$\displaystyle\,\max\\{\deg_{x}(\mathbf{m}^{\mathbf{a}}),\deg_{x}(\mathbf{m}^{\mathbf{b}})\\}.$

  * •

If $x\nmid m_{i}$ and $x\nmid m_{j}$, then

$\deg_{x}({m_{i}}^{\alpha}{m_{j}}^{a_{i}-\alpha}\mathbf{m}^{\mathbf{a}^{\prime}})=\deg_{x}(\mathbf{m}^{\mathbf{a}^{\prime}})=\deg_{x}(\mathbf{m}^{\mathbf{a}})\leq\max\\{\deg_{x}(\mathbf{m}^{\mathbf{a}}),\deg_{x}(\mathbf{m}^{\mathbf{b}})\\}.$

This finishes the proof of $(ii)$.

Finally, $(iii)$ follows directly from the fact that

$\displaystyle\deg({m_{i}}^{\alpha}{m_{j}}^{a_{i}-\alpha}\mathbf{m}^{\mathbf{a}^{\prime}})$
$\displaystyle=\alpha\cdot\deg m_{i}+(a_{i}-\alpha)\cdot\deg
m_{j}+\deg(\mathbf{m}^{\mathbf{a}^{\prime}})$ $\displaystyle=\alpha\cdot\deg
m_{i}+(a_{i}-\alpha)\cdot\deg
m_{j}+\deg(\mathbf{m}^{\mathbf{a}})-a_{i}\cdot\deg m_{i}$
$\displaystyle=\deg(\mathbf{m}^{\mathbf{a}})-(a_{i}-\alpha)(\deg(m_{i})-\deg(m_{j})).$

∎

######  Remark 5.8.

Recall that $s=\left\lceil{\frac{r}{2}}\right\rceil$. When $\alpha=s$, the
condition on $b_{j}$ in Lemma 5.7 can be translated as follows:

$b_{j}>r-s\iff\begin{cases}\text{$b_{j}\geq s$ \quad when $r$ is odd}\\\
\text{$b_{j}>s$ \quad when $r$ is even.}\end{cases}$

Recall that, by Theorem 3.6, if $\Delta$ is a quasi-tree with vertices labeled
with the monomial generating set of a monomial ideal
$I\subseteq\mathsf{k}[x_{1},\ldots,x_{n}]$, then $\Delta$ supports a
resolution of $I$ if and only if $\Delta_{M}$ is connected for every monomial
$M$. It is easy to check that the connectivity need only be verified for $M$
in $\mathrm{LCM}(I)$. We now use this criterion to prove the main result of
the paper.

######  Theorem 5.9 (Main Result).

Let $q\geq 1$ and let $I$ be a monomial ideal minimally generated by square-
free monomials $m_{1},\dots,m_{q}$. By labeling each vertex $\mathbf{a}$ of
the simplicial complexes below with the monomial $\mathbf{m}^{\mathbf{a}}$,
the following hold for any $r\geq 1$:

  1. (1)

$\mathbb{L}^{r}_{q}$ supports a free resolution of $I^{r}$;

  2. (2)

$\mathbb{L}^{r}(I)$ supports a free resolution of $I^{r}$.

###### Proof.

In this proof, we let $\mathbb{L}$ denote either $\mathbb{L}^{r}_{q}$ or
$\mathbb{L}^{r}(I)$. Fix $\mathbb{L}$ as one of these two choices. Let $V$
denote the set of vertices of $\mathbb{L}$ and let $\\{m_{1},\ldots,m_{q}\\}$
be the minimal monomial generating set of $I$. Following 4.1, for
$\mathbf{a}=(a_{1},\ldots,a_{q})\in\mathcal{N}^{r}_{q}$ we let
$\mathbf{m}^{\mathbf{a}}={m_{1}}^{a_{1}}\cdots{m_{q}}^{a_{q}}$ and
$\mathbf{e}_{1},\ldots,\mathbf{e}_{q}$ denote the standard basis vectors for
$\mathbb{R}^{q}$.

We willl show that for every $M$ in $\mathrm{LCM}(I^{r})$, ${\mathbb{L}}_{M}$
is empty or connected, where ${\mathbb{L}}_{M}$ is the induced subcomplex of
the complex $\mathbb{L}$ on the set

$V_{M}=\\{\mathbf{a}\in V\colon\mathbf{m}^{\mathbf{a}}\mid M\\}\,.$

By Proposition 4.6, $\mathbb{L}^{r}_{q}$ is a quasi-tree and by Proposition
3.3, $\mathbb{L}^{r}(I)$ is a quasi-forest, which is connected and thus a
quasi-tree. In view of Theorem 3.6, we can then conclude that $\mathbb{L}$
supports a free resolution of $I^{r}$.

Suppose $M\in\mathrm{LCM}(I^{r})$ and ${\mathbb{L}}_{M}\neq\emptyset$. If
$M={m_{i}}^{r}$ for some $i\in[q]$, then ${\mathbb{L}}_{M}$ is a point by
Proposition 5.6 $(i)$, and hence is connected.

Assume $M\neq{m_{i}}^{r}$ for all $i\in[q]$. Note that this implies $q>1$.

The facets of ${\mathbb{L}}_{M}$ are the maximal sets, under inclusion, among
the sets

$F^{r}_{1}\cap V_{M},\ldots,F^{r}_{q}\cap V_{M},G^{r}_{1}\cap
V_{M},\ldots,G^{r}_{q}\cap V_{M}.$ (5.9.1)

Note that not all these sets are facets of ${\mathbb{L}}_{M}$, but all facets
of ${\mathbb{L}}_{M}$ are among those listed in (5.9.1). We refer to the
facets of ${\mathbb{L}}_{M}$ of form $F^{r}_{i}\cap V_{M}$ as the first layer,
and those of the form $G^{r}_{i}\cap V_{M}$ as the second layer. We refer to
$B^{r}\cap V_{M}$ as the base of ${\mathbb{L}}_{M}$. The base $B^{r}\cap
V_{M}$ could become empty, depending on the choice of $M$.

We use the faces in (5.9.1) to argue the connectedness of ${\mathbb{L}}_{M}$
as follows: Claim 1 below shows that any facet of ${\mathbb{L}}_{M}$ in the
second layer is connected to a facet in the first layer. Claim 2 implies that
any two facets of ${\mathbb{L}}_{M}$ that are in the first layer connect
through the nonempty base. The combination of these two facts implies that
${\mathbb{L}}_{M}$ is connected, which will end our proof.

### Claim 1: (The second layer facets of ${\mathbb{L}}_{M}$ intersect the
first layer facets)

For any $i\in[q]$ we have:

$G^{r}_{i}\cap V_{M}\neq\emptyset\implies G^{r}_{i}\cap F^{r}_{i}\cap
V_{M}\neq\emptyset.$

#### Proof of Claim 1

Assume $G^{r}_{i}\cap V_{M}\neq\emptyset$ for some $i\in[q]$. By Definition
4.2, there exists $a\in[q]$ such that

$\mathbf{d}=(r-1)\mathbf{e}_{i}+\mathbf{e}_{a}\in G^{r}_{i}\cap V_{M}\,.$

If $i\neq a$, then $\mathbf{d}\in F^{r}_{i}$ as well, and hence $G^{r}_{i}\cap
F^{r}_{i}\cap V_{M}\neq\emptyset$ as desired.

Assume $a=i$. Then $\mathbf{d}\in V_{M}$ implies ${m_{i}}^{r}\mid M$. Since
$M\neq{m_{i}}^{r}$, Proposition 5.6 guarantees that there exists $j\in[q]$
with $j\neq i$ and $m_{j}\mid M$ so that
$(r-1)\mathbf{e}_{i}+\mathbf{e}_{j}\in V$.

