---
abstract: |
    Let $(A,{\mathfrak{m} })$ be a Noetherian local ring with infinite residue field and let $I$ be an ideal in $A$ and let $F(I) = 
    \oplus_{n \geq 0}I^n/{\mathfrak{m} }I^n$ be the fiber cone of $I$. We prove certain relations among the Hilbert coefficients $f_0(I),f_1(I), f_2(I)$ of $F(I)$ when the $a$-invariant of the associated graded ring $G(I)$ is negative.
address:
- 'Chennai Mathematical Institute, Plot H1, SIPCOT IT Park Padur PO, Siruseri 603103, India'
- 'Department of Mathematics, IIT Bombay, Powai, Mumbai 400 076, India'
author:
- 'Clare D’Cruz'
- 'Tony J. Puthenpurakal'
title: |
    The Hilbert coefficients of the fiber cone and\
    the $a$-invariant of the associated graded ring
---

[^1]

introduction
============

Let $(A,{\mathfrak{m} })$ be a Noetherian local ring with *infinite residue field* $k = A/{\mathfrak{m} }$. Let $I$ be an ideal in $A$. The *fiber cone* of $I$ is the standard graded $k$-algebra $F(I) = \bigoplus_{n\geq 0}I^n/{\mathfrak{m} }I^{n}.$ Set $l(I) = \dim F(I)$, the *analytic spread* of $I$. The Hilbert polynomial of $F(I)$ is denoted by $f_I(z)$. Write $f_I(z) = \sum_{i = 0}^{l-1}(-1)^i f_i(I)\binom{z+l-1-i}{l-1-i}$ where $l = l(I)$ We call $f_i(I)$ the $i^{th}$ *fiber coefficient* of $I$.

Most recent results in the study of fiber cone involve the depth of $G(I)   = \bigoplus_{n \geq 0}I^n/ I^{n+1} $, the associated graded ring of $I$. When $I$ is ${\mathfrak{m} }$-primary there has been some research relating $f_0(I)$ (the *multiplicity* of $F(I)$) with various other invariants of $I$ (see [@JayV05 4.1], [@CVP032 4.3] and [@Cor05 3.4]). In the case of $G(I)$ the relations among the Hilbert coefficients $e_0(I),e_1(I),e_2(I)$ are well known (see [@VaSix]). However there is no result relating $f_0(I), f_1(I)$ and $f_2(I)$. The reason for this is not difficult to find: any standard $k$-algebra can be thought as a fiber cone of its graded maximal ideal. So any result involving the relation between $f_i(I)$ *would only hold in a restricted class of ideals*. Our paper explores the relation between ${\mathbf{a} }(I)$, *the $a$-invariant of $G(I)$*, and the Hilbert coefficients of $F(I)$. This is a new idea.

We first analyze when $l(I) = 2,3$ as it throws light on the general result.

\[result1\] Let $(A,{\mathfrak{m} })$ be a Noetherian local ring with infinite residue field $k = A/{\mathfrak{m} }$. Let $I$ be an ideal with $l(I) =2$. If ${\mathbf{a} }(I) < 0$ then

$f_1(I) \leq f_0(I) -1. $

Furthermore, equality holds if and only if $F(I^n)$ is [Cohen-Macaulay]{} for all $n \gg 0$. If ${\operatorname{grade}}(I) = 2$ then equality holds.

This result should be compared with a result due to Northcott [@Nor-e1], which in our context states that $f_1({\mathfrak{m} }) \geq f_0({\mathfrak{m} }) -1$ whenever $A$ is [Cohen-Macaulay]{}. In \[2ex1\], we give an example of a two dimensional Noetherian local ring $(A,{\mathfrak{m} })$ with ${\operatorname{depth}}A = 1$ but $f_1(I) < f_0(I) -1$.

To analyze the case when equality holds in Theorem \[result1\] we resolve $F(I^n)$ as a $F(J^{[n]})= k[X^{n}_{1}, X^{n}_{2}]$-module and write it as: $$0 {\longrightarrow}K_n {\longrightarrow}\bigoplus_{i = 1}^{\beta_{1}^{[n]}}F(J^{[n]})( - 1 - \alpha_{i}^{[n]} )  {\longrightarrow}F(J^{[n]})^{\beta_{0}^{[n]}} {\longrightarrow}F(I^n) {\longrightarrow}0$$ Here $\alpha_{i}^{[n]} \geq 0$ for all $i$. As $ {\operatorname{depth}}F(I^n) \geq 1$ for all $n \gg 0$ we get $K_n = 0$ for all $n \gg 0$. We show in Theorem \[sub\] that if ${\mathbf{a} }(I) < 0$ then *for all* $n \gg 0$, $$\begin{aligned}
f_1(I) - f_0(I) + 1 &= - \sum_{i = 1}^{\beta_{1}^{[n]}}\alpha_{i}^{[n]}   \quad \text{and} \\
\beta_{1}^{[n]} &= 0 \ \ \text{if and only if } \alpha_{i}^{[n]} = 0 \ \text{ for all $i$ }.\end{aligned}$$

Our second result Theorem \[sectheorem\] has a noteworthy consequence when $G(I)$ is [Cohen-Macaulay]{}.

\[seccor\] Let $(A,{\mathfrak{m} })$ be [Cohen-Macaulay]{} local ring of dimension $d = 3$. Let $I$ be an ${\mathfrak{m} }$-primary ideal with $G(I)$ [Cohen-Macaulay]{} and ${\operatorname{red}}(I) = 2$. Then

$f_2(I) \geq f_1(I) - f_0(I) + 1$.

We extend our results to higher analytic spread using Rees-superficial sequences (see section $6$ for details), under some mild assumptions on $grade(I)$. We state some of our noteworthy results. The first one (see \[mth1\]) states that if $l(I) \geq 2 $, ${\operatorname{grade}}(I) \geq l(I) -2$ and reduction number of $I \leq 1$ then $f_1(I) \leq f_0(I) - 1$ with equality if ${\operatorname{grade}}(I) = l(I)$. An immediate consequence (see \[NarC\]) is that if $(A,{\mathfrak{m} })$ is [Cohen-Macaulay]{} with $\dim A \geq 2$, $I$ an ${\mathfrak{m} }$-primary ideal and the second Hilbert-Samuel coefficient $e_2(I) = 0$ then $f_1(I) = f_0(I) - 1$ (see 2.7 for definition of $e_2(I)$).

Finally, we show that if $A$ is [Cohen-Macaulay]{} ring of dimension at least three and if $I$ an ${\mathfrak{m} }$-primary ideal of reduction number two whose associated graded ring is [Cohen-Macaulay]{}, then $f_2(I) \geq f_1(I)-f_0(I) +1$ (see \[mth2\]).

Here is an overview of the contents of the paper. In section $1$ we introduce some notations and preliminary facts needed. In section $2$ we introduce two complexes which will be used in the subsequent sections. In section $3$ we prove the main result for $l=2$ (Theorem \[result1\]). In section $4$ we prove our second Theorem and as a consequence obtain Theorem \[seccor\]. In section $5$ we obtain results on the coefficients of the fiber cone for any analytic spread. In the appendix( =section 6) we recall some basic facts regarding minimal reductions and filter-regular elements and prove an elementary result; which is useful in section 3.

*Acknowledgments* The authors thanks the referee for many pertinent comments. The author also thanks Fahed Zulfeqarr and A. V. Jayanthan for help in examples.

preliminaries
=============

From now on $(A,{\mathfrak{m} })$ is a Noetherian local ring of dimension $d$, with infinite residue field. All modules are assumed to be finitely generated. For a finitely generated module $M$, we denote its length by $\ell(M)$.

Let $J = (x_1,\ldots,x_l)$ be a minimal reduction of $I$. We denote by ${\operatorname{red}}_J(I) := \min \{ n | JI^n= I^{n+1} \} $ the *reduction number* of $I$ [with respect to]{} $J$. Let $${\operatorname{red}}(I) = \min \{ {\operatorname{red}}_J(I) \mid J
\text{\ is  a reduction of \ } I \}$$ be the *reduction number* of $I$.

