# Tight closure of powers of parameter ideals in hypersurface rings and their
tight Hilbert polynomials

Saipriya Dubey, Vivek Mukundan and Jugal Verma  Saipriya Dubey
Mathematics Department, Indian Institute of Technology Bombay,
Powai, Mumbai - 400076, India, Email-<EMAIL_ADDRESS>
Vivek Mukundan
Mathematics Department, Indian Institute of Technology Delhi,
Hauz Khas, Delhi - 110016, India, Email-<EMAIL_ADDRESS>
Jugal Verma
Mathematics Department, Indian Institute of Technology Bombay,
Powai, Mumbai - 400076, India, Email-<EMAIL_ADDRESS>Dedicated to Professor
Jürgen Herzog on the occasion of his 80th birthday

###### Abstract.

In this paper we find the tight closure of powers of parameter ideals of
certain diagonal hypersurface rings. In many cases the associated graded ring
with respect to tight closure filtration turns out to be Cohen-Macaulay. This
helps us find the tight Hilbert polynomial in these diagonal hypersurfaces. We
determine the tight Hilbert polynomial in the following cases: (1) $F$-pure
diagonal hypersurfaces where number of variables is equal to the degree of
defining equation, (2) diagonal hypersurface rings where characteristic of the
ring is one less than the degree of defining equation and (3) quartic diagonal
hypersurface in four variables.

###### Key words and phrases:

tight closure, parameter ideals, powers of ideals, tight Hilbert polynomial,
diagonal hypersurface

###### 2010 Mathematics Subject Classification:

Primary: 13A35. Secondary: 13D40

The first author is supported by the Senior Research Fellowship of HRDG, CSIR,
Government of India

##  1\. Introduction

The objective of this paper is to determine the tight Hilbert polynomial of
ideals generated by homogeneous system of parameters in diagonal hypersurface
rings in prime characteristic. The tight Hilbert polynomial was introduced by
K. Goel, V. Mukundan and J. K. Verma in [7]. Let $(R,{\mathfrak{m}})$ be a
$d$-dimensional analytically unramified local ring of prime characteristic
$p>0$ and $I$ be an ${\mathfrak{m}}$-primary ideal. Let $I^{*}$ denote the
tight closure of $I$ (refer Definition 2.5). Let
$H_{I}^{*}(n)=\ell\left(\frac{R}{(I^{n})^{*}}\right)$ where $\ell$ denotes the
length. Then for large $n,$ $H_{I}^{*}(n)$ is given by a polynomial
$P_{I}^{*}(n)$ written as

$P_{I}^{*}(n)=e_{0}^{*}(I)\binom{n+d-1}{d}-e_{1}^{*}(I)\binom{n+d-2}{d-1}+\cdots+(-1)^{d}e_{d}^{*}(I),$

where $e_{i}^{*}(I)\in\mathbb{Z}.$ Recall that $R$ is called an $F$-rational
local ring if the ideals of the principal class, that is, ideals $J$ generated
by height $J$ elements, are tightly closed. It is proved [7] that in a Cohen-
Macaulay local ring $R$, $e_{1}^{*}(I)=0$ if and only if $R$ is $F$-rational.
This indicates that the other tight Hilbert coefficients might play an
important role in detecting other $F$-singularities. They also computed the
tight Hilbert polynomial for two-dimensional diagonal hypersurfaces
$K[X,Y,Z]/(X^{r}+Y^{r}+Z^{r}),$ where $K$ is a field of prime characteristic
$p$ and $(p,r)=1.$ Since there are only a handful of examples of tight Hilbert
polynomials known in the literature, this paper is devoted to computation of
the tight Hilbert polynomial of parameter ideals $I=(x_{1},\ldots,x_{d})$ of
diagonal hypersurfaces in higher dimensions, i.e.
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(X_{1}^{r}+X_{2}^{r}+\cdots+X_{d+1}^{r})$
with the unique homogeneous maximal ideal ${\mathfrak{m}}=(I,x_{d+1})$ in some
special cases. (Throughout, the lower case letters denote the images of the
corresponding variables.)

The tight Hilbert coefficients $e_{i}^{*}(I)$ can be computed via the tight
closure of the powers of the paramter ideals $(I^{n})^{*}$. In general,
computing the tight closure of the powers can be quite hard. In a hypersurface
ring, the tight closure of a homogeneous system of parameters can be computed
using the strong vanishing theorem [12, Theorem 6.4]. This result computes the
tight closure of the homogeneous system of parameters when the characteristic
is “large enough” compared to the degree of the defining ideal of the
hypersurface. It is interesting to note that the tight closure of $I$ equals
$I+{\mathfrak{m}}^{d}$ in this case. Much of this work has been a result of
searching for ideals $I$ whose tight closure of the powers $(I^{n})^{*}$ has a
similar description, such as $I^{n}+{\mathfrak{m}}^{k}$ for some $k$, as that
appearing in the strong vanishing theorem.

This paper is organized as follows. Section 2 contains basic definitions,
notations and results which will provide all the tools necessary in the later
sections. In Section 3, we find the Hilbert polynomial of the filtration
$\\{I_{n}\\}_{n\geq 0},$ where $I_{0}=R$ and
$I_{n}=I^{n}+{\mathfrak{m}}^{n+t}$ for $n\geq 1$ that occur (in the subsequent
sections) as the tight closure filtration of ideals in diagonal hypersurface
rings . In Section 4, we determine the tight Hilbert polynomial in the ring
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(f),$ where degree of $f$ is
$d+1$ and the characteristic $p>(d-1)(d+1)-d.$ Here, the associated graded
ring of the tight closure filtration $\\{(I^{n})^{*}\\}_{n\geq 0}$ is given by
$\operatorname{gr}_{I}^{*}(R)=\bigoplus\limits_{n\geq
0}\frac{(I^{n})^{*}}{(I^{n+1})^{*}}.$

######  Theorem 1.1.

Let $T=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}],$ $R=T/(f)$ where $f\in T$
is a square-free, monic in $X_{d+1}$ homogeneous ploynomial of degree $d+1$
and $p>(d-1)(d+1)-d.$ Let $I=(x_{1},\ldots,x_{d})$ and
${\mathfrak{m}}=(I,x_{d+1}).$ Then $(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+d-1}$
and for $n\geq 1,$

$\ell\left(\frac{R}{(I^{n})^{*}}\right)=(d+1)\binom{n+d-1}{d}-\binom{n+d-2}{d-1}.$

Moreover $\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

One of the important ingredient of the proof is the use of [13] which gives
bounds on the tight closure of $I$ using the betti numbers appearing in a
resolution of $R/I$. Here we use the Eagon-Northcott complex which resolves
$R/I^{n}$ as a means of constructing the tight closure of the powers $I^{n}$.

Recall that, a Noetherian local ring $R$ of prime characteristic $p$ is said
to be $F$-pure if the Frobenius homomorphism $F:M\rightarrow M\otimes_{R}F(R)$
is injective for all $R$-modules $M$ where $F(R)$ is the image of $R$ under
the frobenius endomorphism which takes $r\in R$ to $r^{p}.$ We know that the
test elements play a vital role in the theory of tight closure. In $F$-pure
rings, the test ideal turns out to be a radical ideal. Exploiting this
property of $F$-pure rings, in Theorem 4.3 we determine the tight Hilbert
polynomial in all diagonal hypersurface $F$-pure rings when the number of
variables is equal to the degree of the defining equation of the hypersurface.
Along with results on tight Hilbert polynomial of a minimal reduction of the
maximal homogeneous ideals, we also present the result on F-pure rings.

######  Theorem 1.2.

Let
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{t+2}]/(X_{1}^{r}+X_{2}^{r}+\cdots+X_{t+2}^{r}),$
$I=(x_{1},\ldots,x_{t+1})$ and ${\mathfrak{m}}=(I,x_{t+2}),$ where
$p>tr-(t+1).$ Then
(1) $\operatorname{gr}_{I}^{*}(R)$ is Cohen-Macaulay.
(2) The tight Hilbert polynomial $P_{I}^{*}(n)$ is given by

$P_{I}^{*}(n)=\begin{cases}r\binom{n+t}{t+1},&\text{ if }r\leq t+1,\\\
r\binom{n+t}{t+1}-\binom{r-t}{2}\binom{n+t-1}{t}+\binom{r-t}{3}\binom{n+t-2}{t-1}+\cdots+(-1)^{j}\binom{r-t}{j+1}\binom{n+t-j}{t+1-j},&\text{
if }r\geq t+2.\\\ \end{cases}$

If $r\geq t+2,$ $e_{j}^{*}(I)=0$ for $r-t\leq j\leq t+1.$
(3) If $r=t+2$ and $R$ is F-pure, then for all $n\geq 1$,

$\ell\left(\frac{R}{(I^{n})^{*}}\right)=(t+2)\binom{n+t}{t+1}-\binom{n+t-1}{t}.$

In Section 5, we explore the tight Hilbert polynomial in the diagonal
hypersurface rings where the characteristic of the ring is one less than the
degree of the defining equation.

######  Theorem 1.3.

Let
$R=\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]}{(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1})},$
$I=(x_{1},x_{2},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}).$ Then
(1) $(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+1}$ for all $n\geq 1.$
(2)
$P_{I}^{*}(n)=(p+1)\binom{n+d-1}{d}-\binom{p}{2}\binom{n+d-2}{d-1}+\cdots+(-1)^{j}\binom{p}{j+1}\binom{n+d-j-1}{d-j},$
where $j=\min\\{p-1,d\\}.$
Moreover, $e_{i}^{*}(I)=0$ for $p\leq i\leq d$ and
$\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

Finally in Section 6, we determine the tight Hilbert polynomial for quartic
diagonal hypersurfaces in four variables.