Since ${m_{i}}^{r}\mid M$ and $m_{j}\mid M$, we see that
${m_{i}}^{r-1}m_{j}\mid M$. Indeed, we set $M={m_{i}}^{r}M^{\prime}$ and,
since $m_{j}$ is square-free, we have $m_{j}\mid m_{i}M^{\prime}$, hence
${m_{i}}^{r-1}m_{j}\mid{m_{i}}^{r}M^{\prime}=M$. Thus we have

$(r-1)\mathbf{e}_{i}+\mathbf{e}_{j}\in G^{r}_{i}\cap F^{r}_{i}\cap V_{M}.$

### Claim 2: (The first layer facets connect through the base of
${\mathbb{L}}_{M}$)

For any $i,j\in[q]$ with $i\neq j$ we have:

$\mbox{\rm{if}}\quad(F^{r}_{i}\cup G^{r}_{i})\cap
V_{M}\neq\emptyset\quad\text{and}\quad(F^{r}_{j}\cup G^{r}_{j})\cap
V_{M}\neq\emptyset\quad\mbox{\rm{then}}\quad B^{r}\cap V_{M}\neq\emptyset.$

#### Proof of Claim 2

Assume for some $i,j\in[q]$ with $i\neq j$ that

$(F^{r}_{i}\cup G^{r}_{i})\cap V_{M}\neq\emptyset\quad\mbox{ and
}\quad(F^{r}_{j}\cup G^{r}_{j})\cap V_{M}\neq\emptyset.$

Without loss of generality, assume $\deg(m_{i})\geq\deg(m_{j})$. We choose

$\mathbf{a}=(a_{1},\ldots,a_{q})\in(F^{r}_{i}\cup G^{r}_{i})\cap
V_{M}\quad\mbox{ and }\quad\mathbf{b}=(b_{1},\ldots,b_{q})\in(F^{r}_{j}\cup
G^{r}_{j})\cap V_{M}$

such that

$\deg(\mathbf{m}^{\mathbf{a}})=\min\\{\deg(\mathbf{m}^{\mathbf{d}})\mid\
\mathbf{d}\in(F^{r}_{i}\cup G^{r}_{i})\cap V_{M}\\}\,.$ (5.9.2)

Assume, by way of contradiction, that $B^{r}\cap V_{M}=\emptyset$. Since
$\mathbf{a},\mathbf{b}\in V_{M}$, we have then $\mathbf{a},\mathbf{b}\notin
B^{r}$. Therefore,

$s<a_{i},b_{j}\leq r\quad\mbox{ and
}\quad\mathbf{m}^{\mathbf{a}},\mathbf{m}^{\mathbf{b}}\mid M.$

Set

$\mathbf{a}^{\prime}=\mathbf{a}-a_{i}\mathbf{e}_{i}\in\mathcal{N}^{r-a_{i}}_{q}\quad\mbox{
and
}\quad\mathbf{c}:=(c_{1},\ldots,c_{q})=s\mathbf{e}_{i}+(a_{i}-s)\mathbf{e}_{j}+\sum_{1\leq
k\leq q,k\neq i}a_{k}\mathbf{e}_{k}\in\mathcal{N}^{r}_{q}.$

Since $b_{j}>s\geq r-s$, by Lemma 5.7 $(ii)$ and Remark 5.8,

$\mathbf{m}^{\mathbf{c}}={m_{i}}^{s}{m_{j}}^{a_{i}-s}\mathbf{m}^{\mathbf{a}^{\prime}}\mid\mathrm{lcm}(\mathbf{m}^{\mathbf{a}},\mathbf{m}^{\mathbf{b}})\mid
M.$ (5.9.3)

Moreover $\mathbf{c}\in B^{r}$, because

$\displaystyle c_{i}$ $\displaystyle=s$ $\displaystyle c_{j}$
$\displaystyle=a_{i}-s+a_{j}\leq a_{i}-s+(r-a_{i})=r-s\leq s$
$\displaystyle\mbox{since }a_{i}+a_{j}\leq r$ $\displaystyle c_{k}$
$\displaystyle=a_{k}\leq r-a_{i}<r-s\leq s$ $\displaystyle\mbox{for all }k\neq
i,j.$

If $\mathbb{L}=\mathbb{L}^{r}_{q}$, then $V=\mathcal{N}^{r}_{q}$. Since
(5.9.3) shows $\mathbf{c}\in V_{M}$, we conclude $\mathbf{c}\in B^{r}\cap
V_{M}$, a contradiction. This finishes the proof of part (1).

Assume now $\mathbb{L}=\mathbb{L}^{r}(I)$. Since we assumed $B^{r}\cap
V_{M}=\emptyset$, we must have $\mathbf{c}\notin V$. Definition 5.1 implies
then

$\mathbf{m}^{\mathbf{c}^{\prime}}\mid\mathbf{m}^{\mathbf{c}}\quad\mbox{ for
some }\quad\mathbf{c}^{\prime}:=(c_{1}^{\prime},\ldots,c_{q}^{\prime})\in V.$

Note that $\mathbf{m}^{\mathbf{c}^{\prime}}\mid M$ by (5.9.3) and hence
$\mathbf{c}^{\prime}\in V_{M}$. Since $B^{r}\cap V_{M}=\emptyset$, we must
have $\mathbf{c}^{\prime}\notin B^{r}$. Using the notation in Definition 5.1,
let $V_{a}$ be the equivalence class of which $\mathbf{c}^{\prime}$ is a
representative. If $\mathbf{m}^{\mathbf{c}^{\prime}}=\mathbf{m}^{\mathbf{c}}$,
then $\mathbf{c}\in V_{a}$ as well, and hence $V_{a}\cap B^{r}\neq\emptyset$.
Then Step 1 in Definition 5.1 implies $\mathbf{c}^{\prime}\in B^{r}$, a
contradiction. Hence
$\mathbf{m}^{\mathbf{c}^{\prime}}\neq\mathbf{m}^{\mathbf{c}}$.

Since $\mathbf{c}^{\prime}\notin B^{r}$, there exists $k\in[q]$ such that
$c^{\prime}_{k}>s\geq r-s$. Since $c_{i}=s$ and
$\mathbf{m}^{\mathbf{c}^{\prime}}\mid\mathbf{m}^{\mathbf{c}}$, Lemma 5.7 $(i)$
implies $k=i$. In particular, $c^{\prime}_{i}>s$, and thus

$\mathbf{c}^{\prime}\in(F^{r}_{i}\cup G^{r}_{i})\cap V_{M}\,.$

Since $\mathbf{m}^{\mathbf{c}^{\prime}}$ strictly divides
$\mathbf{m}^{\mathbf{c}}={m_{i}}^{s}{m_{j}}^{a_{i}-s}\mathbf{m}^{\mathbf{a}^{\prime}}$,
using Lemma 5.7 $(iii)$, we have

$\deg(\mathbf{m}^{\mathbf{c}^{\prime}})<\deg(\mathbf{m}^{\mathbf{c}})\leq\deg(\mathbf{m}^{\mathbf{a}}),$

contradicting the assumption made in (5.9.2). This finishes the proof of part
(2). ∎

######  Remark 5.10.

One may feel that part $(1)$ of the statement above is weaker than $(2)$,
since $\mathbb{L}^{r}(I)\subseteq\mathbb{L}^{r}_{q}$. However, the remarkable
aspect of $(1)$ is that, before labeling, $\mathbb{L}^{r}_{q}$ does not depend
on the ideal $I$. Thus the same topological structure, depending only on $r$
and $q$, supports a free resolution of the $r^{th}$ power of any ideal
generated by $q$ square-free monomials. The idea is that $\mathbb{L}^{r}_{q}$
provides an alternative notion to the Taylor complex for powers of $I$: when
$r=1$, $\mathbb{L}^{r}_{q}$ is the Taylor simplex, and when $r>1$,
$\mathbb{L}^{r}_{q}$ is significantly smaller than the Taylor simplex but
still supports a resolution of $I^{r}$.

##  6\. Bounds on Betti numbers

One of the key applications of Theorem 5.9 is that we are able to improve the
bounds on Betti numbers for powers of ideals from that given by the standard
Taylor resolution of $I^{r}$ since

$\mathbb{L}^{r}(I)\subseteq\mathbb{L}^{r}_{q}\subseteq\mathrm{Taylor}(I^{r}).$

This section contains computations that illustrate the extent to which our
results improve the Taylor bounds.

When $r=2$, we were able to provide a concrete formula for the number of faces
of $\mathbb{L}^{r}(I)$ in [4], and as a result we provided bounds for Betti
numbers of $I^{2}$, which are shown in Proposition 7.11 to be sharp. When
$r>2$, however, such formulas are not as easy to write and the numbers are
very large even for small examples. In this case we shift our attention to
$\mathbb{L}^{r}_{q}$. While the bound on the Betti numbers stemming from
$\mathbb{L}^{r}(I)$ is dependent on the relations among the generators of $I$,
one can use the structure of $\mathbb{L}^{r}_{q}$ to get a general, albeit
weaker, bound on $\beta_{t}(I^{r})$.

######  Theorem 6.1 (Bounds on Betti numbers of $I^{r}$).

If $I$ is a square-free monomial ideal with $q$ minimal generators, then the
Betti numbers of $I^{r}$ for $r\geq 2$ are bounded above by

$\beta_{t}(I^{r})\leq
q\left({{q-1}\choose{t}}+{{f}\choose{t+1}}-{{b}\choose{t+1}}\right)+{{b}\choose{t+1}}$

where $t\geq 0$, $b$ is the coefficient of $x^{r}$ in the expansion of

$(1+x+x^{2}+\cdots+x^{\left\lceil{\frac{r}{2}}\right\rceil})^{q},$

and

$f=\frac{{{q+r-1}\choose{r}}-b-q}{q}+b.$

In particular, $\mathrm{pd}(I^{r})\leq\max\\{q-1,f-1\\}.$

###### Proof.

Let $s=\left\lceil{\frac{r}{2}}\right\rceil$. Note that $\beta_{t}(I^{r})$ is
bounded above by the number of faces of dimension $t$ of $\mathbb{L}^{r}_{q}$.
To count the faces of dimension $t$, we use the sets $B^{r},F_{i}^{r},$ and
$G_{i}^{r}$ from Definition 4.2.