As a reference for local cohomology we use [@BSh] (see especially Chapter 18 for relations between local cohomology and reductions). We take local cohomology of $G(I)$ [with respect to]{} $G(I)_+ =  \bigoplus_{n\geq 1}I^n/I^{n+1}$. Set $G = G(I)$ and $G_+ = G(I)_+$. For each $i \geq 0$ the local cohomology modules $H^i_{G_+}(G)$ are graded and furthermore $H^i_{G_+}(G)_n = 0$ for all $n \gg 0$. For each $i \geq 0$ set $$a_i(I) = \max \{ n \mid H^i_{G_+}(G)_n \neq 0 \}.$$

Set $l = l(I)$. Then $H^l_{G_+}(G) \neq 0$ and $H^i_{G_+}(G) = 0 $ for all $i > l$ (see [@Hoa 2.3]). We call $ {\mathbf{a} }(I) = a_l(I)$ to be the *$a$-invariant* of $G(I)$. The *(Castelnuovo-Mumford) regularity* of $G(I)$ is $${\operatorname{reg}}(G(I)) =  \max \{ a_i(G) + i \mid 0 \leq i \leq l \}.$$ The *regularity of $G(I)$ at and above level $r$*, denoted by ${\operatorname{reg}}^r(G(I))$, is $${\operatorname{reg}}^r(G(I)) =  \max \{ a_i(G) + i \mid r \leq i \leq l \}.$$

\[notation\] Let $x \in I\setminus I^2$ be a $I$-superficial element of $I$. For all $r \geq 1$ and $s \geq 0$ it is easy to show $${\operatorname{reg}}^{r}(G(I)) \leq s \implies {\operatorname{reg}}^{r}(G(I/(x)) \leq s.$$

\[asymp\] We will use the following beautiful result due to Hoa([@Hoa 2.6]):

*There exists non-negative integers $n_0, {\mathfrak{r} }(I)$, such that for all $n \geq n_0$ and every minimal reduction $J$ of $I^n$ we have ${\operatorname{red}}_J(I^n)  = {\mathfrak{r} }(I)$. Furthermore* $${\mathfrak{r} }(I) =
\begin{cases}
l(I) - 1 & \text{if } \ {\mathbf{a} }(I) < 0, \\
l(I) & \text{if } \ {\mathbf{a} }(I) \geq 0.
\end{cases}$$

For definition and basic properties of superficial sequences see [@Pu1 p. 86-87] . If $I$ is ${\mathfrak{m} }$-primary then let $p_I(z)$ be the *Hilbert-Samuel polynomial* of $A$ [with respect to]{}  $I$ (so $\ell (A/I^{n+1}) = p_I(n) $, $\forall n \gg 0$). Write $
p_I(z) = \sum_{i = 0}^{d}(-1)^i e_i(I)\binom{z+d -i}{d-i}.
$ For $i \geq 0$ we call $e_i(I)$ the $i^{th}$-*Hilbert coefficient* of $I$.

The Hilbert series of $F(I)$, $G(I)$ is denoted by $H(F(I),z)$, $H(G(I),z)$ respectively, i.e. $$H(F(I),z) = \sum_{n \geq 0} \ell\left( \frac{I^n}{{\mathfrak{m} }I^{n}} \right)z^n\quad \text{and} \quad H(G(I),z) = \sum_{n \geq 0} \ell\left( \frac{I^n}{I^{n+1}}  \right)z^n$$

If $x \in I^j \setminus {\mathfrak{m} }I^j$ then we denote by $x^\circ$ its image in $F(I)_j = I^j/{\mathfrak{m} }I^j$.

Two complexes
=============

Our results are based on analyzing two complexes which we describe in this section. Both the complexes are defined *using the maps in the Koszul complex.* Throughout $I$ is an ideal in $(A, {\mathfrak{m} })$ with a minimal reduction $J = (x_1,\ldots,x_l)$ where $l =l(I)$. Note that $x_1,\ldots,x_l$ are analytically independent ([@NoRees], cf. [@BH 4.6.9]). Using analyticity to detect exactness (at some stage) of a complex

The first complex is when $l = 2$ $$\label{first}
\mathcal{C}_\bullet(I): \quad 0{\rightarrow}\frac{A}{{\mathfrak{m} }} \xrightarrow{\alpha_2} \left(\frac{I}{{\mathfrak{m} }I}\right)^2\xrightarrow{\alpha_1}  \frac{IJ}{{\mathfrak{m} }IJ} {\rightarrow}0, \quad \text{where}$$ $$\alpha_2(a + {\mathfrak{m} }) 
= \begin{pmatrix}-x_2a + {\mathfrak{m} }I \\ x_1a + {\mathfrak{m} }I\end{pmatrix} \ \  \& \ \ 
\alpha_1 \begin{pmatrix}a + {\mathfrak{m} }I \\ b + {\mathfrak{m} }I\end{pmatrix} = x_1a + x_2b + {\mathfrak{m} }IJ.$$

\[firstcomments\]

1.  $H_2({\mathcal{C}_{\bullet} }(I)) = 0$ since $x_1,x_2$ are analytically independent.

2.  If $x_1,x_2$ is a regular sequence in $A$ then clearly $H_1({\mathcal{C}_{\bullet} }(I))  = 0$.

3.  Clearly $\alpha_1$ is surjective. So $H_0({\mathcal{C}_{\bullet} }(I)) = 0$.

The second complex, $\mathcal{D}_\bullet(I)$, is when $l = 3$. $$\label{second}
     0
{\rightarrow}\frac{A}{{\mathfrak{m} }} 
\xrightarrow{\alpha_3} \left(\frac{I}{{\mathfrak{m} }I}\right)^3
\xrightarrow{\alpha_2} \left(\frac{I^2}{{\mathfrak{m} }I^2}\right)^3  
\xrightarrow{\alpha_1} \frac{I^2J}{{\mathfrak{m} }I^2J} 
{\rightarrow}0, \quad \text{where}$$ $$\alpha_3(a + {\mathfrak{m} }) 
= \begin{pmatrix}
    x_3a + {\mathfrak{m} }I \\
   -x_2a + {\mathfrak{m} }I \\ 
    x_1a + {\mathfrak{m} }I
  \end{pmatrix}, \quad 
  \alpha_1
  \begin{pmatrix}
   a + {\mathfrak{m} }I^2 \\ 
   b + {\mathfrak{m} }I^2 \\ 
   c + {\mathfrak{m} }I^2\end{pmatrix} 
= x_1a + x_2b + x_3c + {\mathfrak{m} }I^2J,$$ $$\text{and}\quad \alpha_2\begin{pmatrix}a + {\mathfrak{m} }I \\ b + {\mathfrak{m} }I \\ c + {\mathfrak{m} }I\end{pmatrix} = 
\begin{pmatrix}-x_2a -x_3b + {\mathfrak{m} }I^2 \\ x_1a -x_3c + {\mathfrak{m} }I^2 \\  x_1b + x_2c + {\mathfrak{m} }I^2\end{pmatrix}.$$

\[secondcomments\]

1.  $H_3({\mathcal{D}_{\bullet} }(I)) = 0$ since $x_1,x_2, x_3$ are analytically independent.

2.  In Lemma \[2c1\] we show that if $x_1,x_2,x_3$ is a regular sequence and if $I^2\cap J = JI$ then ${\operatorname{image}}(\alpha_2) = \ker (\alpha_1)$, so $H_1({\mathcal{D}_{\bullet} }(I))  = 0$.

3.  The assumption $I^2\cap J = JI$ holds when the following hold

    1.  $I$ is integrally closed (see[@Hun p. 317] and [@Itoh88 Theorem 1]).

    2.  The initial forms $x_{1}^{*}, x_{2}^{*}, x_{3}^{*}$ in $G(I)_1$ is a regular sequence (see [@VV 2.3]).

4.  Clearly $\alpha_1$ is surjective and so $H_0({\mathcal{D}_{\bullet} }(I)) = 0$.

\[2c1\](with hypothesis as in \[second\]) If $x_1,x_2,x_3$ is a regular sequence and if $I^2\cap J = JI$ then ${\operatorname{image}}(\alpha_2) = \ker (\alpha_1)$.