######  Theorem 1.4.

Let $R=\frac{\mathbb{F}_{p}[X,Y,Z,W]}{(X^{4}+Y^{4}+Z^{4}+W^{4})},$ $I=(x,y,z)$
and ${\mathfrak{m}}=(I,w).$ Then for $n\geq 1,$

$P_{I}^{*}(n)=\begin{cases}4\binom{n+2}{3}-3\binom{n+1}{2}+n,&\text{ if
}p=3,\\\ 4\binom{n+2}{3}-\binom{n+1}{2},&\text{ if }p\geq 5.\end{cases}$

Moreover $\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

### Acknowledgements

We thank Anurag Singh and Karen Smith for discussions about computation of
tight closure of powers of ideals.

##  2\. Preliminaries

We first set up notation, recall certain basic concepts and results needed in
subsequent sections.

######  Definition 2.1.

Let $(R,{\mathfrak{m}})$ be a local ring and $I$ be an
${\mathfrak{m}}$-primary ideal of $R.$ A sequence of ideals
${\mathcal{F}}=\\{I_{n}\\}_{n\in\mathbb{Z}}$ is called an $I$-filtration if
for all $m,n\in\mathbb{Z},$

(1) $I_{n}=R$ for all $n\leq 0,\;\;$ (2) $I_{n}\supseteq I_{n+1},\;\;$ (3)
$I_{m}I_{n}\subseteq I_{m+n}\;\;$ and (4) $I^{n}\subseteq I_{n}.$

The $I$-filtration ${\mathcal{F}}$ is called $I$-admissible if there exists
$r\in\mathbb{N}$ such that $I_{n}\subseteq I^{n-r}$ for all $n\in\mathbb{Z}.$

We associate three blowup algebras with an $I$-admissible filtration
${\mathcal{F}}=\\{I_{n}\\}_{n\geq 0}$. The Rees algebra of ${\mathcal{F}}$ is
denoted by $\mathcal{R}({\mathcal{F}})=\bigoplus_{n\geq 0}I_{n}t^{n}$, the
extended Rees algebra of ${\mathcal{F}}$ is denoted by
$\mathcal{R}^{\prime}({\mathcal{F}})=\bigoplus_{n\in{\mathbb{Z}}}I_{n}t^{n}$
and the associated graded ring of ${\mathcal{F}}$ is denoted by
$\operatorname{gr}_{{\mathcal{F}}}(R)=\bigoplus_{n\geq 0}I_{n}/I_{n+1}.$ If
${\mathcal{F}}=\\{I^{n}\\}_{n\geq 0},$ we denote the blow up algebras by
$\mathcal{R}(I),\mathcal{R}^{\prime}(I)$ and $\operatorname{gr}_{I}(R)$
respectively. Let $I$ be an ${\mathfrak{m}}$-primary ideal. The Hilbert
function of ${\mathcal{F}}$ is defined by
$H_{{\mathcal{F}}}(n)=\ell(R/I_{n}).$ If ${\mathcal{F}}$ is an $I$-admissible
filtration, then for sufficiently large $n,$ $H_{{\mathcal{F}}}(n)$ coincides
with a polynomial $P_{{\mathcal{F}}}(n)$ of degree $d$ for large $n,$ where
$d=\dim R.$ There exist integers
$e_{0}({\mathcal{F}}),e_{1}({\mathcal{F}}),\ldots,e_{d}({\mathcal{F}})$ so
that

$P_{{\mathcal{F}}}(n)=e_{0}({\mathcal{F}})\binom{n+d-1}{d}-e_{1}({\mathcal{F}})\binom{n+d-2}{d-1}+\cdots+(-1)^{d}e_{d}({\mathcal{F}})$

and it is called the Hilbert polynomial of ${\mathcal{F}}.$ The uniquely
determined integers $e_{i}({\mathcal{F}})$ are called the Hilbert coefficients
of ${\mathcal{F}}.$ The coefficient $e_{0}({\mathcal{F}})$ is a positive
integer and it is called the multiplicity of ${\mathcal{F}}$. The coefficient
$e_{1}({\mathcal{F}})$ is called as the Chern number of ${\mathcal{F}}.$ For
${\mathcal{F}}=\\{I^{n}\\}_{n\in\mathbb{N}},$
$e(I):=e_{0}({\mathcal{F}})=e_{0}(I)$ (resp. $e_{1}(I):=e_{1}({\mathcal{F}})$)
is called the multiplicity (resp. the Chern number) of $I.$ The reduction
number of $\mathcal{F}$ with respect to a minimal reduction $J\subseteq I$ is
defined as $r_{J}(\mathcal{F})=\min\\{m|JI_{n}=I_{n+1}\text{ for all }n\geq
m\\}$ and the reduction number of $\mathcal{F}$ is

$r(\mathcal{F})=\min\\{r_{J}(\mathcal{F})|J\text{ is a minimal reduction of
}\mathcal{F}\\}.$

The postulation number of $\mathcal{F}$ is defined as

$\eta(\mathcal{F})=\min\\{n|P_{\mathcal{F}}(m)=H_{\mathcal{F}}(m)\text{ for
all }m>n\\}.$

######  Definition 2.2.

Let $R$ be a commutative ring and $I$ be an ideal of $R.$ An element $x\in R$
is integral over an ideal $I$ if

$x^{n}+a_{1}x^{n-1}+\dots+a_{n}=0$

where $a_{i}\in I^{i}$ for $i=1,2,\dots,n.$

The integral closure $\overline{I}$ of $I$ is the ideal of all $x\in R$ which
are integral over $I.$

A Noetherian local ring $(R,{\mathfrak{m}})$ is said to be analytically
unramified if its completion with respect to the ${\mathfrak{m}}$-adic
topology is reduced.

D. Rees characterised [16, Theorem 1.4] analytically unramified local rings in
terms of the normal filtration $\\{\overline{I^{n}}\\}$ of any
${\mathfrak{m}}$-primary ideal $I.$

######  Theorem 2.3 (D. Rees).

A Noetherian local ring $(R,{\mathfrak{m}})$ is analytically unramified if and
only if the normal filtration $\\{\overline{I^{n}}\\}$ of any
${\mathfrak{m}}$-primary ideal $I$ is $I$-admissible.

Let the normal filtration of $I$ be given by
$\mathcal{N}=\\{\overline{I^{n}}\\}_{n\in\mathbb{Z}}.$ Rees’s Theorem implies
that if $R$ is analytically unramified then
${\mathcal{R}}^{\prime}(\mathcal{N})=\bigoplus_{n\in\mathbb{Z}}\overline{I^{n}}t^{n}$
is a finite module over $\mathcal{R}^{\prime}(I).$ Therefore the normal
Hilbert function $\overline{H}_{I}(n)=\ell(R/\overline{I^{n}})$ coincides with
a polynomial $\overline{P}_{I}(n)$ of degree $d$ for large $n,$ and is known
as the normal Hilbert polynomial of $I$. We write

$\overline{P}_{I}(n)=\overline{e}_{0}(I)\binom{n+d-1}{d}-\overline{e}_{1}(I)\binom{n+d-2}{d-1}+\cdots+(-1)^{d}\overline{e}_{d}(I).$

Here $\overline{e}_{0}(I)=e(I),$ the multiplicity of $I$,
$\overline{e}_{1}(I)$ is known as the normal Chern number of $I$ and the
coefficients $\overline{e}_{i}(I)$ for $i=0,1,\ldots,d$ are called the normal
Hilbert coefficients of $I.$ In this case, we denote the blow up algebras
$\mathcal{R}(\mathcal{N}),$ $\mathcal{R}^{\prime}(\mathcal{N})$ and
$\operatorname{gr}_{\mathcal{N}}(R)$ by $\overline{\mathcal{R}}(I),$
$\overline{\mathcal{R}^{\prime}}(I)$ and $\overline{\operatorname{gr}}_{I}(R)$
respectively.