Observe that the coefficient $b$ of $x^{r}$ in the expansion of

$(1+x+x^{2}+\cdots+x^{s})^{q}$

is exactly the number of $q$-tuples $(a_{1},\ldots,a_{q})$ with $a_{i}\leq s$
and $a_{1}+\cdots+a_{q}=r$, in other words $b=|B^{r}|$.

Note that the vertices of $\mathbb{L}^{r}_{q}$ are formed by selecting $r$ of
the original $q$ generators, so using the standard formula for counting with
repetition, we have

$|V(\mathbb{L}^{r}_{q})|={{q+r-1}\choose{r}}.$

Now $|F_{i}^{r}|=|F_{j}^{r}|$ for all $i,j$, so let $f=|F_{i}^{r}|$. Since
$|G_{i}^{r}\setminus F_{i}^{r}|=1$, there are $|V(\mathbb{L}^{r}_{q})|-q$
vertices in $\cup_{i=1}^{q}F_{i}^{r}$. Also $F_{i}^{r}\cap F_{j}^{r}=B^{r}$
for all $i,j$ such that $i\not=j$, so we have

$f=|F_{i}^{r}|={\frac{1}{q}}\big{(}(|V(\mathbb{L}^{r}_{q})|-q)-|B^{r}|\big{)}+|B^{r}|=\frac{{{q+r-1}\choose{r}}-q-b}{q}+b.$
(6.1.1)

To count the number of faces of dimension $t$, that is, faces with $t+1$
vertices, in $\mathbb{L}^{r}_{q}$, we separate the faces into three distinct
types.

  1. (1)

Faces containing ${m_{i}}^{r}$ for some $i$: By the definition of
$\mathbb{L}^{r}_{q}$, faces of this type must be contained in $G_{i}^{r}$.
Since the vertex corresponding to ${m_{i}}^{r}$ has been fixed, $t$ additional
vertices of $G_{i}^{r}$ are needed. There are $q-1$ such vertices since all
vertices of $G_{i}^{r}$ have the form $(r-1)\mathbf{e}_{i}+\mathbf{e}_{j}$
where $1\leq j\leq q$. Since there are $q$ choices for $i$, there are

$q{{q-1}\choose{t}}$

such faces.

  2. (2)

Faces contained in $B^{r}$: Setting $b=|B^{r}|$ as above, there are

${{b}\choose{t+1}}$

such faces.

  3. (3)

Faces contained in $F_{i}^{r}$ but not in $B^{r}$: Recalling that
$f=|F_{i}^{r}|$ and $B^{r}\subseteq F_{i}^{r}$, there are

$q\left({{f}\choose{t+1}}-{{b}\choose{t+1}}\right)$

such faces.

Combining the three cases, we see that there are

$q\left({{q-1}\choose{t}}+{{f}\choose{t+1}}-{{b}\choose{t+1}}\right)+{{b}\choose{t+1}}$

faces of $\mathbb{L}^{r}_{q}$ of dimension $t$. Thus for $I^{r}$,

$\beta_{t}(I^{r})\leq
q\left({{q-1}\choose{t}}+{{f}\choose{t+1}}-{{b}\choose{t+1}}\right)+{{b}\choose{t+1}}.$

In particular, if $t>q-1$ and $t+1>f$, we must have $\beta_{t}(I^{r})=0$. Thus

$\mathrm{pd}(I^{r})\leq\max\\{q-1,f-1\\}.$ (6.1.2)

∎

######  Corollary 6.2.

If an ideal $I$ is minimally generated by $q$ square-free monomials, then the
Betti numbers of $I^{r}$ are bounded by

$\beta_{t}(I^{r})\leq q{{q-1}\choose{t}}+{{b}\choose{t+1}}$

for $r=2,3$, where $b$ is as defined in Theorem 6.1. In the case where $r=2$,
$b={q\choose 2}$ and the bound reduces to the bound given in [4].

###### Proof.

When $r=2$ or $r=3$, then $F_{i}^{r}=B^{r}$ for all $i$, so $b=f$ and the
result follows immediately from simplifying the formula in Theorem 6.1.
Moreover, when $r=2$, the coefficient of $x^{2}$ in the binomial expansion of
$(1+x)^{q}$ is ${q\choose 2}$ as stated. The reduction is then evident. ∎

Notice that when $r=2$, the bound is known to be sharp; it agrees with the
result in [4]. In Proposition 7.11 we characterize the values of $r$ and $q$
for which these bounds are sharp, i.e. can be realized by some ideal.

######  Example 6.3.

As a first example, we examine cases where $r$ or $q$ is small. We first
consider $b$ and $f$ for small values of $q$.

  * •

For $q=2$, computing $(1+x+x^{2}+\cdots+x^{s})^{2}$ shows that $b=1$ when
$r=2s$ is even and $b=2$ when $r=2s-1$ is odd. If $r$ is even, we then have

$f=\frac{{{2+r-1}\choose{r}}-2-1}{2}+1=\frac{r+1-2-1}{2}+1=\frac{r}{2}=s.$
(6.3.1)

A similar computation shows that when $r$ is odd, $f=s$ as well.

  * •

If $q=3$ and $r=3$ or $r=4$ (so that $s=2$), then computing $(1+x+x^{2})^{3}$
yields $b=7$ and $f=7$ when $r=3$ and $b=6$ and $f=8$ when $r=4$.

Applying the equations from Theorem 6.1 yields:

  * •

For $r=2$, any $q$, and any $t$,

$\beta_{t}(I^{r})\leq q{{q-1}\choose{t}}+{{\frac{1}{2}q(q-1)}\choose{t+1}}.$

  * •

For any $r$, $q=2$, and any $t\geq 2$, if $s=\lceil{r/2\rceil}$,

$\beta_{t}(I^{r})\leq 2{{s}\choose{t+1}}.$

  * •

For $r=q=3$ and any $t$,

$\beta_{t}(I^{r})\leq 3{{2}\choose{t}}+{{7}\choose{t+1}}.$

  * •

For $r=4$, $q=3$ and any $t\geq 3$,

$\beta_{t}(I^{r})\leq 3{{8}\choose{t+1}}-2{{6}\choose{t+1}}.$

######  Remark 6.4.

In (6.1.2) it is useful to understand which of the integers $q-1$, $f-1$
achieves the maximum. For small values of $q$ and $r$, it is possible to have
$q\geq f$. However, the opposite holds in most cases. More precisely, we show
below that if $r$, $q$ satisfy any of the following assumptions:

  1. (1)

$q=2$ and $r\geq 5$;

  2. (2)

$q=3$ and $r\geq 3$;

  3. (3)

$q\geq 4$ and $r\geq 2$,

then $f>q$, and hence

$\mathrm{pd}(I^{r})\leq f-1=\dim\mathbb{L}^{r}_{q}\,.$

In the case $q=2$, (6.3.1) shows $f=s=\lceil{\frac{r}{2}}\rceil$. If $r\geq
5$, we must have $s>2$, thus $f=s>2=q$. This settles (1).

Now suppose $q>2$. If $r=2$, then Corollary 6.2 shows that $f=b={q\choose 2}$.
When $q\geq 4$, we have $f={q\choose 2}>q$. Thus, to show (2) and (3) it
remains to consider the case when $q\geq 3$ and $r\geq 3$. Observe that when
$r\geq 3$, $s\geq 2$. In this scenario, since $b$ is the coefficient of
$x^{r}$ in $(1+x+\dots+x^{s})^{q}$ and $q>2$, then $b\geq 2$ and $bq-b>q$.

To see that $f>q$ holds, we will show $f-q>0$, which by (6.1.1) amounts to

$\frac{\binom{q+r-1}{r}-q-b}{q}+b-q=\frac{\binom{q+r-1}{r}-q-b+bq-q^{2}}{q}>0\,.$

It is sufficient to show the numerator is positive, so we observe the
following, where the first inequality results from $bq-b>q$ and the second
follows since $r\geq 3$:

$\displaystyle\binom{r+q-1}{q-1}-q-b+bq-q^{2}$
$\displaystyle>\binom{r+q-1}{q-1}-q^{2}$
$\displaystyle\geq\binom{q+2}{q-1}-q^{2}$
$\displaystyle=\frac{(q+2)(q+1)q}{6}-q^{2}=\frac{q(q-1)(q-2)}{6}>0\,.$

This ends our argument.

