# Lower bounds on Betti numbers

Adam Boocher  Department of Mathematics, University of San Diego, San Diego,
CA 92110, USA<EMAIL_ADDRESS>and  Eloísa Grifo  Department of
Mathematics, University of Nebraska – Lincoln, Lincoln, NE 68588, USA
<EMAIL_ADDRESS>Dedicated to David Eisenbud on the occasion of his 75th
birthday.

###### Abstract.

We survey recent results on bounds for Betti numbers of modules over
polynomial rings, with an emphasis on lower bounds. Along the way, we give a
gentle introduction to free resolutions and Betti numbers, and discuss some of
the reasons why one would study these.

###### Key words and phrases:

free resolutions, Betti numbers, Buchsbaum–Eisenbud–Horrocks Conjecture, Total
Rank Conjecture

###### 2020 Mathematics Subject Classification:

Primary: 13D02.

##  1\. Introduction

Consider a polynomial ring over a field $k$, say $R=k[x_{1},\ldots,x_{n}]$.
When studying finitely generated graded modules $M$ over $R$, there are many
important invariants we may consider, with the Betti numbers of $M$, denoted
$\beta_{i}(M)$, being among some of the richest. The Betti numbers are defined
in terms of generators and relations (see Section 2), with $\beta_{0}(M)$
being the number of minimal generators of $M$, $\beta_{1}(M)$ the number of
minimal relations on these generators, and so on. Despite this simple
definition, they encode a great deal of information. For instance, if one
knows the Betti numbers111Really, we mean the graded Betti numbers of $M$, to
be defined in Section 3. of $M$, one can determine the Hilbert series,
dimension, multiplicity, projective dimension, and depth of $M$. Furthermore,
the Betti numbers provide even finer data than this, and can often be used to
detect subtle geometric differences (see Example 3.4 for an obligatory example
concerning the twisted cubic curve).

There are many questions one can ask about Betti numbers. What sequences arise
as the Betti numbers of some module? Must the sequence be unimodal? How small,
or how large, can individual Betti numbers be? How large is the sum? Questions
like this are but just a few examples of those that have been studied in the
past decades, and of the flavor we will discuss in this survey. We will focus
on perhaps one of the longest standing open questions in this area, which is
due to Buchsbaum–Eisenbud, and independently Horrocks (BEH). Their conjecture
proposes a lower bound for each $\beta_{i}(M)$ depending only on the
codimension $c$ of $M$: that $\beta_{i}(M)\geqslant{c\choose i}$. While the
conjecture remains widely open in the general setting, there are some special
cases that are known. Moreover, if the conjecture is true, then the total
Betti number of $M$, $\beta(M):=\beta_{0}(M)+\cdots+\beta_{n}(M)$, must
satisfy $\beta(M)\geqslant 2^{c}$. Recently, Mark Walker [65] proved this
bound on the total Betti number — known as the Total Rank Conjecture — in all
cases except when $\textrm{char }k=2$. Walker also showed that equality holds
if and only if $M$ is isomorphic to $R$ modulo a regular sequence — such
modules are called complete intersections.

The Betti numbers of modules that are not complete intersections are quite
interesting. For example, it follows from Walker’s result that if our module
$M$ is not a complete intersection, then $\beta(M)\geqslant 2^{c}+1$, but
there is reason to believe that $\beta(M)$ might be much bigger than $2^{c}$.
Charalambous, Evans, and Miller [31] asked if in fact we must have
$\beta(M)\geqslant 2^{c}+2^{c-1}$, and proved that this holds when $M$ is
either a graded module small codimension ($c\leqslant 4$), or a multigraded
module of finite length (meaning $c=n$) for arbitrary $c$ [30, 29]. More
evidence towards this larger bound for Betti numbers has recently been found,
including [11, 12].

For example, Erman showed [40] that if $M$ is a graded module of small
regularity (in terms of the degrees of the first syzygies), then not only is
the BEH Conjecture 4.1 true, but in fact
$\beta_{i}(M)\geqslant\beta_{0}(M){c\choose i}$. The first author and
Wigglesworth [12] then extended Erman’s work to say that under the same low
regularity hypothesis, $\beta(M)\geqslant\beta_{0}(M)(2^{c}+2^{c-1})$. This
stronger bound asserts that on average, each Betti number $\beta_{i}(M)$ is at
least 1.5 times $\beta_{0}(M){c\choose i}$.

The main goal of this survey is to discuss these lower bounds on Betti numbers
and present some of the motivation for these conjectures. We start with a
short introduction to free resolutions and Betti numbers, why we care about
them, and some of the very rich history surrounding these topics. We also
collect some open questions, discuss some possible approaches, and present
examples that explain why certain hypothesis are important.

##  2\. What is a Free Resolution?

Let $R=k[x_{1},\ldots,x_{n}]$ be a polynomial ring over a field $k$. We will
be primarily concerned with finitely generated graded $R$-modules $M$. One
important invariant of such a module is the minimal number of elements needed
to generate $M$. In fact, this number is the first in a sequence of Betti
numbers that describe how far $M$ is from being a free module. Indeed, suppose
that $M$ is minimally generated by $\beta_{0}$ elements; this means there is a
surjection from $R^{\beta_{0}}$ to $M$, say

$\textstyle{R^{\beta_{0}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{0}}$$\textstyle{M.}$

If $\pi_{0}$ is an isomorphism, then $M\cong R^{\beta_{0}}$ is a free module
of rank $\beta_{0}$. Otherwise, it has a nonzero kernel, which will also be
finitely generated and can be written as the surjective image of some free
module $R^{\beta_{1}}$:

|  |
---|---|---
$\textstyle{R^{\beta_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{\beta_{0}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{0}}$$\textstyle{M.}$$\textstyle{\ker(\pi_{0})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Notice that if $M$ is generated by $m_{1},\ldots,m_{\beta_{0}}$, and $\pi_{0}$
is the map sending each canonical basis element $e_{i}$ in $R^{\beta_{0}}$ to
$m_{i}$, then an element $(r_{1},\ldots,r_{\beta_{0}})^{T}$ in the kernel of
$\pi_{0}$ corresponds precisely to a relation among the $m_{i}$, meaning that

$r_{1}m_{1}+\cdots+r_{\beta_{0}}m_{\beta_{0}}=0.$

Such relations are called syzygies222Fun fact: in astronomy, a syzygy is an
alignment of three or more celestial objects. of $M$ and the module
$\ker{\pi_{0}}$ is called the first syzygy module of $M$.

Continuing this process we can _approximate_ $M$ by an exact sequence

$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{p}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{p}}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{1}}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{0}}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

where each $F_{i}$ is free. Such an exact sequence is called a free resolution
of $M$.

If at each step we have chosen $F_{i}$ to have the minimal number of
generators, then we say the resolution is minimal, and we set $\beta_{i}(M)$
to be the rank of $F_{i}$ in any such minimal free resolution. This is well-
defined, because it is true that two minimal free resolutions of $M$ are
isomorphic as complexes. Furthermore, one has the following,

$\beta_{i}(M)=\operatorname{rk}F_{i}=\operatorname{rk}_{k}\operatorname{Tor}^{R}_{i}(M,k).$

The $i$th syzygy module of $M$, denoted $\Omega_{i}(M)$, is defined to be the
image of $\pi_{i}$, or equivalently the kernel of $\pi_{i-1}$. We note that
$\Omega_{i}(M)$ is defined only up to isomorphism.

If at some point in the resolution we obtain an injective map of free modules,
then its kernel is trivial, and we obtain a finite free resolution, in this
case of length $p$:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{p}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

If a module $M$ has a finite minimal projective resolution, the length of such
a resolution is called the projective dimension of $M$, and we write it
$\operatorname{pdim}M$.

######  Remark 2.1.

We will often implicitly apply the Rank-Nullity Theorem to conclude that

$\beta_{i}(M)=\operatorname{rk}\Omega_{i}(M)+\operatorname{rk}\Omega_{i+1}(M).$

######  Example 2.2.

If $M=R/(f_{1},\ldots,f_{c})$ where the $f_{i}$ form a regular sequence, then
the minimal free resolution of $M$ is given by the Koszul complex. For
instance if $c=4$ then the minimal resolution has the form

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{3}$$\textstyle{R^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{3}$$\textstyle{R^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{M.}$

Note that the numbers over the arrows represent the rank of the corresponding
map, which is equal to the rank of the corresponding syzygy module
$\Omega_{i}(M)$. We will discuss this in more detail in Section 3.2. We will
also see that the ranks occurring in the Koszul complex are conjectured to be
the smallest possible for modules of codimension $c$ (see Conjecture 4.2).

######  Example 2.3.

One of the strongest known bounds on ranks of syzygies is the Syzygy Theorem
3.13 which states that except for the last syzygy module, the rank of
$\Omega_{i}(M)$ is always at least $i$. A typical use of such a result might
be as follows. Suppose we had a rank zero module $M$ with Betti numbers
$\\{1,7,8,8,7,1\\}$. Then we could calculate the ranks of the syzygy modules
by using Remark 2.1 to obtain the ranks labeled in the diagram below:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{6}$$\textstyle{R^{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{2}$$\textstyle{R^{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{6}$$\textstyle{R^{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{M.}$

We would also obtain from Remark 2.1 that $\operatorname{rk}\Omega_{3}(M)=2$,
which we will see violates Theorem 3.13. Therefore, such a module does not
exist! See also Example 5.17.

######  Example 2.4.

In [36], Dugger discusses almost complete intersection ideals and the
tantalizing fact that we currently do not know whether or not there is an
ideal $I$ of height 5 with minimal free resolution

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{6}$$\textstyle{R^{12}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{6}$$\textstyle{R^{10}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{4}$$\textstyle{R^{9}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{5}$$\textstyle{R^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{R/I.}$

David Hilbert, interested in studying minimal free resolutions as a way to
count invariants, was able to prove that finitely generated modules over a
polynomial ring always have finite projective dimension [48].

######  Theorem 2.5 (Hilbert’s Syzygy Theorem, 1890).

Let $R=k[x_{1},\ldots,x_{n}]$ be a polynomial ring in $n$ variables over a
field $k$. If $M$ is a finitely generated graded $R$-module, then $M$ has a
finite free resolution of length at most $n$.

While we are primarily interested in studying polynomial rings over fields,
Hilbert’s Syzygy Theorem is true more generally for any noetherian regular
ring. In fact, if we focus our study on local rings instead, the condition
that every finitely generated module has finite projective dimension
characterizes regular local rings [3, 64]. While we will be working over
polynomial rings throughout the rest of the paper, we point out that the
theory of (infinite) free resolutions over non-regular rings is quite
interesting and rich; [58] and [5] are excellent places to start learning
about this.

The upshot of Hilbert’s Syzygy Theorem is that to each finitely generated
$R$-module $M$ we attach a finite list of Betti numbers
$\beta_{0}(M),\,\ldots,\,\beta_{n}(M)$. Note that while some of these might
vanish, $M$ has at most $n+1$ non-zero Betti numbers.