We now introduce the filtration $\\{(I^{n})^{*}\\}$ of tight closure of powers
an ideal $I.$

######  Definition 2.4.

For a ring $R,$ the subset of $R$ consisting of all the elements which are not
contained in any minimal prime ideals of $R,$ is denoted by $R^{\circ}.$

Let $R$ be a ring of prime characteristic $p.$ Let $q=p^{e}$ for
$e\in\mathbb{N}.$ If $I=(a_{1},\ldots,a_{r})$ is an ideal of $R$ then the
$q^{th}$ Frobenius power of $I$ is the ideal
$I^{[q]}=(a_{1}^{q},\ldots,a_{r}^{q}).$ Notice that for any positive integer
$n,$ $(I^{n})^{[q]}=(I^{[q]})^{n}.$

######  Definition 2.5.

An element $x\in R$ is said to be in the tight closure $I^{*}$ of $I$ if there
exists $c\in R^{\circ}$ such that $cx^{q}\in I^{[q]},$ for all sufficiently
large $q=p^{e}.$ An element $c\in R^{\circ}$ is called a test element for $R$
if for all ideals $I$ in $R$ and $x\in I^{*},$ $cx^{q}\in I^{[q]}$ for all
$q=p^{e}.$
The test ideal of a ring $R,$ denoted by $\tau(R)$ is the ideal generated by
the test elements of $R.$

The tight filtration of $I$ is given by
$\mathcal{T}=\\{(I^{n})^{*}\\}_{n\in\mathbb{Z}}.$ As
$(I^{n})^{*}\subset\overline{I^{n}},$ for all $n\geq 1,$

$\mathcal{R^{\prime}}(I)\hookrightarrow\mathcal{R^{\prime}}(\mathcal{T})\hookrightarrow\mathcal{R^{\prime}}(\mathcal{N}).$

If $R$ is analytically unramified, $\mathcal{R}^{\prime}(\mathcal{N})$ is a
finite $\mathcal{R}^{\prime}(I)$-module and hence it follows that
$\mathcal{R}^{\prime}(\mathcal{T})$ is a finite
$\mathcal{R}^{\prime}(I)$-module. Let $I$ be an ${\mathfrak{m}}$-primary ideal
of a $d$-dimensional analytically unramified local ring $R$ of prime
characteristic $p,$ then the tight Hilbert function defined by
$H_{I}^{*}(n)=\ell(R/(I^{n})^{*})$ coincides with a polynomial,
$P_{I}^{*}(n)$, of degree $d$ for large $n$ with coefficients in $\mathbb{Q}$
and is known as the tight Hilbert polynomial of $I$. We write

$P_{I}^{*}(n)=e_{0}^{*}(I)\binom{n+d-1}{d}-e_{1}^{*}(I)\binom{n+d-2}{d-1}+\cdots+(-1)^{d}e_{d}^{*}(I),$

where $e_{i}^{*}(I)\in\mathbb{Z}.$ Here $e^{*}_{0}(I)=e(I),$ the multiplicity
of $I$, $e^{*}_{1}(I)$ is known as the tight Chern number of $I$ and the
coefficients $e^{*}_{i}(I)$ for $i=0,1,\ldots,d$ are called the tight Hilbert
coefficients of $I.$ The Strong Vanishing theorem for hypersurfaces plays an
important role in computation of the tight closure for “large primes.”

######  Theorem 2.6.

[12, Theorem 6.4] Let $K$ be a field of characteristic $p>0.$ Let $f\in
K[X_{0},X_{1},\ldots,X_{d}]$ is quasihomogeneous and
$R=K[X_{0},X_{1},\ldots,X_{d}]/(f).$ Assume that $R$ is an isolated
singularity and the partial derivatives of $f$ with respect to $X_{i}$ forms a
system of parameters in $R$ where $1\leq i\leq d.$ Further assume that
$p>(d-1)(\deg(f))-\sum_{i=1}^{d}\deg(X_{i}).$ Let $y_{1},y_{2},\ldots,y_{d}$
be a homogeneous system of parameters of degrees $a_{1},a_{2},\ldots,a_{d}.$
Set $A=a_{1}+\cdots+a_{d}$. Then

$(y_{1},y_{2},\ldots,y_{d})^{*}=(y_{1},y_{2},\ldots,y_{d})+R_{\geq A}.$

######  Definition 2.7.

[13] An $\mathbb{N}$-graded Noetherian Cohen-Macaulay ring
$(S,{\mathfrak{m}}),$ over a field $S_{0}$ of characteristic $p>0,$ where
${\mathfrak{m}}$ is the maximal graded ideal and $\dim S=d$ is said to satisfy
the Strong Vanishing Conjecture if

$(x_{1},\ldots,x_{d})^{*}=(x_{1},\ldots,x_{d})+S_{\geq\delta}$

for any homogeneous system of parameters $x_{1},\ldots,x_{d}$ of $S$ where
$\delta$ is the sum of degrees of $x_{1},\ldots,x_{d}.$

One can use the next theorem to determine the tight closure of
${\mathfrak{m}}$-primary ideals.

######  Theorem 2.8.

[13, Theorem 5.11] Let $R=\bigoplus_{n\geq 0}R_{n}$ be a $d$-dimensional
finitely generated $\mathbb{N}$-graded Cohen-Macaulay ring over a field of
prime characteristic $p$. Let ${\mathfrak{m}}$ be its maximal homogeneous
ideal and $I$ be an ${\mathfrak{m}}$-primary ideal such that $R/I$ has finite
projective dimension. Assume that $R$ has an isolated singularity at
${\mathfrak{m}}$ and let

(2.1) $0\to\bigoplus_{i}R(-b_{di})\to\cdots\to\bigoplus_{i}R(-b_{1i})\to R\to
R/I\to 0.$

be a minimal graded free resolution of $R/I.$ Further assume that $R$
satisfies the Strong Vanishing Conjecture. Then

$I+R_{\geq M}\subseteq I^{*}\subseteq I+R_{\geq N}$

where $N=\min_{i}\\{b_{di}\\}$ and $M=\max_{i}\\{b_{di}\\}$.

######  Remark 2.9.

Let
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(X_{1}^{r}+X_{2}^{r}+\cdots+X_{d+1}^{r}),$
$I=(x_{1},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}).$ Then
$\ell\left(\frac{R}{(I^{n})^{*}}\right)=\ell\left(\frac{R_{{\mathfrak{m}}}}{(I^{n}R_{{\mathfrak{m}}})^{*}}\right),$
where the equality follows from the fact that localisation commutes with the
tight closure operation for $I^{n}$ [1, Corollary 3.2]. Therefore the tight
Hilbert polynomial of $I$ is same as that of $I_{{\mathfrak{m}}}.$

######  Theorem 2.10.

[6, Theorem A2.60] Let $R$ be a ring and $\alpha:F=R^{f}\rightarrow G=R^{g}$
where $f\geq g$. The Eagon-Northcott complex of a map $\alpha$ is given by

$\displaystyle
0\rightarrow(\operatorname{Sym}_{f-g}G)^{*}\otimes\wedge^{f}F\xrightarrow{d_{f-g+1}}\cdots\rightarrow(\operatorname{Sym}_{2}G)^{*}\otimes\wedge^{g+2}F\xrightarrow{d_{3}}G^{*}\otimes\wedge^{g+1}F\xrightarrow{d_{2}}\wedge^{g}F\xrightarrow{\wedge^{g}\alpha}\wedge^{g}G.$

The above complex is exact if and only if $\text{grade }I_{g}(\alpha)=f-g+1$.

We refer the reader to [6, Appendix A2H] for information on the above complex
and the description of the maps appearing in the complex. The main feature
that we use in this article is the linearity of the maps $d_{i},i\geq 2$. This
can be realized from the description of the maps (see for example [6, Example
A2.69, Example A2.68] and [5, Appendix A2.6.1]).

In a local ring $(R,{\mathfrak{m}})$ and an ideal $I$ generated by a regular
sequence, the power $I^{t}$ can be resolved by the Eagon-Northcott complex
([4, Page 15, after Remark 2.13] and the above theorem).

##  3\. Hilbert polynomial for the filtration $\\{I_{n}\\}_{n\geq 0},$
where $I_{0}=R$ and $I_{n}=I^{n}+{\mathfrak{m}}^{n+t}$ for $n\geq 1$

In the subsequent sections, we will show that in most of the cases we
consider, the tight closure of powers of a linear system of parameters in
diagonal hypersurfaces takes a special form. Therefore computing the Hilbert
polynomial in such cases becomes crucial.
Let
$R=K[X_{1},X_{2},\ldots,X_{d+1}]/(X_{1}^{r}+X_{2}^{r}+\cdots+X_{d+1}^{r}),$
$I=(x_{1},x_{2},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}),$ where $K$ is
a field. Let $\mathcal{F}=\\{I_{n}\\}_{n\geq 0},$ where
$I_{n}=\\{I^{n}+{\mathfrak{m}}^{n+t}\\}_{n\geq 0}$ for fixed $t\in\mathbb{N}.$
Then we calculate the Hilbert polynomial $P_{\mathcal{F}}(n).$

######  Proposition 3.1.

Let $T=K[X_{1},X_{2},\ldots,X_{d+1}]$ and $R=T/(f)$ where $K$ is an infinite
field and $f\in T$ is a homogeneous square-free polynomial of degree $r\geq
2.$ Let $I$ be a minimal reduction of
${\mathfrak{m}}=(x_{1},\ldots,x_{d+1})R.$ Let $\mathcal{F}=\\{I_{n}\\}_{n\geq
0},$ where $I_{0}=R,$ $I_{n}=I^{n}+{\mathfrak{m}}^{n+t}$ for all $n\geq 1$ and
for some fixed $t\in\mathbb{N}.$ Then
(1) $\operatorname{gr}_{\mathcal{F}}(R)$ is Cohen-Macaulay.
(2) If $r\leq t+1$ then $I_{n}=I^{n}$ for all $n$ and
$P_{\mathcal{F}}(n)=r\binom{n+d-1}{d}.$
(3) If $r\geq t+2$ then

$H(\operatorname{gr}_{\mathcal{F}}(R),\lambda)=\frac{(t+1)+{\lambda}+\cdots+{\lambda}^{r-t-1}}{(1-{\lambda})^{d}},$
$e_{j}(\mathcal{F})=\binom{r-t}{j+1}\text{ for }1\leq j\leq r-t-1\text{ and
}e_{j}(\mathcal{F})=0\text{ for }r-t\leq j\leq d.$

Also $\eta(\mathcal{F})=r-t-d-1$ and $r(\mathcal{F})=r-t-1.$

###### Proof.

We note that $\mathcal{F}$ is an $I$-admissible filtration. Since $f$ is
square-free in $T,$ $R_{{\mathfrak{m}}}$ is reduced and analytically
unramified. Clearly $I_{n}\subseteq\overline{I^{n}}={\mathfrak{m}}^{n}$ which
implies that