######  Example 6.5.

In general, the bounds from Theorem 6.1 will be considerably smaller than
those provided by the Taylor complex, which is a simplex, and the bound on the
projective dimension will also decrease significantly. We display this
phenomenon in the table below.

Bound Comparisons
---
|  $q=2$, $r=3$ |  $q=3$, $r=3$ |  $q=3$, $r=4$
|  Theorem 6.1 |  $3$-simplex |  Theorem 6.1 |  $9$-simplex |  Theorem 6.1 |  $14$-simplex
$\beta_{0}(I^{r})\leq$ |  4 |  4 |  10 |  10 |  15 |  15
$\beta_{1}(I^{r})\leq$ |  3 |  6 |  27 |  45 |  60 |  105
$\beta_{2}(I^{r})\leq$ |  0 |  4 |  38 |  120 |  131 |  455
$\mathrm{pd}(I^{r})\leq$ |  1 |  3 |  6 |  9 |  7 |  14

The bounds on Betti numbers and projective dimension given by the complex
$\mathbb{L}^{r}_{q}$ in Theorem 6.1 are most effective when the generating set
$\\{\mathbf{m}^{\mathbf{a}}\mid\mathbf{a}\in\mathcal{N}^{r}_{q}\\}$ does not
contain redundancies. When there are redundancies, using $\mathbb{L}^{r}(I)$
will yield improved bounds. We illustrate how to use this improvement by
continuing Example 5.2.

######  Example 6.6.

Let $I=(xy,yz,zw,xw)=(m_{1},m_{2},m_{3},m_{4})$ as in Example 5.2. Counting
faces of size $i$ in the complex $\mathbb{L}^{2}(I)$ provides a bound on the
$i^{th}$ Betti number of $I^{2}$. Note that in general, these improvements in
the Betti numbers follow from knowledge of the redundancies in
$\\{\mathbf{m}^{\mathbf{a}}\mid\mathbf{a}\in\mathcal{N}^{r}_{q}\\}$ and can
often be computed from that information using the equivalence classes without
explicitly constructing $\mathbb{L}^{r}(I)$. For instance, a comparison of the
bounds is summarized in the table below for $I^{2}$ and for $I^{3}$ using the
first of the two vertex sets for $\mathbb{L}^{3}(I)$ given in Example 5.2.

Bound Comparisons
---
| $r=2$ | $r=3$
|  $\mathbb{L}^{2}(I)$ |  $8$-simplex |  $\quad\mathbb{L}^{2}_{4}$ |  $9$-simplex |  $\mathbb{L}^{3}(I)$ |  $15$-simplex |  $\quad\mathbb{L}^{3}_{4}$ |  $19$-simplex
|  |  |  (Thm. 6.1) |  |  |  |  (Thm. 6.1) |
$\beta_{0}(I^{r})\leq$ |  9 |  9 |  10 |  10 |  16 |  16 |  20 |  20
$\beta_{1}(I^{r})\leq$ |  20 |  36 |  27 |  45 |  74 |  120 |  132 |  190
$\beta_{2}(I^{r})\leq$ |  18 |  84 |  32 |  120 |  224 |  560 |  572 |  1,140
$\mathrm{pd}(I^{r})\leq$ |  4 |  8 |  5 |  9 |  11 |  15 |  15 |  19

##  7\. Extremal Ideals: When does $\mathbb{L}^{r}_{q}$ support a minimal
resolution?

Based upon the above work, a natural question that arises is the following:
for given $r$ and $q$, can one find ideals $I$ for which $I^{r}$ has a minimal
free resolution supported on $\mathbb{L}^{r}_{q}$ itself? When $r=1$,
$\mathbb{L}^{r}_{q}$ is the Taylor complex, which one can easily see always
supports a minimal free resolution of the ideal generated by $q$ variables
$(x_{1},x_{2},\ldots,x_{q})$. In the case where $r>1$, Proposition 7.11
describes exactly when the bounds for Betti numbers in Theorem 6.1 and
Corollary 6.2 are sharp, in the sense that there exist ideals for which
equality is attained. We call the ideals which realize these bounds
$q$-extremal ideals and denote them by $\mathcal{E}_{q}$. In fact we can show
a much stronger statement in Theorem 7.9: the powers ${\mathcal{E}_{q}}^{r}$
have maximal Betti numbers among the ideals $I^{r}$ where $I$ is generated by
$q$ square-free monomials.

######  Definition 7.1 (Extremal ideals).

Let $q$ be a positive integer. For every set $A$ with $\emptyset\neq
A\subsetneq[q]$, introduce a variable $x_{A}$, and consider the polynomial
ring

$S_{\mathcal{E}}=\mathsf{k}\big{[}x_{A}\colon\emptyset\neq
A\subseteq[q]\big{]}$

over a field $\mathsf{k}$. For each $i\in[q]$ define a square-free monomial
$\epsilon_{i}$ in $S_{\mathcal{E}}$ as

$\epsilon_{i}=\prod_{\begin{subarray}{c}A\subseteq[q]\\\ i\in
A\end{subarray}}x_{A}.$

The square-free monomial ideal
$\mathcal{E}_{q}=(\epsilon_{1},\ldots,\epsilon_{q})$ is called a
$\boldsymbol{q}$-extremal ideal.

When it is unlikely to lead to confusion, we will simplify the notation by
writing $x_{1}$ for $x_{\\{1\\}}$, $x_{12}$ for $x_{\\{1,2\\}}$, etc., and
refer to a $q$-extremal ideal simply as an extremal ideal. The ring
$S_{\mathcal{E}}$ has $2^{q}-1$ variables, corresponding to the power set of
$[q]$ (excluding $\emptyset$), and each $\epsilon_{i}$ is the product of
$2^{q-1}$ variables; that is, those corresponding to the subsets of $[q]$ that
contain $i$. The following example illustrates how this works for
$\mathcal{E}_{4}$.

######  Example 7.2.

When $q=4$, the ideal $\mathcal{E}_{4}$ is generated by the monomials

$\begin{array}[]{ll}\epsilon_{1}&=x_{1}x_{12}x_{13}x_{14}x_{123}x_{124}x_{134}x_{1234};\\\
\epsilon_{2}&=x_{2}x_{12}x_{23}x_{24}x_{123}x_{124}x_{234}x_{1234};\\\
\epsilon_{3}&=x_{3}x_{13}x_{23}x_{34}x_{123}x_{134}x_{234}x_{1234};\\\
\epsilon_{4}&=x_{4}x_{14}x_{24}x_{34}x_{124}x_{134}x_{234}x_{1234}\\\
\end{array}$

in the polynomial ring
$\mathsf{k}[x_{1},x_{2},x_{3},x_{4},x_{12},x_{13},x_{14},x_{23},x_{24},x_{34},x_{123},x_{124},x_{134},x_{234},x_{1234}]$.

Using the terminology of [18], it naturally follows that $\mathcal{E}_{q}$ is
precisely the nearly Scarf ideal of a $(q-1)$-simplex: we see this by matching
the variable $x_{[q]\smallsetminus B}$ with the face $B$ of the simplex.

Following 4.1, let $r$ and $q$ be positive integers and $I$ an ideal generated
by square-free monomials $m_{1},\ldots,m_{q}$. If
$\mathbf{a}=(a_{1},\ldots,a_{q})\in\mathbb{N}^{q}$, set

$\mathbf{m}^{\mathbf{a}}={m_{1}}^{a_{1}}\cdots{m_{q}}^{a_{q}}\quad\mbox{ and
}\quad{\boldsymbol{\epsilon}}^{\mathbf{a}}={\epsilon_{1}}^{a_{1}}\cdots{\epsilon_{q}}^{a_{q}}.$

Observe

${\boldsymbol{\epsilon}}^{\mathbf{a}}=\prod_{j\in[q]}\prod_{\begin{subarray}{c}A\subseteq[q]\\\
j\in A\end{subarray}}{x_{A}}^{a_{j}}=\prod_{\emptyset\neq
A\subseteq[q]}(x_{A})^{\sum_{j\in A}a_{j}}\,.$ (7.2.1)

The $r^{th}$ powers $I^{r}$ and ${\mathcal{E}_{q}}^{r}$ are generated by
monomials of the form $\mathbf{m}^{\mathbf{a}}$ and
${\boldsymbol{\epsilon}}^{\mathbf{a}}$, respectively, with
$\mathbf{a}\in\mathcal{N}^{r}_{q}$. Proposition 7.3 demonstrates that for
${\mathcal{E}_{q}}^{r}$, the elements ${\boldsymbol{\epsilon}}^{\mathbf{a}}$
with $\mathbf{a}\in\mathcal{N}^{r}_{q}$ form a minimal generating set. In
fact, all faces of $\mathbb{L}^{r}_{q}$ are necessary in
$\mathbb{L}^{r}(\mathcal{E}_{q})$; i.e., none may be removed when constructing
$\mathbb{L}^{r}(\mathcal{E}_{q})$.