Let $\mathcal{K}(\mathbf{x})_\bullet$ be the Koszul complex on $x_1,x_2,x_3$. It is acyclic since $x_1,x_2,x_3$ is a regular sequence. Suppose $ \xi 
= \begin{pmatrix}
   {\overline}{a} \\
   {\overline}{b} \\
   {\overline}{c}
\end{pmatrix}
 \in \ker\alpha_1
$. Then $$ax_1 + bx_2 + cx_3 = a'x_1 + b'x_2 + c'x_3 \quad \text{where} \ a',b',c' \in {\mathfrak{m} }I^2.$$ As $\mathcal{K}(\mathbf{x})_\bullet$ is acyclic; there exists $f,g,h \in A $ such that

$$\label{lemmeq}
\begin{pmatrix}a -a' \\ b-b' \\ c-c' \end{pmatrix} = \begin{pmatrix}-x_2f -x_3g \\ x_1f -x_3h \\  x_1g +
 x_2h \end{pmatrix}.$$

We show that $f,g,h$ are in $I$. Using (\[lemmeq\]) we get $$-x_2f -x_3g \in I^2\cap J = JI.$$ So $-x_2f - x_3g = x_1p + x_2q + x_3r$ where $p,q,r \in I$. Again using the fact that $\mathcal{K}(\mathbf{x})_\bullet$ is acyclic we get that there exists $u,v,w \in A$ such that $$\begin{pmatrix}p  \\ q+ f \\ r+ g \end{pmatrix} = \begin{pmatrix}-x_2u -x_3v \\ x_1u -
x_3w \\  x_1v +
 x_2w \end{pmatrix}.$$ So $f,g \in I$. Similarly by using the second row in (\[lemmeq\]) we get that $f,h \in I$. Set $  \eta 
=  \begin{pmatrix}
   {\overline}{f} \\
   {\overline}{g} \\
   {\overline}{h}
\end{pmatrix}
 \in (I/{\mathfrak{m} }I)^3$. Notice $\alpha_2(\eta) = \xi $ since $a', b', c' \in {\mathfrak{m} }I^2$. Thus $\ker \alpha_1 \subseteq {\operatorname{image}}\alpha_2$.

Let us recall the following well-known fact of complexes: Let $$\mathcal{X}_{\bullet}: \quad 0{\rightarrow}X_n {\rightarrow}X_{n-1}{\rightarrow}\ldots {\rightarrow}X_0 {\rightarrow}0,$$ be a complex of $A$-modules with $\ell(X_i)$ finite for all $i$. Then $$\label{hom}
\sum_{i = 0}^{n}(-1)^i\ell(X_i) = \sum_{i = 0}^{n}(-1)^i\ell( H_i\big(\mathcal{X}_{\bullet}\big)).$$

Proof of Theorem \[result1\]
============================

In this section we prove Theorem \[result1\]. The setup below is used throughout. The hypothesis ${\mathbf{a} }(I)$ is crucial for our results. See examples 6.1, 6.2,6.3 for some illustrations of Theorem \[result1\].

**Setup:** Let $J = (x_1,x_2)$ be a minimal reduction of $I$. Notice $ J^{[n]}= (x_{1}^{n},x_{2}^{n})$ is a minimal reduction of $I^n$. If ${\operatorname{grade}}(I) = 2$ then we can take $x_1,x_2$ to be a regular sequence (and so $x_{1}^{n},x_{2}^{n}$ is also a regular sequence). For $i = 1, 2$ set $X_i = x_{i}^{\circ}$, the image of $x_i$ in $I/{\mathfrak{m} }I$. So $F(J^n) = k[X^{n}_{1}, X^{n}_{2}]$ for all $n \geq 1$.

We first prove the inequality stated in Theorem \[result1\].

\[1\] Let $(A,{\mathfrak{m} })$ be a local ring and let $I$ be an ideal with $l(I) =2$. If ${\mathbf{a} }(I) < 0$ then $f_1(I) \leq f_0(I) -1.$ Furthermore if ${\operatorname{grade}}(I) = 2$ then equality holds.

We consider the complex (\[first\]) for the ideal $I^n$ for each $n \geq 0$. Set ${\mathcal{C}_{\bullet} }[n] = {\mathcal{C}_{\bullet} }(I^n)$ for $n \geq 1$. By Remark \[firstcomments\] we have $H_i({\mathcal{C}_{\bullet} }[n]) = 0$ for $i = 0,2$. Using (\[hom\]) for the complex ${\mathcal{C}_{\bullet} }[n]$ for each $n$ we get an equation $$\label{r1}
1 - 2\ell(I^n/{\mathfrak{m} }I^n) + \ell(I^nJ^{[n]}/{\mathfrak{m} }I^nJ^{[n]}) = -\ell(H_1({\mathcal{C}_{\bullet} }[n])).$$

Since ${\mathbf{a} }(I) < 0$, by \[asymp\] we have ${\operatorname{red}}_{J^{[n]}}(I^n) =1$ for all $n \gg 0$. So $I^nJ^{[n]} = I^{2n}$ for all $n \gg 0$. Also for all $n \gg 0$ we have $f_I(n) = \ell(I^n/{\mathfrak{m} }I^n)$. Setting these in (\[r1\]) we get for all $n \gg 0$, $$1 - 2f_I(n) + f_I(2n) = -\ell(H_1({\mathcal{C}_{\bullet} }[n])).$$ Write $f_I(n) = f_0(n+1) -f_1$. Therefore $$1 - 2\{f_0(n+1) -f_1 \} + f_0(2n+1) -f_1 = -\ell(H_1({\mathcal{C}_{\bullet} }[n])).$$ $$\text{Thus} \quad 1 - f_0 + f_1  =  -\ell(H_1({\mathcal{C}_{\bullet} }[n])).$$ Hence $$\text \quad 1 - f_0 + f_1  \leq 0.$$ By our assumption on $J$, $H_1({\mathcal{C}_{\bullet} }[n]) = 0$ for each $n$ if ${\operatorname{grade}}(I) = 2$. Hence equality holds in the above equation. This proves the result.

The following example shows that if ${\mathbf{a} }(I) < 0$ but ${\operatorname{grade}}(I) \neq 2$ then $f_1(I) < f_0(I) -1 $ is possible.

\[2ex1\] Set $A = k[[X_1,X_2, X_3]]/(X_{1}^{2}, X_1X_2) = k[[x_1,x_2,x_3]]$. Set $I = {\mathfrak{m} }= (x_1,x_2,x_3)$. and $J = (x_2,x_3)$. Then $J$ is a reduction of ${\mathfrak{m} }$ and ${\mathfrak{m} }^2 = J{\mathfrak{m} }$. By [@TrungProc 3.2] we get ${\mathbf{a} }({\mathfrak{m} })< 0$.

It can be checked that ${\operatorname{grade}}({\mathfrak{m} }) = 1$ and $x_3$ is a non-zero divisor. Using COCOA [@cocoa] it can verified that the Hilbert-series of $F({\mathfrak{m} }) = G({\mathfrak{m} })$ is $$\frac{1+z-z^2}{(1-z)^2}.$$ So $f_1({\mathfrak{m} }) = -1$ but $f_0({\mathfrak{m} }) = 1$.

The next example shows that Theorem \[1\] need not hold when $  {\mathbf{a} }(I)> 0$.

\[2ex2\] Let $(A,{\mathfrak{m} })$ be a two dimensional [Cohen-Macaulay]{} local ring with ${\operatorname{red}}({\mathfrak{m} }) = 2$. Then will have $G({\mathfrak{m} }) = F({\mathfrak{m} })$ is [Cohen-Macaulay]{}, ([@Sa2 2.1]) and it’s Hilbert-series is $$\frac{1+z+ cz^2}{(1-z)^2} \quad \text{where} \ c > 0.$$ So $f_1({\mathfrak{m} })-f_0({\mathfrak{m} }) + 1 = c > 0$.

