidi= bi. Thus in the

expression for the multiplicity from lemmamultiplicity5.9

h!e(R/I) =

∑mi≤ji≤mi

∏hi=1 ji[

∏βijiV (j1, · · · , jn)]∑

mi≤ji≤mi

∏hi=1 βijiV (j1, · · · , jh)

,

there is only one sum. Thus h!e(R/I) =Q

i diQ

i biV (d1,··· ,dh)Qi biV (d1,··· ,dh)

=∏

i di. So, we have

CorPure Corollary 5.13.HM[13] If R/I is Cohen Macaulay with a pure resolution,

0→ Rbh(−dh)→ · · · → Rb1(−d1)→ R→ R/I,

then the multiplicity of R/I isQ

i di

h!.

Proof. We will give a complete proof. Let R/I be CM with a pure resolution

0→ Rbh(−dh)→ · · · → Rb1(−d1)→ R→ R/I.

Consider the determinant

Q =

∣∣∣∣∣∣∣b1d1 b2d2 · · · bhdh

......

......

b1dh1 b2d

h2 · · · bhd

hh

∣∣∣∣∣∣∣ =∏i

di

∣∣∣∣∣∣∣b1 b2 · · · bh...

......

...

b1dh−11 b2d

h−12 · · · bhd

h−1h

∣∣∣∣∣∣∣ .13

Now we replace the last column by the alternating sums of the columns,∑

i(−1)iCi =

Ch, to see that

Q =∏i

di(−1)h

∣∣∣∣∣∣∣b1 b2 · · ·

∑i(−1)ibi

......

......

b1dh−11 b2d

h−12 · · ·

∑i(−1)ibid

h−1i

∣∣∣∣∣∣∣ .By applying the equations∑

i(−1)ibi = −1∑i(−1)ibid

ti = 0 for 1 ≤ t < h∑

i(−1)ibidhi = (−1)hh!e

we get

Q =∏i

di

∣∣∣∣∣∣∣b1d1 b2d2 · · · bh−1dh−1

......

......

b1dh−11 b2d

h−12 · · · bh−1d

h−1h−1

∣∣∣∣∣∣∣ .On the other hand, we can replace the last column by the alternating sums of the

columns∑

i(−1)iCi = Ch. Then we get the determinant

Q = (−1)h

∣∣∣∣∣∣∣b1d1 b2d2 · · ·

∑i(−1)ibidi

......

......

b1dh1 b2d

h2 · · ·

∑i(−1)ibid

hi

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣b1d1 b2d2 · · · 0

......

... 0

b1dh1 b2d

h2 · · · h!e

∣∣∣∣∣∣∣ = h!e

∣∣∣∣∣∣∣b1d1 b2d2 · · · bh−1dh−1

......

......

b1dh1 b2d

h2 · · · bh−1d

h−1h

∣∣∣∣∣∣∣ .Comparing the two expressions for Q and noting that the determinant∣∣∣∣∣∣∣

b1d1 b2d2 · · · bh−1dh−1...

......

...

b1dh1 b2d

h2 · · · bh−1d

h−1h

∣∣∣∣∣∣∣ =∏i

bi∏i

diV (d1, · · · dh−1) 6= 0

because d1 < d2 · · · dh, we get that h!e =∏i = 1hdi as desired. �

Remark 5.14. We note that if the resolution is quasi pure but not pure, then there

is at least one i for which mi 6= mi. Then there are more than one (j1, · · · , jh) and

the product of one of these is strictly bigger than the other. Hence we cannot have

equality for the multiplicity with either the minimum or the maximum value. Thus

we see that if R/I is Cohen Macaulay with a quasi pure resolution, the equality for

the upper or the lower bound for the multiplicity implies that the resolution is pure.

In fact there is an addition to the conjecture to this effect by Herzog and Zheng that14

the equality holds at either end if only if R/I is CM with pure resolution. The proof

for the quasi pure case above demonstrates it for CM quasi pure resolutions.

M. Boij and J. Soderberg observed that if the resolution F is the sum of two graded

resolutions G1 and G2, say of same codimension, then the multiplicity e(R/I) is cal-

culated as e(F) = e(G1) + e(G2). Then the bounds conjectured for the multiplicity

by Huneke and Srinivasan for the CM case would be true if we can write all CM

(perfect) resolutions as the sum of CM quasi pure resolutions. In fact, they conjec-

tured,

Conjecture 5.15. (Boij and SoderbergBS1[1])

• 1. The minimal graded resolution of a graded module M can be written as

the sum of positive rational multiples of pure resolutions.

• 2. For every sequence d = (d1 < · · · < dt) of positive integers, there is a CM

module with a pure resolution given by the shifts d.

Thus our discussion above says that

Theorem 5.16. The Boij and Soderberg conjecture (1) implies the multiplicity con-

jecture.

6. Some Free Resolutions

In this section, we will briefly outline some of the standard results on free resolu-

tions.

Let a1, · · · an be a set of elements in R and I = (a1, · · · , an) be the ideal generated

by the a′is. They may not be minimal generators. The Koszul complex associated

to the sequence a1, · · · , an, denoted by K = K(a1, · · · , an), is as follows:

Let F =∑n

i=1Rei be a free R module of rank n.

K = 0→n∧Fδn→

n−1∧Fδn−1→ · · · →

i∧F

δi→i−1∧

Fδi−1→ · · · → F

δ1→R→ R/(a1, · · · an)→ 0,

where the maps δi :∧Fi →

∧Fi−1 are given by δi(ej1∧· · ·∧eji) =

∑it=1(−1)t−1at(· · ·∧

et−1 ∧ et+1 · · · ).This complex K is exact if and only if the sequence (a1, · · · , an) is a regular

sequence or equivalently, the ideal (a1, · · · , an) = I is of height n.

If a1, · · · , an are all homogeneous with degree ai = di, then K is the graded

complex with shifts∧i F =

∑1≤j1<j2···ji≤nR(−(dj1 + · · ·+ dji)). Hence the minimal

and maximal shifts are

mi =i∑

j=1

dj and mi =n∑

j=n−i+1

dj.

The multiplicity of R/I is∏di, if a− 1, . . . , an form a regular sequence.

15

To verify the conjectured bounds, we note that mi =∑i

j=1 dj ≤ idi and mi =∑nj=n−i+1 dj ≥ idn−i+1. So,

∏mi ≤

∏idi ≤ n!