######  Proposition 7.3.

Let $r$ and $q$ be positive integers, and
$\mathbf{a}=(a_{1},\ldots,a_{q}),\mathbf{b}=(b_{1},\ldots,b_{q})\in\mathbb{N}^{q}$.
Then

  1. (i)

${\boldsymbol{\epsilon}}^{\mathbf{b}}\mid{\boldsymbol{\epsilon}}^{\mathbf{a}}$
$\iff$ $b_{i}\leq a_{i}$ for every $i\in[q]$.

  2. (ii)

If $\mathbf{a},\mathbf{b}\in\mathcal{N}^{r}_{q}$, then
${\boldsymbol{\epsilon}}^{\mathbf{b}}\mid{\boldsymbol{\epsilon}}^{\mathbf{a}}$
$\iff$ $\mathbf{b}=\mathbf{a}$.

In particular, if $\mathbf{a},\mathbf{b}\in\mathcal{N}^{r}_{q}$, then then
$\mathbb{L}^{r}_{q}=\mathbb{L}^{r}(\mathcal{E}_{q})$.

###### Proof.

To prove $(i)$, use (7.2.1) to justify the first equivalence below.

${\boldsymbol{\epsilon}}^{\mathbf{b}}\mid{\boldsymbol{\epsilon}}^{\mathbf{a}}\iff\sum_{j\in
A}b_{j}\leq\sum_{j\in A}a_{j}\quad\mbox{ for each }\quad\emptyset\neq
A\subseteq[q]\iff b_{j}\leq a_{j}\quad\mbox{ for each }\quad j\in[q].$

Now $(ii)$ follows from $(i)$ and the added assumption

$b_{1}+\cdots+b_{q}=a_{1}+\cdots+a_{q}=r.$

The final claim follows directly from Definition 5.1. ∎

In general, given the discussions above (see also [13]), for every $r$ and $q$
we have

$\mathbb{L}^{r}(\mathcal{E}_{q})=\mathbb{L}^{r}_{q}\subseteq\mathrm{Taylor}({\mathcal{E}_{q}}^{r}),$
(7.3.1)

where we assume that every vertex $\mathbf{a}$ of $\mathbb{L}^{r}_{q}$ is
labeled with a generator ${\boldsymbol{\epsilon}}^{\mathbf{a}}$ of
${\mathcal{E}_{q}}^{r}$, and each face $\sigma$ is labeled with the lcm of the
labels of its vertices, denoted $M_{\sigma}$. The following observation will
be useful when working with the monomial label $M_{\sigma}$ of
$\sigma\in\mathrm{Taylor}({\mathcal{E}_{q}}^{r})$.

######  Remark 7.4.

Let
$\sigma=\\{{\boldsymbol{\epsilon}}^{\mathbf{a}_{1}},\dots,{\boldsymbol{\epsilon}}^{\mathbf{a}_{t}}\\}\in\mathrm{Taylor}({\mathcal{E}_{q}}^{r})$
and set ${\mathbf{a}_{i}}=(a_{i1},a_{i2},\dots,a_{iq})$ for $i\in[t]$. Using
(7.2.1), we then have

$M_{\sigma}=\mathrm{lcm}\Big{(}\prod_{\emptyset\neq
A\subseteq[q]}(x_{A})^{\sum_{j\in
A}a_{ij}}:i\in[t]\Big{)}=\prod_{\emptyset\neq
A\subseteq[q]}(x_{A})^{\underset{1\leq i\leq t}{\max}\sum_{j\in A}a_{ij}}\,.$
(7.4.1)

Furthermore, if $\mathbf{c}=(c_{1},\dots,c_{q})\in\mathcal{N}^{r}_{q}$,
(7.2.1) and (7.4.1) give

${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid M_{\sigma}\iff\sum_{j\in
A}c_{j}\leq\underset{1\leq i\leq t}{\max}\sum_{j\in A}a_{ij}\qquad\text{for
all}\quad\emptyset\neq A\subseteq[q].$ (7.4.2)

We show in Theorem 7.9 that powers of extremal ideals attain maximal Betti
numbers among powers of all square-free monomial ideals with the same number
of generators. To prove this, we first define a ring homomorphism
$S_{\mathcal{E}}\to S$ and discuss its properties.

######  Definition 7.5 (The ring homomorphism $\psi_{I}$).

Let $I$ be an ideal of the polynomial ring $S=\mathsf{k}[x_{1},\ldots,x_{n}]$
minimally generated by square-free monomials $m_{1},\ldots,m_{q}$. For each
$k\in[n]$, set

$A_{k}=\\{j\in[q]\colon x_{k}\mid m_{j}\\}\,.$

We have thus

$j\in A_{k}\iff x_{k}\mid m_{j}\,.$

Define $\psi_{I}$ to be the ring homomorphism

$\psi_{I}\colon S_{\mathcal{E}}\to S\quad\mbox{ where
}\quad\psi_{I}(x_{A})=\begin{cases}\displaystyle\prod_{\begin{subarray}{c}k\in[n]\\\
A=A_{k}\end{subarray}}x_{k}&\mbox{if }A=A_{k}\mbox{ for some }k\in[n],\\\
1&\mbox{otherwise}.\end{cases}$

Before proceeding directly with our work on extremal ideals, we illustrate how
the map $\psi_{I}$ works in a sample case where there are four generators and
seven variables.

######  Example 7.6.

Let $I$ be the ideal of $\mathsf{k}[x_{1},\ldots,x_{7}]$ generated by the
square-free monomials

$m_{1}=x_{1}x_{2}x_{5}x_{7},\quad m_{2}=x_{2}x_{3}x_{7},\quad
m_{3}=x_{3}x_{4}x_{6},\quad m_{4}=x_{1}x_{4}.$

Since $n=7$ and $q=4$, it follows that

$A_{1}=\\{1,4\\},\quad A_{2}=\\{1,2\\},\quad A_{3}=\\{2,3\\},\quad
A_{4}=\\{3,4\\},\quad A_{5}=\\{1\\},\quad A_{6}=\\{3\\},\quad
A_{7}=\\{1,2\\}.$

The function

$\psi_{I}:\mathsf{k}\big{[}x_{A}\colon\emptyset\neq
A\subseteq[q]\big{]}\to\mathsf{k}[x_{1},\ldots,x_{7}]$

maps

$\begin{array}[]{llll}\psi_{I}(x_{\\{1,4\\}})=x_{1},&\psi_{I}(x_{\\{1,2\\}})=x_{2}x_{7},&\psi_{I}(x_{\\{2,3\\}})=x_{3},&\psi_{I}(x_{\\{3,4\\}})=x_{4},\\\
&&&\\\
\psi_{I}(x_{\\{1\\}})=x_{5},&\psi_{I}(x_{\\{3\\}})=x_{6},&\mbox{and}&\psi_{I}(x_{A})=1\,\,\mbox{otherwise}.\end{array}$

Observe that

$\psi_{I}(\epsilon_{1})=\psi_{I}\Big{(}\prod_{\begin{subarray}{c}A\subseteq[q]\\\
1\in A\end{subarray}}x_{A}\Big{)}=\prod_{\begin{subarray}{c}A\subseteq[q]\\\
1\in
A\end{subarray}}\psi_{I}(x_{A})=\psi_{I}(x_{\\{1,4\\}})\psi_{I}(x_{\\{1,2\\}})\psi_{I}(x_{\\{1\\}})=x_{1}(x_{2}x_{7})x_{5}=m_{1}\,.$

Generalizing the properties of $\psi_{I}$ from the example, we arrive at the
following lemma.