Our main goal in this paper is to discuss the following question:

######  Question A.

If $M$ is a finitely generated graded module over $R=k[x_{1},\ldots,x_{n}]$,
where $k$ is a field, can we bound the Betti numbers of $M$, either from above
or below?

As we will see, there are many results and conjectures relevant to the answer
to this question. Feel free to skip the next section if you can’t handle the
suspense!

##  3\. Why Study Resolutions?

Before getting to the heart of the matter in Section 4, we would first like to
offer some motivation as to why one might care about Betti numbers at all.

###  3.1. Betti Numbers Encode Geometry

In a sense, a minimal free resolution of $M$ contains redundant information —
after all, the first map $\pi_{1}\\!\\!:F_{1}\to F_{0}$ is a presentation of
$M$. However, suppose we do not know the maps in the resolution, but just the
numerical data of the resolution, namely the numbers $\\{\beta_{i}\\}$.
Surprisingly, this coarse invariant encodes much geometric and algebraic
information about $M$. First of all, the Betti numbers $\beta_{i}$ tell us
that $M$ has $\beta_{0}$ generators, that there are $\beta_{1}$ relations
among those generators, and $\beta_{2}$ relations among those relations, and
so on. But the Betti numbers also encode more sophisticated information about
$M$. For instance, since rank is additive across exact sequences, we have

$\operatorname{rk}M=\beta_{0}-\beta_{1}+\cdots+(-1)^{n}\beta_{n}.$

Moreover, if we have a graded module $M$, we can take the resolution of $M$ to
be a graded resolution, and if among the $\beta_{i}$ generators of
$\Omega_{i}(M)$, exactly $\beta_{ij}$ of them live in degree $j$, then the
following formula gives the Hilbert series for $M$:

(3.1)
$HS(M)=\displaystyle\frac{\,\,\displaystyle\sum_{i=0}^{d}(-1)^{i}\beta_{ij}t^{j}\,\,}{(1-t)^{d}}.$

We recall that the Hilbert series of $M$ is a power series that encodes the
$k$-vector space dimension of each graded piece $M_{i}$ of $M$, as follows:

$HS(M)=\displaystyle\sum_{i=0}^{\infty}\dim_{k}(M_{i})t^{i}.$

This is a classical tool that contains important algebraic and geometric
information about our module. For example, once we write
$HS(M)=p(t)/(1-t)^{m}$ with $p(1)\neq 0$, we have $\dim(M)=m$ and $p(1)$ is
equal to the degree of $M$. So just by knowing its (graded) Betti numbers, we
can then determine the multiplicity (i.e. degree), dimension, projective
dimension, Cohen-Macaulayness, and other properties and invariants of a module
$M$.

The following example gives the spirit of these ideas:

######  Example 3.1.

Suppose that $R=k[x,y,z]$ and that $M=R/(xy,xz,yz)$ corresponds to the affine
variety defining the union of the  three coordinate lines in $k^{3}$. This
variety has dimension one and degree  three. Let us illustrate how the
(graded) Betti numbers communicate this. The minimal free resolution for $M$
is

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\displaystyle{\psi}=\begin{bmatrix}z&0\\\
-y&y\\\
0&-x\end{bmatrix}}$$\textstyle{R^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\displaystyle{\phi}=\begin{bmatrix}xy&xz&yz\end{bmatrix}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M.}$

From this minimal resolution, we can read the Betti numbers of $M$:

  * •

$\beta_{0}=1$, since $M$ is a cyclic module;

  * •

$\beta_{1}={\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}3}$,
and these three quadratic generators live in degree 2;

  * •

$\beta_{2}={\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}2}$,
and these represent linear (degree 1) syzygies on quadrics (degree 2), and
thus live in degree  3 $(=1+2)$ .

We can include this _graded_ information in our resolution, and write a
_graded_ free resolution of $M$:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-3})^{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\displaystyle{\psi}=\begin{bmatrix}z&0\\\
-y&y\\\
0&-x\end{bmatrix}}$$\textstyle{R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-2})^{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color<EMAIL_ADDRESS>
The
$R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-2})^{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}3}}$
indicates that we have three generators of degree 2. Formally, the $R$-module
$R(-a)$ is one copy of $R$ whose elements have their degrees shifted by $a$:
the polynomial $1$ lives in degree $0$ in $R$ and degree $a$ in $R(-a)$, and
in general the degree $d$ piece of $R(-a)$ consists of the elements of $R$ of
degree $d-a$. With this convention, the map $\phi$ keeps degrees unchanged —
we say it is a degree $0$ map: for example, it takes the vector $[1,1,1]^{T}$,
which lives in degree $2$, to the element $xy+xz+yz$, which is an element of
degree $2$. When we move on to the next map, $\psi$, we only need to shift the
degree of each generator by 1, but since $\psi$ now lands on
$R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-2})^{3}$,
we write
$R({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-3})^{2}$.

The graded Betti number $\beta_{ij}(M)$ of $M$ counts the number of copies of
$R(-j)$ in homological degree $i$ in our resolution. So we have

$\beta_{00}=1,\beta_{1{{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}2}}}={{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}3}},\textrm{and
}\beta_{2{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}}}={{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color<EMAIL_ADDRESS>
We can collect the graded Betti numbers of $M$ in what is called a _Betti
table_ :

$\begin{array}[]{cc}\begin{array}[]{r|ccccc}\beta(M)&0&1&2&\\\ \hline\cr
0&\beta_{00}&\beta_{11}&\beta_{22}&\\\ 1&\beta_{01}&\beta_{12}&\beta_{23}&\\\
&\end{array},&\begin{array}[]{r|ccccc}\beta(M)&0&1&2&\\\ \hline\cr 0&1&-&-&\\\
1&-&{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}3}}&{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}2}}&\\\
&\end{array}.\end{array}$

######  Remark 3.2.

To the reader who is seeing Betti tables for the first time, we point out that
although we will write resolutions so that the maps go from left to right, and
thus the Betti numbers appear from right to left
$\\{\ldots,\beta_{2},\beta_{1},\beta_{0}\\}$ in a Betti table, the opposite
order is used. Furthermore, by convention, the entry corresponding to $(i,j)$
in the Betti table of $M$ is $\beta_{i,i+j}(M)$, and not $\beta_{ij}(M)$.

Finally, we can use this information to calculate the Hilbert series of $M$:

$HS(M)=\frac{1t^{0}-{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}3}}t^{{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}2}}+{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}2}}t^{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}}}{(1-t)^{3}}=\frac{1+2t}{(1-t)^{1}},$

and since this last fraction is in lowest terms, we see that the dimension of
$M$ is
${\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}1}$
(the degree of the denominator) and that the degree of $M$ is equal to
$p(1)=1+2\cdot
1={\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}3}$.
Recall that $M$ corresponded to the union of 3 lines. Notice that in this
example, the projective dimension of $M$ is $2$, which is equal to the
codimension
$3-{\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}1}=2$
of $M$. Hence, $M$ is Cohen-Macaulay. In summary, we can get lots of
information about $M$ from its (graded) Betti numbers.

######  Example 3.3.

(The Hilbert series doesn’t determine the Betti numbers) Let $k$ be a field,
$R=k[x,y]$, and consider the two ideals

$I=(x^{2},xy,y^{3})\quad\textrm{ and }\quad J=(x^{2},xy+y^{2}).$

One can check that both $R/I$ and $R/J$ have the same Hilbert series:

$HS(R/I)=HS(R/J)=1+2t+1t^{2}.$

However, these modules have different Betti numbers. We work out the minimal
free resolution and Betti numbers for $R/I$. Since $I$ has two generators of
degree 2 and one of degree 3, there are graded Betti numbers $\beta_{12}$ and
$\beta_{13}$. Similarly, the two minimal syzygies of $R/I$ correspond to the
relations

$\displaystyle y(x^{2})-x(xy)=0$ $\displaystyle\mbox{ \ which has degree 3, so
}\beta_{23}=1$ and $\displaystyle y^{2}(xy)-x(y^{3})=0$ $\displaystyle\mbox{\
which has degree 4, so }\beta_{24}=1.$

Continuing this process, we find the following minimal free resolutions and
graded Betti numbers for $R/I$ and $R/J$, respectively:

---
$\textstyle{{\begin{array}[]{c}R({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-3})^{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}\\\
\bigoplus\\\
R({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-4})^{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}\\\
\end{array}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{bmatrix}y&0\\\
-x&y^{2}\\\
0&-x\end{bmatrix}}$$\textstyle{{\begin{array}[]{c}R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-2})^{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}2}}\\\
\bigoplus\\\
R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-3})^{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}1}}\end{array}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{bmatrix}x^{2}&xy&y^{3}\end{bmatrix}}$$\textstyle{R}$$\textstyle{\begin{array}[]{c}\beta_{2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}}(R/I)={{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}\\\
\beta_{2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}}(R/I)={{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}\end{array}}$$\textstyle{{\begin{array}[]{c}\beta_{1{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}2}}(R/I)={{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}2}}\\\
\beta_{1{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}3}}(R/I)={{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}1}}\end{array}}}$

---
$\textstyle{R({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-4})^{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{bmatrix}xy+y^{2}\\\
-x^{2}\end{bmatrix}}$$\textstyle{R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}-2})^{{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{bmatrix}x^{2}&xy+y^{2}\end{bmatrix}}$$\textstyle{R.}$$\textstyle{{\beta_{2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}}(R/J)={{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}1}}}}$$\textstyle{{\beta_{1{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}2}}(R/J)={{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}2}}}}$

$\begin{array}[]{cc}\begin{array}[]{r|ccccc}\beta(R/I)&0&1&2&\\\ \hline\cr
0&1&-&-&\\\
1&-&2&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}&\\\
2&-&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}&1&\\\
\end{array}&\hskip 60.00009pt\begin{array}[]{r|ccccc}\beta(R/J)&0&1&2&\\\
\hline\cr 0&1&-&-&\\\ 1&-&2&-&\\\ 2&-&-&1&\\\ \end{array}\end{array}$

Finally, if we calculate the Hilbert series from Equation 3.1, we notice that
the calculation is the same for $R/I$ and $R/J$:

$HS(R/I)=\frac{1-2t^{2}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}-t^{3}+t^{3}}+t^{4}}{(1-t)^{3}}=\frac{1-2t^{2}+t^{4}}{(1-t)^{3}}=HS(R/J).$

The cancellation of the
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}t^{3}}$
terms is known as a consecutive cancellation, and one can see the two
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$s on the
diagonal in the Betti table for $R/I$. For the reader who knows about Gröbner
degenerations, $I$ is the initial ideal of $J$ coming from a Lex term-order.
Any such degeneration will preserve the Hilbert series, but not necessarily
the Betti numbers. For results concerning the relationship between the Betti
numbers of ideals and those of their initial ideals, see [2, 10, 34, 32, 33,
60].

######  Example 3.4.

We would be remiss if, in this article dedicated to David Eisenbud on his
birthday, we didn’t also mention that the connection between graded Betti
numbers and geometry is a rich and beautiful story. In his book [37], he
paints a story that begins with the following surprising fact from geometry.
If $X$ is a set consisting of seven general333this means that no more than 3
lie on a plane and no more than 5 on a conic. points in $\mathbb{P}^{3}$, then
the Hilbert series of the coordinate ring for $X$ is completely determined by
this data. However, this is not sufficient to determine the Betti numbers of
the coordinate ring of $X$. Indeed, these numbers are either $\\{1,4,6,3\\}$
or $\\{1,6,8,3\\}$ depending on whether or not the points lie on a curve of
degree 3.

###  3.2. Resolutions for Ideals with Few Generators

Over a polynomial ring $R=k[x_{1},\ldots,x_{n}]$, calculating a free
resolution is tantamount to producing the sets of dependence relations among
the generators of a module. In simple cases this is straightforward, as the
following example shows:

######  Example 3.5.

Consider the module $M=R/(f)$, where $f$ is a homogeneous polynomial in $R$.
Then

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\begin{bmatrix}f\end{bmatrix}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M}$

is a minimal free resolution of length $1$, since over our polynomial ring
$R$, $f$ is a _regular_ element and cannot be killed by multiplication by any
nonzero element.

If $I$ is an ideal minimally generated by two polynomials $f$ and $g$, then
the minimal free resolution of $R/I$ has length two. Indeed, if $c=\gcd(f,g)$,
then the following is a minimal free resolution:

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137.64029pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 167.64029pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
205.31042pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
205.31042pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R/I}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

This example can be summarized by the following result:

######  Proposition 3.6.

If $I$ is an ideal in a polynomial ring $R$ that is minimally generated by one
or two homogeneous polynomials, then the projective dimension of $R/I$ is
equal to the minimal number of generators, and the Betti numbers are either
$\\{1,1\\}$ or $\\{1,2,1\\}$.

Whatever optimistic generalization of this proposition one might have in mind
for ideals with $3$ or more generators will certainly fail to be true, as we
have the following astonishing results of Burch and Bruns:

######  Theorem 3.7 (Burch, 1968 [20]).

For each $N\geqslant 2$, there exists a three-generated ideal $I$ in a
polynomial ring $R=k[x_{1},\ldots,x_{N}]$ such that
$\operatorname{pdim}(R/I)=N$.

So we can always find free resolutions of maximal length by simply using $3$
generated ideals. In fact, in some sense “every” free resolution is the free
resolution of a $3$-generated ideal:

######  Theorem 3.8 (Bruns, 1976 [15]).

Let $R=k[x_{1},\ldots,x_{n}]$ and

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{d-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M}$

be a minimal free resolution of a finitely generated graded $R$-module $M$.
Then there exists a $3$-generated ideal $I$ in $R$ with minimal free
resolution

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
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148.8002pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
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0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
229.8898pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
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267.55994pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
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0.0pt\hbox{$\textstyle{R/I}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

######  Remark 3.9.

Note that the rank of $F_{2}^{\prime}$ may be different than that of $F_{2}$,
but a rank calculation yields that

$\operatorname{rk}F_{2}^{\prime}=3-1+\operatorname{rk}F_{3}-\operatorname{rk}F_{4}+\cdots\pm\operatorname{rk}F_{n}=2+\operatorname{rk}F_{2}-\operatorname{rk}F_{1}+\operatorname{rk}F_{0}.$

From this, it follows that $\beta_{2}$ can be arbitrarily large for
3-generated ideals.

Our point in presenting these results is to make plain that free resolutions
are complicated — even for ideals with 3 generators! However, if in Example
3.5 we add a further restriction for the ideal $I=(f,g)$ and require that $f$
and $g$ have no common factors (meaning that $g$ is a _regular_ element modulo
$f$), then the only relations between $f$ and $g$ are given by the “obvious”
relation that $gf-fg=0$. This fact _does_ generalize nicely to any set
$\\{f_{1},\ldots,f_{c}\\}$ of homogeneous polynomials provided $f_{i}$ is a
regular element modulo the previous $f_{j}$. Such elements form what is called
a regular sequence, and the ideal they generate is resolved by the Koszul
complex. Rather than introducing the topic here, we point the reader to some
of the many nice references for learning about the Koszul complex, such as
[37, Chapter 17], [16, Section 1.6], or [5, Example 1.1.1].

The most important fact we will need about the Koszul complex is that it is a
resolution (of $R/(f_{1},\ldots,f_{c})$) if and only if the
$f_{1},\ldots,f_{c}$ form a regular sequence, and that the Betti numbers (and
ranks of syzygy modules) of the Koszul complex are given by binomial
coefficients.

######  Theorem 3.10.

If $I$ is an ideal generated by a regular sequence of $c$ homogeneous
polynomials, then

$\operatorname{rk}\Omega_{i}(R/I)={c-1\choose i-1},$

and therefore

$\beta_{i}(R/I)={c\choose i}.$

######  Remark 3.11.

To the reader not familiar with Koszul complexes, it might be instructive to
carefully write out the maps involved to get a feel for how resolutions are
constructed. Essentially, the point is that the generating $i$th syzygies are
built from using $i$ generators and the fact that $f_{j}f_{i}=f_{i}f_{j}$.
Alternatively, perhaps the quickest way to define the Koszul complex is just
to take the tensor product of the $c$ minimal free resolutions of $R/(f_{i})$:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Since multiplication by $f_{i}$ has rank one, if one calculates the ranks in
the tensor product inductively, one will see Pascal’s Triangle appearing,
providing a justification of the claims in Theorem 3.10.

###  3.3. How Small Can the Ranks of Syzygies Be?

If $I$ is an ideal that is generated by a regular sequence then as we saw in
the previous section, the minimal free resolution for $R/I$ is given by the
Koszul complex. For instance, if $I$ has height $8$, then $\beta_{4}(R/I)$
will be equal to ${8\choose 4}=70$, and the syzygy module $\Omega_{4}(R/I)$
will have rank ${7\choose 3}=35$. We will see in the next section (Conjectures
4.1 and 4.2) that among all ideals of height $8$ these numbers are conjectured
to be the smallest possible values for $\beta_{4}$ and
$\operatorname{rk}\Omega_{4}$ respectively. In short, these conjectures assert
a relationship between the ranks of syzygies and the height (or codimension)
of the ideal. Before we present these conjectures, which will occupy the
remainder of the paper, we close with an example and theorem that give the
sharpest possible bound for ranks of syzygies if one does not refer to
codimension.

######  Example 3.12 (Bruns, 1976 [15]).

Let $R=k[x_{1},\ldots,x_{n}]$. There is a finitely generated module $M$ over
$R$ with the following resolution:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{2n-3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

In other words, the $i$th Betti number is $2i+1$ except for the last two Betti
numbers. This is the case for an even nicer reason: if one calculates the
ranks of each syzygy module (which can be read off as the rank of the $i$th
map $\pi_{i}$ in the resolution) one sees that the ranks are:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{n-1}$$\textstyle{R^{2n-3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{n-2}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{3}$$\textstyle{R^{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{2}$$\textstyle{R^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

In other words, in this example the $i$th syzygy module has rank equal to $i$,
except for the last one. This bound holds for any module, which is the content
of the great Syzygy Theorem.

######  Theorem 3.13 (Syzygy Theorem, Evans–Griffith, 1981 [43]).

Let $M$ be a finitely generated module over a polynomial ring $R$. If
$\Omega_{i}(M)$ is not free, then $\operatorname{rk}\Omega\geqslant i$. Hence,
if $\operatorname{pdim}M=p$, then

$\operatorname{rk}\Omega_{i}(M)\geqslant i,\ \mbox{for $i<p$.}$

Moreover,

$\beta_{i}(M)=\operatorname{rk}\Omega_{i}(M)+\operatorname{rk}\Omega_{i+1}(M)\geqslant\begin{cases}2i+1&\mbox{if
$i<p-1$}\\\ p&\mbox{if $i=p-1$}\\\ 1&\mbox{if $i=p$}\end{cases}$

where $\Omega_{i}(M)$ denotes the $i$th syzygy module of $M$.

The Syzygy Theorem together with Bruns’ example provides a sharp lower bound
for $\beta_{i}(M)$. Without further conditions on $M$, there is not much more
we can say. However, if we add additional hypotheses on $M$ — for instance,
requiring $M$ to be Cohen-Macaulay, or of a fixed codimension $c$ — then the
bounds above appear to be far from sharp. Indeed, we will discuss a conjecture
that states that in fact $\beta_{i}(M)\geqslant{c\choose i}$; when $c$ is
large, this conjecture is much stronger than the Syzygy Theorem’s bound of
$2i+1$. Note that the ideal in Example 3.12 is of codimension $2$.

###  3.4. Other Possible Directions

Before we begin to focus on codimension, we want to say that there are many
distinct and interesting alternative questions on bounds for Betti numbers
that have been considered. We present some possibilities below.

One could decide to study ideals and then fix the number of generators of $I$;
for example, one could study the sets of Betti numbers of ideals defined by
$5$ homogeneous polynomials. Theorem 3.8 shows that this approach will not
allow for any upper bounds, except in trivial cases.

Refining this idea, one could add a condition on the _degrees_ of the
generators of these ideals, and for example ask what the maximal Betti numbers
for an ideal with $3$ quadratic generators might be. This question is
tractable, though incredibly difficult. Note that here we are not saying how
many variables are in the ring $R$. For instance, the largest Betti numbers
possible for an ideal generated by $3$ quadrics is $\\{1,3,5,4,1\\}$; note
that the projective dimension is $4$. More generally, the question of whether
there exists an upper bound on the projective dimension of an ideal defined by
$r$ forms of degree $d_{1},\ldots,d_{r}$ depending only on $r$ and
$d_{1},\ldots,d_{r}$, and not on the number of variables, is known as
Stillman’s Conjecture, and has been solved by Ananyan and Hochster [1] in
general. The question of providing effective upper bounds is much harder, and
some of the efforts in this direction can be found in [41]. See [42] for an
exposition on some of the followup results that expanded on the ideas
initiated by Ananyan and Hochster in their proof of Stillman’s conjecture; see
also [59] for a survey and [50, 25] for related work on the subject.

We saw in sections 2 and 3 that the (graded) Betti number determine the
Hilbert series; however, there can be many distinct sets $\\{\beta_{ij}(M)\\}$
for $R$-modules $M$ all with the same given Hilbert series. If one fixes a
Hilbert series, what are the possible sets $\\{\beta_{ij}(M)\\}$ for modules
$M$ with Hilbert series $h(t)$? The following theorems give a beautiful answer
that provides an upper bound for the Betti numbers.

