**Proof.** It is clear that the resolution of

I is as given in the assertion of the theorem. Since

I is the cover ideal of

${K}_{3}$,

${I}^{s}$ has linear minimal free resolution for all

$s\ge 1$. Note also that by ([

16] Lemma 3.1),

$depthR/{I}^{s}=0$ for all

$s\ge 2$ so that

$pdimR/{I}^{s}=3$ for all

$s\ge 2$. Therefore, the minimal free resolution of

$R/{I}^{s}$ is of the form

Now we describe the syzygies and Betti numbers of $R/{I}^{s}$ for $s\ge 2$. Let ${g}_{1}={x}_{1}{x}_{2},{g}_{2}={x}_{1}{x}_{3}$ and ${g}_{3}={x}_{2}{x}_{3}$. The generators of ${I}^{s}$ are of the form ${g}_{1}^{{\ell}_{1}}{g}_{2}^{{\ell}_{2}}{g}_{3}^{{\ell}_{3}}$, where $0\le {\ell}_{1},{\ell}_{2},{\ell}_{3}\le s$ and ${\ell}_{1}+{\ell}_{2}+{\ell}_{3}=s$. Set ${f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}={g}_{1}^{{\ell}_{1}}{g}_{2}^{{\ell}_{2}}{g}_{3}^{{\ell}_{3}}={x}_{1}^{s-{\ell}_{3}}{x}_{2}^{s-{\ell}_{2}}{x}_{3}^{s-{\ell}_{1}}$. It is easy to see that $\mu \left({I}^{s}\right)=\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)$, as the cardinality of a minimal generating set of ${I}^{s}$ is same as the total number of non-negative integral solution of ${\ell}_{1}+{\ell}_{2}+{\ell}_{3}=s$, which is $\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)$.

Let $\{{e}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}\phantom{\rule{3.33333pt}{0ex}}:\phantom{\rule{3.33333pt}{0ex}}0\le {\ell}_{1},{\ell}_{2},{\ell}_{3}\le s;{\ell}_{1}+{\ell}_{2}+{\ell}_{3}=s\}$ denote the standard basis for ${R}^{\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)}$ and consider the map ${\partial}_{1}:{R}^{\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)}\to R$ defined by ${\partial}_{1}\left({e}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}\right)={f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}$. As ${f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}$’s are monomials, the kernel is generated by binomials of the form ${m}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}{f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}-{m}_{{t}_{1},{t}_{2},{t}_{3}}{f}_{{t}_{1},{t}_{2},{t}_{3}}$, where ${m}_{i,j,k}$’s are monomials in R. Also, as the minimal free resolution is linear, the kernel is generated in degree 1. Note that $\frac{{f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}}{{f}_{{t}_{1},{t}_{2},{t}_{3}}}={x}_{1}^{{t}_{3}-{\ell}_{3}}{x}_{2}^{{t}_{2}-{\ell}_{2}}{x}_{3}^{{t}_{1}-{\ell}_{1}}$. Therefore, for $\frac{{f}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}}{{f}_{{t}_{1},{t}_{2},{t}_{3}}}$ to be a linear fraction, $|{t}_{i}-{\ell}_{i}|\le 1$ for $i=1,2,3$.

Let

${t}_{3}={\ell}_{3}$,

${t}_{2}={\ell}_{2}+1$ and

${t}_{1}={\ell}_{1}-1$. The corresponding linear syzygy relation is

where

$1\le {\ell}_{1}\le s$, and

$0\le {\ell}_{2},{\ell}_{3}\le s-1$. Note that the number of such relations is equal to the number of integral solution to

$({\ell}_{1}-1)+{\ell}_{2}+{\ell}_{3}=s$, i.e.,

$\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)$. Similarly, if

${t}_{3}={\ell}_{3}+1$,

${t}_{2}={\ell}_{2}-1$ and

${t}_{1}={\ell}_{1}$, then we get the corresponding linear syzygy relation as

where

$0\le {\ell}_{1},{\ell}_{3}\le s-1$ and

$1\le {\ell}_{2}\le s$. Note that the linear syzygy relation obtained by fixing

${\ell}_{2}$ and taking

$|{\ell}_{i}-{t}_{i}|=1$ for

$i=1,2$
can be obtained from Equations (

2) and (

3) by setting the

${\ell}_{i}$’s appropriately. Therefore,

As there are

$\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)$ minimal generators of type (

2) and (

3),

$\mu (ker{\partial}_{1})={\beta}_{2}=2\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)$. Write the standard basis of