Next we analyze the case when $f_1(I) = f_0(I) -1$. \[veronese\]Observe that $F(I^n) = F(I)^{<n>}$, the *$n^{th}$ Veronese* subring of $F(I)$. In particular $l(I^n) = l(I)$ for each $n \geq 1$. Local cohomology commutes with the Veronese functor, cf. [@HMc 2.5]. In [@Pu3 2.8] it is proved that ${\operatorname{depth}}F(I^n)$ is constant for all $n \gg 0$. We prove

\[veronese-th\] Let $(A,{\mathfrak{m} })$ be a local ring with infinite residue field and let $I$ be an ideal with $s = l(I) > 0$. Then

1.  ${\operatorname{depth}}F(I^n) > 0$ for all $n \gg 0$.

2.  There exists a minimal reduction $J = (x_1,\ldots,x_s)$ of $I$ such that $(x_1^n)^\circ$ is $F(I^n)$ regular for all $n \gg 0$.

\(1) Set $E = H^{0}_{F(I)_+}(F(I))$. Clearly $\ell(E) < \infty$. Say $E = \bigoplus_{i = 0}^{r}E_i$. Notice $E$ is an ideal of $F(I)$ with finite length. If $E_0 \neq 0$ then $1_{F(I)} \in E$. So $E = F(I)$ will have finite length, a contradiction since $\dim F(I) = l(I) \geq 1$. Therefore $E_0 = 0$ and as a consequence we have $$H^{0}_{F(I^n)_+} F(I^n) =  \left( H^{0}_{F(I)_+}(F(I) \right)^{<n>} = 0  \quad \text{for all} \ \ n > r.$$ Thus ${\operatorname{depth}}F(I^n) > 0$ for all $n \gg 0$.

\(2) By [@TrungProc 3.8] we get that there exists a minimal reduction $J = (x_1,\ldots, x_s)$ of $I$ such that $x_1^\circ, \ldots, x_s^\circ \in F(I)_1$ is a $F(I)$-filter regular sequence. Set $x = x_1$. Since $x^\circ$ is $F(I)$-filter regular, by \[filt-ideal-cor\] we get that $(x^n)^\circ$ is $F(I^n)$-filter regular for each $n \geq 1$. By (1) ${\operatorname{depth}}F(I^n) > 0$ for all $n \gg 0$. So by \[filtISreg\] we get that $(x^n)^\circ$ is $F(I^n)$-regular for all $n \gg 0$.

\[rossi\] As $l(I) = 2$, by computing the Hilbert polynomial of $F(I)$ and $F(I^n)$ we obtain $$f_1(I^n) - f_0(I^n) + 1 = f_1(I) - f_0(I) + 1 \quad \text{for all } n \geq 1.$$

\[temp\] Let $J = (x_1,x_2)$ be a minimal reduction of $I$ as constructed in Theorem \[veronese-th\]. In particular $(x^{n}_{1})^\circ$ is $F(I^n)$-regular for all $n \gg 0$. Set $X^{n}_{j} = (x^{n}_{1})^\circ$ for $j = 1,2$. We resolve $F(I^n)$ as a $F(J^{[n]}) =  k[X^{n}_{1}, X^{n}_{2}]$ module and write it as: $$0 {\longrightarrow}K_n {\longrightarrow}\bigoplus_{i = 1}^{\beta_{1}^{[n]}}F(J^{[n]})( - 1 - \alpha_{i}^{[n]} )  {\longrightarrow}F(J^{[n]})^{\beta_{0}^{[n]}} {\longrightarrow}F(I^n) {\longrightarrow}0$$ Here $\alpha_{i}^{[n]} \geq 0$ for all $i$. As $ {\operatorname{depth}}F(I^n) \geq 1$ for all $n \gg 0$ we get $K_n = 0$ for all $n \gg 0$. We prove

\[sub\] (with assumptions as in \[temp\]) If ${\mathbf{a} }(I) < 0$ then *for all* $n \gg 0$, $$\begin{aligned}
f_1(I) - f_0(I) + 1 &= - \sum_{i = 1}^{\beta_{1}^{[n]}}\alpha_{i}^{[n]}   \quad \text{and} \\
\beta_{1}^{[n]} &= 0 \ \ \text{if and only if } \alpha_{i}^{[n]} = 0 \ \text{ for all $i$ }.\end{aligned}$$

For the proof of this theorem we need the following:

\[euclid\] Let $R = k[X]$ and let $S = \oplus_{n \geq 0}S_n$ be a standard $k$-algebra of dimension $1$ and multiplicity $p+1$. Assume $S = Ru_1 + \ldots +Ru_m$ where degree $u_i \leq 1$ and $u_1 = 1_{S}$. Then $S$ has the following resolution over $R$ $$0 {\longrightarrow}\bigoplus_{i = 1}^{q}R( -1  - \alpha_i  ){\longrightarrow}R \oplus R(-1)^{p+ q} {\longrightarrow}S {\longrightarrow}0.$$ Also $S$ is free if and only if all $\alpha_i = 0$.

By the hypothesis on $S$ and as $R$ is an Euclidean domain we get $$S = R \bigoplus R(-1)^{p} \bigoplus \left(  \bigoplus_{i = 1}^{q}\frac{R}{(X^{\alpha_i}) } (-1)    \right), \quad  \text{where} \  \alpha_i \geq 0.$$ The result follows.

We choose $n_0$ such that for all $n \geq n_0$ we have ${\operatorname{depth}}F(I^n) \geq 1$ and ${\operatorname{red}}_{J^{[n]}}(I^n) = 1$. Fix $n \geq n_0$. Set $\alpha_i = \alpha_{i}^{[n]} $. Set $F(J^{[n]}) = k[X^{n}_{1}, X^{n}_{2}]$. Since ${\operatorname{red}}_{J^{[n]}}(I^n) = 1$ it follows that $F(I^n)$ is generated as a $F(J^{[n]})$ in degrees $\leq 1$.

Notice that by construction, $X_1^{n}$ is a non-zero divisor on $F(I^n)$ (see \[temp\]). Set $R = F(J^{[n]})/(X^{n}_{1}) =  k[X^{n}_{2}]$ and $S = F(I^n)/X^{n}_{1}F(I^n)$. Note that $S$ is generated as a $R$ module in degrees $\leq 1$. By Lemma \[euclid\] the resolution of $S$ as a $R$-module is $$0 {\longrightarrow}\bigoplus_{i=1}^{q}R( -1  - \alpha_i)   {\longrightarrow}R \oplus R(-1)^{p+ q} {\longrightarrow}S {\longrightarrow}0.$$ Since $X_{1}^{n}$ is a non-zero divisor on $F(I^n)$ and $F(J^{[n]})$, we get that the resolution of $F(I^n)$ as $F(J^{[n]})$-module is $$0 {\longrightarrow}\bigoplus_{i=1}^{q} F(J^{[n]})( -1 - \alpha_i) {\longrightarrow}F(J^{[n]}) \oplus F(J^{[n]})(-1)^{p+ q} {\longrightarrow}F(I^n) {\longrightarrow}0.$$ Thus $q = \beta_{1}^{[n]} $. Set $\phi(z) = \sum_{i = 1}^{q} z^{\alpha_i + 1}$. Therefore the Hilbert series of $ F(I^n)$ is $$\frac{ 1 + (p+q)z -\phi(z)}{(1-z)^2}.$$ So $f_0(I^n) = 1 + p$. Notice $$f_1(I^n) = p + q - \sum_{i = 1}^{q}(\alpha_i + 1)  = p - \sum_{i = 1}^{q}{\alpha_i}.$$ Using \[rossi\] we obtain $$\sum_{i = 1}^{q}{\alpha_i} = f_0(I^n) - f_1(I^n) - 1 = f_0(I) - f_1(I) - 1.$$ Also by Lemma \[euclid\], $q = 0$ if and only if all $\alpha_i = 0$.

In view of this result we are tempted to ask

(with notation as in \[temp\]) Let $(A,{\mathfrak{m} })$ be a local ring and let $I$ be a proper ideal with $l(I) = 2$. Is $ \sum_{i = 1}^{\beta_{1}^{[n]}}{\alpha_i}^{[n]}$ constant for all $n \gg 0$?

Finally we prove

This follows from Theorems \[1\], \[sub\].