######  Theorem 3.14 (Bigatti, 1993 [9], Hulett, 1993 [49], Pardue, 1996
[62]).

Let $I$ be a homogeneous ideal in $R=k[x_{1},\ldots,x_{n}]$. Consider the set

$\mathcal{H}=\\{J\subseteq R\textrm{ an ideal }\ |\ HS(R/J)=HS(R/I)\\}.$

There exists an ideal $L\in\mathcal{H}$ with the property that among all
ideals in $H$, the Betti numbers of $L$ are the largest:

$\beta_{ij}(R/J)\leqslant\beta_{ij}(R/L)\ \mbox{for all $i,j$ and for all
$J\in\mathcal{H}$.}$

The ideal $L$ that achieves the largest Betti numbers in the Theorem can be
described explicitly, and goes back 100 years to work of Macaulay [57]; it is
the known as the Lex-segment ideal. To construct $L$, we start by going over
each degree $D$ and ordering all the monomials in $R_{D}$ lexicographically.
Then we collect the first $\dim_{k}(J_{D})$ monomials in degree $D$, for all
$D$. Macaulay showed the ideal $L$ generated by all these monomials has the
same Hilbert function as our original ideal $J$; in other words, it is an
ideal in $\mathcal{H}$. Bigatti, Hulett, and Pardue then showed that this
special ideal has in fact the largest possible Betti numbers with the same
Hilbert function as $I$. Moreover, if we fix a Hilbert polynomial, and
consider all the saturated ideals $I$ with that fixed Hilbert polynomial,
there is also a particular lex-segment ideal that maximizes the Betti numbers
[26].

Finally, we remark that while this paper is devoted to the ranks of modules
appearing in a minimal resolution — that is, the study of acyclic complexes.
There has been much work devoted more generally to complexes, or even more
generally to differential modules. For instance, it was conjectured in [7,
Conjecture 5.3] that if $F_{\bullet}$ is any complex over a $d$-dimensional
local ring, and if the homology $H(F)$ has finite length, then
$\sum_{i}\operatorname{rk}F_{i}\geqslant 2^{d}$. This was shown in [7] for the
case when $d\leqslant 3$, and in [35] in the multigraded setting (for all
$d$). However, the conjecture is false in general. Indeed, in [52], an example
is given of a complex of $R$-modules such that $H(F)$ has length $2$ but
$\sum_{i}\operatorname{rk}F_{i}<2^{d}$ for all $d\geqslant 8$. See also [22,
24, 23, 13].

In the remainder of the paper we will state several conjectures concerning
lower bounds for the $\beta_{i}(R/I)$ in terms of $c=\operatorname{codim}R/I$.
As an appetizer, notice that the Krull altitude theorem asserts that the
codimension of $R/I$ must be at most the minimal number of generators, i.e.
$\beta_{1}(R/I)\geqslant c.$ Meanwhile, the Auslander-Buchsbaum formula above
guarantees that the length of the resolution of $R/I$ is at least the
codimension $c$, which implies that $\beta_{c}(R/I)\geqslant 1$. With these
two classical results giving us information about Betti numbers in terms of
codimension, we now proceed to the main conjecture we want to focus on.

##  4\. The Buchsbaum–Eisenbud–Horrocks Conjecture and the Total Rank
Conjecture

In the late 1970s, Buchsbaum and Eisenbud [18], and independently Horrocks
[46, Problem 24], conjectured that the Koszul complex is the smallest free
resolution possible; more precisely, that the Betti numbers of any finitely
generated module are at least as large as those of a complete intersection of
the same codimension as given in Theorem 3.10:

######  Conjecture 4.1 (BEH Conjecture).

Let $R=k[x_{1},\ldots,x_{n}]$, where $k$ is a field, and $M$ be a nonzero
finitely generated graded $R$-module of codimension $c$, meaning that
$\operatorname{ht}\textrm{ann}(M)=c$. Then

$\beta_{i}(M)\geqslant{c\choose i}$

for all $0\leqslant i\leqslant\operatorname{pdim}_{R}M$.

Actually, both Buchsbaum and Eisenbud [18] and Horrocks [46, Problem 24]
propose the following stronger conjecture:

######  Conjecture 4.2 (Stronger BEH Conjecture for the ranks of the
syzygies).

Let $R=k[x_{1},\ldots,x_{n}]$, where $k$ is a field, and $M$ be a nonzero
finitely generated graded $R$-module of codimension $c$. Then

$\operatorname{rk}(\Omega_{i}(M))\geqslant{c-1\choose i-1}.$

Originally, Horrocks’ problem was stated for finite length modules over a
regular local ring, i.e., the case that $\operatorname{codim}M$ was as large
as possible, and equal to the dimension of the ring. On the other hand,
Buchsbaum and Eisenbud were interested in resolutions of $R/I$ for a general
local ring $R$. They conjectured that the minimal free resolution of $R/I$
possessed the structure of a commutative associative differential graded
algebra; they then showed that if this held, and $I$ had grade $c$, then the
corresponding inequalities (which they independently attribute to Jürgen
Herzog) on the ranks above would hold:

######  Theorem 4.3 (Buchsbaum–Eisenbud, 1977, Proposition 1.4 in [18]).

If $R/I$ has codimension $c$ and the minimal free resolution of $R/I$
possesses the structure of an associative commutative differential graded
algebra, then $\beta_{i}(R/I)\geqslant{c\choose i}$ for all $i$. Furthermore,
the rank of the $i$th syzygy module is at least ${c-1\choose i-1}$.

For some time it was open whether or not all resolutions could be given such a
DGA structure. It turns out that this is not necessarily the case [4, Example
5.2.2], though notably any algebra $R/I$ of projective dimension at most $3$
or of projective dimension $4$ that is Gorenstein will have such a resolution
[18, 55, 54]. See also [8] for more on the $\operatorname{pdim}(R/I)\leqslant
3$ case.

######  Remark 4.4.

Throughout, we will adopt the convention that ${n\choose k}$ is zero unless
$0\leqslant k\leqslant n$.

As a motivating example, let $R/I$ be a cyclic module of codimension $c$.

  * •

The principal ideal theorem guarantees that $I$ must be generated by at least
$c$ elements, so $\beta_{1}(R/I)\geqslant{c\choose 1}$.

  * •

The Auslander–Buchsbaum formula implies that
$\operatorname{pdim}(R/I)\geqslant c$, which implies that
$\beta_{c}(R/I)\geqslant{c\choose c}.$

  * •

If $I$ is generated by exactly $c$ elements, then $R/I$ is resolved by the
Koszul complex, and then $\beta_{i}(R/I)={c\choose i}$ for all $i$.

If $I$ has more than $c$ generators, then $I$ will not be a complete
intersection, and in general there is no structural result concerning its
minimal free resolution. However, it stands to reason (at least for optimists)
that perhaps the Betti numbers can only increase as the number of generators
grows and grows.

In the rest of this paper we have two goals. First, we want to survey various
generalizations of the BEH Conjecture and give the state of the art for each
of these. Second, we want to include a few basic constructions and techniques
that could be helpful to those who want to work in this field. For a more
thorough treatment, we refer the reader to the book [44] and survey article
[28].

We have opted to give a summary of classical results on the BEH Conjecture
first, but we want to point out right away that an immediate consequence of
the BEH Conjecture is that if the conjecture is true, then the sum of the
Betti numbers will be at least $2^{c}$. This weaker conjecture, known as the
Total Rank Conjecture, was proven by Walker in 2018 [65]. Since then, there
has been increasing evidence that apart from complete intersections, which are
resolved by the Koszul complex, it may be true that in fact the sum of the
Betti numbers is always at least $2^{c}+2^{c-1}$. In the final section of the
survey, we present the case for this stronger conjecture.

###  4.1. General Purpose Tools

The BEH Conjecture is known in a surprisingly small number of cases. Indeed,
as a first challenge, it is open an open question whether
$\beta_{2}(R/I)\geqslant{5\choose 2}$ whenever $I$ is an ideal of codimension
$5$. In this section, we present a collection of general purpose tools and use
them to show that if $c\leqslant 4$ then the conjecture holds. We also
carefully describe how localization can reduce the conjecture to the finite
length case, provided we work over arbitrary regular local rings.

######  Proposition 4.5 (Buchsbaum–Eisenbud, 1973, Theorem 2.1 in [17], see
also [57]).

Suppose that $M$ is a module of codimension $c$. Then

$\beta_{1}(M)-\beta_{0}(M)+1\geqslant c.$

If equality holds, then $M$ is resolved by the Buchsbaum–Rim complex.

Note that this result includes both the Principal Ideal Theorem (when $M=R/I$
and thus $\beta_{0}(M)=1$) and the fact that the Koszul complex (a special
instance of the Buchsbaum–Rim complex [19]) resolves complete intersections.
Below is a version of this result in terms of Betti numbers:

######  Corollary 4.6.

If $M$ is a module of codimension $c$, then

$\beta_{1}(M)\geqslant\beta_{0}(M)+c-1.$

If equality holds, then for all $i\geqslant 2$

$\beta_{i}(M)={\beta_{0}(M)+i-3\choose
i-2}{\beta_{1}(M)\choose\beta_{0}(M)+i-1}.$

As an exercise, the reader can prove that if $\beta_{1}(M)=\beta_{0}(M)+c-1$
then the BEH conjecture holds, by the equality of binomial coefficients above.

Discounting cases when equality holds, this lower bound
$\beta_{1}(M)>\beta_{0}(M)+c-1$ might not at first glance seem very useful,
since it only gives information about $\beta_{1}(M)$. However, when $M$ is
Cohen-Macaulay we can use this result to also gain information about
$\beta_{c-1}(M)$ as well by appealing to duality. Indeed, if $M$ is a Cohen-
Macaulay module, meaning that the codimension $c$ of $M$ is equal to its
projective dimension, then applying $\operatorname{Hom}(-,R)$ to a resolution
of $M$ will yield a resolution of $\operatorname{Ext}^{c}_{R}(M,R)$, another
Cohen-Macaulay module. This yields the following observation:

######  Proposition 4.7.

If $\\{\beta_{0},\ldots,\beta_{c}\\}$ is the Betti sequence for a Cohen-
Macaulay module, then so is the reverse sequence
$\\{\beta_{c},\ldots,\beta_{0}\\}$.

As an application of these ideas, let us use these results to prove Conjecture
4.1 for $c\leqslant 4$. We focus on $c=3$ and $c=4$, as the smaller cases
follow immediately from the principal ideal theorem.

When $\mathbf{c=3}$, Corollary 4.6 and Proposition 4.7 imply that

$\\{\beta_{0},\beta_{1},\beta_{2},\beta_{3}\\}\geqslant\\{\beta_{0},3+\beta_{0}-1,3+\beta_{3}-1,\beta_{3}\\}\geqslant\\{1,3,3,1\\}$

where the inequalities are interpreted entry by entry.