######  Lemma 7.7.

Let $I$ be an ideal of the polynomial ring $S=\mathsf{k}[x_{1},\ldots,x_{n}]$,
minimally generated by square-free monomials $m_{1},\ldots,m_{q}$. Then

  1. (i)

$\psi_{I}({\boldsymbol{\epsilon}}^{\mathbf{a}})=\mathbf{m}^{\mathbf{a}}$ for
each $\mathbf{a}\in\mathcal{N}^{r}_{q}$;

  2. (ii)

$\psi_{I}({\mathcal{E}_{q}}^{r})S=I^{r}$ for every $r>0$;

  3. (iii)

$\psi_{I}$ preserves least common multiples, that is:

$\psi_{I}(\mathrm{lcm}({\boldsymbol{\epsilon}}^{\mathbf{a}_{1}},\dots,{\boldsymbol{\epsilon}}^{\mathbf{a}_{t}}))=\mathrm{lcm}(\mathbf{m}^{\mathbf{a}_{1}},\dots,\mathbf{m}^{\mathbf{a}_{t}})\quad\mbox{
for all }\quad\mathbf{a}_{1},\ldots,\mathbf{a}_{t}\in\mathcal{N}^{r}_{q},\quad
t\geq 1.$

###### Proof.

By Definitions 7.1 and 7.5, for every $j\in[q]$

$\psi_{I}(\epsilon_{j})=\psi_{I}\Big{(}\prod_{\begin{subarray}{c}A\subseteq[q]\\\
j\in A\end{subarray}}x_{A}\Big{)}=\prod_{\begin{subarray}{c}A\subseteq[q]\\\
j\in
A\end{subarray}}\psi_{I}(x_{A})=\prod_{\begin{subarray}{c}A\subseteq[q]\\\
j\in A\end{subarray}}\prod_{\begin{subarray}{c}k\in[n]\\\
A=A_{k}\end{subarray}}x_{k}=\prod_{\begin{subarray}{c}k\in[n]\\\ j\in
A_{k}\end{subarray}}x_{k}=\prod_{\begin{subarray}{c}k\in[n]\\\ x_{k}\mid
m_{j}\end{subarray}}x_{k}=m_{j}.$

It follows that
$\psi_{I}({\boldsymbol{\epsilon}}^{\mathbf{a}})=\mathbf{m}^{\mathbf{a}}$ for
all $\mathbf{a}\in\mathcal{N}^{r}_{q}$, which establishes $(i)$ and also
$(ii)$. It remains to show $(iii)$.

Set ${\mathbf{a}_{i}}=(a_{i1},a_{i2},\dots,a_{iq})$, for $i\in[t]$. Using
(7.4.1) in the first equality below, we have:

$\displaystyle\psi_{I}(\mathrm{lcm}({\boldsymbol{\epsilon}}^{\mathbf{a}_{1}},\dots,{\boldsymbol{\epsilon}}^{\mathbf{a}_{t}}))$
$\displaystyle=\psi_{I}\Big{(}\prod_{\emptyset\neq
A\subseteq[q]}(x_{A})^{\underset{1\leq i\leq t}{\max}\sum_{j\in
A}a_{ij}}\Big{)}$ $\displaystyle=\prod_{\emptyset\neq
A\subseteq[q]}\Big{(}\prod_{\begin{subarray}{c}k\in[n]\\\
A=A_{k}\end{subarray}}x_{k}\Big{)}^{\underset{1\leq i\leq t}{\max}\sum_{j\in
A}a_{ij}}$
$\displaystyle=\prod_{k\in[n],A_{k}\neq\emptyset}(x_{k})^{\underset{1\leq
i\leq t}{\max}\sum_{j\in A_{k}}a_{ij}}$
$\displaystyle=\mathrm{lcm}\Big{(}\prod_{k\in[n],A_{k}\neq\emptyset}(x_{k})^{\sum_{j\in
A_{k}}a_{1j}},\dots,\prod_{k\in[n],A_{k}\neq\emptyset}(x_{k})^{\sum_{j\in
A_{k}}a_{tj}}\Big{)}$
$\displaystyle=\mathrm{lcm}\Big{(}\prod_{j\in[q]}\prod_{\begin{subarray}{c}k\in[n]\\\
j\in
A_{k}\end{subarray}}(x_{k})^{a_{1j}},\dots,\prod_{j\in[q]}\prod_{\begin{subarray}{c}k\in[n]\\\
j\in A_{k}\end{subarray}}(x_{k})^{a_{tj}}\Big{)}$
$\displaystyle=\mathrm{lcm}\Big{(}\prod_{j\in[q]}m_{j}^{a_{1j}},\dots,\prod_{j\in[q]}m_{j}^{a_{tj}}\Big{)}$
$\displaystyle=\mathrm{lcm}(\mathbf{m}^{\mathbf{a}_{1}},\dots,\mathbf{m}^{\mathbf{a}_{t}}).$

∎

Theorem 7.9 below demonstrates why the ideals from Definition 7.5 are called
extremal: they have the greatest Betti numbers among all ideals minimally
generated by $q$ square-free monomials. The following lemma provides the
technical preliminaries necessary for the proof of Theorem 7.9.

######  Lemma 7.8.

Let $I$ be an ideal minimally generated by $q$ square-free monomials in a
polynomial ring $\mathsf{k}[x_{1},\ldots,x_{n}]$. With notation as in
Definition 7.5, if $S$ is viewed as an $S_{\mathcal{E}}$-module via the ring
homomorphism $\psi_{I}\colon S_{\mathcal{E}}\to S$, then

$S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r}\otimes_{S_{\mathcal{E}}}S\cong
S/I^{r}\quad\text{and}\quad\mathrm{Tor}_{i}^{S_{\mathcal{E}}}(S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r},S)=0\quad\text{for
all $i>0$}.$

###### Proof.

First, note that

$\frac{S_{\mathcal{E}}}{{\mathcal{E}_{q}}^{r}}\otimes_{S_{\mathcal{E}}}S\cong\frac{S}{({\mathcal{E}_{q}}^{r})S}=\frac{S}{\psi_{I}({\mathcal{E}_{q}}^{r})S}=\frac{S}{I^{r}}\,.$

To compute
$\mathrm{Tor}_{i}^{S_{\mathcal{E}}}(S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r},S)$,
use the Taylor complex $\mathrm{Taylor}({\mathcal{E}_{q}}^{r})$, which
supports a free resolution $\mathbb{F}$ of
$S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r}$; see (2.4.1) for a description of the
differentials of $\mathbb{F}$. Since the homomorphism $\psi_{I}$ changes the
labels ${\boldsymbol{\epsilon}}^{\mathbf{a}}$ to the labels
$\mathbf{m}^{\mathbf{a}}$ and preserves least common multiples, the chain
complex $\mathbb{F}\otimes_{S_{\mathcal{E}}}S$ is isomorphic to a
homogenization of the chain complex associated to the simplex with vertices
corresponding to $\mathbf{a}\in\mathcal{N}^{r}_{q}$ and labeled with the
monomials $\mathbf{m}^{\mathbf{a}}$. This is a version of the Taylor
resolution of $S/I^{r}$ defined starting with a possibly non-minimal
generating set of $I^{r}$. Such a non-minimal version is a free resolution of
$S/I^{r}$ as well, hence
$\mathrm{Tor}_{i}^{S_{\mathcal{E}}}(S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r},S)=0$
when $i>0$. (See Remark 2.6.) ∎

######  Theorem 7.9 (Powers of extremal ideals have maximal Betti numbers).

Given positive integers $r$ and $q$,

$\beta_{i}^{S}(I^{r})\leq\beta_{i}^{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})$

for any $i\geq 0$ and any ideal $I$ minimally generated by $q$ square-free
monomials in a polynomial ring $S$.

###### Proof.

Let $\mathbb{F}$ be a minimal free resolution of
$S_{\mathcal{E}}/{\mathcal{E}_{q}}^{r}$ over $S_{\mathcal{E}}$. Then Lemma 7.8
shows that $\mathbb{F}\otimes_{S_{\mathcal{E}}}S$ is a free resolution of
$S/I^{r}$ over $S$. Consequently,

$\beta_{i}^{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=\text{rank}_{S_{\mathcal{E}}}(\mathbb{F}_{i+1})=\text{rank}_{S}(\mathbb{F}_{i+1}\otimes_{S_{\mathcal{E}}}S)\geq\beta_{i}^{S}(I^{r}).$

∎

In view of Theorem 7.9, the homogenized chain complex of any (cell) complex
that supports a minimal free resolution of ${\mathcal{E}_{q}}^{r}$ can be
thought of as an upper bound for the minimal free resolution of the $r^{th}$
power of any ideal minimally generated by $q$ square-free monomials.
Proposition 7.10 establishes when our simplicial complex $\mathbb{L}^{r}_{q}$
supports a minimal free resolution of ${\mathcal{E}_{q}}^{r}$.

######  Proposition 7.10 (When $\mathbb{L}^{r}_{q}$ supports a minimal free
resolution of ${\mathcal{E}_{q}}^{r}$).

Let $r$ and $q$ be positive integers. The following statements are equivalent:

  1. (1)

The simplicial complex $\mathbb{L}^{r}_{q}$ supports a minimal free resolution
of the ideal ${\mathcal{E}_{q}}^{r}$.

  2. (2)

$\mathrm{pd}_{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=\dim\mathbb{L}^{r}_{q}$.

  3. (3)

One of the following conditions holds:

     * •

$q=1$ and $r\geq 1$;

     * •

$q=2$ and $1\leq r\leq 4$;

     * •

$q\geq 3$ and $1\leq r\leq 2$.

###### Proof.

(1)$\implies$(2): This implication is clear.