$\mathcal{R^{\prime}}(I)\hookrightarrow\mathcal{R^{\prime}}(\mathcal{F})\hookrightarrow\overline{\mathcal{R^{\prime}}}(I).$

As $R_{{\mathfrak{m}}}$ is analytically unramified,
$\overline{\mathcal{R^{\prime}}}(I)$ is a finite
$\mathcal{R^{\prime}}(I)$-module so that $\mathcal{R^{\prime}}(\mathcal{F})$
is a finite $\mathcal{R^{\prime}}(I)$-module making $\mathcal{F}$ an
$I$-admissible filtration.
(1) Since $R$ is Cohen-Macaulay, by Valabrega-Valla Theorem [10, Proposition
3.5], it is enough to show that $I\cap I_{n}=II_{n-1}$ for $n\geq 1.$ Notice
that $I\cap{\mathfrak{m}}^{n}=I{\mathfrak{m}}^{n-1}$ for $n\geq 1$ as
$\operatorname{gr}_{{\mathfrak{m}}}(R)\cong R$ is Cohen-Macaulay. This implies
that for all $n\geq 1,$

$I\cap
I_{n}=I^{n}+I\cap{\mathfrak{m}}^{n+t}=I^{n}+I{\mathfrak{m}}^{n+t-1}=II_{n-1}.$

Therefore $\operatorname{gr}_{\mathcal{F}}(R)$ is Cohen-Macaulay.
(2) Let $r\leq t+1.$ We will show that $r({\mathfrak{m}})=r-1.$ The Hilbert
series of $R$ is given by

$H(R,{\lambda})=\frac{1-{\lambda}^{r}}{(1-{\lambda})^{d+1}}=\frac{1+{\lambda}+\cdots+{\lambda}^{r-1}}{(1-{\lambda})^{d}}.$

Since $\operatorname{gr}_{{\mathfrak{m}}}(R)$ is Cohen-Macaulay and
$\eta({\mathfrak{m}})=r-1-d,$ we have $r({\mathfrak{m}})=r-1$ [15, Corollary
5.7].
Let $r\leq t+1.$ Since $r({\mathfrak{m}})=r-1,$ for $n\geq 1,$

$I_{n}=I^{n}+{\mathfrak{m}}^{n+t}=I^{n}+I^{n}{\mathfrak{m}}^{t}=I^{n}.$

This implies that $\mathcal{F}$ is the $I$-adic filtration and for $n\geq 1,$

$H_{\mathcal{F}}(n)=P_{\mathcal{F}}(n)=r\binom{n+d-1}{d}.$

(3) Let $r\geq t+2.$ Apply induction on $d.$ If $d=0$ then
$R=\frac{K[X_{1}]}{(X_{1}^{r})},$ $I=(0),$ ${\mathfrak{m}}=(x_{1})$ and
$\mathcal{F}=\\{I_{n}\\}_{n\geq 0}$ where $I_{0}=R$ and
$I_{n}={\mathfrak{m}}^{n+t}$ for $n\geq 1.$ In this case
$\operatorname{gr}_{\mathcal{F}}(R)=\frac{R}{{\mathfrak{m}}^{t+1}}\bigoplus_{n\geq
1}\frac{{\mathfrak{m}}^{n+t}}{{\mathfrak{m}}^{n+t+1}}.$ This implies that

$\displaystyle H(\operatorname{gr}_{\mathcal{F}}(R),\lambda)$
$\displaystyle=\sum_{n\geq
0}\ell([\operatorname{gr}_{\mathcal{F}}(R)]_{n}){\lambda}^{n}$
$\displaystyle=\ell\left(\frac{R}{{\mathfrak{m}}^{t+1}}\right)+\ell\left(\frac{{\mathfrak{m}}^{t+1}}{{\mathfrak{m}}^{t+2}}\right){\lambda}+\cdots+\ell\left(\frac{{\mathfrak{m}}^{r-1}}{{\mathfrak{m}}^{r}}\right){\lambda}^{r-1-t}.$

Note that $\frac{R}{{\mathfrak{m}}^{t+1}}\cong\frac{K[X_{1}]}{(X_{1}^{t+1})}$
since $r\geq t+2\geq t+1.$ Therefore
$\ell\left(\frac{R}{{\mathfrak{m}}^{t+1}}\right)=t+1.$ If $1\leq i\leq r-1-t$
then

$\frac{R}{{\mathfrak{m}}^{t+i}}\cong\frac{K[X_{1}]}{(X_{1}^{t+i})}\text{ and
}\frac{R}{{\mathfrak{m}}^{t+i+1}}\cong\frac{K[X_{1}]}{(X_{1}^{t+i+1})}.$

This implies that
$\ell\left(\frac{{\mathfrak{m}}^{t+i}}{{\mathfrak{m}}^{t+i+1}}\right)=1$ for
$1\leq i\leq r-t-1.$ Therefore

$H(\operatorname{gr}_{\mathcal{F}}(R),\lambda)=(t+1)+{\lambda}+\cdots+{\lambda}^{r-t-1}.$

Let $d\geq 1.$ After a linear change of coordinates, we may assume that $f$ is
monic in $X_{d+1}.$ Since $f$ is monic in $X_{d+1},$ $(x_{1},\ldots,x_{d})$ is
a minimal reduction of ${\mathfrak{m}}.$ Thus we may assume that
$I=(x_{1},\ldots,x_{d}).$ Clearly $x_{1}^{*}$ is a nonzerodivisor on
$\operatorname{gr}_{\mathcal{F}}(R)$ so that

$\frac{\operatorname{gr}_{\mathcal{F}}(R)}{x_{1}^{*}\operatorname{gr}_{\mathcal{F}}(R)}\cong\operatorname{gr}_{\mathcal{F}/(x_{1})}(R/(x_{1})).$

Let $S=R/(x_{1}),$ $J=IS$ and $\Gamma=\\{J_{n}\\}_{n\geq 0}$ where $J_{0}=S$
and $J_{n}=J^{n}+\mathfrak{n}^{n+t}$ for $n\geq 1$ where ${\mathfrak{n}}$ is
the unique maximal homogeneous ideal in the $(d-1)$-dimensional ring $S.$
Notice that the filtration $\Gamma$ is same as the filtration
$\mathcal{F}/(x_{1})=\\{L_{n}\\}_{n\geq 0}$ in $S.$ Indeed for $n\geq 1,$
$J_{n}=J^{n}+{\mathfrak{n}}^{n+t}=\frac{I^{n}+{\mathfrak{m}}^{n+t}+(x_{1})}{(x_{1})}$
and
$L_{n}=\frac{I_{n}+(x_{1})}{(x_{1})}=\frac{I^{n}+{\mathfrak{m}}^{n+t}+(x_{1})}{(x_{1})}.$
Also $J_{0}=S$ and $L_{0}=\frac{I_{0}+(x_{1})}{(x_{1})}=R/(x_{1})=S.$ By
induction hypothesis,

$H(\operatorname{gr}_{\Gamma}(S),\lambda)=\frac{(t+1)+{\lambda}+\cdots+{\lambda}^{r-t-1}}{(1-{\lambda})^{d-1}}.$

Since $x_{1}^{*}$ is a nonzerodivisor of $\operatorname{gr}_{\mathcal{F}}(R),$
it follows that

$H(\operatorname{gr}_{\mathcal{F}}(R),\lambda)=\frac{(t+1)+{\lambda}+\cdots+{\lambda}^{r-t-1}}{(1-{\lambda})^{d}}.$

Let $h({\lambda})=(t+1)+{\lambda}+\cdots+{\lambda}^{r-t-1}.$ Then by [3,
Proposition 4.1.9] $e_{j}(\mathcal{F})=\frac{h^{(j)}(1)}{j!}.$ Note that if
$r-t\leq j\leq d$ then $e_{j}(\mathcal{F})=0.$ Let $1\leq j\leq r-t-1$ then
$h^{(j)}({\lambda})=\sum_{i\geq
j}^{r-t-1}i(i-1)\cdots(i-(j-1)){\lambda}^{i-j}.$ Therefore for $1\leq j\leq
r-t-1,$

$e_{j}(\mathcal{F})=\sum_{i\geq
j}^{r-t-1}\frac{i(i-1)\cdots(i-(j-1))}{j!}=\sum_{i\geq
j}^{r-t-1}\binom{i}{j}=\binom{r-t}{j+1}.$

Since $\operatorname{gr}_{\mathcal{F}}(R)$ is Cohen-Macaulay,
$\eta(\mathcal{F})=r-t-d-1$ and $r(\mathcal{F})=r-t-1.$ ∎

##  4\. The Tight Hilbert polynomial for diagonal hypersurfaces

The purpose of this section is to find the tight Hilbert polynomial in certain
diagonal hypersurface rings. As evidenced in [7], this will enable us to
detect $F$-rationality of such rings. The first proposition helps us to
compute the tight Hilbert polynomial of a system of parameters generated by
test elements in excellent reduced Cohen-Macaulay rings. In fact, it also
provides information about the various blow up algebras associated with the
tight closure filtration in this case. It is an easy consequence of [2, Lemma
3.1].