∏di = dn2dn−1 · · ·nd1 =

∏idn−i+1 ≤

∏mi.

Thus the multiplicity bounds are natural for the Koszul complex and like the

shifts in the Koszul complex this inequality also depends only on the numbers di.

In fact, if di = d for all i, we get the pure resolution with equality everywhere. As

a matter of fact, the Koszul complex is pure if and only if di = d for all i.

An easy way to construct resolutions and betti diagrams to do some of the com-

putations suggested in these lectures is to look at the koszul complex with various

d′is. The multiplicity conjecture or the bounds are trivially true in these cases since

we know the actual multiplicity as the product of the di’s. However, they are still

helpful in making the computations. For instance, with a Koszul complex generated

by two quadratic and one quintic will have the betti diagram

K(2, 2, 5) =

1 − − −− 2 − −− − 1 −− − − −− 1 − −− − 2 −− − − 1

Here K(a, b, c) represent the betti diagram of the Koszul complex associated with

three generators in degrees a, b and c. The minimal shifts are (0,2,4,9) and the max-

imal shifts are (0,5,7,9). To illustrate Boij-Soderberg conjectures, let us write this as

the sum of pure sequences. Let P (0, 2, 4, 9) represent the normalized betti diagram

with the pure sequence (0,2,4,9). That is the unique betti diagram associated with

a pure sequence with b0 = 1.

Then K(2, 2, 5) = (5/9)P (0, 2, 4, 9) + (20/63)P (0, 2, 7, 9) + (8/63)P (0, 5, 7, 9).

Now we know the minimal resolution of the ideal of the n×n minors of an m×nmatrix with m ≥ n given by the Eagon-Northcott complex. Again, if the minors

are homogeneous, that is the entries xij of the m × n matrix are homogeneous of

degree ri + sj for nonnegative integers 1 ≤ r1 ≤ · · · ≤ rm and s1 ≤ · · · ≤ sn, then

the minors will be of different degrees. The resolution is pure and linear if they

are all generated in the same degree. These are another type of resolutions to get

examples. For example consider the module defined by the ideal of 2× 2 minors of

a 4× 2 matrix with degrees of the entries as follows:

16

E(X) Example 6.1.

X =

1 2

1 2

2 3

2 3

The Eagon Northcott resolutions of this CM module has the betti diagram

E(X) =

1 − − −0 0 − −− 1 − −− 4 2 −− 1 4 1

− − 2 1

− − − 1

.

What is the representation of this in terms of the pure diagrams? We will come

back to this later in section 8. Using the fact that B(a,b,c,d) represent the normalized

pure betti sequence with possibly rational betti numbers it can be verified that

E(X) = 835P (0, 3, 5, 7)+ 2

35P (0, 4, 5, 7)+ 1

21P (0, 4, 6, 7)+1

3P (0, 4, 6, 8)+1

9P (0, 4, 6, 9)+

221P (0, 4, 7, 9) + 8

63P (0, 5, 7, 9)

7. Monomial Ideals

In this section we will talk about the special case of monomial ideals. When I

is an ideal generated by monomials there are several aspects of the resolution that

are better understood. For the stable monomial ideals there is the resolution given

by Eliahou and Kervaire which is minimal and for the general monomial ideals,

the resolution of Taylor is very specific without being minimal except when it is

generated by a regular sequence. The multiplicity conjecture for the monomial

ideals case has been studied beginning with the first paper of Herzog and SrinivasanHS1[11], where they prove it for stable and square free stable monomial ideals. Later,

they extend the problem to one in graph theory by looking of a weaker upper bound

given by Taylor resolutionHS2[12].

We refer to the paper by Herzog and SrinivasanHS2[12] for the details of the stable

and square free strongly stable resolutions and the proof of the multiplicity bounds

in those cases. The proof in both cases is essentially by establishing the inequality

numerically.

We will describe the Taylor resolution for any monomial ideal.

Let [n] denote the set of first n positive integers. Let I be a monomial ideal

generated by the monomials {f1, · · · , fn}. If σ is a subset of [n], let fσ denote the

lcm of fi, i ∈ σ and |σ| denote the size of σ. Then the shifts in the Taylor resolution

of R/I at the j-th place are the numbers {degree(fσ) : |σ| = j} . The maximal shift

in the j-th place is Lj = max{degree(fσ) : |σ| = j}.17

Thus, the Taylor resolution is a generalization of the Koszul complex and as such

it is minimal exactly when it is the Koszul complex. That is, if the generators

f1, · · · , fn form a regular sequence, as square free monomials, they must be pairwise

relatively prime. So, the Taylor resolution is exactly the Koszul complex with the

shifts

degree(∏

i∈σ,|σ|=j

fi) =∑

i∈σ,|σ|=j

degree(fi).

In any case, by taking the product of the maximal shifts in the resolution, we get a

possible weaker upper bound for the multiplicity which we call the Taylor bound.

Then the conjecture of Herzog and Srinivasan is that

Taylorconj Conjecture 7.1. (Herzog and SrinivasanHS2[12]) Suppose that I is a monomial ideal

of height h and is minimally generated by f1, · · · , fn. Let Li be the maximal shifts

in the Taylor resolution of R/I. Then the multiplicity e of R/I satisfies

1

h!

h∏i=1

Li ≥ e.

Clearly, if I is a complete intersection ideal, this resolution is the same as the

minimal resolution and all the bounds hold. As a first reduction:

Proposition 7.2. Let I be a monomial ideal with a minimal generating set con-

taining a regular sequence of length h = height(I). Then the Taylor bound holds for

the multiplicity of R/I.

Proof. Let K be the ideal generated by a regular sequence of length h that form

a part of the minimal generating set of I. Then since K ⊂ I, e(R/I) ≤ e(R/K)

and Li(I) ≥ Li(K). But the Taylor bound holds for e(R/K) since K is a complete

intersection. So we get that, e(R/I) ≤ e(R/K) ≤∏i=h

i=1 Li. �

Herzog and Srinivasan prove:

hs21 Theorem 7.3. (Corollary 4.3HS2[12]) For a monomial ideal of height 2, the Taylor

bound holds.

hs22 Theorem 7.4. (Theorem 5.3HS2[12]) Let I be an almost complete intersection mono-

mial ideal. Then the Taylor bound holds for I.