Similarly, for $\mathbf{c=4}$ we obtain

$\\{\beta_{0},\beta_{1},\beta_{2},\beta_{3},\beta_{4}\\}\geqslant\\{\beta_{0},4+\beta_{0}-1,\beta_{2},4+\beta_{4}-1,\beta_{4}\\}\geqslant\\{1,4,\beta_{2},4,1\\}.$

From here, we can apply the Syzygy Theorem (3.13) and notice that in a minimal
free resolution

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{\beta_{4}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{\beta_{3}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{3}}$$\textstyle{R^{\beta_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{2}}$$\textstyle{R^{\beta_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{\beta_{0}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M,}$

the image of $\pi_{3}$ is equal to $\Omega_{3}(M)$, and thus the rank of
$\pi_{3}$ is at least $3$ by the Syzygy Theorem; here we used that $c=4$, so
that $\Omega_{3}(M)$ is not free.

Similarly, working now on the resolution of the dual
$\operatorname{Ext}^{4}_{R}(M,R)$, we can see that the rank of $\pi_{2}$ must
be at least $3$ as well. Hence

$\beta_{2}=\operatorname{rk}\pi_{2}+\operatorname{rk}\pi_{3}\geqslant 6,$

as required.

However, if we try the same tricks with $\mathbf{c=5}$, the best we can get is
that

$\\{\beta_{0},\beta_{1},\beta_{2},\beta_{3},\beta_{4},\beta_{5}\\}\geqslant\\{1,5,7,7,5,1\\}.$

There are, however, other techniques one could use to try and complete this
case:

  * •

Suppose $M$ is cyclic, that is, $\beta_{0}=1$. Then one may assume that
$\beta_{1}>5$. Indeed, if $\beta_{1}=5$, then $M$ is a complete intersection
and the Koszul complex is a resolution. Surprisingly, Conjecture 4.1 is still
open even if we assume $c=5$ and that $M$ is cyclic. More precisely, it is
still open whether or not $\beta_{2}\geqslant{5\choose 2}=10$.

  * •

One could suppose further that $\beta_{1}=6$, so $M=R/I$ is an almost complete
intersection. A result of Kunz [53] guarantees that $R/I$ is not Gorenstein,
and thus $\beta_{5}\geqslant 2$. Using linkage, Dugger [36] was able to show
in this case that $\beta_{2}\geqslant 9$.

  * •

In general, for cyclic modules, the rank of $\pi_{1}$ will be 1, and thus the
Syzygy Theorem implies that

$\beta_{2}=\operatorname{rk}\pi_{2}+\operatorname{rk}\pi_{3}\geqslant\operatorname{rk}\pi_{2}+3=(\beta_{1}-\operatorname{rk}\pi_{1})+3=\beta_{1}+2,$

so whenever $\beta_{0}=1$ and $\beta_{1}\geqslant 8$ we will have the BEH
bound for $\beta_{2}$.

We close out this section with another general technique and an application.
Let $M$ be a graded module and $P$ be a prime ideal in its support. Since
localization is exact, any minimal free resolution of $M$ over $R$ will remain
exact upon localization at $P$. Hence, over the local ring $R_{P}$, the
minimal free resolution of $M_{P}$ must be a direct summand of this
resolution. In other words,

$\beta_{i}^{R_{P}}(M_{P})\leqslant\beta_{i}^{R}(M).$

We now give two applications of this idea. The first shows that if we wanted
to prove a stronger version of the BEH conjecture, we could restrict to finite
length modules.

######  Conjecture 4.8 (Local BEH Conjecture).

Let $R$ be a local ring and $M$ a finitely generated $R$-module of codimension
$c$. Then for all $i$,

$\beta_{i}(M)\geqslant{c\choose i}.$

######  Lemma 4.9.

To prove Conjecture 4.8, it suffices to prove it for modules of finite length.

###### Proof.

Let $M$ be an arbitrary module, not necessarily of finite length. Say that $M$
has codimension $c$, and note that there must be a minimal prime $P$ of $M$ of
height $c$. Then $M_{P}$ is a finite length module over $R_{P}$, and

$\beta_{i}^{R}(M)\geqslant\beta_{i}^{R_{P}}(M_{P})$

by our localization argument. Since $M_{P}$ must then have codimension $c$,
the result follows. ∎

We apply this idea to the case of monomial ideals and present a short proof
that the BEH conjecture holds for monomial ideals. As we will see in Section
5, there are in fact stronger bounds that hold in the monomial case.

######  Theorem 4.10.

Let $I$ be a monomial ideal of height $c$ in a polynomial ring $R$. Then the
BEH conjecture holds and $\beta_{i}(R/I)\geqslant{c\choose i}.$

###### Proof.

Our first step is to reduce to the case that $I$ is squarefree. Indeed, if $I$
is a monomial ideal, then there is a squarefree monomial ideal (perhaps in a
larger number of variables) called the _polarization_ of $I$ which has the
same codimension and Betti numbers as $I$.

So consider the primary decomposition of a squarefree monomial ideal. It
consists entirely of minimal primes that are generated by subsets of the
variables, and all must have height at least $c$. Choose any one you like and
call it $P$. Note that $I_{P}=P_{P}$, since $P$ is minimal. Without loss of
generality, we can assume $P=(x_{1},\ldots,x_{r})$ for some $r\geqslant c$.
Then upon localizing $R/I$ at $P$, it is easy to see that

$(R/I)_{P}\cong
R[x_{1},\ldots,x_{r}]_{(x_{1},\ldots,x_{r})}/(x_{1},\ldots,x_{r}),$

whose Betti number are obtained from the Koszul complex on
$x_{1},\ldots,x_{r}$. Thus

$\beta_{i}(R/I)\geqslant\beta_{i}(R_{P}/I_{P})={r\choose i}\geqslant{c\choose
i}.$

The reader will note that if we choose $r$ as large as possible, then $r$
would be the _big height_ of the squarefree monomial ideal $I$, that is, the
largest height of an associated prime. ∎

######  Remark 4.11.

Notice that it is _not_ clear that to prove the original BEH Conjecture (which
was stated over a polynomial ring) one can simply study finite length modules.
Indeed, this localization argument might require one to work over
localizations of polynomial rings, which despite being regular will not be
polynomial rings.

Finally, we include another important general result that comes up frequently.
As motivation we refer to Example 3.1 with $I=(xy,xz,yz)$. Notice that the
element $\ell=x-y-z$ is a regular element on $M=R/I$, for instance by looking
at a primary decomposition. If we work over $\overline{R}=R/(\ell)\cong
k[y,z]$, then $\overline{M}\cong\overline{R}/(y^{2},yz,z^{2})$, which is a
module of finite length. Standard arguments show that when we go modulo a
regular element like this, the homological invariants (including the Betti
numbers) do not change. One application of this is the fact that the Betti
numbers of Cohen-Macaulay modules are the same as those of finite length
modules. We make this sentence precise in the following:

######  Proposition 4.12.

Let $M$ be a Cohen-Macaulay module of codimension $c$ over the polynomial ring
$R=k[x_{1},\ldots,x_{n}]$ where $k$ is any field. There exist a field
$k^{\prime}$and a finite length module $M^{\prime}$ over the polynomial ring
$R^{\prime}=k^{\prime}[y_{1},\ldots,y_{c}]$ such that the Betti numbers of $M$
and $M^{\prime}$ coincide. Thus the following sets are equal:

$\\{\beta_{i}(M):M\textrm{ Cohen-Macaulay of codimension }c\textrm{ over
}k[x_{1},\ldots,x_{n}]\textrm{ for some }k\\}$

$\mathrel{\rotatebox{90.0}{$=$}}$

$\\{\beta_{i}(M):M\textrm{ is finite length over
}k[x_{1},\ldots,x_{c}]\textrm{ for some }k\\}.$

###### Proof.

Let $M$ be a Cohen-Macaulay module of codimension $c$ over
$k[x_{1},\ldots,x_{n}]$. If $k$ is infinite, set $k^{\prime}=k$. If $k$ is
finite, then we may enlarge the field, say to the algebraic closure
$k^{\prime}=\overline{k}$, since flat base change will not affect the Betti
numbers of $M$. Set $\overline{R}=k^{\prime}[x_{1},\ldots,x_{n}]$ and
$\overline{M}=M\otimes_{R}\overline{R}$, where $\overline{M}$ is regarded as
an $\overline{R}$-module. Note that $\beta_{i}(M)=\beta_{i}(\overline{M})$.
Now, since we are working over an infinite field, there is a sequence of
linear forms $\ell_{1},\ldots,\ell_{n-c}\in\overline{R}$ that is a maximal
regular sequence on $\overline{M}$. Now let
$R^{\prime}=\overline{R}/(\ell_{1},\ldots,\ell_{n-c})$ and set

$M^{\prime}=\overline{M}\otimes_{\overline{R}}R^{\prime}.$

Then since we have gone modulo a regular sequence,
$\beta_{i}(M^{\prime})=\beta_{i}(M)$, and since the $\ell_{i}$ were linear
forms, $R^{\prime}$ is isomorphic to a polynomial ring
$k^{\prime}[y_{1},\ldots,y_{c}]$. ∎

###  4.2. Other Results

As we mentioned in the previous section, the BEH conjecture 4.1 remains open
for modules of codimension $c\geqslant 5$ except in a small collection of
cases. There are, however, some classes of modules for which the BEH
Conjecture is known.

A deformation argument was used in [51, Remark 4.14] to show that the
conjecture holds for arbitrary $c$ when $M=R/I$ and $I$ is in the linkage
class of a complete intersection. Additionally, in [40] it was shown that if
the regularity of $M$ is small relative to the degrees of the first syzygies
of $M$, meaning the entries in a presentation matrix for $M$, then the BEH
conjecture holds. This will be discussed more carefully in Section 5.

The conjecture holds also when $M$ is multigraded, meaning that $M$ remains
graded no matter what weights the generators $x_{i}$ are given. In fact, there
are several proofs of this fact, for example [29, 30, 63], but perhaps the
strongest version is the result due to Brun and Römer, [14] which shows that
if $M$ is multigraded, then in fact $\beta_{i}(M)\geqslant{p\choose i}$, where
$p$ is the projective dimension of $M$. Since the projective dimension can
exceed the codimension, this is a much stronger result. Such a result cannot
hold more generally — after all, there are 3-generated ideals $I$ with
projective dimension $1000$, by Theorem 3.8, and in that case
$\beta_{1}(R/I)=3<{1000\choose 1}$. Nevertheless, it would be interesting to
know if there are other classes where ${p\choose i}$ is a lower bound for the
Betti numbers. We know of at least one other case, when the resolution of
$R/I$ is linear, which we present in Theorem 5.14. We will discuss the
multigraded case in more detail in Section 5, when we discuss stronger bounds
on Betti numbers.

Finally, the BEH conjecture 4.1 also holds for finite length modules of Loewy
length 2 over any regular local ring $(R,\mathfrak{m})$, meaning modules $M$
satisfying $\mathfrak{m}^{2}M=0$ [27, 21].

###  4.3. The Total Rank Conjecture

If the Buchsbaum–Eisenbud–Horrocks Rank Conjecture is true, an immediate
corollary would be the Total Rank Conjecture, which is obtained by adding the
individual inequalities:

######  Conjecture 4.13 (Total Rank Conjecture).

If $M\neq 0$ is a finitely generated graded module over
$R=k[x_{1},\ldots,x_{n}]$ of codimension $c$, then

$\sum_{i=0}^{c}\beta_{i}(M)\geqslant 2^{c}.$

This Conjecture was settled in 2018 by Walker [65], except in the case that
$k$ has characteristic $2$. In fact, Walker’s result also applies to finitely
generated modules over an arbitrary local ring $R$ containing a field of odd
characteristic. This result truly was a breakthrough in the field.

Even though the Total Rank Conjecture is settled (except in characteristic
two), we cannot resist sharing some of the beautiful historical results in
this story and compare them with the modern treatment. For example, the odd
length case has a simple solution via elementary methods:

######  Lemma 4.14.

Suppose that $M$ is a finitely generated $R$-module of (finite) odd length
over $R=k[x_{1},\ldots,x_{n}]$. Then

$\sum_{i=0}^{n}\beta_{i}(M)\geqslant 2^{n}.$

###### Proof.

The Hilbert series $h_{M}(t)$ of $M$ is a polynomial in $t$, say
$h_{M}(t)=h_{0}+h_{1}t+\cdots+h_{r}t^{r}$. We can also write it as

$h_{M}^{R}(t)=\frac{\sum_{i,j}(-1)^{i}\beta_{i,j}(M)t^{j}}{(1-t)^{n}}.$

Plugging in $t=-1$, we obtain

$2^{n}h_{M}^{R}(-1)=\sum_{i,j}(-1)^{i+j}\beta_{i,j}(M),$

so

$2^{n}\left|h_{0}-h_{1}+\cdots+(-1)^{r}h_{r}\right|=\left|\sum_{i,j}(-1)^{i+j}\beta_{i,j}(M)\right|\leqslant\sum_{i}\beta_{i}(M).$

On the other hand, $h_{0}+h_{1}+\cdots+h_{r}$ is the rank of $M$, which we
assumed to be odd. Therefore, $h_{0}-h_{1}+\cdots+(-1)^{r}h_{r}$ is also odd,
and thus nonzero. In particular,

$2^{n}\leqslant\sum_{i=0}^{n}\beta_{i}(M).\qed$

In other words, for modules of finite odd length, the Total Rank Conjecture
holds simply due to constraints on its Hilbert function. In 1993, Avramov and
Buchweitz were able to obtain a generalization of this fact in [6]. Their most
general bound was that if $d\geqslant 5$ and $M$ is a module of finite length
over $R$, then

$\sum_{i=0}^{d}\beta_{i}(M)\geqslant\frac{3}{2}(d-1)^{2}+8.$

In particular, this shows that when $d=5$ the lower bound of $32=2^{5}$ in the
Total Rank Conjecture does hold. Their results were in fact much finer,
depending on the prime factors of the length of $M$, $\ell(M)$. For instance,
they show that

  * •

If $\ell(M)$ is odd, then $\sum\beta_{i}(M)\geqslant 2^{d}$, so they recover
the above result.

  * •

If $\ell(M)$ is even but not divisible by 6, then $\sum\beta_{i}(M)\geqslant
3^{d/2}\geqslant 2^{0.79d}$.

  * •

If $\ell(M)$ is divisible by 6 but not by 30, then $\sum\beta_{i}(M)\geqslant
5^{d/4}\geqslant 2^{0.58d}$.

  * •

If $\ell(M)$ is divisible by 30 but not by 60, then $\sum\beta_{i}(M)\geqslant
2^{(d+1)/2}$.

If we move forward 25 years, the following is a summary of Walker’s results:

######  Theorem 4.15 (Walker, 2018 [65, 66]).

Let $M$ be a finitely generated module of codimension $c$ over
$k[x_{1},\ldots,x_{n}]$.

  * •

If $\operatorname{char}{k}\neq 2$, then $\sum\beta_{i}(M)\geqslant 2^{c}.$

  * •

If $\operatorname{char}(k)=2$, then $\sum\beta_{i}(M)\geqslant
2{(\sqrt{3})}^{c-1}>2^{0.79c+0.208}.$

While the Total Rank Conjecture remains open in characteristic $2$, for that
case Walker [66, Theorem 5] did give the above bound of $2{(\sqrt{3})}^{d-1}$,
which improves the previous bounds by Avramov and Buchweitz [6]. We also
remark that the Total Rank Conjecture is related to the Toral Rank Conjecture
of Halperin [45]. For a survey on this and related results, see [61, 22, 24,
23].

In the following table, we indicate the current status (as of the writing of
this survey) of both the Total Rank Conjecture 4.13 and the BEH Conjecture
4.1. The reader may want to refer to Table 2 at the end of the paper to see
what the case for stronger bounds is.

| $c\leqslant 4$ | $c\geqslant 5$
---|---|---
$\beta_{i}\geqslant{c\choose i}$ | follows from the |
Syzygy Theorem | Open
(Evans–Griffith 1981) [43] |
$\sum_{i}\beta_{i}\geqslant 2^{c}$ | follows from box above ✓ |  $c=5$ (Avramov–Buchweitz, 1993) [6]
all $c$ char$(k)\neq 2$ (Walker, 2018) [65]
Table 1. Status of the BEH and Total Rank Conjectures for a module of
codimension $c$

##  5\. Stronger bounds

We now turn to the question of whether there are larger bounds for Betti
numbers and whether or not these bounds are achieved, starting with Walker’s
result [65].

######  Theorem 5.1 (Walker, 2018, Theorem 1 in [65]).

Suppose that $\operatorname{char}k\neq 2$, and let $M$ be a finitely generated
graded $k[x_{1},\ldots,x_{n}]$-module of codimension $c$. Then

$\sum_{i=0}^{c}\beta_{i}(M)\geqslant 2^{c}$

with equality if and only if $M$ is not a complete intersection.

######  Remark 5.2.

The situation where we have a module $M\cong R/I$ with $I$ an ideal generated
by a regular sequence is very important, and we will want to distinguish it
from any other kind of module; we will abuse notation444This is an abuse of
notation since the expression “complete intersection” typically refers to a
ring, not a module. and say that $M$ is a complete intersection. We will say
that a module $M$ is not a complete intersection whenever $M$ is not
isomorphic to any quotient of $R$ by a regular sequence; in particular, when
we refer to modules $M$ that are not a complete intersection, we will include
any non-cyclic module.

Notice that this theorem says that the only time that the Betti numbers sum to
$2^{c}$ is in the case of a complete intersection. Surprisingly, the next
smallest value for the sum of the Betti numbers that we know of is
$2^{c}+2^{c-1}$, which is 50% larger than the bound of $2^{c}$. The next two
examples show how to achieve this value. Notice that in one example this stems
from the fact that $1+3+2=6$, whereas in the other it is because $1+5+5+1=12$.

######  Example 5.3.

Let $I$ be the ideal $(x^{2},xy,y^{2})$ in $R=k[x,y]$. Then $R/I$ is a finite
length module of codimension $c=2$ with Betti numbers $\\{1,3,2\\}$. Notice
that these sum to $6$ which is $2^{2}+2^{2-1}$.

By adding new variables (to $R$ and also to $I$) we can extend this example to
any $c\geqslant 2$. Indeed, set $R=k[x,y,z_{1},\ldots,z_{c-2}]$, and let
$I=(x^{2},xy,y^{2},z_{1}^{2},z_{2}^{2},\ldots,z^{2}_{c-2})$. Then the minimal
free resolution of $R/I$ is obtained by tensoring the Koszul complex on
$\\{z_{1},\ldots,z_{c-2}\\}$ with the minimal free resolution of
$R/(x^{2},xy,y^{2})$. Thus

$\beta_{i}(R/I)={c-2\choose i}+3{c-2\choose i-1}+2{c-2\choose i-2}$

and we see that $\sum\beta_{i}(R/I)=2^{c}+2^{c-1}.$ We chose to adjoin
$z_{i}^{2}$ just so that our generators were all in the same degree, but one
could choose these additional generators to be of any degree. Note that in all
of these examples $I$ is monomial and $R/I$ is of finite length.

######  Example 5.4.

Consider the ideal

$G=(x^{2},y^{2},z^{2},xy-yz,yz-xy)$

in the ring $R=k[x,y,z]$. The height of $G$ is 3 and the Betti numbers of
$R/I$ are $\\{1,5,5,1\\}$. Note that $1+5+5+1=2^{3}+2^{2}$.

Let $c\geqslant 3$. Then as in Example 5.3, we can just add generators in new
variables, say

$I=G+(z_{1}^{2},\ldots,z_{c-3}^{2})$

and after tensoring with a Koszul complex we have that

$\beta_{i}(R/I)={c-3\choose i-3}+5{c-3\choose i-2}+5{c-3\choose
i-1}+{c-3\choose i}.$

Therefore,

$\sum\beta_{i}(R/I)=2^{c}+2^{c-1}.$

Note that all of the modules $R/I$ in this example are of finite length.

######  Example 5.5.

If one repeats Example 5.4 with $R=k[x,y,z,u,v]$ and $J=(xy,yz,zu,uv,vx)$
playing the role of $G$, then the numerics are exactly the same. $R/J$ has
codimension $3$ and the Betti numbers are $\\{1,5,5,1\\}$. The analogous
examples obtained by adding new generators will all be monomial but not of
finite colength. This distinction is important, because we will later see in
Corollary 5.8 that there are bounds on the individual Betti numbers for
monomial ideal of finite colength that do not hold for monomial ideals more
generally, nor for general ideals of finite colength.

The following result in [31] shows that for modules that are not complete
intersections, this behavior of Betti numbers adding up to “50% more than
$2^{c}$” does hold for $c\leqslant 4$:

######  Theorem 5.6 (Charalambous–Evans–Miller, 1990, Theorem 3 in [31]).

Let $M$ be a finitely generated graded module of height $c$ over a polynomial
ring. Suppose $M$ is not a complete intersection. If $c\leqslant 4$, then
$\sum\beta_{i}(M)\geqslant 2^{c}+2^{c-1}$.