######  Proposition 4.1.

Let $(R,{\mathfrak{m}})$ be $d$-dimensional excellent reduced Cohen-Macaulay
local ring. Let $x_{1},x_{2},\ldots,x_{d}$ be test elements and
$Q=(x_{1},x_{2},\ldots,x_{d})$ be ${\mathfrak{m}}$-primary.Then
(1) $(Q^{n})^{*}=Q(Q^{n-1})^{*}$ for all $n\geq 2.$
(2) $e_{0}(Q)-e_{1}^{*}(Q)=\ell(R/Q^{*}).$
(3)
$\ell(R/(Q^{n})^{*})=e_{0}(Q)\binom{n+d-1}{d}-e_{1}^{*}(Q)\binom{n+d-2}{d-1}$
for all $n\geq 1.$
(4)  $G^{*}(Q)=\bigoplus_{n\geq 0}(Q^{n})^{*}/(Q^{n+1})^{*}$ and
$\mathcal{R}^{*}(Q)=\bigoplus_{n\geq 0}(Q^{n})^{*}t^{n}$ are Cohen-Macaulay.

###### Proof.

By [2, Lemma 3.1], $(Q^{n})^{*}=Q^{n-1}Q^{*}.$ This implies that

$(Q^{n})^{*}\subseteq
Q((Q^{n-2})^{*}Q^{*})^{*}=Q(Q^{n-1})^{*}\subseteq(Q^{n})^{*}\text{ for all
}n\geq 2.$

This proves (1). The remaining parts follow by Huneke-Ooishi Theorem [11]. ∎

In the next theorem, we demonstrate the use of Strong Vanishing Theorem for
computing the tight Hilbert polynomial. It is well known that the Strong
Vanishing Theorem helps in computation of the tight closure of an ideal
generated by system of parameters for rings with “large” prime characteristic.
However to find the tight Hilbert polynomial, we need to compute the tight
closure of powers of such ideals. For this, we use Theorem 2.8 which ensures a
bound for tight closure of ${\mathfrak{m}}$-primary ideals once we have its
projective resolution. This resolution in our case will be given by the Eagon-
Northcott complex.

######  Theorem 4.2.

Let $T=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}],$ $R=T/(f)$ where $f\in T$
is a square-free, monic in $X_{d+1}$ homogeneous ploynomial of degree $d+1$
and $p>(d-1)(d+1)-d.$ Let $I=(x_{1},\ldots,x_{d})$ and
${\mathfrak{m}}=(I,x_{d+1}).$ Then $(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+d-1}$
and for $n\geq 1,$

$\ell\left(\frac{R}{(I^{n})^{*}}\right)=(d+1)\binom{n+d-1}{d}-\binom{n+d-2}{d-1}.$

Moreover $\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

###### Proof.

Since $p>(d-1)(d+1)-d,$ it follows from Theorem 2.6 that Strong Vanishing
holds for $R.$ Therefore we have $I^{*}=I+{\mathfrak{m}}^{d}.$ Using the fact
that the Eagon-Northcott complex resolves $I^{n}$ and that the differentials
$d_{i},i\geq 2$ of this complex is linear, it follows from Theorem 2.8 that
$(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+d-1}.$ We claim that $r^{*}(I)\leq 1.$
Indeed, for $n\geq 1,$
$I(I^{n})^{*}=I^{n+1}+I{\mathfrak{m}}^{n+d-1}=I^{n+1}+{\mathfrak{m}}^{n+d}$
where the last equality follows as $r({\mathfrak{m}})=d.$ Therefore by Huneke-
Ooishi Theorem (Proposition 4.1) we have for $n\geq 1,$

$\displaystyle\ell\left(\frac{R}{(I^{n})^{*}}\right)$
$\displaystyle=e_{0}(I)\binom{n+d-1}{d}-e_{1}^{*}(I)\binom{n+d-2}{d-1}$
$\displaystyle=(d+1)\binom{n+d-1}{d}-e_{1}^{*}(I)\binom{n+d-2}{d-1}$

Put $n=1$ to obtain
$\ell\left(\frac{R}{I+{\mathfrak{m}}^{d}}\right)=(d+1)-e_{1}^{*}(I).$ Since

$\frac{R}{I+{\mathfrak{m}}^{d}}\cong\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(f)}{\left((X_{1},X_{2},\ldots,X_{d})+(X_{1},X_{2},\ldots,X_{d+1})^{d}+(f)\right)/(f)}\cong\frac{\mathbb{F}_{p}[X_{d+1}]}{({X_{d+1}})^{d}},$

it follows that $e_{1}^{*}(I)=1.$ ∎

Due to [9, Proposition 5.21 (c)], many diagonal hypersurface rings are
$F$-pure, namely,
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(X_{1}^{d+1}+X_{2}^{d+1}+\cdots+X_{d+1}^{d+1}),$
where $p\equiv 1(\operatorname{mod}d+1).$ In such cases, the test ideal turns
out to be a radical ideal. Using this, we find the tight Hilbert polynomial in
$F$-pure diagonal hypersurface rings where the number of variables is equal to
the degree of the defining equation.

######  Theorem 4.3.

Let
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]/(X_{1}^{d+1}+X_{2}^{d+1}+\cdots+X_{d+1}^{d+1})$
be F-pure, $I=(x_{1},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1})$. Then for
all $n\geq 1,$

$\ell\left(\frac{R}{(I^{n})^{*}}\right)=(d+1)\binom{n+d-1}{d}-\binom{n+d-2}{d-1}.$

Moreover $\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

###### Proof.

Since $R$ is F-pure, it implies that ${\mathfrak{m}}\subseteq\tau(R).$
Therefore by Proposition 4.1 it follows that for $n\geq 1,$

$\ell\left(\frac{R}{(I^{n})^{*}}\right)=(d+1)\binom{n+d-1}{d}-e_{1}^{*}(I)\binom{n+d-2}{d-1},$

where $e_{1}^{*}(I)=\ell\left(\frac{I^{*}}{I}\right).$ Using Briançon-Skoda
Theorem $x_{d+1}^{d}\in I^{*}\setminus I$ which implies that
$\tau(R)={\mathfrak{m}}.$ We claim that $I^{*}=I+{\mathfrak{m}}^{d}.$ Since
$I:\tau(R)=I^{*}$ [12, Corollary 4.2], it is enough to show that
$x_{d+1}^{d-1}\notin I^{*}.$ If possible, let $x_{d+1}^{d-1}\in I^{*}$ then
$x_{d+1}^{d}\in I,$ a contradiction. Hence the claim. Since

$\frac{R}{I+{\mathfrak{m}}^{d}}\cong\frac{\mathbb{F}_{p}[X_{1},\ldots,X_{d+1}]}{(X_{1},\ldots,X_{d},X_{d+1}^{d})}\cong\frac{\mathbb{F}_{p}[X_{d+1}]}{(X_{d+1}^{d})},$

it follows that $e_{1}^{*}(I)=1.$ ∎

As promised in section 3, we now give a class of rings where the tight closure
filtration turns out to be the filtration $\\{I_{n}\\}_{n\geq 0},$ where
$I_{0}=R$ and $I_{n}=I^{n}+{\mathfrak{m}}^{n+t}$ for $n\geq 1.$

######  Theorem 4.4.

Let
$R=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{t+2}]/(X_{1}^{r}+X_{2}^{r}+\cdots+X_{t+2}^{r}),$
with $p\nmid r,$ $I=(x_{1},\ldots,x_{t+1})$ and ${\mathfrak{m}}=(I,x_{t+2}),$
where $p>tr-(t+1).$ Then
(1) $\operatorname{gr}_{I}^{*}(R)$ is Cohen-Macaulay.
(2) The tight Hilbert polynomial $P_{I}^{*}(n)$ is given by

$P_{I}^{*}(n)=\begin{cases}r\binom{n+t}{t+1},&\text{ if }r\leq t+1,\\\
r\binom{n+t}{t+1}-\binom{r-t}{2}\binom{n+t-1}{t}+\binom{r-t}{3}\binom{n+t-2}{t-1}+\cdots+(-1)^{j}\binom{r-t}{j+1}\binom{n+t-j}{t+1-j},&\text{
if }r\geq t+2.\\\ \end{cases}$

If $r\geq t+2,$ $e_{j}^{*}(I)=0$ for $r-t\leq j\leq t+1.$

###### Proof.

Since $p>tr-(t+1),$ it follows from Theorem 2.6 that Strong Vanishing
Conjecture holds for $R.$ Therefore we have $I^{*}=I+{\mathfrak{m}}^{t+1}.$
Using the Eagon-Northcott complex, it follows from Theorem 2.8 that
$(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+t}.$ The result follows from Proposition
3.1. ∎

##  5\. The Tight Hilbert polynomial for degree $p+1$ hypersurface ring of
characteristic $p$

The goal of this section is to compute the tight Hilbert polynomial of linear
system of parameters in diagonal hypersurface rings where the characteristic
of the ring is one less than the degree of the defining equation. In order to
find the tight Hilbert polynomial, we calculate the initial ideals of certain
ideals in $K[X_{1},\ldots,X_{n}].$

######  Lemma 5.1.

Let $A=K[X_{1},X_{2},\ldots,X_{m}]$ be a polynomial ring over field $K$. Let
$k\in\mathbb{N}$ and $q=p^{e}$ for some $e.$ Let
$J=(X_{1}^{N}+X_{2}^{N}+\cdots+X_{m}^{N},(X_{2}^{q},X_{3}^{q},\ldots,X_{m}^{q})^{k}).$
If $>$ denotes the graded reverse lex ordering with
$X_{1}>X_{2}>\cdots>X_{m},$ then

$\operatorname{in}_{>}(J)=(X_{1}^{N},(X_{2}^{q},X_{3}^{q},\ldots,X_{m}^{q})^{k}).$