An important aspect of this is the translation to graphs. As a first reduction, Her-

zog and Srinivasan show that if the Taylor bound holds for all square free monomial

ideals, then it holds for all monomial ideals. Thus the problem is one for square free

monomial ideals. If I is a square free monomial ideal, then its multiplicity can also

be computed as the number of primes of height h in an irredundant primary decom-

position of I. Then they estimate this number from the primary decomposition of

I to arrive at the Taylor bound.18

In fact, in light of the first reduction to square free monomial ideals, the conjecture

can be stated as a problem in combinatorics. An anti-chain A on n vertices is a

collection of subsets of [n] such that none of the sets in A contains another set in

A. A subset B ⊆ [n] is a minimal vertex cover of A, if B ∩ A 6= ∅ for all A ∈ A,

and for any proper subset C ⊂ B there exists A′ ∈ A such that C ∩ A′ = ∅. Let

M(A) = {B1, · · · , Bt} be the set of distinct minimal vertex covers of A and let

Li = max{| ∪k∈σ Bk| : σ ⊂ [t], |σ| = i}. If h = h(A) denotes the least cardinality of

an element of A, and e(A) equals the number of subsets in A of cardinality h, then

conjectureTaylorconj7.1 states that h!e(A) ≤ L1L2 · · ·Lh.

A simple reduction one can make to prove conjectureTaylorconj7.1 is

Theorem 7.5. Suppose the Taylor bound holds for monomial ideals of height h

generated by f1, f2, · · · , ft such that no t− 1 of them will generate an ideal of height

h. Then the Taylor bound holds for all monomial ideals of height h. That is, the

conjectured bound for the antichains A holds if it holds for those antichains A such

that every proper subset B of M(A) has a minimal vertex cover of cardinality h− 1

where h = h(A).

As a consequence of the translation to graphs the proof of Taylor bounds has a

dual. Since it can be easily proved that M(M(A)) = A for any anti chain A, a

theorem for A has a dual theorem for M(A). This is also called Alexander Duality

in Graph theory. Recall that Supheight is the maximal height of a minimal prime of

I. Thus the theorem in codimension two tells us that if I is a square free monomial

ideal in a polynomial ring of dimension n, then the number of generators of I in

degree two is bounded above by (n/2)Supheight I; that is, theoremhs217.3 has the

following dual.

Theorem 7.6. Let I be a monomial ideal generated in degrees d and above. Then

the number of generators of degree d is bounded above by (nSupheight I)/2.

A discussion of the conversion to graph theory and the problem is given in the

appendix: Ramanujan’s day talk.

8. Boij and Soderberg Conjectures

In this section, we will outline the ideas of Boij and Soderberg and their theorems

which led to the final proof of the multiplicity conjectures. Let M be a graded

R-module with a minimal resolution F:

0→ms⊕j=ms

Rβsj (−j) δs→· · · →mi⊕j=mi

Rβij (−j) δi→· · · →m1⊕j=m1

Rβ1j (−j) δ1→m0⊕j=m0

Rβ0j (−j)→M → 0

Consider all the Betti diagrams with the given set of minimal and maximal shifts.

These form a matrix with s+ 1 columns, numbered 0,1,...s and r+1 rows numbered19

0,1,..r where the element in the ith row and jth column is the number βj,i+j =

dimk(Torj(M,k))i+j.

From the equationsPSModule5.1, associated to such a Betti diagram, there is a “height” h,

and a multiplicity e.

If B1 and B2 are two matrices with the same “height” h, then the multiplicity

e(B1 +B2) = e(B1) + e(B2). Let B(M) = (bij) be the Betti diagram of M . If there

exist Betti diagrams Bi such that B(M) =∑

i riBi such that Bi are Betti diagrams

of pure sequences and ri are positive rational numbers, which sum to 1, we see that

e(M) =∑

i rie(Bi) and e(M) satisfy the multiplicity bounds provided Bi satisfy the

bounds for all i.

Consider the set of matrices

Vm,m = {B | B = (bij),∑

(−1)ibijjt = 0, 0 ≤ t < h, bij = 0, mi < j < mi}.

Note that Vm,m is contained in the vector space of matrices over the rational

numbers Q, with n rows and max{Mi | 0 ≤ i ≤ n} columns.

Let Bm,m be the subset of Vm,m consisting of Betti diagrams of graded modules

and Bm,m be the subset consisting of all the B in Bm,m after dividing by β0 =∑

j β0j.

Let πm,m be the set of all pure diagrams in Vm,m. By a pure diagram, we mean

there exists a sequence d0 ≤ d1 ≤ · · · ≤ dn such that bij = 0, j 6= di, 0 ≤ i ≤ n.

The Betti numbers of a Cohen Macaulay pure resolution are determined by the

equationsPSModule5.1.

Lemma 8.1. Let P = (d0, d1, · · · , dh) be a CM pure diagram. Then for 1 ≤ i ≤ h,

bi = bi,di= b0(−1)i

∏i>0

dk − d0

dk − di= b0

V (d0, · · · , di−1, di+1, · · · )V (d1, · · · , dh)

Proof. We have the h+1 equations∑

i(−1)ibidti = 0, 0 ≤ i ≤ h−1 and

∑i(−1)ibid

hi =

(−1)hh!e. In other words, (b0, · · · , bh)B = (0, · · · , (−1)hh!e), where B is the square

matrix B = ((−1)idji ). Solving for the variables we get them as quotients of two

determinants. Note that the determinant of B is (−1)h(h+ 1)/2V (d0, d1, · · · , dh).So,

bi = (−1)h+1+i+1(−1)hh!e(−1)iV (· · · , di−1di+1, · · · )

V (d0, · · · , dh)= h!e(−1)i

1∏k 6=i(dk − di)

.

But for i = 0, h!e = b0∏

k 6=0(dk − d0). So,

bi = b0(−1)i∏k 6=0,i

dk − d0

dk − di.

�

Define a partial order on CM pure diagrams as follows:20

po Definition 8.2. For pure diagrams, d = (d0, · · · , dh), and c = (c0 · · · , ch), define

(1) d ≤ c if di ≤ ci for all i.

(2) let bi(d) denote the i-th Betti number of the resolution with the pure diagram

as the shifts.

(3) π(d) = (−1)i∏

k 6=0,idk−d0dk−di

= bib0

(4) π(d) ≤ π(c) if d ≤ c

(5) e(π(d)) =Qh

i=1(di−d0)

h!