(2)$\implies$(3): Assume that
$\mathrm{pd}_{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=\dim\mathbb{L}^{r}_{q}$
and the values of $r$ and $q$ do not satisfy the conditions in (3). In
particular, we must have $q\geq 2$ and $r\geq 5$, or $q\geq 3$ and $r=3,4$. By
Remark 6.4 and Equation 6.1.1, for all $i,j\in[q]$,

$|F_{i}^{r}|=f>q=|G_{j}^{r}|$ (7.10.1)

and so $F_{1}^{r},\ldots,F_{q}^{r}$ are the facets of $\mathbb{L}^{r}_{q}$ of
highest dimension, with the caveat that $F_{1}^{r}=\cdots=F_{q}^{r}$ when
$r=3=q$, and thus $\dim\mathbb{L}^{r}_{q}=f-1$. Since we assumed
$\mathrm{pd}_{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=\dim\mathbb{L}^{r}_{q}$,
we have $\mathrm{pd}_{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=f-1$ as well.
Let $\mathbb{F}$ denote the free resolution supported on $\mathbb{L}^{r}_{q}$,
and let $\partial$ denote its differential, which is described in (2.4.1).
Since $\mathrm{pd}_{S_{\mathcal{E}}}({\mathcal{E}_{q}}^{r})=f-1$ and a minimal
free resolution of ${\mathcal{E}_{q}}^{r}$ is isomorphic to a direct summand
of $\mathbb{F}$, there must exist a basis element
$e\in\mathbb{F}_{f-1}\smallsetminus(S_{\mathcal{E}})_{\geqslant
1}\mathbb{F}_{f-1}$ such that

$\partial(e)\in(S_{\mathcal{E}})_{\geqslant 1}\mathbb{F}_{f-2}\,.$ (7.10.2)

As in (2.4.1), let $e_{\sigma}$ denote the basis element in $\mathbb{F}$
corresponding to $\sigma\in\mathbb{L}^{r}_{q}$. We write

$e=\sum_{F\in\mathbb{L}^{r}_{q},|F|=f}\alpha_{F}e_{F}$

with $\alpha_{F}\in S_{\mathcal{E}}$. The assumption
$e\notin(S_{\mathcal{E}})_{\geqslant 1}\mathbb{F}_{f-1}$ implies that
$\alpha_{F}$ is a unit for some $F\in\mathbb{L}^{r}_{q}$ with $|F|=f$. By
Equation 7.10.1, we see that $F=F_{i}^{r}$ for some $i\in[q]$. Without loss of
generality, assume $i=1$.

Recall that $M_{\sigma}$ denotes the $\mathrm{lcm}$ of the monomial labels of
the vertices in $\sigma\in\mathbb{L}^{r}_{q}$.

######  Claim.

There exists $\mathbf{c}\in F_{1}^{r}\smallsetminus G_{1}^{r}$ such that
$M_{F_{1}^{r}}=M_{F_{1}^{r}\smallsetminus\\{\mathbf{c}\\}}$.

Proof of Claim. Assume first $r>4$, and $q\geq 2$, so that
$s=\left\lceil{\frac{r}{2}}\right\rceil\geq 3$. Let
$\mathbf{a},\mathbf{b},\mathbf{c}\in\mathcal{N}^{r}_{q}$ be such that

${\boldsymbol{\epsilon}}^{\mathbf{a}}={\epsilon_{1}}^{r-s}{\epsilon_{2}}^{s},\quad{\boldsymbol{\epsilon}}^{\mathbf{b}}={\epsilon_{1}}^{r-s+2}{\epsilon_{2}}^{s-2}\quad\mbox{
and
}\quad{\boldsymbol{\epsilon}}^{\mathbf{c}}={\epsilon_{1}}^{r-s+1}{\epsilon_{2}}^{s-1},$

so that $\mathbf{a},\mathbf{b},\mathbf{c}$ are distinct vertices of
$\mathbb{L}^{r}_{q}$. Note that
${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid\mathrm{lcm}({\boldsymbol{\epsilon}}^{\mathbf{a}},{\boldsymbol{\epsilon}}^{\mathbf{b}})$.
Indeed, after removing the common factors, this divisibility is equivalent to

$\epsilon_{1}\epsilon_{2}\mid\mathrm{lcm}(\epsilon_{1}^{2},\epsilon_{2}^{2})\,,$

and can be verified using (7.4.2).

Now let $r=3$ or $r=4$, and $q\geq 3$. Then $s=2$, and if one sets

${\boldsymbol{\epsilon}}^{\mathbf{a}}={\epsilon_{1}}^{r-2}{\epsilon_{2}}^{2},\quad{\boldsymbol{\epsilon}}^{\mathbf{b}}={\epsilon_{1}}^{r-1}{\epsilon_{3}}\quad\mbox{
and
}\quad{\boldsymbol{\epsilon}}^{\mathbf{c}}={\epsilon_{1}}^{r-2}{\epsilon_{2}}\epsilon_{3},$

then we see that
${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid\mathrm{lcm}({\boldsymbol{\epsilon}}^{\mathbf{a}},{\boldsymbol{\epsilon}}^{\mathbf{b}})$.
Indeed, after removing the common factors, this divisibility is equivalent to

$\epsilon_{2}\epsilon_{3}\mid\mathrm{lcm}(\epsilon_{2}^{2},\epsilon_{1}\epsilon_{3})\,,$

and can be verified using (7.4.2).

In both cases, we have $\mathbf{a},\mathbf{b},\mathbf{c}\in F_{1}^{r}$ and
$\mathbf{c}\notin G_{1}^{r}$. The divisibility
${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid\mathrm{lcm}({\boldsymbol{\epsilon}}^{\mathbf{a}},{\boldsymbol{\epsilon}}^{\mathbf{b}})$
establishes the conclusion of the Claim.

We finish the proof of (2)$\implies$(3) as follows. We have

$\partial(e)=\sum_{F\in\mathbb{L}^{r}_{q},|F|=f}\alpha_{F}\partial(e_{F})=\sum_{F\in\mathbb{L}^{r}_{q},|F|=f}\alpha_{F}\left(\sum_{\mathbf{c}^{\prime}\in
F}\pm\frac{M_{F}}{M_{F\smallsetminus\\{\mathbf{c}^{\prime}\\}}}e_{F\smallsetminus\\{\mathbf{c}^{\prime}\\}}\right)\,.$
(7.10.3)

Let $\mathbf{c}$ be as in the Claim and let $F\in\mathbb{L}^{r}_{q}$ with
$|F|=f$. By Equation 7.10.1, we have $F=F_{j}^{r}$ for some $j$. If
$\mathbf{c}^{\prime}\in F$, observe

$F_{1}^{r}\smallsetminus\\{\mathbf{c}\\}=F\smallsetminus\\{\mathbf{c}^{\prime}\\}\iff\text{
$F_{1}^{r}=F$ and $\mathbf{c}=\mathbf{c}^{\prime}$.}$ (7.10.4)

To prove this statement, we refer to Section 4 for basic properties of the
sets $F_{i}^{r}$, $G_{i}^{r}$ and $B^{r}$. Indeed, (7.10.4) is clear when
$r=3$, since $F_{1}^{r}=B^{r}=F_{j}^{r}$ for all $j\in[q]$ in this case.
Assume now $r>3$, and recall that $q\geq 2$. Assuming
$F_{1}^{r}\smallsetminus\\{\mathbf{c}\\}=F_{j}^{r}\smallsetminus\\{\mathbf{c}^{\prime}\\}$
and $j\neq 1$, we use $F_{1}^{r}\cap F_{j}^{r}=B^{r}$, and conclude
$F_{1}^{r}=B^{r}\cup\\{\mathbf{c}\\}$. Since $B^{r}\cap G_{1}^{r}=\emptyset$
and $\mathbf{c}\notin G_{1}^{r}$, this contradicts the fact that
$F_{1}^{r}\cap G_{1}^{r}\neq\emptyset$ when $r>3$ and $q\geq 2$. Thus,
(7.10.4) must hold.

In view of the Claim, (7.10.4) and (7.10.3), we see that the coefficient of
$e_{F_{1}^{r}\smallsetminus\\{\mathbf{c}\\}}$ in $\partial(e)$ is equal to
$\pm\alpha_{F_{1}^{r}}$ hence it is a unit. This contradicts (7.10.2).

(3)$\implies(1)$: Theorem 5.9 shows that $\mathbb{L}^{r}_{q}$ supports a free
resolution of ${\mathcal{E}_{q}}^{r}$ for all $r,q\geq 1$. To show minimality,
according to Theorem 3.1, it suffices to show that $M_{\sigma}\neq
M_{\sigma^{\prime}}$ for any faces
$\sigma,\sigma^{\prime}\in\mathbb{L}^{r}_{q}$ with
$\sigma\neq\sigma^{\prime}$, or, in other words, that each monomial label
appears only once.

If $q=1$, then $I^{r}=({m_{1}}^{r})$ for all $r$, and all complexes in (7.3.1)
are one point, so each supports a minimal resolution by default.