###### Proof.

It suffices to show that

$G=\Bigg{\\{}\sum_{i=1}^{m}X_{i}^{N},\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]}\text{,
}r_{j}\in\\{0,1,2,\ldots,k\\}\text{ for }2\leq j\leq m-1\Bigg{\\}}$

is a Gröbner basis of $J$. For $r_{j}\in\\{0,1,\ldots,k\\},$ consider the
$S$-polynomial of $f=X_{1}^{N}+X_{2}^{N}+\cdots+X_{m}^{N}$ and
$g=X_{2}^{qr_{2}}X_{3}^{qr_{3}}\cdots
X_{m-1}^{qr_{m-1}}X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]},$

$\displaystyle S(f,g)$
$\displaystyle=\frac{X_{1}^{N}\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]}}{X_{1}^{N}}f-\frac{X_{1}^{N}\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]}}{\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]}}g$
$\displaystyle=\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]}\left(\sum_{i=2}^{m}X_{i}^{N}\right).$

By division algorithm, it follows that $\overline{S(f,g)}^{G}=0$ for
$f=X_{1}^{N}+X_{2}^{N}+\cdots+X_{m}^{N}$ and
$g=X_{2}^{qr_{2}}X_{3}^{qr_{3}}\cdots
X_{m-1}^{qr_{m-1}}X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]},$ where
$r_{j}\in\\{0,1,\ldots,k\\}.$ Now for
$\overline{r}=(r_{2},r_{3},\ldots,r_{m-1})$ and
$\overline{s}=(s_{2},s_{3},\ldots,s_{m-1})$ with
$\overline{r}\neq\overline{s},$ we have $S$-polynomial,

$\displaystyle
S\left(\left[\prod_{j=2}^{m-1}X_{j}^{qr_{j}}\right]X_{m}^{q[k-(r_{2}+r_{3}+\cdots+r_{m-1})]},\left[\prod_{j=2}^{m-1}X_{j}^{qs_{j}}\right]X_{m}^{q[k-(s_{2}+s_{3}+\cdots+s_{m-1})]}\right)=0.$

It follows that $G$ is a Gröbner basis of $J$ and hence
$\operatorname{in}_{>}(J)=(X_{1}^{N},(X_{2}^{q},X_{3}^{q},\ldots,X_{m}^{q})^{k}).$
∎

######  Theorem 5.2.

Let $p$ be an odd prime,
$R=\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]}{(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1})},$
$I=(x_{1},x_{2},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}).$ Then

$I^{*}=I+{\mathfrak{m}}^{2}.$

###### Proof.

First we will show that $x_{d+1}^{2}\in I^{*}.$ Clearly,

(5.2)
$(x_{d+1}^{p+1})^{p^{e}}=-(x_{1}^{p+1})^{p^{e}}-(x_{2}^{p+1})^{p^{e}}-\cdots-(x_{d}^{p+1})^{p^{e}}\in
I^{[p^{e+1}]}.$

Therefore
$\underbrace{(x_{d+1}^{2})^{p^{e}}\cdots(x_{d+1}^{2})^{p^{e}}\,}_{\text{$m$
times}}\in I^{[p^{e+1}]}$ where $p+1=2m.$ Since $p>m,$ it follows that

$(x_{d+1}^{2})^{p^{e+1}}=\underbrace{(x_{d+1}^{2})^{p^{e}}\cdots(x_{d+1}^{2})^{p^{e}}\,}_{\text{$p$
times}}\in I^{[p^{e+1}]}.$

Hence $x_{d+1}^{2}\in I^{*}.$ It is enough to show that $x_{d+1}\notin I^{*}.$
Since $x_{d}^{p}$ is a test element, we show that
$x_{d}^{p}x_{d+1}^{p^{3}}\notin I^{[p^{3}]}.$ If possible, let
$x_{d}^{p}x_{d+1}^{p^{3}}\in I^{[p^{3}]}.$ Note that
$x_{d}^{p}x_{d+1}^{p^{3}}\in I^{[p^{3}]}$ if and only if
$X_{d}^{p}X_{d+1}^{p^{3}}\in J,$ where
$J=(X_{1}^{p+1}+\cdots+X_{d+1}^{p+1},X_{1}^{p^{3}},\ldots,X_{d}^{p^{3}})$ in
$\mathbb{F}_{p}[X_{1},\ldots,X_{d+1}].$ Let $>$ denote the graded reverse lex
ordering with $X_{d+1}>X_{1}>X_{2}>\cdots>X_{d}$ in
$\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}].$ Let $p^{3}=(p+1)u+i,$ where
$1\leq i\leq p.$ Note that $i\neq 0.$ Clearly,

$X_{d}^{p}X_{d+1}^{p^{3}}=X_{d}^{p}X_{d+1}^{(p+1)u+i}\equiv f(\text{mod }J),$

where $f=(-1)^{u}X_{d}^{p}(X_{1}^{p+1}+\cdots+X_{d}^{p+1})^{u}X_{d+1}^{i}.$
Note that $\operatorname{in}_{>}(f)=X_{d}^{p}X_{1}^{(p+1)u}X_{d+1}^{i}$ and by
Lemma 5.1 it follows that
$\operatorname{in}_{>}(J)=(X_{d+1}^{p+1},X_{1}^{p^{3}},\ldots,X_{d}^{p^{3}}).$
The exponent of $X_{1}$ in $\operatorname{in}_{>}(f)$ is
$(p+1)u=p^{3}-i<p^{3},$ exponent of $X_{d}$ and $X_{d+1}$ in
$\operatorname{in}_{>}(f)$ is $p$ and $i$ respectively. This implies that
$\operatorname{in}_{>}(f)\notin\operatorname{in}_{>}(J).$ Hence $x_{d+1}\notin
I^{*}$ so that $I^{*}=I+{\mathfrak{m}}^{2}.$ ∎

######  Lemma 5.3.

Let
$R=\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]}{(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1})}$
and $I=(x_{1},x_{2},\ldots,x_{d})$ where $p$ is prime.

$(x_{d+1}^{n})^{[p^{e+1}]}\in\begin{cases}(I^{n-1})^{[p^{e+1}]},&\text{ if
}n\not\equiv 0\pmod{p+1},\\\ (I^{n})^{[p^{e+1}]},&\text{ if }n\equiv
0\pmod{p+1}.\end{cases}$

###### Proof.

Let $n\equiv 0\pmod{p+1}.$ Then $n=(p+1)m$ for some $m\in\mathbb{N}.$ Then

$(x_{d+1}^{n})^{p^{e+1}}=(x_{d+1}^{(p+1)m})^{p^{e+1}}=[(x_{d+1}^{p+1})^{p^{e+1}}]^{m}\in\left((I^{p+1})^{[p^{e+1}]}\right)^{m}=(I^{n})^{[p^{e+1}]}.$

Next let $n\not\equiv 0\pmod{p+1}.$ Write $n=(p+1)m+r$ where $1\leq r\leq p.$
If $m=0$ then we may assume that $r>1.$

$\displaystyle(x_{d+1}^{r})^{p^{e+1}}$
$\displaystyle=\underbrace{(x_{d+1}^{r})^{p^{e}}\cdots(x_{d+1}^{r})^{p^{e}}\,}_{\text{$p$
times}}=\underbrace{(x_{d+1}^{p})^{p^{e}}\cdots(x_{d+1}^{p})^{p^{e}}\,}_{\text{$r$
times}}$
$\displaystyle=\underbrace{(x_{d+1}^{p+1})^{p^{e}}\cdots(x_{d+1}^{p+1})^{p^{e}}\,}_{\text{$r-1$
times}}(x_{d+1}^{p-(r-1)})^{p^{e}}\in(I^{r-1})^{[p^{e+1}]},$

where the last statement follows from equation 5.2 in Theorem 5.2. This
implies that

$(x_{d+1}^{n})^{[p^{e+1}]}=(x_{d+1}^{m(p+1)}x_{d+1}^{r})^{p^{e+1}}\in(I^{(m(p+1))}I^{r-1})^{[p^{e+1}]}=(I^{n-1})^{[p^{e+1}]}.$

∎

######  Theorem 5.4.

Let
$R=\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]}{(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1})},$
$I=(x_{1},x_{2},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}).$ Then
(1) $(I^{n})^{*}\supseteq I^{n}+{\mathfrak{m}}^{n+1};$
(2) If $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}}\in{\mathfrak{m}}^{n}\setminus I^{n}$ then
$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}.$
(3) $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}$ for
$\sum_{j=1}^{d+1}a_{j}<n.$

###### Proof.

(1) Consider $x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots
x_{d+1}^{b_{d+1}}\in{\mathfrak{m}}^{n+1}$ where $\sum_{j=1}^{d+1}b_{j}=n+1$.
Using Lemma 5.3, observe that

$(x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots
x_{d+1}^{b_{d+1}})^{p^{e+1}}=(x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots
x_{d}^{b_{d}})^{p^{e+1}}(x_{d+1}^{b_{d+1}})^{p^{e+1}}$
$\in\begin{cases}(I^{n+1-b_{d+1}})^{[p^{e+1}]}(I^{b_{d+1}-1})^{[p^{e+1}]},&\text{if
}b_{d+1}\not\equiv 0\pmod{p+1},\\\
(I^{n+1-b_{d+1}})^{[p^{e+1}]}(I^{b_{d+1}})^{[p^{e+1}]},&\text{if
}b_{d+1}\equiv 0\pmod{p+1}.\end{cases}$

In either case, $(x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots
x_{d+1}^{b_{d+1}})^{p^{e+1}}\in(I^{n})^{[p^{e+1}]}.$ This implies that
$I^{n}+{\mathfrak{m}}^{n+1}\subseteq(I^{n})^{*}.$
(2) Note that $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}}\in{\mathfrak{m}}^{n}\setminus I^{n}$ implies that
$a_{d+1}\geq 1$ and that $p+1\nmid a_{d+1}.$ Indeed, if $a_{d+1}=(p+1)u$ for
some $u$ then

$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}}=(-1)^{u}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d}^{a_{d}}(x_{1}^{p+1}+\cdots+x_{d}^{p+1})^{u}\in I^{n},$

a contradiction to the assumption. Hence $p+1\nmid a_{d+1}.$ We use the fact
that $x_{d}^{p}$ is a test element. Observe that
$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}$ if and
only if $x_{d}^{p}(x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}})^{q}\notin(I^{n})^{[q]}$ for some $q.$ For $e\in\mathbb{N}$
set $q=p^{e+1}>p+1$ and $a_{d+1}q=(p+1)u+i$ where $1\leq i\leq p.$ Let $>$ be
the graded reverse lex ordering with $X_{d+1}>X_{1}>X_{2}>\cdots>X_{d}$ in
$A=\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d},X_{d+1}].$
Let
$J=(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1},(X_{1}^{q},X_{2}^{q},\ldots,X_{d}^{q})^{n})$
be an ideal in $A.$ Using Lemma 5.1, it follows that
$\operatorname{in}_{>}(J)=(X_{d+1}^{p+1},(X_{1}^{q},X_{2}^{q},\ldots,X_{d}^{q})^{n}).$
Now

$\displaystyle{X_{d}}^{p}({X_{1}}^{a_{1}q}\cdots{X_{d+1}}^{a_{d+1}q})$
$\displaystyle={X_{1}}^{a_{1}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}{X_{d+1}^{(p+1)u+i}}$
$\displaystyle\equiv(-1)^{u}X_{1}^{a_{1}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}(X_{1}^{p+1}+\cdots+X_{d}^{p+1})^{u}X_{d+1}^{i}\;(\text{mod
}J).$

Let $f=(-1)^{u}X_{1}^{a_{1}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}(X_{1}^{p+1}+\cdots+X_{d}^{p+1})^{u}X_{d+1}^{i}.$
Clearly

$X_{d}^{p}(X_{1}^{a_{1}q}X_{2}^{a_{2}q}\cdots X_{d+1}^{a_{d+1}q})\in
J\Leftrightarrow f\in J.$

Note that $\operatorname{in}_{>}(f)=X_{1}^{a_{1}q+(p+1)u}X_{2}^{a_{2}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}X_{d+1}^{i}.$ The exponent of $X_{1}$ in
$\operatorname{in}_{>}(f)$ is

$a_{1}q+a_{d+1}q-i=(a_{1}+a_{d+1}-1)q+(q-i)=[n-(\sum_{j=2}^{d}a_{j})-1]q+(q-i).$

The exponents of $X_{2},\ldots,X_{d}$ in $\operatorname{in}_{>}(f)$ are
$a_{2}q,\ldots,a_{d-1}q,a_{d}q+p$ respectively. Since $p+1<q,$ the sum of
multiples of $q$ in exponents of $X_{1},X_{2},\ldots,X_{d}$ is $(n-1)q.$ We
have $\operatorname{in}_{>}(f)\notin\operatorname{in}_{>}(J).$ This implies
that
${x_{d}}^{p}({x_{1}}^{a_{1}}{x_{2}}^{a_{2}}\cdots{x_{d+1}}^{a_{d+1}})^{q}\notin(I^{n})^{[q]}.$
Hence $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}.$
(3) Now we show that $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}$ for $\sum_{j=1}^{d+1}a_{j}<n.$ Observe
that

$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}\Leftrightarrow
x_{d}^{p}(x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}})^{q}\notin(I^{n})^{[q]}=(x_{1}^{q},x_{2}^{q},\ldots,x_{d}^{q})^{n}$

for some $q.$ For $e\in\mathbb{N}$ set $q=p^{e+1}>p+1$ and $a_{d+1}q=(p+1)u+i$
where $0\leq i\leq p.$ Let $>$ be the graded reverse lex ordering with
$X_{d+1}>X_{1}>X_{2}>\cdots>X_{d}$ in
$A=\mathbb{F}[X_{1},X_{2},\ldots,X_{d}].$ Let
$J=(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1},(X_{1}^{q},X_{2}^{q},\ldots,X_{d}^{q})^{n})$
be an ideal in $A.$ Using Lemma 5.1, it follows that
$\operatorname{in}_{>}(J)=(X_{d+1}^{p+1},(X_{1}^{q},X_{2}^{q},\ldots,X_{d}^{q})^{n}).$
Using the arguments as above, we get

$\displaystyle X_{d}^{p}(X_{1}^{a_{1}q}\cdots
X_{d+1}^{a_{d+1}q})\equiv(-1)^{u}{X_{1}}^{a_{1}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}(X_{1}^{p+1}+\cdots+X_{d}^{p+1})^{u}X_{d+1}^{i}\;(\text{mod
}J).$

Let $f=(-1)^{u}X_{1}^{a_{1}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}(X_{1}^{p+1}+\cdots+X_{d}^{p+1})^{u}X_{d+1}^{i}.$
Clearly

$X_{d}^{p}(X_{1}^{a_{1}q}X_{2}^{a_{2}q}\cdots X_{d+1}^{a_{d+1}q})\in
J\Leftrightarrow f\in J.$

Note that $\operatorname{in}_{>}(f)=X_{1}^{a_{1}q+(p+1)u}X_{2}^{a_{2}q}\cdots
X_{d-1}^{a_{d-1}q}X_{d}^{a_{d}q+p}X_{d+1}^{i}.$ The exponent of $X_{1}$ in
$\operatorname{in}_{>}(f)$ is

$a_{1}q+a_{d+1}q-i=(a_{1}+a_{d+1}-1)q+(q-i).$

The exponents of $X_{2},\ldots,X_{d}$ in $\operatorname{in}_{>}(f)$ is
$a_{2}q,\ldots,a_{d-1}q,a_{d}q+p$ respectively. The sum of multiples of $q$ in
exponents of $X_{1},X_{2},\ldots,X_{d}$ is

$(a_{1}+a_{d+1})q+a_{2}q+\cdots+a_{d}q\text{ or
}(a_{1}+a_{d+1}-1)q+a_{2}q+\cdots+a_{d}q$

depending on $i=0$ or $i\neq 0$ respectively. Since $p+1<q,$ in either case
the sum of multiples of $q$ in exponents of $X_{1},X_{2},\ldots,X_{d}$ in
$\operatorname{in}_{>}(f)<nq.$ Therefore
$\operatorname{in}_{>}(f)\notin\operatorname{in}_{>}(J).$ Thus
$x_{d}^{p}(x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{d+1}^{a_{d+1}})^{q}\notin(I^{n})^{[q]}$ for some $q$ which implies that
$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{d+1}^{a_{d+1}}\notin(I^{n})^{*}.$ ∎

######  Theorem 5.5.

Let
$R=\frac{\mathbb{F}_{p}[X_{1},X_{2},\ldots,X_{d+1}]}{(X_{1}^{p+1}+X_{2}^{p+1}+\cdots+X_{d+1}^{p+1})},$
$I=(x_{1},x_{2},\ldots,x_{d})$ and ${\mathfrak{m}}=(I,x_{d+1}).$ Then
(1) $(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+1}.$
(2)
$P_{I}^{*}(n)=(p+1)\binom{n+d-1}{d}-\binom{p}{2}\binom{n+d-2}{d-1}+\cdots+(-1)^{j}\binom{p}{j+1}\binom{n+d-j-1}{d-j},$
where $j=\min\\{p-1,d\\}.$
Moreover, $e_{i}^{*}(I)=0$ for $p\leq i\leq d$ and
$\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

###### Proof.

(1) and (2) are immediate consequences of Theorem 5.4 and Proposition 3.1
respectively. ∎

##  6\. Quartic diagonal hypersurfaces

Let $R=\frac{\mathbb{F}_{p}[X,Y,Z]}{(X^{r}+Y^{r}+Z^{r})}$ and $I=(y,z).$ Let
$p$ be a prime number such that $p\nmid r.$ Let
$\mathcal{T}=\\{(I^{n})^{*}\\}_{n\geq 0}$ be tight Hilbert filtration. Since
$I$ is a reduction of ${\mathfrak{m}}$ and
$\operatorname{gr}_{{\mathfrak{m}}}(R)$ is a domain, we have
$\overline{I^{n}}=\overline{{\mathfrak{m}}^{n}}={\mathfrak{m}}^{n}.$ Since $I$
is generated by a regular sequence, from Huneke-Itoh intersection theorem [11,
14], it follows that for all $n\geq 0,$

$I^{n}\cap\overline{I^{n+1}}=I^{n}\cap\overline{{\mathfrak{m}}^{n+1}}={\mathfrak{m}}I^{n}.$

Using Theorem 2.6 it follows that for $p>r-2,$ we have
$I^{*}=I+{\mathfrak{m}}^{2}.$ By Theorem 2.8, it follows that
$(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+1}.$ Using Proposition 3.1, we get