In fact, [31] actually provides minimal Betti sequences for each codimension.
For example, in codimension $c=4$ they show that
$\\{\beta_{0},\ldots,\beta_{4}\\}$ must be bigger (entry by entry) than at
least one of the following:

$\begin{array}[]{cccc}\\{1,5,9,7,2\\},&\\{1,6,10,6,1\\},&\\{2,6,8,6,2\\},&\\{1,6,9,6,2\\}\\\
\\{2,7,9,5,1\\},&&&\\{2,6,9,6,1\\}\end{array}.$

Note that the entries on the bottom row are the reverse of those directly
above. The proof of this result uses techniques of linkage and relies on the
classification [56] of the possible algebra structures on
$\operatorname{Tor}_{\bullet}^{R}(R/I,k)$. This result led the authors to ask
the following question:

######  Question B (Charalambous–Evans–Miller, 1990).

If $M$ is finitely generated graded module over $k[x_{1},\ldots,x_{n}]$ of
codimension $c$ that is not a complete intersection, is

$\sum\beta_{i}(M)\geqslant(1.5)2^{c}=2^{c}+2^{c-1}?$

We will now discuss several instances where we have an affirmative answer to
this question. We remark, however, that the techniques — and indeed the
underlying reasons — in each instance are completely different! Here are some
natural follow-up questions.

######  Question C.

What other modules $M$ of codimension $c$ satisfy
$\sum\beta_{i}(M)=(1.5)2^{c}$?

######  Question D.

What are the smallest Betti sequences in a given codimension $c$, when we
range over all finitely generated modules of codimension $c$ over a polynomial
ring on any number of variables?

###  5.1. The Multigraded Case

Let $R=k[x_{1},\ldots,x_{n}]$ and let $M$ be a finitely generated graded-
module over $R$. We say that $M$ is multigraded if it remains graded with
respect to any grading of the variables. For example, when $I$ is a monomial
ideal, $I$ and $R/I$ are multigraded (each).

######  Example 5.7.

Let $R=k[x,y,z]$. Consider

$M=\operatorname{Coker}\begin{pmatrix}y&0&z\\\ -x&z&0\\\
0&-y&-x\end{pmatrix}.$

The module $M$ is generated by $3$ elements; for example, since $M$ is a
quotient of $R^{3}$, we can take the images of the canonical basis elements
$e_{1},e_{2},e_{3}$ for generators of $M$. Then $M$ has relations

$ye_{1}=xe_{2},\ \ ze_{2}=ye_{3},\ \ ze_{1}=xe_{3}.$

Suppose we are given any weights on the variables. Then the module $M$ will be
graded as well by setting $\deg e_{1}:=\deg(x),\deg e_{2}:=\deg(y),\deg
e_{3}:=\deg(z)$.

In contrast, consider

$N=\operatorname{Coker}\begin{pmatrix}x&y&z&0\\\ 0&x&y&z\end{pmatrix},\hskip
40.00006pt\begin{array}[]{r|cccc}\beta(N)&0&1&2&3\\\ \hline\cr 0&2&4&-&-\\\
1&-&-&4&2\\\ \end{array}.$

Then $N$ is generated by $e_{1},e_{2}$, and the relations

$ye_{1}+xe_{2}=0,\quad ze_{1}+ye_{2}=0$

imply that

$\deg(y)+\deg(e_{1})=\deg(x)+\deg(e_{2})$
$\deg(z)+\deg(e_{1})=\deg(y)+\deg(e_{2})$

which has no solution for example when $\deg(y)=\deg(x)=0$ and $\deg(z)=1$. In
other words, $N$ is not multigraded. Notice that $N$ is finite length as an
$R$-module, and thus has codimension $3$.

The following theorem gives strong bounds on the individual Betti numbers of
modules that are multigraded and of finite-length. For instance they imply
that a module of codimension $c$ must have Betti numbers that either exceed
$\\{1,4,5,2\\}$ or $\\{2,5,4,1\\}$. Noting that the Betti numbers of $N$ in
the previous example violate both of these bounds provides yet another reason
why it is not multigraded.

######  Theorem 5.8 (Charalambous–Evans, 1991 [30]).

Let $M$ be a multigraded module of finite length and let $\gamma_{i}(M)$
denote the rank of the $i$th syzygy module of $M$. Then for all $i$

$\gamma_{i}(M)\geqslant{n-1\choose i-1}\mbox{ and therefore
}\beta_{i}(M)\geqslant{n\choose i}.$

Further if $M$ is not a complete intersection, then at least one of the
following holds:

  1. (a)

for all $i$, $\gamma_{i}(M)\geqslant{n-1\choose i-1}+{n-2\choose i-2}$, and
therefore $\beta_{i}(M)\geqslant{n\choose i}+{n-1\choose i-1}$;

  2. (b)

for all $i$, $\gamma_{i}(M)\geqslant{n-1\choose i-1}+{n-2\choose i-1}$, and
therefore $\beta_{i}(M)\geqslant{n\choose i}+{n-1\choose i}$.

######  Remark 5.9.

We want to emphasize that without the assumption that $M$ is multigraded _and_
of finite length, Theorem 5.8 is false if $c\geqslant 3$. Indeed, Examples 5.4
(respectively 5.5) give families of modules $R/I$ that are finite length
(respectively multigraded) but with Betti numbers that violate the bounds in
Theorem 5.8. This is essentially due to the fact that $R/I$ is Gorenstein in
both cases. Indeed, since Theorem 5.8 implies that either
$\beta_{0}(M)\geqslant 2$ or $\beta_{c}(M)\geqslant 2$, any Gorenstein algebra
$R/I$ that is not a complete intersection will violate the bounds in Corollary
5.10. In fact, we can use this to deduce the following classical fact: if $I$
is a monomial ideal in a polynomial ring $R$ such that $R/I$ is of finite
length and Gorenstein, then $R/I$ is a complete intersection.

Summing the inequalities for the Betti numbers in Theorem 5.8 yields the
following result, which is a special case of Question B.

######  Corollary 5.10.

If $M$ is a multigraded module of finite length then

$\sum\beta_{i}(M)\geqslant 2^{n}.$

Further if $M$ is not a complete intersection, then

$\sum_{i=0}^{n}\beta_{i}(M)\geqslant 2^{n}+2^{n-1}.$

We remark that in this case $n=\operatorname{codim}M$.

Notice that one of the examples in Remark 5.9 has Betti numbers
$\\{1,5,5,1\\}$, and although this violates the bounds in Theorem 5.8, they
nonetheless add up to $2^{3}+2^{2}$. Recently, the first author and Seiner
were able to show that one can remove the finite length assumption, provided
one works with multigraded cyclic modules:

######  Theorem 5.11 (Boocher–Seiner, 2018 [11]).

Let $I\subseteq R=k[x_{1},\ldots,x_{n}]$ be a monomial ideal of any
codimension $c\geqslant 2$. If $R/I$ is not a complete intersection, then

$\sum_{i=0}^{c}\beta_{i}(R/I)\geqslant 2^{c}+2^{c-1}.$

Unlike the proofs in the finite length case, this theorem does not apparently
follow from a bound on the individual Betti numbers. Indeed, the argument
follows via a degeneration argument that reduces everything to either a Betti
sequence $\\{1,3,2\\}$ with $c=2$ or a Betti sequence $\\{1,5,5,1\\}$ with
$c=3$. Perhaps it is a coincidence that these Betti numbers sum to
$(1.5)2^{c}$.

######  Question E.

Examples 5.4 and 5.5 are both examples of Gorenstein algebras where the sum of
the Betti numbers is equal to $2^{c}+2^{c-1}$. What other Gorenstein algebras
$R/I$ of codimension $c$ have this sum?

######  Question F.

In Examples 5.3 and 5.4, we saw two distinct families of Betti numbers whose
Betti numbers sum to $2^{c}+2^{c-1}$. Are there other examples of Betti
numbers that achieve this sum?

######  Question G.

If $M$ is a multigraded $k[x_{1},\ldots,x_{n}]$-module of codimension $c<n$
that is not a complete intersection, then does

$\sum\beta_{i}(M)\geqslant 2^{c}+2^{c-1}$

hold?

As a partial answer to this, we have the following:

######  Theorem 5.12 (Brun–Romer, 2004 [14]).

If $M$ is multigraded $k[x_{1},\ldots,x_{n}]$-module of projective dimension
$p$, then $\beta_{i}(M)\geqslant{p\choose i}.$

Since $p\geqslant c$ with equality only in the case that $M$ is Cohen-
Macaulay, we see that Question G can be reduced to the Cohen-Macaulay case.

Finally, we cannot resist including the following beautiful result of
Charalambous and Evans, which gives a sharp strong bound for monomial ideals
of finite colength:

######  Theorem 5.13 (Charalambous–Evans, 1991 [30]).

Let $R=k[x_{1},\ldots,x_{n}]$ and $M=R/I$, where the ideal $I$ is minimally
generated by $n$ pure powers of the variables and one additional generator
$m=x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$. Suppose that $\ell$ is the number of
nonzero $a_{i}$’s. Then for all $i$, we have

$\beta_{i}(M)={n\choose i}+{n-1\choose i-1}+\cdots+{n-(\ell-1)\choose
i-(\ell-1)}.$

For instance, this says that the Betti numbers of the ideal
$I=(x^{2},y^{2},z^{2},w^{2},xywz)$ must sum to at least
$2^{4}+2^{3}+2^{2}+2=30$. Indeed, the Betti numbers are $\\{1,5,10,10,4\\}$.

######  Question H.

Can this theorem be extended outside of the case of finite colength monomial
ideals? Is there a version for general monomial ideals? Is there a version for
multigraded modules? For general ideals?

###  5.2. Low Regularity Case

We finish this survey with some of the most recent results on larger lower
bounds for Betti numbers. So far we have not paid much attention to the
degrees of the syzygies. After all, our bounds are in terms of the Betti
numbers $\beta_{i}$, which count the number of generators, but not their
degrees, of the $i$th syzygy module. But since we are working with graded
modules, we will now actually look at $\beta_{ij}$.

In terms of degrees, the simplest resolutions are those whose matrices all
have linear entries. Such resolutions are called linear.

######  Theorem 5.14 (Herzog–Kühl, 1984 [47]).

If $M$ is a graded $R$-module of projective dimension $p$ with a linear
resolution, then $\beta_{i}(M)\geqslant{p\choose i}$.

######  Remark 5.15.

Notice that this is the same bound given by Brun and Römer for multigraded
modules in Theorem 5.12. In the same paper, Herzog and Kühl show that apart
from this bound, linear resolutions can behave quite wildly.555This is the
term used by Herzog and Kühl. Indeed, they show how to produce squarefree
monomial ideals with a linear resolution such that the Betti numbers form a
non-unimodal sequence with arbitrarily many extrema.

Linear resolutions have the property that each matrix has entries all of which
are linear. This is a particular case of what is called a pure resolution. We
say that a module $M$ is pure if it is Cohen-Macaulay and each map has entries
all of the same degree. Equivalently, in the free resolution
$F_{\bullet}\rightarrow M$, each $F_{i}$ is generated in a single degree
$d_{i}$. This sequence of numbers $\\{d_{0},\ldots,d_{c}\\}$ is called the
degree sequence of $M$.

######  Example 5.16.

The module $M$ given in Example 5.7 with Betti table

$\begin{array}[]{r|cccc}\beta(M)&0&1&2&3\\\ \hline\cr 0&2&4&-&-\\\
1&-&-&4&2\\\ \end{array}$

is pure with degree sequence $0,1,3,4$. The module $R/G$ in Example 5.4 is an
example of a pure module with degree sequences $\\{0,2,3,5\\}$ and Betti table

$\begin{array}[]{r|cccc}\beta(M)&0&1&2&3\\\ \hline\cr 0&1&-&-&-\\\
1&-&5&5&-\\\ 2&-&-&-&1\\\ \end{array}.$

In [47], Herzog and Kühl showed that if $M$ is a pure module with degree
sequence $\\{d_{0},\ldots,d_{c}\\}$, then for all $i\geqslant 1$ we have

$\beta_{i}(M)=\beta_{0}(M)\prod_{\stackrel{{\scriptstyle 1\leqslant j\leqslant
c}}{{j\neq i}}}\frac{|d_{j}-d_{0}|}{|d_{i}-d_{j}|}.$

Quite surprisingly, given any degree sequence $d_{0}<d_{1}<\cdots<d_{p}$,
there exists a Cohen-Macaulay module $M$ whose resolution is pure with this
degree sequence. This was proven in [39, 38] as part of the resolution of the
Boij-Söderberg conjectures.

######  Question I.

If $M$ is a pure module of codimension $c$, is $\beta_{i}(M)\geqslant{c\choose
i}$?

Given the Herzog-Kühl equations, one might expect that this question is
numerical in nature, and in a sense it is. However, the following example
shows a major obstacle:

######  Example 5.17.

Let $M$ be a pure module with degree sequence $\\{0,2,3,7,8,10\\}$. Such a
module has codimension $5$ and its Betti table is

$\begin{array}[]{r|cccccc}\beta(M)&0&1&2&3&4&5\\\ \hline\cr
0&\beta_{0}(M)&-&-&-&-&-\\\ 1&-&7\beta_{0}(M)&8\beta_{0}(M)&-&-&-\\\
2&-&-&-&-&-&-\\\ 3&-&-&-&-&-&-\\\ 4&-&-&-&8\beta_{0}(M)&7\beta_{0}(M)&-\\\
5&-&-&-&-&-&\beta_{0}(M)\\\ \end{array}.$

Notice that if $\beta_{0}(M)=1$, then this would give an example of a module
with $\beta_{2}(M)<{5\choose 2}$. So part of answering Question I involves
showing that $\beta_{0}(M)\geqslant 2$. One way to prove this is to apply a
big hammer — the Total Rank Conjecture, now Walker’s Theorem [65]. Using
Walker’s Theorem, we notice that if $\beta_{0}(M)=1$, then the sum of the
Betti numbers would be equal to $2^{c}$, but evidently $M$ is not a complete
intersection, which contradicts Walker’s result. Alternatively, one could note
that from the Betti table, the rank of $\Omega_{3}(M)$ would be
$2\beta_{0}(M)$, which would violate the Syzygy Theorem 3.13 when
$\beta_{0}(M)=1$.

Extending this sort of argument to general degree sequences will present many
challenges. In fact, we need only to turn to the degree sequence
$\\{0,1,2,3,5,7,8,9,10\\}$ to see to limits of this argument. A module $M$
possessing a pure resolution with this degree sequence would necessarily be of
codimension $8$ and would have Betti table

$\begin{array}[]{r|ccccccccc}\beta(M)&0&1&2&3&4&5&6&7&8\\\ \hline\cr
0&4N&25N&60N&60N&-&-&-&-&-\\\
1&-&-&-&-&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42N}&-&-&-&-\\\
2&-&-&-&-&-&60N&60N&25N&4N\\\ \end{array}$

for some positive integer $N$. Boij-Söderberg Theory guarantees that such a
module exists, but $N$ may be large. Note that if $N=1$ then
$\beta_{4}<{8\choose 4}$, and the sum of the Betti numbers would be
$340<2^{8}+2^{7}$ which would violate both the BEH Conjecture 4.1 and provide
a negative answer to Question B. Notice that the Betti sequence is non-
unimodal, regardless of $N$.

The numerical behavior resulting from the Herzog-Kühl equations is nontrivial
to analyze, but is slightly manageable in the case where the last degree
$d_{c}$ is small relative to $d_{1}$. Note that $d_{1}$ and $d_{c}$ are
esssentially degrees of the first syzygies of $M$ and the Castelnuovo-Mumford
Regularity. This insight was first noticed by Erman in [40]. Coupling this
observation with the full force of the newly proven Boij-Söderberg Theory
allowed him to prove the BEH Conjecture for those graded modules whose
regularity is low relative to the degrees of the first syzygies.

######  Theorem 5.18 (Erman, 2010 [40]).

Let $M$ be a graded $R$-module of codimension $c\geqslant 3$ generated in
degree $0$ and let $a\geqslant 2$ be the minimal degree of a first syzygy of
$M$. If $\operatorname{reg}(M)\leqslant 2a-2$, then

$\beta_{i}(M)\geqslant\beta_{0}(M){c\choose i}.$

In particular the sum of the Betti numbers is at least $\beta_{0}(M)2^{c}$.

To put the regularity bound into perspective, if $M$ is $R/I$ for some ideal
$I$ generated by quadrics, then the above theorem would apply to any $M$ with
regularity at most $2$, which means the Betti table has at most 2 rows. The
regularity condition is relaxed enough to include, for example, the coordinate
rings of smooth curves embedded by linear systems of high degree, those of
toric surfaces, as well as any finite length module whose socle degree is
relatively low. In Example 5.17 the two Betti tables do not obey the low
regularity bound. In the first, $a=2$ and $\operatorname{reg}(M)=5$; in the
second, $a=1$ and $\operatorname{reg}(M)=2$.

Erman’s proof uses general Boij-Söderberg techniques to reduce studying the
Betti tables of arbitrary modules to the study of pure modules and then use a
degeneration argument to supply the required numerical bound. These techniques
were pushed even further in [12], where it is shown that in fact the sum of
the Betti numbers is 50% larger:

######  Theorem 5.19 (Boocher–Wigglesworth, 2020 [12]).

Let $M$ be a graded $R$-module of codimension $c\geqslant 3$ generated in
degree $0$ and let $a\geqslant 2$ be the minimal degree of a first syzygy of
$M$. If $\operatorname{reg}(M)\leqslant 2a-2$, then

$\beta(M)\geqslant\beta_{0}(M)(2^{c}+2^{c-1}).$

If moreover $c\geqslant 9$, then

$\beta_{i}(M)\geqslant 2\,\beta_{0}(M){c\choose i}$

for the first half of the Betti numbers, meaning for $1\leqslant
i\leqslant\lceil c/2\rceil$.

Essentially, this says that if the regularity is “low”, then for $c\geqslant
9$, the _first half_ of the Betti numbers are at least _double_ the
conjectured Buchsbaum–Eisenbud–Horrcks bounds. Then on average the Betti
numbers, will be at least $1.5$ times the BEH bounds, and thus the sum of all
the Betti numbers needs to be at least $1.5(2^{c})$. The authors deal with the
cases $c\leqslant 8$ separately. Again it seems almost miraculous that the
bound of $2^{c}+2^{c-1}$ pops up — in this case aided by the fact that the
first half of the Betti numbers are twice as large as excepted.

######  Remark 5.20.

Notice that if $R=k[x_{1},\ldots,x_{c}]$ with $c\geqslant 2$, then any ideal
$I$ generated by $c+1$ generic quadrics will be an ideal of height $c$, and
$\beta_{1}(R/I)=c+1<2{c\choose 1}$. So without some other condition, for
example on the regularity, there is no hope of finding a stronger bound for
the first Betti number.

As a corollary of Theorem 5.19, for ideals generated by quadrics with
$c\geqslant 9$ we have

$\operatorname{reg}(R/I)<3,\mbox{ and $R/I$ is not a CI
}\implies\beta_{1}(R/I)\geqslant 2c,\,\beta_{2}(R/I)\geqslant 2{c\choose
2},\,\ldots$

In other words, low regularity forces this rather specific bound for the
number of generators.

We end with a table summarizing the results concerning these stronger bounds
(each). We remind the reader that these entries all concern modules that are
not complete intersections.

| $c\leqslant 4$ | $c\geqslant 5$ | multigraded | low regularity
---|---|---|---|---
$\begin{array}[]{c}\beta_{i}\geqslant{c\choose i}+{c-1\choose i-1}\textrm{ for all }i\\\ \textrm{ or }\\\ \beta_{i}\geqslant{c\choose i}+{c-1\choose i}\textrm{ for all }i\end{array}$ | $\begin{array}[]{c}\textrm{{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}False}}\\\ \textrm{\ref{remark that shows how we violate stronger bounds}}\end{array}$ | $\begin{array}[]{c}\textrm{{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}False}}\\\ \textrm{\ref{remark that shows how we violate stronger bounds}}\end{array}$ | $\begin{array}[]{c}c=n\\\ \textrm{(CE, 1991) }\\\ \textrm{\cite[cite]{[\@@bibref{}{CE}{}{}]}}\end{array}$ | $\begin{array}[]{c}c<n\\\ \textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{\ref{remark that shows how we violate stronger bounds}}\end{array}$ | $\begin{array}[]{c}\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{Rk \ref{remark that shows how we violate stronger bounds}}\end{array}$
$\sum_{i}\beta_{i}\geqslant$ $(1.5)$ $2^{c}$ | $\begin{array}[]{c}\textrm{(CEM,}\\\ \textrm{ 1990)}\\\ \textrm{\cite[cite]{[\@@bibref{}{CEM}{}{}]}}\end{array}$ | Open | $\begin{array}[]{c}M\cong R/I\\\ \textrm{(BS, '18 \cite[cite]{[\@@bibref{}{BS}{}{}]}) }\end{array}$ | True$\textstyle{\textrm{True if }c\geqslant 9\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{5\leqslant c\leqslant 8}$check(BW, ’20 [12])
$\begin{array}[]{c}\textrm{general }M\\\
\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{Open}}\end{array}$
$\beta_{i}\geqslant 2{c\choose i}$ for $i<\frac{c}{2}$ | $\begin{array}[]{c}\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{\ref{rmk: c+1 vs 2c} }\end{array}$ | $\begin{array}[]{c}\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{\ref{rmk: c+1 vs 2c}}\end{array}$ | $\begin{array}[]{c}\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{\ref{rmk: c+1 vs 2c}}\end{array}$ | $\begin{array}[]{c}\textrm{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}{False}}\\\ \textrm{\ref{rmk: c+1 vs 2c}}\end{array}$
Table 2. $M$ is a module of codimension $c$ that is not a complete
intersection

## Acknowledgements

We thank Daniel Erman and Josh Pollitz for their detailed comments on a
previous version of this survey. We also thank Mark Walker for helpful
correspondence. The second author was partially supported by NSF grant
DMS-2001445, now DMS-2140355.

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