We give an example which shows that in the case $l(I) = {\operatorname{grade}}(I) = 2$ and ${\mathbf{a} }(I) < 0$ it is possible for $F(I)$ to be not [Cohen-Macaulay]{} even though $f_1(I) = f_0(I) -1$ and $F(I^n)$ is [Cohen-Macaulay]{} for all $n \gg 0$. The example below was constructed by Marley [@Mar 4.1] in his study of the associated graded ring $G(I)$.

Let $A = k[X,Y]_{(X,Y)}$ and let $I = (X^7, X^6Y, XY^6, Y^7) $. Using COCOA one verifies that $e_2(I) = 0$ and that the Hilbert series of the fiber cone $F(I)$ is $$\frac{1 + 2z + 2z^2  + 2z^3 + 2z^4 + 2z^5 - 4z^6 }{(1-z)^2}$$ From the Hilbert series it’s clear that $f_0(I) = 7$, $f_1(I) = 6$ but $F(I)$ is not [Cohen-Macaulay]{}.

It is also possible to prove Theorem \[result1\] directly from Theorem \[sub\], Observation \[rossi\] and by using a result of Kishor Shah [@shah91 Theorem 1]. We kept Theorem \[1\] in this section since it gives the inequality $f_1(I) \leq f_0(I) - 1$ very easily and more importantly gives us a natural way of trying to relate $f_i(I) $ for $i = 0,1,2$ which is new and important. This is done in our next section.

Results when analytic spread is three
=====================================

In this section we assume that $l(I) = 3$. If $J = (x_1,x_2,x_3)$ is a reduction of $I$, we also assume that $x_1,x_2,x_3$ is a regular sequence. The goal of this section is to prove $f_2(I) \geq f_1(I) - f_0(I) + 1$ under suitable conditions on $I$.

\[scrap\] We consider the complex (\[second\]) for the ideal $I^n$ for each $n \geq 0$. Set ${\mathcal{D}_{\bullet} }[n] = {\mathcal{D}_{\bullet} }(I^n)$ for $n \geq 1$. By Remark \[firstcomments\] we have $H_i({\mathcal{D}_{\bullet} }[n]) = 0$ for $i = 0,3$.

To analyze the case when $H_1({\mathcal{D}_{\bullet} }[n]) $ is zero we make the following:

Let $I$ be an ideal and let $J = (x_1,x_2,x_3)$ be a minimal reduction of $I$. We say the pair $(I,J)$ satisfy $V_{2}^{\infty}$ if $$I^{2n}\cap J^{[n]} = J^{[n]}I^n \quad \text{for all} \ n\gg 0.$$ This condition has been studied by Elias in [@Elasymp].

\[scrap2\] By Lemma \[2c1\], $H_1({\mathcal{D}_{\bullet} }[n]) = 0$ for all $n \gg 0$ *if* the pair $(I,J)$ satisfy $V_{2}^{\infty}$.

\[hyp-satisfy\] When ${\operatorname{grade}}(I) = l(I)$ then using \[secondcomments\](iii) the hypothesis $V_{2}^{\infty}$ holds when

1.  $I$ is asymptotically normal i.e., $I^n$ is integrally closed for all $n \gg 0$.

2.  The initial forms $x_{1}^{*},\ldots, x_{l(I)}^{*}$ in $G(I)_1$ form a regular sequence.

We now state our second main theorem.

\[sectheorem\] Let $(A,{\mathfrak{m} })$ be local and let $I$ be an ideal in $A$ with $l(I) = {\operatorname{grade}}(I) =3$. Let $J = (x_1,x_2,x_3)$ be a minimal reduction of $I$ and assume the pair $(I,J)$ satisfy $V_{2}^{\infty}$. If ${\mathbf{a} }(I) < 0$ then $f_2(I) \geq f_1(I) -f_0(I) + 1$.

The hypothesis of the theorem above is quite stringent. However they are necessary (see Examples \[ex1\], \[ex2\], \[ex3\]). Also note that if $F(I)$ is [Cohen-Macaulay]{} then inequality holds (infact it holds for any [Cohen-Macaulay]{} $k$-algebra).

We give two examples where the condition of Theorem \[sectheorem\] holds.

\[pos1\] Let $(A,{\mathfrak{m} })$ be a three dimensional [Cohen-Macaulay]{} local ring with $I$ an ${\mathfrak{m} }$-primary ideal with reduction number two and $G(I)$ [Cohen-Macaulay]{}. Then the hypothesis of Theorem \[sectheorem\] holds by \[hyp-satisfy\](2). This is shown in Theorem \[as=2\]. Using Example 6.1 from [@jpv] we can construct an interesting example of this kind as follows:

Let $T = k[[t^6, t^{11}, t^{15}, t^{31}]]$, $K = (t^6, t^{11},
t^{31})$ and $L = (t^6)$. Then, it can easily be verified that $K^3 = LK^2$. Since $K^2 \cap L = LK$, $G(K)$ is Cohen-Macaulay by a result of Valabrega and Valla [@VV 2.3]. It can also be seen that $t^{37} \in {\mathfrak{m} }K^2$, but $t^{37} \notin {\mathfrak{m} }LK$. Therefore $F(K)$ is not Cohen-Macaulay by a criterion due to Cortadellas and Zarzuela [@CZ 3.2]. One can verify that the Hilbert-Series of $F(K)$ is $(1 + 2z)/(1-z)$.

Let $R = k[[X,Y,Z,W]]$ and ${\mathfrak{q} }= (y^2z-xw,x^2z^2-yw,x^3z-y^3,x^3yw-z^4,z^5-y^4w,xyz^3-w^2,y^5-wx^4,x^2y^3-z^3,x^5-z^2,x^4y^2-zw)$. Set $B = R/{\mathfrak{q} }= k[[x,y,z,w]]$. Using COCOA, one can verify that $B \cong T$. Under this isomorphism the ideal $(x,y,w)$ maps to $K$ and $(x)$ goes to $L$

Set $A = B[[U,V]]$. Clearly $A$ is [Cohen-Macaulay]{} of dimension $3$. Set $I = (x,y,w, U, V)$. and $J = (x, U, V)$. Clearly $J$ is a minimal reduction of $I$ and $I^3 = JI^2$. Furthermore $G(I) \cong G(K)[U, V]$ and $F(I) \cong F(K)[U, V]$. So $G(I)$ is [Cohen-Macaulay]{} while $F(I)$ has depth $2$.

Be giving the next example we make the following:

If $A$ satisfies the condition of the theorem then as ${\mathbf{a} }(I) < 0$ we get by \[asymp\] that ${\operatorname{red}}_{J^{[n]}}(I^n) = 2$ for all $n \gg 0$. So by using the Valabrega-Valla criterion [@VV 2.3] it follows that ${\operatorname{grade}}G(I^n) \geq 3$ for all $n \gg 0$.

\[pos2\] Let $A = k[[X,Y,Z]]$ and let ${\mathfrak{m} }= (X,Y,Z)$. Let $I$ is an ${\mathfrak{m} }$-primary ideal with $I^r = {\mathfrak{m} }^s$ for some $s> r$ and $G(I)$ not [Cohen-Macaulay]{} ( for a specific example see [@CPR 4.3] also see [@CPR 3.8]). If $J$ is a reduction of $I$ then the pair $(I,J)$ satisfy $V_{2}^{\infty}$, by \[hyp-satisfy\](1). Also by Hoa’s result it follows that $a(I) < 0$. Any such example is different from Example \[pos1\]; since $G(I)$ is [Cohen-Macaulay]{} in Example \[pos1\].

Before proving the theorem we give a proof of Theorem \[seccor\]. We restate it for the convenience of the reader.

\[as=2\] Let $(A,{\mathfrak{m} })$ be [Cohen-Macaulay]{} local ring of dimension $d = 3$. Let $I$ be an ${\mathfrak{m} }$-primary ideal with $G(I)$ [Cohen-Macaulay]{} and ${\operatorname{red}}(I) \leq 2$. Then

$f_2(I) \geq f_1(I) - f_0(I) + 1$.

As $G(I)$ is [Cohen-Macaulay]{} we have that ${\mathbf{a} }(I)  =  {\operatorname{red}}(I) -3 \leq -1$. Let $J$ be a minimal reduction of $I$ such that ${\operatorname{red}}_J(I) = {\operatorname{red}}(I)$. As $G(I)$ is [Cohen-Macaulay]{} the pair $(I,J)$ satisfy $V_{2}^{\infty}$. So the result follows from Theorem \[sectheorem\].

We now prove Theorem \[sectheorem\]. The notation is as in \[scrap\].

We use the complex ${\mathcal{D}_{\bullet} }[n]$. By \[scrap2\] and (\[hom\]) we get $  $ $$\label{r2}
-1 + 3\ell\left(\frac{I^n}{{\mathfrak{m} }I^n}  \right) - 3\ell\left( \frac{I^{2n}}{{\mathfrak{m} }I^{2n}} \right) +
 \ell\left( \frac{I^{2n}J^{[n]}}{{\mathfrak{m} }I^{2n}J^{[n]} } \right) = \ell\left( H_2({\mathcal{D}_{\bullet} }[n]) \right), \ \  \forall n \gg 0.$$ Since ${\mathbf{a} }(I) < 0$ we have by (\[asymp\]) ${\operatorname{red}}_{J^{[n]}}(I^n) = 2$ for all $n \gg 0$. So $I^{2n}J^{[n]} = I^{3n}$ for all $n \gg 0$. Also for all $n \gg 0$ we have $f_I(n) = \ell(I^n/{\mathfrak{m} }I^n)$. Setting these in (\[r2\]) we get $$-1 + 3f_I(n) - 3f_I(2n) + f_I(3n) \geq 0.$$ Since, $$\quad f_I(n) = f_0\binom{n+2}{2} - f_1(n+1) + f_2,$$ an easy computation yields $$-1 + 3f_I(n) - 3f_I(2n) + f_I(3n) = -1 + f_0 - f_1 + f_2$$ and the result follows.

Our second result Theorem \[sectheorem\] has three hypothesis, namely \[hypoth\]

1.  ${\operatorname{grade}}(I) = l(I)$.

2.  The pair $(I,J)$ satisfies $V_{2}^{\infty}$.

3.  ${\mathbf{a} }(I) < 0.$

We show that if any of the hypothesis in \[hypoth\] is not satisfied then the conclusion of Theorem \[sectheorem\] (i.e., $f_2(I) \geq f_1(I) -f_0(I) + 1 $) need not hold.

In the first example only hypothesis \[hypoth\] (a) is not satisfied, infact we have $l(I) - {\operatorname{grade}}(I) = 1$.

\[ex1\] Let $A = k[[X,Y,U,V]]/(XY, Y^3) = k[[x,y,u,v]]$ and $I = {\mathfrak{m} }= (x,y,u,v)$. One can readily see that ${\operatorname{grade}}({\mathfrak{m} }) = 2$ while $l({\mathfrak{m} }) =  \dim A = 3$. Set $J = (x,u,v)$. Then $I^3 = JI^2$. So by [@TrungProc 3.2] we get ${\mathbf{a} }(I) <0$. The pair $(I,J)$ satisfy $V_{2}^{\infty}$ by Proposition \[ex1J\]. In fact in \[ex1J\] we show ${\mathfrak{m} }^{2n} \cap J^{[n]} = J^{[n]} {\mathfrak{m} }^n$ for all $n \geq 1$. However Hilbert series of $F({\mathfrak{m} }) = G({\mathfrak{m} })$ (by COCOA) is $$\frac{1+ z -z^3}{(1-z)^3}$$ So $f_0(I) = 1, f_1(I) = -2$ and $ f_2(I) = -3$. Thus $f_2(I) \ngeq f_1(I) -f_0(I) + 1$.

\[ex1J\]\[With $A$, ${\mathfrak{m} }$, $J$ as in \[ex1\]\] $${\mathfrak{m} }^{2n} \cap J^{[n]} = J^{[n]} {\mathfrak{m} }^n \quad \text{for all }  \ n \geq 1.$$

Fix $n \geq 1$. We claim $$\text{if}  \  x^nc \in {\mathfrak{m} }^{2n}  \quad  \text{ then} \  c \in {\mathfrak{m} }^n.  \tag{$\dagger$}$$

Let us assume ($\dagger$) and prove our result. Fix $n \geq 1$. Let $\xi \in {\mathfrak{m} }^{2n} \cap J^{[n]}$. Write $\xi = \alpha x^n + \beta u^n + \gamma v^n$. Set $(T,{\mathfrak{q} }) = (A/(x^n), {\mathfrak{m} }/(x^n))$. Note $$T = \frac{k[[x,y,u,v]]}{(xy,y^3, x^n)} = \left(\frac{k[[x,y]]}{(xy,y^3, x^n)}\right)[[u,v]]$$ Thus $u^*, v^*$ are $G({\mathfrak{q} })$-regular. So $${\overline}{\beta}{\overline}{ u}^n + {\overline}{\gamma} {\overline}{v}^n \in  {\overline}{{\mathfrak{m} }}^{2n} \cap ({\overline}{ u}^n,{\overline}{v}^n) = {\overline}{{\mathfrak{m} }}^{n}({\overline}{ u}^n,{\overline}{v}^n).$$ It follows that ${\overline}{\beta}, {\overline}{\gamma} \in {\overline}{{\mathfrak{m} }}^{n}$. Thus $$\xi = \alpha x^n + \beta u^n + \gamma v^n + \theta x^n \quad \text{where} \ \beta,\gamma  \in {\mathfrak{m} }^n \ \& \ \theta \in A.$$ Notice $(\alpha + \theta )x^n \in {\mathfrak{m} }^{2n}$. So by ($\dagger$) we get $(\alpha + \theta ) \in {\mathfrak{m} }^n$. It follows that $\xi \in {\mathfrak{m} }^{n} J^{[n]}$.

We now prove ($\dagger$). Let $S =k[[x,u,v]]$ considered as a subring of $A$. Any element $a \in A$ can be written as $$\begin{aligned}
 \label{AS}
 a &= \phi_{0}^{(a)}(x,u,v) + \phi_{1}^{(a)}(x,u,v)y + \phi_{2}^{(a)}(x,u,v)y^2  \\
\text{where} \ &\phi_{i}^{(a)}(x,u,v) \in S \ \text{for} 
\ i = 0,1,2.\end{aligned}$$ So $A$, as a $S$-module, is generated by $1,y,y^2 $. Thus $\dim S = \dim A = 3$. It follows that $S \cong k[[X,U,V]]$.

Notice $A/(y) = S$ and the natural map $\pi \colon A {\rightarrow}S$ is a splitting (as $S$-modules) of the inclusion $\imath \colon S {\rightarrow}A$. Set $L = Sy + Sy^2$. Then $A = S \oplus L$ as a $S$-module. It follows that $\phi_{0}^{(a)}(x,u,v)$ in (\[AS\]) is uniquely determined by $a$.

Let ${\mathfrak{n} }$ be the unique maximal ideal of $S$. Then $$\label{AS2}
 {\mathfrak{m} }^i = {\mathfrak{n} }^i \oplus ({\mathfrak{m} }^i \cap L) \quad \text{for all } \ i \geq 1.$$

Set $$c = \phi_{0}^{(c)}(x,u,v) + \phi_{1}^{(c)}(x,u,v)y + \phi_{2}^{(c)}(x,u,v)y^2.$$ *Notice* $x^nc = x^n\phi_{0}^{(c)}(x,u,v)$. By uniqueness of $\phi_{0}^{*}$ we get $$\phi_{0}^{(x^nc)}(x,u,v) = x^n\phi_{0}^{(c)}(x,u,v).$$ Since $x^nc \in {\mathfrak{m} }^{2n}$, we get by (\[AS2\]) that $x^n\phi_{0}^{(c)}(x,u,v) \in {\mathfrak{n} }^{2n}$. Clearly $x^*$ is $G_{{\mathfrak{n} }}(S)$-regular. So $ \phi_{0}^{(c)}(x,u,v) \in {\mathfrak{n} }^{n}$. By \[AS2\] again we get that $c \in {\mathfrak{m} }^n$.

This proves $(\dagger)$. As stated earlier this finishes the proof of the proposition.

In the second example only hypothesis \[hypoth\] (b) is not satisfied, We adapt an example from [@CJ3 6.2]. If $K$ is an ideal in $A$ let ${\widetilde }{K} = \cup_{n\geq 1} (K^{n+1} \colon K^n)$ be the Ratliff-Rush closure of $K$.

\[ex2\] Let $ A = \mathbb{Q}[[X,Y,Z]]$. Let $I = (X^4, X^3Y, XY^3, Y^4, Z)$. The ideal $J = (X^4, Y^4, Z )$ is a minimal reduction of $I$, infact $I^3 = JI^2$. So by [@TrungProc 3.2] we get ${\mathbf{a} }(I) < 0$. Set $B = \mathbb{Q}[[X,Y]]$ and $q = (X^4, X^3Y, XY^3, Y^4)$. One can show ${\widetilde }{q} \neq q$. However notice $G(I) = G(q)[Z^*]$. So $Z^*$ is $G(I)$-regular. In particular ${\widetilde }{I} = I$. By [@Pu3 7.9] we get ${\operatorname{depth}}G(I^n) = 1$ for all $n \gg 0$. The Hilbert series of $F(I)$ is $$\frac{1 + 2z + 2z^2 - z^3}{(1-z)^3}.$$ So $f_0(I) = 4$, $f_1(I) = 3$ and $f_2(I) = -1$. Thus $f_2(I) \ngeq f_1(I) -f_0(I) + 1$.

In the third example hypothesis \[hypoth\](c) is not satisfied. Instead of \[hypoth\](b) the hypothesis $$I^{2n}\cap (x_{1}^{n}, x_{2}^{n}) = I^n(x_{1}^{n}, x_{2}^{n}) \quad \text{for all} \ n \gg 1. \tag{$\dagger$}$$ is satisfied. Recall $J = (x_1,x_2,x_3)$ is a minimal reduction of $I$. The hypothesis ($\dagger$) is equivalent to ${\operatorname{depth}}G(I^n) \geq 2$ for all $n \gg 0$ [@Elasymp 2.4].

\[ex3\] Let $A = k[X,Y,Z]_{(X,Y,Z)}$ and $I = (X^3, XY^4Z, XY^5, Z^5, Y^7)$. Set $u = Z^5, V = 5X^3 + 3Y^7$ and $w=  X^3 -3XY^4Z  +2Z^5$. Set $J = (u,v,w)$. Using COCOA we can check that $I^6 = JI^5$. So $J$ is a minimal reduction of $I$. The Hilbert Series of $G(I)$, $G(I/(u))$, $G(I/(u,v))$ is $$\begin{aligned}
 H(G(I),z) &= (1-z)H(G(I/(u)),z)  = (1-z)^2H(G(I/(u,v)),z), \\
           &= \frac{77 + 15z + 8z^2 + 2z^3 + 2z^4 + z^5}{(1-z)^3}, \   \text{and} \\
 H(G(I/(u,v,w)),z) &= 77 + 28z\end{aligned}$$ So $u^*, v^*$ is a $G(I)$-regular sequence. Note that ${\operatorname{depth}}G(I) = 2$. It follows that ${\operatorname{depth}}G(I^n) \geq 2$ for all $n \geq 1$. So hypothesis ($\dagger$) is satisfied.

We prove that ${\mathbf{a} }(I) \geq 0$. Set $G = G(I)$. $$a_{i}(I) = \max\{ n \mid H^i(G)_n \neq 0 \} \quad i = 0,1,2,3$$ \[compt\] Note that $a_0(G) = a_1(G) = - \infty$. As ${\operatorname{red}}_J(I) = 5$ we get by [@TrungProc 3.2] that $a_3(G) \leq 2$. By [@Mar 2.1(a)] we have $a_2(G) < a_3(G) = 2$.

By [@Hoa 2.4] we have $$a_{i}(I^n)  \leq \left[ \frac{a_i(I)}{n} \right] \quad \text{for} \ i = 0,1,2.  \quad \& \quad
 a_{3}(I^n) = \left[ \frac{a_3(I)}{n} \right]$$ Notice by \[compt\] we get $$\begin{aligned}
 a_{i}(I^3) &\leq 0 \quad \text{for} \ i = 0,1,2. \\
a_3(I^3) &\leq -1 \quad \text{if }  \ a_3(I) < 0 \quad \text{and} \quad 
a_3(I^3) = 0  \quad \text{if }  \ a_3(I) \geq 0. \end{aligned}$$ By [@TrungProc 3.2] it follows that $${\operatorname{red}}_{J^{[3]}}(I^3) = 2  \quad \text{if }  \ a_3(I) < 0   \quad \text{and} \quad 
{\operatorname{red}}_{J^{[3]}}(I^3) = 3  \quad \text{if }  \ a_3(I) \geq 0.$$ However using COCOA we have verified that $I^9 \neq I^6J^{[3]}$. Thus ${\mathbf{a} }(I) = a_3(I) \geq 0$. The fiber coefficients are $ f_0(I) = 17, f_1(I) = 34, f_2(I) = 17.$ So $$f_{2}(I)- f_{1}(I) + f_{0}(I) - 1 = -1$$

Results when analytic spread is high 
=====================================

In this section we extend our results Theorem \[result1\] and Theorem \[sectheorem\] to the case when $l(I) \geq 3 $ and $l(I) \geq 4 $ respectively. The main tool is the use of Rees-superficial sequences (see \[def\]). The utility of a Rees-superficial element in the study of fiber cone was first demonstrated in [@jpv].

\[def\] An element $x \in I$ is said to be *Rees superficial* if there exists $r_0 \geq 1$ such that $$(x) \cap I^r{\mathfrak{m} }^s = xI^{r-1}{\mathfrak{m} }^s \quad \text{for all} \ r \geq r_0 \ \& \ s \geq 0.$$

The following was proved in [@jpv 2.8] for ${\mathfrak{m} }$-primary $I$ in a local ring $A$. The same proof works in general.

\[diff\] Let $(A,{\mathfrak{m} })$ be a local ring and $I$ an ideal. Let $x \in I$ be a nonzero divisor in $R$ which is also Rees-superficial for $I.$ Set $(B,{\mathfrak{n} }) = (A/(x), {\mathfrak{m} }/(x))$ and $K = I/(x)$. Then $$\ell\left(\frac{I^n}{{\mathfrak{m} }I^n} \right) -  \ell\left(\frac{I^{n-1}}{{\mathfrak{m} }I^{n-1}} \right) = \ell\left(\frac{K^n}{{\mathfrak{n} }K^n} \right)  \quad \text{for all} \ n\gg 0.$$ In particular $f_i(K) = f_i(I)$ for $i = 0,\ldots, l(I)-2$.

The existence of a Rees-superficial element which is also regular follows from the the following special case of a Lemma (see [@re 1.2]) due to Rees.

\[reeslemma\] Let $(A,{\mathfrak{m} })$ be local and let $I$ be an ideal in $A$. Let $\mathcal{P}$ be a finite set of primes not containing $I{\mathfrak{m} }$. Then there exists $x \in I$ and $r_0 \geq 1$ such that

1.  $x \notin \mathfrak{P}$ for all $\mathfrak{P} \in \mathcal{P}$

2.  $\displaystyle{(x) \cap I^r{\mathfrak{m} }^s = xI^{r-1}{\mathfrak{m} }^s \quad \text{for all} \ r \geq r_0 \ \& \ s \geq 0.}$

\[ReCsup\] If $x \in I$ is Rees-superficial and a non-zero divisor then it is easy to check that $x$ is $I$-superficial.

We say $x_1,\ldots,x_r \in I$ is a *Rees-superficial* sequence if $x_i$ is Rees superficial for the $A/(x_1,\ldots,x_{i-1})$-ideal $I/(x_1,\ldots,x_{i-1})$ for $i = 1,\ldots, r$.

\[rseq\] **a**. If ${\operatorname{grade}}(I) \geq r$ then using \[reeslemma\] we can show that there exists a Rees-superficial sequence $x_1,\ldots,x_r$ in $I$ which is also a regular sequence.