$P_{I}^{*}(n)=r\binom{n+1}{2}-\binom{r-1}{2}n+\binom{r-1}{3}.$

When $r=3,$

$P_{I}^{*}(n)=3\binom{n+1}{2}-n.$

When $r=4,$

$P_{I}^{*}(n)=4\binom{n+1}{2}-3n+1.$

These agree with the formulas derived in [7]. We discuss about tight closure
of parameter ideals in

$R=\frac{\mathbb{F}_{p}[X,Y,Z,W]}{(X^{4}+Y^{4}+Z^{4}+W^{4})}.$

The Hilbert series and the Hilbert polynomial of $R$ are given by

$\displaystyle H(R,\lambda)$ $\displaystyle=\sum_{n=0}^{\infty}\dim
R_{n}\lambda^{n}=\frac{1-\lambda^{4}}{(1-\lambda)^{4}},$ $\displaystyle
P_{R}(n)$
$\displaystyle=e_{0}\binom{n+2}{2}-e_{1}({\mathfrak{m}})(n+1)+e_{2}({\mathfrak{m}})=4\binom{n+2}{2}-6(n+1)+4.$

Since, $\deg H(R,\lambda)=0,$ $H_{R}(0)\neq P_{R}(0)\text{ and
}H_{R}(n)=P_{R}(n)\text{ for all }n\geq 1.$

######  Theorem 6.1.

Let $R=\frac{\mathbb{F}_{p}[X,Y,Z,W]}{(X^{4}+Y^{4}+Z^{4}+W^{4})},$ $I=(x,y,z)$
and ${\mathfrak{m}}=(I,w).$ Then

$I^{*}=\begin{cases}{\mathfrak{m}},&\text{ if }p=2,\\\
I+{\mathfrak{m}}^{2},&\text{ if }p=3,\\\ I+{\mathfrak{m}}^{3},&\text{ if
}p\geq 5.\end{cases}$

###### Proof.

By the Briançon-Skoda Theorem, we have, $I+\mathfrak{m}^{3}\subseteq I^{*}.$
We check if $w,w^{2}\in I^{*}.$ We note that $x^{3},y^{3},z^{3},w^{3}$ are
test elements of $R.$ We find $I^{*}$ for various primes.
Case (i) Suppose $p=2$. Clearly, $w^{4}=-x^{4}-y^{4}-z^{4}.$ We have

$w^{2^{e}}=(x^{4}+y^{4}+z^{4})^{2^{e-2}}.$

For $q=2^{e},$ $w^{q}=x^{q}+y^{q}+z^{q}\in I^{[q]}.$ Hence $w\in I^{*}.$ This
implies that $I^{*}=\mathfrak{m}.$
Case (ii) Suppose $p=3$. Clearly,

(6.3) $(w^{4})^{3^{e}}=-(x^{4})^{3^{e}}-(y^{4})^{3^{e}}-(z^{4})^{3^{e}}\in
I^{[3^{e+1}]}$

Hence $(w^{2})^{3^{e}}(w^{2})^{3^{e}}(w^{2})^{3^{e}}=(w^{2})^{3^{e+1}}\in
I^{[3^{e+1}]}$ for all $e.$ Therefore $w^{2}\in I^{*}.$ We have
$w^{3}w^{27}\notin I^{[27]}.$ If possible let $w^{30}\in I^{[27]}.$ Then

(6.4) $W^{30}\in(X^{27},Y^{27},Z^{27})+(X^{4}+Y^{4}+Z^{4}+W^{4})$

in $\mathbb{F}_{p}[X,Y,Z,W].$ Let $>$ denote the graded reverse lex ordering
with $W>X>Y>Z$ in $\mathbb{F}_{p}[X,Y,Z,W].$ Note that Lemma 5.1 yields a
contradiction to equation 6.4. Since $w^{3}$ is a test element, it follows
that $w\notin I^{*}.$ Therefore, $I^{*}=(x,y,z,w^{2})=I+\mathfrak{m}^{2}.$
Case (iii) Suppose $p=5$. Since $p\equiv 1\pmod{4}$ by [9, Proposition 5.21
(c)], it follows that $R$ is F-pure. Hence by [8, Proposition 2.5], the test
ideal is a radical ideal of $R$ and since $x^{3},y^{3},z^{3},w^{3}\in\tau(R),$
it follows that $x,y,z,w\in\tau(R).$ We will show that
$I^{*}=I+\mathfrak{m}^{3}.$ If possible let $w^{2}\in I^{*}$ then $ww^{2}\in
I$ as $w$ is a test element. This implies that

$W^{3}\in(X,Y,Z)+(X^{4}+Y^{4}+Z^{4}+W^{4}).$

Consider the graded reverse lex ordering $>$ with $W>X>Y>Z$ in
$\mathbb{F}_{p}[X,Y,Z,W]$ and use Lemma 5.1 to obtain a contradiction. Hence
it follows that $w^{2}\notin I^{*}.$ Hence $I^{*}=(x,y,z,w^{3}).$
Case (iv) Suppose $p>5$. It follows from Theorem 2.6 that,

$I^{*}=I+\mathfrak{m}^{3}.$

Hence $R$ is not $F$-rational. ∎

In order to compute the tight Hilbert polynomial, we find $(I^{n})^{*}.$

######  Proposition 6.2.

Let $R=\frac{\mathbb{F}_{p}[X,Y,Z,W]}{(X^{4}+Y^{4}+Z^{4}+W^{4})},$ $I=(x,y,z)$
and ${\mathfrak{m}}=(I,w).$ Then for $n\geq 1,$

$(I^{n})^{*}=\begin{cases}I^{n}+(x+y+z+w),&\text{ if }p=2,\\\
I^{n}+{\mathfrak{m}}^{n+1},&\text{ if }p=3,\\\
I^{n}+{\mathfrak{m}}^{n+2},&\text{ if }p\geq 5.\end{cases}$

###### Proof.

Case (i) Suppose $p=2.$ It is clear that the only minimal prime of $R$ is
$(x+y+z+w)=\mathfrak{p},$ say. Since $R/\mathfrak{p}$ is a polynomial ring,
every ideal in $R/\mathfrak{p}$ is tightly closed. We know that for any ideal
$J$ in $R$ and $r\in R,$ $r\in J^{*}$ if and only if
$\overline{r}\in\left(J\frac{R}{\mathfrak{p}}\right)^{*}$ for $\overline{r}\in
R/\mathfrak{p}.$
Since

$\left(\frac{I^{n}+\mathfrak{p}}{\mathfrak{p}}\right)^{*}=\frac{I^{n}+(x+y+z+w)}{(x+y+z+w)}=\frac{I^{n}+\mathfrak{p}}{\mathfrak{p}}\text{
in }\frac{R}{\mathfrak{p}},$

it implies that $I^{*}=I+(x+y+z+w)={\mathfrak{m}}$ and
$(I^{n})^{*}=I^{n}+(x+y+z+w).$ Note that $\\{(I^{n})^{*}\\}_{n\geq 0}$ is not
an $I$-admissible filtration.
Case (ii) Suppose $p=3.$ From Theorem 5.5 (1), it follows that
$(I^{n})^{*}=I^{n}+{\mathfrak{m}}^{n+1}.$
Case (iii) Suppose $p=5.$ If $p\equiv 1(\operatorname{mod}4),$ then $R$ is
F-pure. It follows from [2, Lemma 3.1] that

$(I^{n})^{*}=I^{n-1}(I+\mathfrak{m}^{3})=I^{n}+{\mathfrak{m}}^{n+2}\text{ as
}r({\mathfrak{m}})=3.$

Case (iv) Suppose $p>5.$ We use Theorem 2.8 to find out the tight closure of
powers of ideal $I$ in $R.$ We show that $pd(R/I^{n})<\infty.$ The Koszul
complex gives a projective resolution in the case when $n=1.$ By using
induction on $n$ and the following short exact sequence one can easily
conclude that the projective dimension of $R/I$ and $R/I^{n}$ are equal for
all $n.$

$0\longrightarrow I^{n-1}/I^{n}\longrightarrow R/I^{n}\longrightarrow
R/I^{n-1}\longrightarrow 0$

The Eagon-Northcott complex for $R/I^{n}$ gives a minimal resolution which is
linear. The degree of all generators of $I^{n}$ is $n.$ When $p>5,$ $R$
satisfies Strong Vanishing Conjecture by Theorem 2.6. In this case
$b_{3i}=n+2$ (refer Theorem 2.8). Therefore it follows that
$(I^{n})^{*}=I^{n}+\mathfrak{m}^{n+2}.$ ∎

######  Theorem 6.3.

Let $R=\frac{\mathbb{F}_{p}[X,Y,Z,W]}{(X^{4}+Y^{4}+Z^{4}+W^{4})},$ $I=(x,y,z)$
and ${\mathfrak{m}}=(I,w).$ Then for $n\geq 1,$

$P_{I}^{*}(n)=\begin{cases}4\binom{n+2}{3}-3\binom{n+1}{2}+n,&\text{ if
}p=3,\\\ 4\binom{n+2}{3}-\binom{n+1}{2},&\text{ if }p\geq 5.\end{cases}$

Moreover $\operatorname{gr}^{*}_{I}(R)$ is Cohen-Macaulay.

###### Proof.

Case (i) Suppose $p=3$. Use Theorem 5.5 (2).
Case (ii) Suppose $p=5$. Since $p\equiv 1\pmod{4},$ the result follows from
Theorem 4.3.
Case (iii) Suppose $p>5$. It follows as a consequence of Theorem 4.2. ∎

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