In particular, if the Betti diagram of the pure diagram is B, then π(d) is in B.

Note that this partial order simply amounts to partially ordering the Vandermonde

determinants, V (d0, · · · , dh) by the partial ordering of the entries. Also, if d ≤ c,

then e(π(d)) ≤ e(c).

With this partial order on the pure diagrams, Boij and Soderberg prove that any

totally ordered set of pure diagrams is linearly independent. The next theorem shows

that any set of pure diagrams in a totally ordered subset is linearly independent.

The maximal chain will then be a basis for Vm,m over Q. Thus it remains to see if

the vectors in Bm,m can be positive rational combinations of the elements in πm,m.

Remark 8.3. If d = (d0, · · · , dn), then B = π(d) is the matrix with exactly one non

zero entry in every column and the non zero entry in the kth column is bkdk.

Theorem 8.4. With the notations as above,

(1) Any set of pure diagrams in a totally ordered subset is linearly independent.

(2) The pure diagrams in any maximal chain in πm,m will form a Q basis for

Vm,m.

Proof. Let P1 < P2 < · · · < Pt be a chain of pure diagrams in πm,m. Suppose∑ti=1 riPi = 0. Let Pi = π(di). P1 < P2 means d1 < d2. Hence there is a k such

that d1k < d2k. So the entry in the k-th column of the matrix of P2 is zero in the

d1k-th “row” and the same is true for all Pi, i ≥ 2. So, looking at the entries in the

dik-th row, we get, r1b1k = 0. Since, b1k 6= 0, r1 = 0.

So,∑t

i=2 riPi = 0. We repeat the argument starting with P2 < P3 and get r2 = 0.

Continuing this, we get ri = 0, 1 ≤ i ≤ t. Thus, Pi are all linearly independent as

desired.

To establish that the pure diagrams in any maximal chain of pure diagrams form

a basis for V , we just need to show that the dimension of V equals the length of the

maximal chain. In fact, since the pure diagrams in the maximal chain are linearly

independent, it suffices to show that

dim V ≤ t = length of a maximal chain of pure diagrams.

To compute the dimension of V , we take the dimension of the vector space of full

matrices and subtract the number of places that must be zeros. The matrices have21

h+ 1 columns and M = 1 + maxi{mi} rows. Further, the non zero entries can only

occur in rows r, with mi ≤ r ≤ mi, 0 ≤ i ≤ h. So,

dim Vm,m ≤ (h+ 1)(M)−∑i

(mi + (M −mi − 1)) =∑i

(mi −mi) + h+ 1.

If P = P1 < · · · < Pt is a maximal chain of pure diagrams, then any two con-

secutive pure diagrams will differ in exactly one place and by exactly 1; that is,

dik < d(i+1)k for exactly one k and d(i+1)k = dik + 1. Further, P1 = (m0, · · · ,mh)

and Pt = (m0, · · · ,mh). Thus,

t =∑

(mi −mi + 1) =∑

(mi −mi) + h+ 1.

This proves that t = dim Vm,m and hence the pure diagrams in any maximal chain

of pure diagrams form a basis for Vm,m. �

Corollary 8.5. If mi > mi for all i, then the multiplicity conjecture holds.

Conjecture 8.6. (Boij and Soderberg): Bm,m is the convex hull of πm,m.

The example we saw inE(X)6.1 shows that if I is the ideal of the 2× 2 minors of the

4× 2 matrix

X =

1 2

1 2

2 3

2 3

, then the minimal resolution of R/I has the betti diagram which decomposes

as the sum of the pure diagrams 835P (0, 3, 5, 7) + 2

35P (0, 4, 5, 7) + 1

21P (0, 4, 6, 7) +

13P (0, 4, 6, 8) + 1

9P (0, 4, 6, 9) + 2

21P (0, 4, 7, 9) + 8

63P (0, 5, 7, 9)

From the discussions before in section 5, we immediately get,

Theorem 8.7. The Boij and Soderberg conjecture implies the multiplicity conjec-

ture.

Proof. m0 = m0 = 1, for any B ∈ B. So, if B =∑riPi, with

∑ri = 1, and ri > 0,

let Pi be the pure diagram with shifts (di0, · · · , dij). Then,

e(B) =∑

rie(Pi) =∑i

rih!

∏j

dij.

But∏

jmj ≤∏

j dij ≤∏

jmj. So, we get∏j

mj ≤ h!e(B) ≤∏j

mj.

�22

Now, associated to Πm,m, there is a simplex ∆(Πm,m) consisting of all the chains

of pure diagrams in Πm,m. Vertices of ∆ are the diagrams, faces are chains and the

maximal chains are facets. A chain is in the boundary of ∆ if it is contained in

exactly one maximal chain.

Proposition 8.8. Let C = π(d1) < · · · < π(dt) be a maximal chain and F =

π(d1) < · · · < π(di−1) < π(di+1) · · · < π(dt) be a facet. F is in the boundary if and

only if it satisfies one of the following.

(1) i = 1 or i = t,

(2) di−1 and di+1 differ in only one place,

(3) di−1 and di+1 differ in 2 adjacent places k and k+1 and di−1,k+1 = di−1,k+1.

Proof. Suppose i 6= 1 or t. Let di−1 = (di−1,1, · · · , di−1,h). di−1 and di+1 can differ

in at most two places for just the addition of di in between them makes the chain

maximal. If they differ in only one place, say the k-th place, then di,k = di−1,k+1 and

di+1,k = di−1,k+2. If they differ in two places, say, k and l > k, then di+1,k = di−1,k+1

and di+1,l = di−1,l + 1. If l > k + 1 then di can differ from di−1 at either the k-th

place or the l-th place. Thus F can be in two maximal chains which makes it not

in the boundary. So, l = k + 1. Now, again, if there is to be only one possible di,

then it can differ from di−1 in either the k-th or the (k + 1)-th place only. Thus

di−1,k+1 = di−1,k + 1 and di,k = di−1,k + 1 and di+1,k+1 = di,k+1 + 1. �

9. On the proof Boij and Soderberg Conjectures

We will briefly outline some of the ideas in the proof of the conjectures.

It is clear from the proof of the multiplicity conjecture for the quasi pure resolu-

tions, that in order to settle the multiplicity conjecture all we need is that the betti

diagrams from actual modules is a positive rational multiple of the pure ( or quasi

pure) resolutions. This is the statement of the second part of the Boij-Soderberg

conjecture. The question comes as to from where does the first part come? We will

see that the first part is indeed needed if the multiplicity conjecture is to be proved

by this.

Translating it to the language of cones, the set of all Betti diagrams of CM

modules with minimal and maximal shifts bounded by m and m do generate a

positive rational cone. In order to prove the multiplicity conjecture, one does need

to show that this is indeed contained in the rational cone generated by the pure

resolutions. Call this the Boij-Soderberg cone. If the supporting hyperplanes of the

boundary facet of the Boij-Soderberg cone is also the boundary facet of the betti

cone, it will be enough. In order to do that, we need to establish that at least one

of the betti diagrams in the betti cone is on this boundary. That is, upto a rational

multiple any pure diagram is an actual betti diagram of a CM module. This is the

conjecture 1 of Boij and Soderberg . Thus, the proof of the multiplicity conjecture23

along these lines, will indeed require both the 1 and the 2 part of the Boij-Soderberg

conjecture.

Developing on the theme of looking at the Betti diagrams of CM modules as

forming a cone, Eisenbud, Schreyer and Weymen proved that indeed the cone of

Betti diagrams is contained in the rational convex cone generated by the pure Betti

diagrams. Further they also proved that given any pure diagram, there is a CM

module whose Betti diagram is a (positive) rational multiple of the pure diagram.

Boij and SoderbergBS2[2] extended this idea to the non CM case, by extending the

partial order to the non CM case.

Theorem 9.1.ES[6] Let ∆ be a maximal chain of pure sequences in Πm,m. Let N =

∆di be a facet of ∆. Then the supporting hyperplane of N is

(1) βj,mj= 0 if di = m and d2,j = mj + 1.

(2) βj,mj= 0 if di = m and dt−1,,j = mj − 1.

(3) βj,k = 0 if di−1,k = di,k − 1 = di+1,k − 2

Further, in all these three cases, the Betti diagrams with the minimal and maximal

shifts bounded by m and m respectively lie on the positive side of the supporting

hyperplane.

Theorem 9.2.ES[6] Suppose ∆ is a maximal chain of pure diagrams and a facet of

∆ is given by F = ∆di such that di−1,k+1 = di,k+1 − 1 and di,k = di+1,k − 1. Then

there are unique hyperplanes U and L containing F such that Betti diagrams in the

cone bounded by the minimal and maximal shifts m and m that are ≥ di+1 are in

the hyperplane U and those that are ≤ di−1 are in L.

The final theorems which prove the conjectures of Boij and Soderberg are sum-

marized in the next statement.

Bettidiagram Theorem 9.3. (ES[6],

BS2[2])

• Let B be the Betti diagram of an R-module of codimension h. Then B is

a non negative rational linear combination of the pure Betti diagrams of

codimension at least h.

• Further, if B is the Betti diagram of a CM R-module, then it is a non negative

rational linear combination of CM pure diagrams of the same codimension

and this representation is unique.

• Let π be a pure diagram that satisfies the equations (**2). Then there exists a

positive integer d such that dπ is the Betti diagram of a CM R-module.

As a result of the theorems inES[6] and

BS2[2] they give an algorithm for writing a

resolution as a rational multiple of CM pure resolutions.24

Consider the free resolution 0 → R(−8) → R10(−6) → R16(−5) → R16(−3) →R10(−2)→ R.

This is the minimal resolution of the ideal of the 4 × 4 pfaffians of a 5 × 5 skew

symmetric matrix Y together with the entries of Y X where X is a generic column

matrix. This has beens studied inS[16]. In fact, it is exact if and only if the ideal

so formed has height 5. This resolution has an interesting fact that it is different

in characteristic 2. If the underlying field is characteristic 2, then the quadratic

syzygies in degree 3 . This is a pure resolution with degree sequence (0,2,3,5,6,8).

Now consider the resolution Bt : 0 → R(−8) → R10(−6) → R16(−5)⊕

Rt(−4) →Rt(−4)

⊕R16(−3) → R10(−2) → R. This is quasi pure and even a virtual pure

resolution. As such, there is no problem in checking the multiplicity conjecture for

this. However, to illustrate the theorem of Boij-Soderberg , let us write it as the

sum of the positive rational multiples of pure resolutions not exceeding the minimal

and maximal shifts.

The betti diagram we want is

Bt =

1 − − − − −− 10 16 t − −− − t 16 10 −− − − − − 1

Here is the method. Say t=11. Let us consider the minimal shifts in the betti

diagram B11. They are (0,2,3,4,6,8). Now the pure resolution with these minimal

shifts has a betti diagram, a rational multiple of

B′ =

5 − − − − −− 60 128 90 − −− − − − 20 −− − − − − 3

.

Now we need to find the largest possible rational number r such that B11 − rB′is non negative. This is 11/90.

B11 − (11/90)B′ =

35/90 − − − − −− 8/3 32/90 0 − −− − 11 16 68/9 −− − − − − 19/30

We now have one extra zero. The minimal shifts are (0,2,3,5,6,8) which correspond

to the pure resolution, a multiple of25

B0 =

1 − − − − −− 10 16 − −− − − 16 10 −− − − − − 1

The highest r such that subtracting rB0 will still leave B11 − (11/90)B′ − rB0

positive is r = 1/45.

Thus,

B11 − (11/90)B′ − (1/45)B0 =

35/90− 2/90 − − − − −− 8/3− 10/45 32/90− 32/90 − − −− − 11 16− 16/45 68/9− 10/45 −− − − − − 19/30− 1/45

=

11/30 − − − − −− 22/9 0 − − −− − 11 704/45 22/3 −− − − − − 55/90

.

The minimal shifts are (0,2,4,5,6,8) and the corresponding pure betti sequence is

B′′ =

3 − − − − −− 20 − − − −− − 60 128 60 −− − − − − 5

We see that 11/90B′′ = B11−(11/90)B′−(1/45)B0. ThusB11 = (11/90)π(0, 2, 3, 4, 6, 8)+

(1/45)(0, 2, 3, 5, 6, 8) + (11/90)(0, 2, 4, 5, 6, 8).

The plan is to get at least one extra zero in the betti diagram with every subtrac-

tion of a multiple of pure diagram. As a consequence of Eisenbud-Shreyer theorem

this will certainly terminate.

As a result of the theorems of Eisenbud, Schreyer, Weymen, Boij and Soderberg

the multiplicity conjectures of Herzog, Huneke and Srinivasan are established, For

the proof of the conjectures of Boij and Soderberg, we refer to the papers{EFS}, {BS2}[5, 2] and

we will not repeat them here.

Since the Hilbert series is also strictly increasing as a function of the pure dia-

grams, with respect to the partial order, as a corollary to the proof of TheoremBettidiagram9.3,

we obtain26

Theorem 9.4. (Boij, SoderbergBS2[2]) Let M be a finitely generated graded R-module

of projective dimension r and codimension s and generated in degree zero. Then

β0(M)H(π(0,m1,m2 · · ·mr), t) ≤ H(M, t) ≤ β0(M)H(π(0,m1, · · · ,ms), t).

Further, equality on either side implies that M has a pure resolution.

One of the natural questions arising from the study of multiplicity conjectures and

the various attempts to prove them is what sort of “Betti diagrams” are possible

Betti diagrams. We saw earlier, that a Betti diagram such as n = 2, β0 = 0, β12 =

3; β21 = β25 = 1 is a codimension 2 diagram which will not satisfy the upper bound

and hence cannot be a Betti diagram of a module. In fact, it also fails to be a Betti

diagram for any module because β21 6= 0 and β11 = 0 for all j. In codimension three

and higher, it is easy to create such examples by moving the shifts without changing

the equations governing them.

Even if a diagram satisfies all the conditions of the equations as well as the multi-

plicity bounds, it may not be a Betti diagram of a finitely generated module. Thus,

the question of determining the equations that will determine if a potential Betti

diagram is a Betti diagram of a module is still open. Some partial results on this

problem can be found in the preprintEr[7].

One other natural question is, given a pure diagram π, what is the smallest possible

positive integer d such that there is a CM module with Betti diagram dπ?

It is also of interest to find some idea as to what the pure resolutions are that

occur in the decomposition of the minimal resolution of a module and interpret the

coefficients of the pure diagrams that occur in the decompositions. For instance,

in the example above, B11 = (11/90)π(0, 2, 3, 4, 6, 8) + (1/45)π(0, 2, 3, 5, 6, 8) +

(11/90)π(0, 2, 4, 5, 6, 8) interpret the numbers 11/90 and 1/45.

A comprehensive list of references on the work on multiplicity conjectures prior

to the proofs of the complete conjecture are inFS[8].

References

BS1 [1] M. Boij and J. Soderberg: Graded Betti Numbers of Cohen-Macaulay modules and the Mul-tiplicity conjectures, math. AC/0611081, (2) 78, (2008),(1), 85-106.

BS2 [2] M. Boij and Soderberg: Betti Numbers of Graded Modules and the Multiplicity Conjecturein the Non-Cohen-Macaulay Case, preprint, posted on the Arxiv.

BE1 [3] D. Buchsbaum and D. Eisenbud: Algebra Structures for Finite Free Resolutions, and SomeStructure Theorems for Ideals of Codimension 3, Amer. J. Math., 25(1973), 259-268.

BE2 [4] D. Buchsbaum and D. Eisenbud: What Makes a Complex Exact?, J. Algebra (99) (1977),447-485.

EFS [5] D. Eisenbud, G. Floystad and J. Weyman: The Existence of Pure Free Resolutions,arXiv.0709.1529

ES [6] D. Eisenbud and F. Schreyer: Betti Numbers of Graded Modules and Cohomology of Homo-geneous Vector Bundles, preprint. posted on the Arxiv

Er [7] D. Erman: The Semigroup of Betti Diagrams, preprint, Arxiv, December 2008.27

FS [8] C. Francisco and H.Srinivasan: Multiplicity Conjectures, Syzygies and Hilbert functions, 145–178, Lect. Notes Pure Appl. Math., 254, Chapman & Hall/CRC.

LHH [9] L. Gold, H. Schenck and H.Srinivasan: Linkage, Betti Numbers and Degree Bounds for ZeroDimensional Schemes, preprint.

HK [10] J. Herzog and M. Kuhl: On the Betti Numbers of Finite Pure Resolutions, Comm. Algebra,12. 1627-1646 (1984).

HS1 [11] J. Herzog and H. Srinivasan: Bounds for Multiplicities, Trans. Amer. Math. Soc. 350 (1998),no. 7, 2879–2902

HS2 [12] J. Herzog and H. Srinivasan: Multiplicities of Monomial ideals, J. Algebra, 274 (2004). 230-244.

HM [13] C. Huneke and M. Miller: A note on the multiplicity of Cohen Macaulay algebras with pureresolutions, Canadian J. Math, 37, (1985),no. 6, 1149-1162.

MN [14] J. Migliore, U. Nagel and T. Romer: Extensions of the multiplicity conjecture. Trans. Amer.Math. Soc. 360 (2008), no. 6, 2965–2985.

PS [15] C. Peskine and L. Szpiro: Syzygies et Multiplicities, C.R. Acad. Sci. Paris Ser. A 278 (1974),no.1, 163-194.

S [16] H. Srinivasan : A Grade Free Gorenstein Algebra with No Minimal Algebra Resolutions, J.Algebra 179, (1996) 362-379.

T [17] D. Taylor: Ideals Generated by Monomials in an R-Sequence, Thesis, University of Chicago,1966.

28

10. Appendix

Free Resolutions and Graph Theory

These are the notes from the talk given on the Ramanujan′s Day ( December

22, 2008), Ramanujan’s birth day at the Indian Institute of Technology, Madras.

The talk was part of the Golden Jubilee National Symposium on Mathematics,

Methods and Applications.

10.1. Background. Let R denote the ring of polynomials with coefficients in a

field k. If we pick a set of polynomials, f1, f2, . . . , fn, any common solution to these

polynomials will also be a solution to any sum of multiples of them. Thus the set

of sums of multiples of these polynomials is a natural object to study when one

studies polynomials and their solutions. This is called the ideal generated by these

polynomials and is written

I = (f1, f2, · · · , fn).

For example:

R = k[x1, x2, x3],

I = (x21 − x2x3, x

22, x1x2).

I is generated by three homogeneous polynomials of degree 2.

Much of commutative algebra is devoted to studying these objects and their struc-

tures from varied perspectives. Ideals are in some sense a generalization of the notion

of vector spaces. That is, they are non empty sets that are closed with respect to

addition and multiplication by elements of R. The main difference is that the ring

R, unlike the real numbers, is not a field. There are elements in R, such as any

non-constant polynomial, with no multiplicative inverses.

In our example, the ideal I = (x21 − x2x3, x

22, x1x2) has a ”basis” of three poly-

nomials in the sense that no one of these polynomials can be written as a sum

of multiples of the other two. Thus, we have a map, ϕ1 : R3 → R given by

ϕ1

abc

= ax21 − x2x3 + bx2

2 + cx1x2

where the image of ϕ1 is precisely the ideal I.

10.2. Ideals. Now going from an ideal to its set of solutions, we can also go back

from a set of solutions to finding all polynomials which have these as solutions.

Clearly the ideal generated by the polynomials will be contained in them but then

there may be more.

For instance, if f vanishes at a set of points, it is also true that square root of f ,

if it exists, will vanish at these points. Thus the set of polynomials vanishing at a

given set of points will include all the square roots, cube roots etc. of the elements

in the ideal. This is called the radical of the ideal I. So, I must have some part29

of its structure that is shared with its radical. Whatever it is, it must come from

the geometry. Thus the notion of dimension or codimension is preserved when one

looks at the radical. That is dim R/I = dim R/√I. dim R− dim R/I is called the

height of I. Thus the height of I is the same as the height of the radical of I. Later

we will see another, numerical method for determining the height of an ideal.

In our example I = (x21 − x2x3, x

22, x1x2), the set of points where these three

equations vanish is {(0, 0, a) | a ∈ k}.

Thus if we collect all polynomials vanishing at these points, we get all sums of

multiples of x1 and x2. Thus, the radical of I is the ideal√I = (x1, x2).

Recall R = k[x1, x2, x3]. dim R is 3. R/√I = R/(x1, x2) ∼= k[x3] has dimension

1.

Thus the height of I and the height of the radical of I are both 2.

10.3. Homogeneous ideals. We will concentrate on homogeneous ideals, that is

ideals generated by homogeneous polynomials such as our example ideal I.

Now, these polynomials have degree 2. We encode the fact that the ideal I is

generated by three homogeneous polynomials of degree 2 by

R(−2)3 ϕ1→ R

This does not tell you what the polynomials are but that the ideal is generated

by three polynomials each of degree 2.

Consider the map ϕ : R → R given by ϕ(g(x)) = x21g(x). This map takes every

polynomial of degree d to a polynomial of degree d+2. Homogenizing the map means

we want the degree to be preserved by the map. So, we look at R(−2), the ring R

where we simply increase the degrees of all polynomials by 2. So, ϕ : R(−2)→ R is

now degree preserving or homogeneous. The −2 is called the shift.

10.4. Free Resolutions. In general, a homogeneous ideal I has a presentation

t⊕i=1

R(−di)ϕ1→ R

when I is generated by t homogeneous polynomials of degree d1, d2, . . . , dt. This is

the beginning of a free resolution of the ideal. These generators may not really be a

”basis” that is, they are not exactly ”linearly independent” over R. That is, there

are likely to be relations among the generators and these relations can be captured

in the following manner.30

Example 10.1. Returning to our example, I = (x21 − x2x3, x

22, x1x2) = (f1, f2, f3)

with

f1 = x21 − x2x3, f2 = x2

2, f3 = x1x2.

We see that

x2f1 + x3f2 − x1f3 = 0, x1f2 − x2f3 = 0.

It turns out that these are precisely the two ”linearly independent” relations on

f1, f2, f3. That is all other relations are linear combinations of these.

We encode this as

0→⊕

R(−3)2 ϕ2→⊕

R(−2)3 ϕ1→ R→ R/I → 0.

In general, a homogeneous ideal I or R/I, has a homogeneous resolution such as:

0→bt⊕j=1

R(−dtj)ϕt→ · · ·

bi⊕j=1

R(−dij)ϕi→

bi−1⊕j=1

R(−d(i−1)j)→

· · · →b1⊕j=1

R(−d1j)ϕ1→ R→ R/I → 0.

The maps ϕi can be represented by matrices of size bi × bi−1 the composition of

two consecutive ϕ’s is zero.

The composition is zero exactly: i.e If ϕi(a1, a2, · · · , abi) is zero, then it must be

equal to ϕi+1(y1, y2, · · · , ybi+1).

This is called a free resolution of R/I and is of length t.

10.5. Numerical information on the resolution. The shifts in the resolution

have a lot of information and it is sometimes enough to tell the structure of the

resolution. Let us look at our example: I = (x21 − x2x3, x

22, x1x2) has the resolution

0→ R(−3)2 ϕ2→ R(−2)3 ϕ1→ R→ R/I → 0.

Without even knowing anything except this picture of the resolution, one can say

that up to multiplication by an element of R, the ideal I is indeed generated by the

2 by 2 determinants of a 2 by 3 matrix.

In fact, the matrix [x1 x2 x3

x2 0 x1

]is the matrix whose 2 by 2 minors will be precisely the generators of I we mentioned

earlier.31

10.6. Relation among shifts. We notice some relations among the shifts in the

resolution.

0→ R(−3)2 ϕ2→ R(−2)3 ϕ1→ R→ R/I → 0

such as the alternating sum of the ranks 2 − 3 + 1− 0, and the alternating sum of

the shifts 3 + 3− (2 + 2 + 2) = 2× 3− 3× 2 = 0.

Indeed the shifts in any homogeneous resolution satisfy a set of equations (Herzog-

Kuhl, Peskine-Szpiro:) Letting b0 = 1,∑i

(−1)ibi = 0

∑i

∑j

(−1)idkij = 0, 0 ≤ k < h

The first time this alternating sum is not zero, is naturally an important invariant

of the ideal, called the height.

Thus the height of the ideal in our example is 2.

When this sum is not zero, the nonzero value must also be important. It is:∑i

∑j

(−1)idhij = (−1)hh!e,

where e is called the multiplicity of R/I or loosely I.

When the length t = h, the ideal is called perfect.

10.7. Multiplicity conjecture. The multiplicity of the homogeneous ideal is the

focus of the multiplicity conjecture.

Specifically, we ( Herzog, Huneke and Srinivasan) conjectured in 1995 that∏i

minj(dij) ≤ h!e,

if t = h

and

h!e ≤∏i

Maxj(dij)

in general.

There is a long history of various people working on this conjecture and its many

variants and modifications. Herzog and Srinivasan who originally proposed the

conjecture (Transactions of AMS, 1997, [4]) solved it for various important cases.

The conjecture is now completely solved by a combination of three recent papers by

Eisenbud, Schreyer, Weyman and Boij and Soderberg . This is one of the exciting

events in commutative algebra this year.32

10.8. Monomial Ideals. A monomial is just a polynomial with only one term and

is thus necessarily homogeneous. An ideal which is generated by some monomials

in the ring is called a monomial ideal. For a monomial ideal I, the minimal set of

monomials which generate I is unique. As with any ideal, there is a free resolution

for the monomial ideal.

• ¡1-¿ I = (f1, f2, · · · , fm) is generated by m monomials of degree d1, d2, · · · , dmrespectively.

• ¡2-¿ A resolution of R/I starts as⊕R(−di)

ϕ1→ R→ R/I → 0.

In fact there is an even easier way to construct a free resolution for a monomial

ideal which we will call the Taylor Resolution [6]. It is a natural generalization of

the idea of Koszul complex. The resolution is minimal if and only if it is the Koszul

complex or equivalently that the ideal is generated by a regular sequence. This is

called the Taylor Resolution [6].

10.9. Taylor Resolution. Denote by f1∨f2∨f3 the LCM of the monomials f1, f2, f3

and let deg(f) denote the degree of f .

Then the resolution looks like

0→⊕

R(−deg(f1 ∨ f2 ∨ · · · ∨ fm))ϕm→ · · · →

m⊕i,j,k=1

R(−deg(fi ∨ fj ∨ fk))

ϕ1→m⊕

i,j=1

R(−deg(fi ∨ fj))ϕ2→⊕

R(−di)ϕ1→ R→ R/I → 0.

Thus the resolution is necessarily of length m. This resolution is not minimal in

the sense that there are many relations that will not be linearly independent. For

instance, whenever the LCM of two different set of monomials is the same, there

will be an extra relation.

Nevertheless, the numerical equations will have to be still satisfied and we there-

fore can compute both the height and the multiplicity from these shifts. The mul-

tiplicity conjecture from this is weaker than the original one proposed;

Namely:

h!e ≤h∏i=1

Max{deg(fj1 ∨ fj2 ∨ · · · ∨ fji) | 1 ≤ j1 < j2 · · · < ji ≤ m}

10.10. Graph theory problem. What is a graph? It is a picture with a set of

vertices V and edges or faces. Normally one would think of an edge as something

connecting just two vertices or a loop or something like that.

We can simply extend our imagination to think of edges or faces consisting of any

number of vertices.33

Then a graph G on a vertex set V is a set of subsets of V . To avoid degeneracy,

we may as well say that no subset can be contained in another. Thus a graph G is

a set of subsets of V such that no set in G is contained in another set in G. Thus a

loop at a point will be a singleton set.

For a subset A of V , we will denote the cardinality of A by |A|.The height of a graph G is simply the minimal cardinality of a subset in G.

e(G) = Number of subsets in G with cardinality h.

A cover of G is a subset T of V such that T ∩ A 6= ϕ for any A in G. That is T

meets every subset A in G. Naturally, if T is a cover, then any subset containing T

must also be a cover. So, we will only collect the minimal covers. Thus G∗ = C(G)

denotes the set of minimal covers of G. By the very method of definition, C(G) or

G∗ itself is a graph and if we apply C again, we get back G. That is the cover of

the cover of G is itself.

For any graph G, let Mi be the largest cardinality of a union of i distinct sets in

G. Thus, M1 is the largest size of a set in G.

For example the triangle ABC with three edges AB, BC and CA is represented

by the set of subsets

{{A,B}, {B,C}, {C,A}}

This graph has height 2 and multiplicity 3. It’s M1 = 2 and M2 = 3 = M3.

This graph is special for G∗ = G.

10.11. Problem. Let G be a graph on a vertex set V . Prove that the following

equalities hold.

h!e ≤h∏i=1

Mi(G∗)

and

h(G∗))!e(G∗) ≤h(G∗)∏i=1

Mi(G∗).

The Taylor bound for the multiplicity conjecture translates to this problem. This

conjecture ( problem) was stated in Herzog-Srinivasan in 2003, in the Journal of Pure

and Applied Algebra [5] In the same paper we give prove it when h = 2 using graph

theoretic and combinatorial computations. Since then there were also other special

cases by many others essentially using structures of free resolutions or techniques

from combinatorics. The final solution which is a corollary of the result mentioned

for the free resolutions is as follows:

Theorem 10.2. The above inequalities are true for any graph G.34

Proof: This is a corollary of the theorem on multiplicities bounds by Eisenbud,

Schreyer, Boij and Soderberg . They prove the upper bound for the multiplicity

for minimal resolution. Taylor resolution is a nonminimal resolution in general and

hence the Taylor bound holds.

However, there is still no proof that uses mainly graph theoretic notions to address

this weaker inequality on the invariants of a graph and it’s vertex covers.

Referencesbib1 [1] M. Boij and Soderberg : Betti Numbers of Graded Modules and the Multiplicity Cinjecture

in the NOn-Cohen-Macaulay Case, Preprint.2 [2] D. Eisenbud and F. Schreyer: Betti Numbers of Graded Modules and Cohomology of Homo-

geneous Vector Bundles, preprint.3 [3] C. Francisco and H.Srinivasan; Multiplicity Conjectures, Syzygies and Hilbert functions, 145–

178, Lect. Notes Pure Appl. Math., 254, Chapman & Hall/CRC4 [4] J. Herzog and H. Srinivasn: Bounds for Multiplicities, Trans. Amer. Math. Soc. 350 (1998),

no. 7, 2879–29025 [5] J. Herzog and H. Srinivasan: Multiplicities of Monomial ideals, J. Algebra, 274 (2004). 230-

244.5 [6] D.Taylor, Ideals generated by monomials in an R-sequence, Thesis, Chicago University.

Hema Srinivasan, Mathematics Department, 202 Mathematical Sciences Bldg,

University of Missouri, Columbia, MO 65211 USA

E-mail address: srinivasanh@missouri.edu <srinivasanh@missouri.edu>

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