**b.** In this case we can further prove (by using Proposition \[diff\] repeatedly) that if $K = I/(x_1,\ldots,x_r)$ then $$\label{seq}
f_i(K) = f_i(I) \quad \text{for} \ i =  0,\ldots, l(I)-r -1.$$

We state our main results.

\[mth1\] Let $(A,{\mathfrak{m} })$ be local and let $I$ be an ideal in $A$ with $l = l(I) \geq 2$ and ${\operatorname{grade}}(I) \geq  l(I)-2$. Assume either ${\operatorname{reg}}^2(G(I)) \leq 1$ or ${\operatorname{red}}(I) \leq 1$. Then $f_1(I) \leq f_0(I) -1$. Furthermore equality holds if ${\operatorname{grade}}(I) =  l(I)$.

If $l(I) = 2$ then we are done by Theorem \[result1\]. If $l \geq 3$ and as ${\operatorname{grade}}(I) \geq  l(I)-2$ by \[rseq\].a we can choose a Rees-superficial sequence $x_1,\ldots,x_{l-2}$ which is also a regular sequence (and so an $I$-superficial sequence). Set $K = I/(x_1,\ldots,x_{l-2})$. Then $l(K) = 2$.

If ${\operatorname{reg}}^2(G(I)) \leq 1$ then by \[notation\] we get ${\mathbf{a} }(K) + 2 = {\operatorname{reg}}^2(G(K)) \leq 1$. So ${\mathbf{a} }(K) < 0$. If ${\operatorname{red}}(I) \leq 1$ then ${\operatorname{red}}(K) \leq 1$. So by [@TrungProc 3.2] we get ${\mathbf{a} }(K) < 0$. Thus at any rate ${\mathbf{a} }(K) < 0$.

By Theorem \[result1\] we get $f_1(K) \leq f_0(K) -1$. The result follows since by \[rseq\].b we have $f_i(I) = f_i(K)$ for $i = 0,1$.

\[NarC\] Let $(A,{\mathfrak{m} })$ be [Cohen-Macaulay]{} local ring of dimension $d \geq 2$. If $I$ is ${\mathfrak{m} }$-primary and $e_{2}(I) = 0$ then $f_1(I) = f_0(I) -1$.

First assume that $d = 2$. By Narita’s result [@Nar] we get that if $J$ is any reduction of $I$ then ${\operatorname{red}}_{J^{[n]}}(I^n) = 1$ for all $n \gg 0$. Also as ${\operatorname{grade}}(I) = 2$ by Theorem \[result1\] we get $f_1(I) =  f_0(I) -1$.

When $d \geq 3$ we choose a Rees-superficial sequence $x_1,\ldots,x_{d-2}$ which is also an $A$-regular sequence (and so an $I$-superficial sequence). Set $K = 
I/(x_1,\ldots,x_{d-2})$. Note that $e_2(K) =  e_2(I) = 0$. Also $f_i(I) = f_i(K)$ for $i = 0,1$.

Next we give an application of Theorem \[as=2\].

\[mth2\] Let $(A,{\mathfrak{m} })$ be [Cohen-Macaulay]{} local ring of dimension $d \geq 3$ and let $I$ be an ${\mathfrak{m} }$-primary ideal. If $G(I)$ is [Cohen-Macaulay]{} and ${\operatorname{red}}(I) \leq 2$ then $f_2(I) \geq f_1(I) -f_0(I) +1$.

When $d = 3$ then the result follows from Theorem \[as=2\]. If $d \geq 4$ we choose a Rees-superficial sequence $x_1,\ldots,x_{d-3}$ which is also an $A$-regular sequence (and so an $I$-superficial sequence). Set $K = I/(x_1,\ldots,x_{d-3})$. Note that $G(K)$ is [Cohen-Macaulay]{} and reduction number of $K$ is $\leq 2$. By Theorem \[as=2\] we have $f_2(K) \geq f_1(K) -f_0(K) +1$. Also as $f_i(I) = f_i(K)$ for $i = 0,1,2$ (\[rseq\].b) we get the result.

appendix: Minimal reductions and Filter-regular elements
========================================================

In this section we prove that if $x^\circ  \in F(I)_1$ is $F(I)$ filter-regular then $(x^n)^\circ  \in F(I^n)_1$ is $F(I^n)$ filter-regular. This is used in proof of Theorem \[veronese-th\].

The relation between minimal reductions and Filter-regular sequences first appeared in the work of Trung [@TrungProc]. We state one of his results ([@TrungProc 3.8]) in the form we need it.

Let $I$ be an ideal with $s = l(I) > 0$. Then there exists a minimal reduction $J = (x_1,\ldots, x_s)$ of $I$ such that $x_1^\circ, \ldots, x_s^\circ \in F(I)_1$ is a $F(I)$-filter regular sequence.

For definition of filter-regular sequence see [@BH 18.3.7]. We however are only interested in a filter-regular element.

\[filter-defn\] Recall that the following assertions are equivalent

-   $x^\circ \in F(I)_1$ is $F(I)$-filter regular

-   $(0 \colon_{F(I)} \ x^\circ)_n = 0$ for all $n \gg 0$.

-   $x^\circ$ is $F(I)/H^{0}_{ F(I)_{+}}(F(I))$ regular.

For proof of the above equivalence see [@TrungProc 2.1]. It is perhaps better to see Exercise [@BSh 18.3.8].

\[filtISreg\] By \[filter-defn\] we get that if ${\operatorname{depth}}F(I) > 0$ and $x^\circ$ is $F(I)$ filter-regular then $x^\circ$ is $F(I)$ regular.

We give an ideal-theoretic criterion for an element $x^\circ \in F(I)_1$ to be $F(I)$-filter regular.

\[filt-ideal-th\] Let $I$ be an ideal with $s = l(I) > 0$ and let $x \in I \setminus {\mathfrak{m} }I$. [The following conditions are equivalent:]{}

1.  $x^\circ $ is $F(I)$ filter-regular.

2.  $({\mathfrak{m} }I^{j+1} \colon x )\cap I^j = {\mathfrak{m} }I^j$ for all $j \gg 0$.

\(i) $\implies$ (ii) We assume $(0 \colon x^\circ)_n = 0$ for all $n \geq c$. Clearly ${\mathfrak{m} }I^j \subseteq ({\mathfrak{m} }I^{j+1} \colon x )\cap I^j $ for all $j$. If $a \in I^j \setminus {\mathfrak{m} }I^j$ and $xa \in {\mathfrak{m} }I^{j+1} $ then we have $x^\circ \bullet a^\circ = 0$. It follows that $j < c$.

\(ii) $\implies$ (i) Coversely assume $({\mathfrak{m} }I^{j+1} \colon x )\cap I^j = {\mathfrak{m} }I^j$ for all $j \geq c$. Say $a^\circ \in F(I)_j$ is non-zero and $x^\circ \bullet a^\circ = 0$. Then $a \in ({\mathfrak{m} }I^{j+1} \colon x )\cap I^j$. By hypothesis. it follows that $j < c$. So $x^\circ $ is $F(I)$ filter-regular.

\[filt-ideal-cor\]\[with hypothesis as in \[filt-ideal-th\]\] If $x^\circ  \in F(I)_1$ is $F(I)$ filter-regular then $(x^n)^\circ  \in F(I^n)_1$ is $F(I^n)$ filter-regular.

Since $x^\circ$ is $F(I)$ filter-regular, by \[filt-ideal-th\], there exists $c > 0$ such that $$({\mathfrak{m} }I^{j+1} \colon x )\cap I^j = {\mathfrak{m} }I^j \quad \text{ for all  }  \  j \geq  c.$$ So for $j \geq c $ we have $ ({\mathfrak{m} }I^{j+n} \colon x^n )\cap I^j = {\mathfrak{m} }I^j$. Therefore for $j \geq c $ we obtain $$({\mathfrak{m} }I^{n(j+1)} \colon x^n )\cap I^{nj} = ({\mathfrak{m} }I^{(nj+n)} \colon x^n )\cap I^{nj} = {\mathfrak{m} }I^{nj}$$ Thus by \[filt-ideal-th\] we get that $(x^n)^\circ$ is $F(I^n)$ filter-regular.

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[^1]: The second author was partly supported by IIT Bombay seedgrant 03ir053