If $q=2$, then $\mathcal{E}_{2}=(x_{1}x_{12},x_{2}x_{12})$, and
$\mathbb{L}^{2}(\mathcal{E}_{2})$, $\mathbb{L}^{3}(\mathcal{E}_{2})$, and
$\mathbb{L}^{4}(\mathcal{E}_{2})$ are shown below. Observe that each monomial
label appears once in each complex, hence $\mathbb{L}^{2}(\mathcal{E}_{2})$,
$\mathbb{L}^{3}(\mathcal{E}_{2})$, and $\mathbb{L}^{4}(\mathcal{E}_{2})$
support a minimal free resolution of ${\mathcal{E}_{2}}^{2}$,
${\mathcal{E}_{2}}^{3}$ and ${\mathcal{E}_{2}}^{4}$, respectively.

${x_{1}}^{2}{x_{12}}^{2}$${x_{1}}{x_{2}}{x_{12}}^{2}$${x_{2}}^{2}{x_{12}}^{2}$${x_{1}}^{2}{x_{2}}{x_{12}}^{2}$${x_{1}}{x_{2}}^{2}{x_{12}}^{2}$ |  ${x_{1}}^{3}{x_{12}}^{3}$${x_{1}}^{2}{x_{2}}{x_{12}}^{3}$${x_{1}}{x_{2}}^{2}{x_{12}}^{3}$${x_{2}}^{3}{x_{12}}^{3}$${x_{1}}{x_{2}}^{3}{x_{12}}^{3}$${x_{1}}^{3}{x_{2}}{x_{12}}^{3}$${x_{1}}^{2}{x_{2}}^{2}{x_{12}}^{3}$ |  ${x_{1}}^{4}{x_{12}}^{4}$${x_{1}}^{3}{x_{2}}{x_{12}}^{4}$${x_{1}}{x_{2}}^{3}{x_{12}}^{4}$${x_{2}}^{4}{x_{12}}^{4}$${x_{1}}{x_{2}}^{4}{x_{12}}^{4}$${x_{1}}^{4}{x_{2}}{x_{12}}^{4}$${x_{1}}^{2}{x_{2}}^{2}{x_{12}}^{4}$${x_{1}}^{3}{x_{2}}^{2}{x_{12}}^{4}$${x_{1}}^{2}{x_{2}}^{3}{x_{12}}^{4}$
---|---|---
$\mathbb{L}^{2}(\mathcal{E}_{2})$ | $\mathbb{L}^{3}(\mathcal{E}_{2})$ | $\mathbb{L}^{4}(\mathcal{E}_{2})$

Now assume $q\geq 3$ and $1\leq r\leq 2$. We will show that, for every
$\mathbf{c}\in\mathcal{N}^{r}_{q}$ and $\sigma\in\mathbb{L}^{r}_{q}$,

${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid M_{\sigma}\iff\mathbf{c}\in\sigma.$
(7.10.5)

When $r=1$ observe that for every $i\in[q]$,

$\epsilon_{i}\nmid\mathrm{lcm}(\epsilon_{k_{1}},\dots,\epsilon_{k_{p}})\quad\mbox{
for all }\quad i\in[q],\quad k_{1},\dots,k_{p}\in[q]\smallsetminus\\{i\\}.$
(7.10.6)

This can be seen by noting that the right-hand side of (7.4.2) becomes $1\leq
0$ for $A=\\{i\\}$, ${\boldsymbol{\epsilon}}^{\mathbf{c}}=\epsilon_{i}$ and
$\sigma=\\{\epsilon_{k_{1}},\dots,\epsilon_{k_{p}}\\}$. Therefore
$\mathbb{L}^{1}_{q}=\mathbb{L}^{1}(\mathcal{E}_{q})=\mathrm{Taylor}(\mathcal{E}_{q}),$
and (7.10.5) follows immediately.

Now let $r=2$, $q\geq 3$, $\sigma\in\mathbb{L}^{2}_{q}$ and
$\mathbf{c}\in\mathcal{N}_{q}^{2}$, with
${\boldsymbol{\epsilon}}^{\mathbf{c}}\mid M_{\sigma}$. Since $r=2$,
${\boldsymbol{\epsilon}}^{\mathbf{c}}=\epsilon_{i}\epsilon_{j}$ for some
$i,j\in[q]$. Pick $A=\\{i,j\\}$ in the right-hand side of (7.4.2).

  * •

If $i=j$, then ${\boldsymbol{\epsilon}}^{\mathbf{c}}={\epsilon_{i}}^{2}$ and
$A=\\{i\\}$. By (7.4.2) we must have $\mathbf{c}\in\sigma$.

  * •

If $i\neq j$, then by (7.4.2) there exists $\mathbf{b}\in\sigma$ with

${\boldsymbol{\epsilon}}^{\mathbf{b}}=\epsilon_{i}\epsilon_{j},\
{\epsilon_{i}}^{2}\mbox{ or }{\epsilon_{j}}^{2}.$

If ${\boldsymbol{\epsilon}}^{\mathbf{b}}=\epsilon_{i}\epsilon_{j}$ then
$\mathbf{c}=\mathbf{b}\in\sigma$, as desired. Suppose
${\boldsymbol{\epsilon}}^{\mathbf{c}}\notin\sigma$, so without loss of
generality ${\boldsymbol{\epsilon}}^{\mathbf{b}}={\epsilon_{i}}^{2}$. As
$\sigma\in\mathbb{L}^{2}_{q}$, we must have $\sigma\subseteq G_{i}^{2}$. So
there exist $k_{1},\dots,k_{p}$ in $[q]\smallsetminus\\{i,j\\}$ such that

$M_{\sigma}=\mathrm{lcm}({\epsilon_{i}}^{2},\epsilon_{i}\epsilon_{k_{1}},\dots,\epsilon_{i}\epsilon_{k_{p}})=\epsilon_{i}\mathrm{lcm}(\epsilon_{i},\epsilon_{k_{1}},\dots,\epsilon_{k_{p}})\,.$

Since $\epsilon_{i}\epsilon_{j}\mid M_{\sigma}$, it follows that
$\epsilon_{j}\mid\mathrm{lcm}(\epsilon_{i},\epsilon_{k_{1}},\dots,\epsilon_{k_{p}})$,
which contradicts (7.10.6). Thus $\mathbf{c}\in\sigma$.

We have now proved the statement in (7.10.5), and by Theorem 3.1 we conclude
that $\mathbb{L}^{2}_{q}$ supports a minimal free resolution of
${\mathcal{E}_{q}}^{2}$ for every $q\geq 3$.

∎

A direct consequence of our work in Proposition 7.10 is the following
statement.

######  Proposition 7.11 (When $\mathbb{L}^{r}_{q}$ supports a minimal
resolution of some $I^{r}$).

If $r$ and $q$ are positive integers, then $\mathbb{L}^{r}_{q}$ supports a
minimal free resolution of $I^{r}$ for some ideal $I$ minimally generated by
$q$ square-free monomials if and only if one of the following holds:

  * •

$q=1$ and $r\geq 1$;

  * •

$q=2$ and $1\leq r\leq 4$;

  * •

$q\geq 3$ and $1\leq r\leq 2$.

###### Proof.

For any square-free monomial ideal $I$ with $q$ minimal generators the Betti
numbers of $I^{r}$ are bounded above by the Betti numbers of
${\mathcal{E}_{q}}^{r}$ by Theorem 7.9. So the question is reduced to when
$\mathbb{L}^{r}_{q}=\mathbb{L}^{r}(\mathcal{E}_{q})$ supports a minimal free
resolution of ${\mathcal{E}_{q}}^{r}$. The rest follows from Proposition 7.10.
∎

When $r=2$, the fact that these bounds are sharp had been previously announced
in [4]. The search for sharp(er) bounds when $r>2$ is continued in [5].

#### Acknowledgments

The research leading to this paper was initiated during the week long workshop
“Women in Commutative Algebra” (19w5104) which took place at the Banff
International Research Station (BIRS). The authors would like to thank the
organizers and acknowledge the hospitality of BIRS and the additional support
provided by the National Science Foundation, DMS-1934391.

The authors are extremely grateful to the anonymous referee for their careful
reading of this paper, excellent suggestions, as well as detecting and
correcting a mistake in the proof of Proposition 5.6.

For this work Liana Şega was supported in part by a grant from the Simons
Foundation (#354594), and Susan Cooper and Sara Faridi were supported by the
Natural Sciences and Engineering Research Council of Canada (NSERC).

The computations for this project were done using the computer algebra program
Macaulay2 [11].

For the last author, this material is based upon work supported by and while
serving at the National Science Foundation. Any opinion, findings, and
conclusions or recommendations expressed in this material are those of the
authors and do not necessarily reflect the views of the National Science
Foundation.

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