${R}^{2\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)}$ as

and define

${\partial}_{2}:{R}^{2\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)}\to {R}^{\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)}$ by

From the equation,

${\beta}_{3}-2\left(\genfrac{}{}{0pt}{}{(s+1)}{2}\right)+\left(\genfrac{}{}{0pt}{}{s+2}{2}\right)-1=0$, it follows that

${\beta}_{3}=\left(\genfrac{}{}{0pt}{}{s}{2}\right)$. Now, we describe

$ker{\partial}_{2}$. For

$2\le {\ell}_{1}\le s\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}0\le {\ell}_{2},{\ell}_{3}\le s-2$, set

Let $N=ker{\partial}_{2}$. We claim that ${\overline{B}}_{3}=\left\{{\overline{E}}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}\right\}$ is a basis for $N/\mathfrak{m}N$. It is enough to prove that ${\overline{B}}_{3}$ is $R/\mathfrak{m}$-linearly independent. Suppose $\sum _{{\ell}_{1},{\ell}_{2},{\ell}_{3}}{\alpha}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}{\overline{E}}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}=\overline{0}$. Since ${B}_{3}\subset {N}_{1}$ and $\mathfrak{m}N\subset {\oplus}_{r\ge 2}{N}_{r}$, the above equation implies that $\sum {\alpha}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}{E}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}=0$. In the above equation, the coefficient of ${e}_{2,(i,j),k}=-{x}_{3}{\alpha}_{i+1,j,i-1}+{x}_{2}{\alpha}_{i+1,j-1,k}$ and coefficient of ${e}_{1,(i,j),k}={x}_{2}{\alpha}_{k+2,j-2,i}-{x}_{3}{\alpha}_{k+1,j-1,i}$. As the set ${B}_{2}$ is linearly independent in ${R}^{2\left(\genfrac{}{}{0pt}{}{s+1}{2}\right)}$, all of these coefficients have to be zero, i.e., ${\alpha}_{i,j,k}=0$ for all $i,j,k$. This proves our claim and hence $N=\langle {B}_{3}\rangle $. Let $\{{H}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}\phantom{\rule{3.33333pt}{0ex}}:\phantom{\rule{3.33333pt}{0ex}}2\le {\ell}_{1}\le s,\phantom{\rule{0.277778em}{0ex}}0\le {\ell}_{2},{\ell}_{3}\le s-2\}$ denote the standard basis of ${R}^{\left(\genfrac{}{}{0pt}{}{s}{2}\right)}$. Define ${\partial}_{3}:{R}^{\left(\genfrac{}{}{0pt}{}{s}{2}\right)}\to {R}^{2\left(\genfrac{}{}{0pt}{}{(s+1)}{2}\right)}$ by ${\partial}_{3}\left({H}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}\right)={E}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}.$ As $pdim(R/{I}^{s})=3$, this map is injective and the resolution is complete.

It can also be seen that

$deg{e}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}=2s,\phantom{\rule{3.33333pt}{0ex}}deg{e}_{1,({\ell}_{2}+1,{\ell}_{3}),{\ell}_{1}-1}=deg{e}_{2,({\ell}_{1},{\ell}_{2}),{\ell}_{3}}=2s+1$ and

$deg{H}_{{\ell}_{1},{\ell}_{2},{\ell}_{3}}=2s+2$. Therefore, we get the minimal graded free resolution of

$R/{I}^{s}$ as

Therefore, the Hilbert series of

$R/{I}^{s}$ is given by

By expanding ${(1-t)}^{-2}$ in the power series form and multiplying with the numerator, we get the required expression. □

We now proceed to compute the minimal graded free resolution of powers of complete tripartite graphs.

Observe that the above expression for

$reg\left({J}_{G}^{s}\right)$ is not really in the form

$ds+e$. Given a graph, one can compute

d and

e by studying the interplay between the cardinality of the partitions. For example, suppose the graph is unmixed, i.e., all the partitions are of same cardinality. Then,

$reg\left({J}_{G}^{s}\right)=2\ell s+(2\ell -2)$ for all

$s\ge 2$, where

$\ell ={m}_{1}={m}_{2}={m}_{3}$. Note also that the stabilization index in this case is 2. As in

Section 3.1, one can derive various expressions for

$reg\left({J}_{G}^{s}\right)$ for different cases as well. Consider the arithmetic progression

${m}_{1}=m+2r$,

${m}_{2}=m$, and

${m}_{3}=m+r$: