1 Introduction and notation
Throughout this paper, let
$A$
denote a Cohen–Macaulay local ring with maximal ideal
$\mathfrak{m}$
and positive dimension
$d.$
Let
$I$
be an
$\mathfrak{m}$
-primary ideal in
$A$
and, for simplicity, we assume the residue class field
$A/\mathfrak{m}$
is infinite. Let
$\ell _{A}(N)$
denote, for an
$A$
-module
$N$
, the length of
$N$
. The integers
$\{\text{e}_{i}(I)\}_{0\leqslant i\leqslant d}$
such that the equality
$$\begin{eqnarray}\displaystyle \ell _{A}(A/I^{n+1})=\text{e}_{0}(I)\binom{n+d}{d}-\text{e}_{1}(I)\binom{n+d-1}{d-1}+\cdots +(-1)^{d}\text{e}_{d}(I) & & \displaystyle \nonumber\end{eqnarray}$$
holds true for all integers
$n\gg 0$
, are called the Hilbert coefficients of
$A$
with respect to
$I$
. This polynomial, known as the Hilbert–Samuel polynomial of
$I$
and denoted by
$HP_{I}(n),$
encodes the asymptotic information coming from the Hilbert function
$H_{I}(t)$
of
$I$
which is defined as
$$\begin{eqnarray}H_{I}(t)=\ell _{A}(I^{t}/I^{t+1}).\end{eqnarray}$$
The generating function of the numerical function
$H_{I}(t)$
is the power series
$$\begin{eqnarray}~~HS_{I}(z)=\mathop{\sum }_{t\geqslant 0}H_{I}(t)z^{t}.\end{eqnarray}$$
This series is called the Hilbert series of
$I.$
It is well known that this series is rational and that, even more, there exists a polynomial
$h_{I}(z)$
with integer coefficients such that
$h_{I}(1)\not =0$
and
$$\begin{eqnarray}HS_{I}(z)=\frac{h_{I}(z)}{(1-z)^{d}}.\end{eqnarray}$$
Notice that for all
$i\geqslant 0,$
the Hilbert coefficients can be computed as it follows
$$\begin{eqnarray}\text{e}_{i}(I):=\frac{h_{I}^{(i)}(1)}{i!}\end{eqnarray}$$
where
$h_{I}^{(i)}(1)$
denotes the
$i$
th derivative of
$h_{I}(z)$
evaluated at
$1$
(
$h^{(0)}=h_{I}$
).
Choose a parameter ideal
$Q$
of
$A$
which forms a reduction of
$I$
and let
$$\begin{eqnarray}R=\text{R}(I):=A[It]\qquad \operatorname{and}\qquad T=\text{R}(Q):=A[Qt]\subseteq A[t]\end{eqnarray}$$
denote, respectively, the Rees algebras of
$I$
and
$Q$
. Let
$$\begin{eqnarray}\displaystyle & & \displaystyle R^{\prime }=\text{R}^{\prime }(I):=A[It,t^{-1}]\subseteq A[t,t^{-1}]\quad \operatorname{and}\quad \nonumber\\ \displaystyle & & \displaystyle G=\text{G}(I):=R^{\prime }/t^{-1}R^{\prime }\cong \bigoplus _{n\geqslant 0}I^{n}/I^{n+1}.\nonumber\end{eqnarray}$$
Following Vasconcelos [Reference VasconcelosV], we consider
$$\begin{eqnarray}S=\text{S}_{Q}(I)=IR/IT\cong \bigoplus _{n\geqslant 1}I^{n+1}/Q^{n}I\end{eqnarray}$$
the Sally module of
$I$
with respect to
$Q$
.
The notion of filtration of the Sally module was introduced by Vaz Pinto [Reference Vaz PintoVP] as follows. We denote by
$E(\unicode[STIX]{x1D6FC})$
, for a graded
$T$
-module
$E$
and each
$\unicode[STIX]{x1D6FC}\in \mathbb{Z}$
, the graded
$T$
-module whose grading is given by
$[E(\unicode[STIX]{x1D6FC})]_{n}=E_{\unicode[STIX]{x1D6FC}+n}$
for all
$n\in \mathbb{Z}$
.
Definition 1.1. [Reference Vaz PintoVP]
We set, for each
$i\geqslant 1$
,
$$\begin{eqnarray}C^{(i)}=(I^{i}R/I^{i}T)(-i+1)\cong \bigoplus _{n\geqslant i}I^{n+1}/Q^{n-i+1}I^{i},\end{eqnarray}$$
and let
$L^{(i)}=T[C^{(i)}]_{i}$
. Then, because
$L^{(i)}\cong \bigoplus _{n\geqslant i}Q^{n-i}I^{i+1}/Q^{n-i+1}I^{i}$
and
$C^{(i)}/L^{(i)}\cong C^{(i+1)}$
as graded
$T$
-modules, we have the following natural exact sequences of graded
$T$
-modules
$$\begin{eqnarray}0\rightarrow L^{(i)}\rightarrow C^{(i)}\rightarrow C^{(i+1)}\rightarrow 0\end{eqnarray}$$
for every
$i\geqslant 1.$
We notice that
$C^{(1)}=S$
, and
$C^{(i)}$
are finitely generated graded
$T$
-modules for all
$i\geqslant 1$
, since
$R$
is a module-finite extension of the graded ring
$T$
.
So, from now on, we set
$$\begin{eqnarray}C=\text{C}_{Q}(I)=C^{(2)}=(I^{2}R/I^{2}T)(-1)\end{eqnarray}$$
and we shall explore the structure of
$C.$
Assume that
$I$
is integrally closed. Then, by [Reference Elias and VallaEV, Reference Guerrieri and RossiGR], the inequality
$$\begin{eqnarray}\text{e}_{1}(I)\geqslant \text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)\end{eqnarray}$$
holds true and the equality
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)$
holds if and only if
$I^{3}=QI^{2}$
. When this is the case, the associated graded ring
$G$
of
$I$
is Cohen–Macaulay and the behavior of the Hilbert–Samuel function
$\ell _{A}(A/I^{n+1})$
of
$I$
is known (see [Reference Elias and VallaEV], Corollary 2.10). Thus the integrally closed ideal
$I$
with
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)$
enjoys nice properties and it seems natural to ask what happens to the integrally closed ideal
$I$
which satisfies the equality
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
. The problem is not trivial even if we consider
$d=1.$
We notice here that
$\ell _{A}(I^{2}/QI)=\text{e}_{0}(I)+(d-1)\ell _{A}(A/I)-\ell _{A}(I/I^{2})$
holds true (see for instance [Reference Rossi and VallaRV3]), so that
$\ell _{A}(I^{2}/QI)$
does not depend on a minimal reduction
$Q$
of
$I$
.
Let
$B=T/\mathfrak{m}T\cong (A/\mathfrak{m})[X_{1},X_{2},\ldots ,X_{d}]$
which is a polynomial ring with
$d$
indeterminates over the field
$A/\mathfrak{m}$
. The main result of this paper is stated as follows.
Theorem 1.2. Assume that
$I$
is integrally closed. Then the following conditions are equivalent:
-
(1)
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
, -
(2)
$\mathfrak{m}C=(0)$
and
$\text{rank}_{B}C=1$
, -
(3)
$C\cong (X_{1},X_{2},\ldots ,X_{c})B(-1)$
as graded
$T$
-modules for some
$1\leqslant c\leqslant d$
, where
$X_{1},X_{2},\ldots ,X_{d}$
are linearly independent linear forms of the polynomial ring
$B$
.
When this is the case,
$c=\ell _{A}(I^{3}/QI^{2})$
and
$I^{4}=QI^{3}$
, and the following assertions hold true:
-
(i)
$\text{depth}~G\geqslant d-c$
and
$\text{depth}_{T}C=d-c+1$
, -
(ii)
$\text{depth}~G=d-c$
, if
$c\geqslant 2$
. -
(iii) Suppose
$c=1<d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 0$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)+1 & \quad \text{if }i=2,\\ 1 & \quad \text{if }i=3\text{ and }d\geqslant 3,\\ 0 & \quad \text{if }4\leqslant i\leqslant d.\end{array}\right.\end{eqnarray}$$
-
(iv) Suppose
$2\leqslant c<d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 0$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I) & \quad \text{if }i=2,\\ 0 & \quad \text{if }i\neq c+1,c+2,\quad 3\leqslant i\leqslant d\\ (-1)^{c+1} & \quad \text{if }i=c+1,c+2,\quad 3\leqslant i\leqslant d\end{array}\right.\end{eqnarray}$$
-
(v) Suppose
$c=d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 2$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I) & \quad \text{if }i=2\text{ and }d\geqslant 2,\\ 0 & \quad \text{if }3\leqslant i\leqslant d\end{array}\right.\end{eqnarray}$$
-
(vi) The Hilbert series
$HS_{I}(z)$
is given by
$$\begin{eqnarray}HS_{I}(z)=\frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-\ell _{A}(I^{2}/QI)-1\}z+\{\ell _{A}(I^{2}/QI)+1\}z^{2}+(1-z)^{c+1}z}{(1-z)^{d}}.\end{eqnarray}$$
Let us briefly explain how this paper is organized. We shall prove Theorem 1.2 in Section 3. In Section 2 we will introduce some auxiliary results on the structure of the
$T$
-module
$C=\text{C}_{Q}(I)=(I^{2}R/I^{2}T)(-1)$
, some of them are stated in a general setting. Our hope is that this information will be successfully applied to give new insight to problems related to the structure of the Sally module. In Section 4 we will introduce some consequences of Theorem 1.2. In particular, we shall explore the integrally closed ideals
$I$
with
$\text{e}_{1}(I)\leqslant \text{e}_{0}(I)-\ell _{A}(A/I)+3.$
In Section 5 we will construct a class of Cohen–Macaulay local rings satisfying condition (1) in Theorem 1.2.
2 Preliminary steps
The purpose of this section is to summarize some results on the structure of the graded
$T$
-module
$C=\text{C}_{Q}(I)=(I^{2}R/I^{2}T)(-1)$
, which we need throughout this paper. Remark that in this section
$I$
is an
$\mathfrak{m}$
-primary ideal not necessarily integrally closed.
Lemma 2.1. The following assertions hold true.
-
(1)
$\mathfrak{m}^{\ell }C=(0)$
for integers
$\ell \gg 0$
; hence
$\dim _{T}C\leqslant d$
. -
(2) The homogeneous components
$\{C_{n}\}_{n\in \mathbb{Z}}$
of the graded
$T$
-module
$C$
are given by
$$\begin{eqnarray}C_{n}\cong \left\{\begin{array}{@{}rl@{}}(0) & \quad \text{if }n\leqslant 1,\\ I^{n+1}/Q^{n-1}I^{2} & \quad \text{if }n\geqslant 2.\end{array}\right.\end{eqnarray}$$
-
(3)
$C=(0)$
if and only if
$I^{3}=QI^{2}$
. -
(4)
$\mathfrak{m}C=(0)$
if and only if
$\mathfrak{m}I^{n+1}\subseteq Q^{n-1}I^{2}$
for all
$n\geqslant 2$
. -
(5)
$C=TC_{2}$
if and only if
$I^{4}=QI^{3}$
.
Proof. (1) Let
$u=t^{-1}$
and
$T^{\prime }=\text{R}^{\prime }(Q)$
. Notice that we have
$C=(I^{2}R/I^{2}T)(-1)\cong (I^{2}R^{\prime }/I^{2}T^{\prime })(-1)$
as graded
$T$
-modules. We then have
$u^{\ell }\cdot (I^{2}R^{\prime }/I^{2}T^{\prime })=(0)$
for some
$\ell \gg 0$
, because the graded
$T^{\prime }$
-module
$I^{2}R^{\prime }/I^{2}T^{\prime }$
is finitely generated and
$[I^{2}R^{\prime }/I^{2}T^{\prime }]_{n}=(0)$
for all
$n\leqslant 0$
. Therefore,
$\mathfrak{m}^{\ell }C=(0)$
for
$\ell \gg 0$
, because
$Q^{\ell }=(Qt^{\ell })u^{\ell }\subseteq u^{\ell }T^{\prime }\cap A$
and
$\mathfrak{m}=\sqrt{Q}$
.
(2) Since
$[I^{2}R]_{n}=(I^{n+2})t^{n}$
and
$[I^{2}T]_{n}=(I^{2}Q^{n})t^{n}$
for all
$n\geqslant 0$
, assertion (2) follows from the definition of the module
$C=(I^{2}R/I^{2}T)(-1)$
.
Assertions (3), (4), and (5) readily follow from assertion (2). ◻
In the following result we need that
$Q\cap I^{2}=QI$
holds true. This condition is automatically satisfied in the case where
$I$
is integrally closed (see [Reference HunekeH, Reference ItohI2]).
Proposition 2.2. Suppose that
$Q\cap I^{2}=QI$
. Then we have
$\text{Ass}_{T}C\subseteq \{\mathfrak{m}T\}$
so that
$\dim _{T}C=d$
, if
$C\neq (0)$
.
Let
$Q=(a_{1},\ldots ,a_{d})$
be a minimal reduction of
$I.$
In the proof of Proposition 2.2 we need the following lemmata.
Lemma 2.3. Suppose that
$Q\cap I^{2}=QI$
. Then we have the equality
$(a_{1},a_{2},\ldots ,a_{i})\cap Q^{n+1}I^{2}=(a_{1},a_{2},\ldots ,a_{i})Q^{n}I^{2}$
for all
$n\geqslant 0$
and
$1\leqslant i\leqslant d$
. Therefore,
$a_{1}t\in T$
is a nonzero divisor on
$T/I^{2}T$
.
Proof. We have only to show that
$(a_{1},a_{2},\ldots ,a_{i})\cap Q^{n+1}I^{2}\subseteq (a_{1},a_{2},\ldots ,a_{i})Q^{n}I^{2}$
holds true for all
$n\geqslant 0$
. We proceed by induction on
$n$
and
$i$
. We may assume that
$i<d$
and that our assertion holds true for
$i+1$
. Suppose
$n=0$
and take
$x\in (a_{1},a_{2},\ldots ,a_{i})\cap QI^{2}$
. Then, by the hypothesis of induction on
$i$
, we have
$$\begin{eqnarray}\displaystyle (a_{1},a_{2},\ldots ,a_{i})\cap QI^{2} & \subseteq & \displaystyle (a_{1},a_{2},\ldots ,a_{i},a_{i+1})\cap QI^{2}\nonumber\\ \displaystyle & = & \displaystyle (a_{1},a_{2},\ldots ,a_{i},a_{i+1})I^{2}.\nonumber\end{eqnarray}$$
Then we write
$x=y+a_{i+1}z$
with
$y\in (a_{1},a_{2},\ldots ,a_{i})I^{2}$
and
$z\in I^{2}$
. Therefore,
$z\in [(a_{1},a_{2},\ldots ,a_{i}):a_{i+1}]\cap I^{2}=(a_{1},a_{2},\ldots ,a_{i})\cap I^{2}\subseteq (a_{1},a_{2},\ldots ,a_{i})\cap QI$
. Notice that, since
$Q/QI\cong (A/I)^{d}$
,
$(a_{1},a_{2},\ldots ,a_{i})\cap QI=(a_{1},a_{2},\ldots ,a_{i})I$
holds true. Hence, we get
$x=y+a_{i+1}z\in (a_{1},a_{2},\ldots ,a_{i})I^{2}$
. Therefore, we have
$(a_{1},a_{2},\ldots ,a_{i})\cap QI^{2}=(a_{1},a_{2},\ldots ,a_{i})I^{2}.$
Assume that
$n\geqslant 1$
and that our assertion holds true for
$n-1$
. Take
$x\in (a_{1},a_{2},\ldots ,a_{i})\cap Q^{n+1}I^{2}$
. Then, by the hypothesis of induction on
$i$
, we have
$$\begin{eqnarray}\displaystyle (a_{1},a_{2},\ldots ,a_{i})\cap Q^{n+1}I^{2} & \subseteq & \displaystyle (a_{1},a_{2},\ldots ,a_{i},a_{i+1})\cap Q^{n+1}I^{2}\nonumber\\ \displaystyle & = & \displaystyle (a_{1},a_{2},\ldots ,a_{i},a_{i+1})Q^{n}I^{2}.\nonumber\end{eqnarray}$$
Write
$x=y+a_{i+1}z$
with
$y\in (a_{1},a_{2},\ldots ,a_{i})Q^{n}I^{2}$
and
$z\in Q^{n}I^{2}$
. By the hypothesis of induction on
$n$
, we have
$z\in [(a_{1},a_{2},\ldots ,a_{i}):a_{i+1}]\cap Q^{n}I^{2}=(a_{1},a_{2},\ldots ,a_{i})\cap Q^{n}I^{2}=(a_{1},a_{2},\ldots ,a_{i})Q^{n-1}I^{2}$
. Therefore, we get
$x=y+a_{i+1}z\in (a_{1},a_{2},\ldots ,a_{i})Q^{n}I^{2}$
. Thus,
$(a_{1},a_{2},\ldots ,a_{i})\cap Q^{n+1}I^{2}=(a_{1},a_{2},\ldots ,a_{i})Q^{n}I^{2}$
as required.◻
Lemma 2.4. Suppose that
$Q\cap I^{2}=QI$
. Then
$Q^{n+1}\cap Q^{n}I^{2}=Q^{n+1}I$
for all
$n\geqslant 0$
.
Proof. We have only to show that
$Q^{n+1}\cap Q^{n}I^{2}\subseteq Q^{n+1}I$
for
$n\geqslant 0$
. Take
$f\in Q^{n+1}\cap Q^{n}I^{2}$
and write
$f=\sum _{|\unicode[STIX]{x1D6FC}|=n}x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}$
with
$x_{\unicode[STIX]{x1D6FC}}\in I^{2}$
, where
$\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{d})\in \mathbb{Z}^{d}$
and
$|\unicode[STIX]{x1D6FC}|=\sum _{i=1}^{d}\unicode[STIX]{x1D6FC}_{i}$
. Then, we have
$\overline{ft^{n}}=\sum _{|\unicode[STIX]{x1D6FC}|=n}\overline{x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{n}}=0$
in
$\text{G}(Q)=\text{R}^{\prime }(Q)/t^{-1}\text{R}^{\prime }(Q)$
, where
$\overline{ft^{n}}$
and
$\overline{x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{n}}=0$
denote the images of
$ft^{n}$
and
$x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{n}$
in
$\text{G}(Q)$
, respectively. Because
$\text{G}(Q)\cong (A/Q)[\overline{a_{1}t},\overline{a_{2}t},\ldots ,\overline{a_{d}t}]$
is the polynomial ring over the ring
$A/Q$
, we have
$x_{\unicode[STIX]{x1D6FC}}\in Q\cap I^{2}=QI$
. Thus
$f\in Q^{n+1}I$
so that
$Q^{n+1}\cap Q^{n}I^{2}=Q^{n+1}I$
holds true.◻
Lemma 2.5. Suppose that
$Q\cap I^{2}=QI$
. Then the following sequences
-
(1)
$0\rightarrow T/IT\overset{a_{1}}{\rightarrow }T/I^{2}T\rightarrow T/[I^{2}T+a_{1}T]\rightarrow 0$
, and -
(2)
$0\rightarrow T/[I^{2}T+QT](-1)\overset{a_{1}t}{\rightarrow }T/[I^{2}T+a_{1}T]\rightarrow T/[I^{2}T+\,a_{1}tT+a_{1}T]\rightarrow \,0$
of graded
$T$
-modules are exact.
Proof. (1) Let us consider the homomorphism
$$\begin{eqnarray}\unicode[STIX]{x1D719}:T\rightarrow T/I^{2}T\end{eqnarray}$$
of graded
$T$
-modules such that
$\unicode[STIX]{x1D719}(f)=\overline{a_{1}f}$
for
$f\in T$
where
$\overline{a_{1}f}$
denotes the image of
$a_{1}f$
in
$T/I^{2}T$
. Because
$\unicode[STIX]{x1D719}(IT)=(0)$
and
$\text{Coker}\unicode[STIX]{x1D719}=T/[I^{2}T+a_{1}T]$
, we have only to show that
$\ker \unicode[STIX]{x1D719}\subseteq IT$
. Take
$f\in [\ker \unicode[STIX]{x1D719}]_{n}$
and write
$f=xt^{n}$
with
$n\geqslant 0$
and
$x\in Q^{n}$
. Then we have
$a_{1}f=a_{1}xt^{n}\in I^{2}T$
so that
$a_{1}x\in Q^{n}I^{2}\cap Q^{n+1}=Q^{n+1}I$
by Lemma 2.4. Because
$a_{1}t\in T$
forms a nonzero divisor on
$T/IT$
(notice that
$T/IT\cong (A/I)[\overline{a_{1}t},\overline{a_{2}t},\ldots ,\overline{a_{d}t}]$
is a polynomial ring over the ring
$A/I$
), we have
$(a_{1})\cap Q^{n+1}I=a_{1}Q^{n}I$
so that
$x\in Q^{n}I$
. Therefore,
$f\in IT$
, and hence
$\ker \unicode[STIX]{x1D719}\subseteq IT$
. Thus, we get the first required exact sequence.
(2) Let us consider the homomorphism
$$\begin{eqnarray}\unicode[STIX]{x1D711}:T(-1)\rightarrow T/[I^{2}T+a_{1}T]\end{eqnarray}$$
of graded
$T$
-modules such that
$\unicode[STIX]{x1D711}(f)=\overline{a_{1}tf}$
for
$f\in T$
where
$\overline{a_{1}tf}$
denotes the image of
$a_{1}tf$
in
$T/[I^{2}T+a_{1}T]$
. Because
$\unicode[STIX]{x1D711}(I^{2}T+QT)=(0)$
and
$\text{Coker}\unicode[STIX]{x1D719}=T/[I^{2}T+a_{1}tT+a_{1}T]$
, we need to show that
$[\ker \unicode[STIX]{x1D711}]_{n}\subseteq I^{2}T_{n-1}+QT_{n-1}$
for all
$n\geqslant 1$
. Take
$f\in [\ker \unicode[STIX]{x1D711}]_{n}$
and write
$f=xt^{n-1}$
with
$n\geqslant 1$
and
$x\in Q^{n-1}$
. Then we have
$a_{1}tf=a_{1}xt^{n}\in I^{2}T+a_{1}T$
so that
$a_{1}x\in Q^{n}I^{2}+a_{1}Q^{n}$
. Write
$a_{1}x=y+a_{1}z$
with
$y\in Q^{n}I^{2}$
and
$z\in Q^{n}$
. Then have
$$\begin{eqnarray}a_{1}(x-z)=y\in (a_{1})\cap Q^{n}I^{2}=a_{1}Q^{n-1}I^{2}\end{eqnarray}$$
by Lemma 2.3. Hence
$x-z\in Q^{n-1}I^{2}$
so that
$x\in Q^{n-1}I^{2}+Q^{n}$
. Therefore,
$f\in I^{2}T_{n-1}+QT_{n-1}$
and hence
$[\ker \unicode[STIX]{x1D711}]_{n}\subseteq I^{2}T_{n-1}+QT_{n-1}$
. Consequently, we get the second required exact sequence.◻
The following Lemma 2.6 is the crucial fact in the proof of Proposition 2.2.
Lemma 2.6. Assume that
$Q\cap I^{2}=QI$
. Then we have
$\text{Ass}_{T}(T/I^{2}T)=\{\mathfrak{m}T\}$
.
Proof. Take
$P\in \text{Ass}_{T}(T/I^{2}T)$
; then we have
$\mathfrak{m}T\subseteq P$
, because
$\mathfrak{m}^{\ell }(T/I^{2}T)=(0)$
for
$\ell \gg 0$
. Assume that
$\mathfrak{m}T\subsetneq P$
; then
$\text{ht}_{T}P\geqslant 2$
, because
$\mathfrak{m}T$
is a height one prime ideal in
$T$
(notice that
$\dim T=d+1$
). Consider the following exact sequences
$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle T_{P}/IT_{P}\rightarrow T_{P}/I^{2}T_{P}\rightarrow T_{P}/[I^{2}T_{P}+a_{1}T_{P}]\rightarrow 0\quad (\ast _{1})\quad \text{and}\nonumber\\ \displaystyle 0 & \rightarrow & \displaystyle T_{P}/[I^{2}T_{P}+QT_{P}]\rightarrow T_{P}/[I^{2}T_{P}+a_{1}T_{P}]\rightarrow T_{P}/[I^{2}T_{P}+a_{1}tT_{P}+a_{1}T_{P}]\rightarrow 0\quad (\ast _{2})\nonumber\end{eqnarray}$$
of
$T_{P}$
-modules, which follow from the exact sequences in Lemma 2.5. Then, since
$\text{depth}_{T_{P}}T_{P}/IT_{P}>0$
(notice that
$T/IT\cong (A/I)[X_{1},X_{2},\ldots ,X_{d}]$
is the polynomial ring with
$d$
indeterminates over the ring
$A/I$
) and
$\text{depth}_{T_{P}}T_{P}/I^{2}T_{P}=0$
, we have
$\text{depth}_{T_{P}}T_{P}/[I^{2}T_{P}+a_{1}T_{P}]=0$
by the exact sequence
$(\ast _{1})$
. Notice that
$T/[I^{2}T+QT]\cong (A/[I^{2}+Q])[X_{1},X_{2},\ldots ,X_{d}]$
is the polynomial rings with
$d$
indeterminates over the ring
$A/[I^{2}+Q]$
. Hence,
$\text{depth}_{T_{P}}T_{P}/[I^{2}T_{P}+a_{1}tT_{P}+a_{1}T_{P}]=0$
by the exact sequence
$(\ast _{2})$
. Then, because
$a_{1}t\in \text{Ann}_{T}(T/[I^{2}T+a_{1}tT+a_{1}T])$
, we have
$a_{1}t\in P$
. Therefore,
$\text{depth}_{T_{P}}T_{P}/I^{2}T_{P}>0$
by Lemma 2.3; however, it is impossible. Thus,
$P=\mathfrak{m}T$
as required.◻
Let us now give a proof of Proposition 2.2.
Proof of Proposition 2.2.
Take
$P\in \text{Ass}_{T}C$
. Then we have
$\mathfrak{m}T\subseteq P$
, because
$\mathfrak{m}^{\ell }C=(0)$
for some
$\ell \gg 0$
by Lemma 2.1(1). Suppose that
$\mathfrak{m}T\subsetneq P$
; then
$\text{ht}_{T}P\geqslant 2$
, because
$\mathfrak{m}T$
is a height one prime ideal in
$T$
. We look at the following exact sequences
$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow I^{2}T_{P}\rightarrow I^{2}R_{P}\rightarrow C_{P}\rightarrow 0\quad (\ast _{3})\quad \text{and}\quad \nonumber\\ \displaystyle & & \displaystyle 0\rightarrow I^{2}T_{P}\rightarrow T_{P}\rightarrow T_{P}/I^{2}T_{P}\rightarrow 0\quad (\ast _{4})\nonumber\end{eqnarray}$$
of
$T_{P}$
-modules which follow from the canonical exact sequences
$$\begin{eqnarray}0\rightarrow I^{2}T(-1)\rightarrow I^{2}R(-1)\rightarrow C\rightarrow 0\quad \text{and}\quad 0\rightarrow I^{2}T\rightarrow T\rightarrow T/I^{2}T\rightarrow 0\end{eqnarray}$$
of
$T$
-modules. We notice here that
$\text{depth}_{T_{P}}I^{2}R_{P}>0$
, because
$a_{1}\in A$
is a nonzero divisor on
$I^{2}R$
. Thanks to the depth lemma and the exact sequence
$(\ast _{3})$
, we get
$\text{depth}_{T_{P}}I^{2}T_{P}=1$
, because
$\text{depth}_{T_{P}}I^{2}R_{P}>0$
and
$\text{depth}_{T_{P}}C_{P}=0$
. Then, since
$\text{depth}_{T_{P}}T_{P}\geqslant 2$
, we have
$\text{depth}_{T_{P}}T_{P}/I^{2}T_{P}=0$
by the exact sequence
$(\ast _{4})$
. Therefore, we have
$P=\mathfrak{m}T$
by Lemma 2.6, which is impossible. Thus,
$P=\mathfrak{m}T$
as required.◻
The following techniques are due to Vaz Pinto [Reference Vaz PintoVP, Section 2].
Let
$L=L^{(1)}=TS_{1}$
then
$L\cong \bigoplus _{n\geqslant 1}Q^{n-1}I^{2}/Q^{n}I$
and
$S/L\cong C$
as graded
$T$
-modules. Then there exists a canonical exact sequence
$$\begin{eqnarray}0\rightarrow L\rightarrow S\rightarrow C\rightarrow 0\quad (\dagger )\end{eqnarray}$$
of graded
$T$
-modules (Definition 1.1). We set
$D=(I^{2}/QI)\otimes _{A}(T/\text{Ann}_{A}(I^{2}/QI)T)$
. Notice here that
$D$
forms a graded
$T$
-module and
$T/\text{Ann}_{A}(I^{2}/QI)T\cong (A/\text{Ann}_{A}(I^{2}/QI))[X_{1},X_{2},\ldots ,X_{d}]$
is the polynomial ring with
$d$
indeterminates over the ring
$A/\text{Ann}_{A}(I^{2}/QI)$
. Let
$$\begin{eqnarray}\unicode[STIX]{x1D703}:D(-1)\rightarrow L\end{eqnarray}$$
denote an epimorphism of graded
$T$
-modules such that
$\unicode[STIX]{x1D703}(\sum _{\unicode[STIX]{x1D6FC}}\overline{x_{\unicode[STIX]{x1D6FC}}}\otimes X_{1}^{\unicode[STIX]{x1D6FC}_{1}}X_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FC}_{d}})=\sum _{\unicode[STIX]{x1D6FC}}\overline{x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{|\unicode[STIX]{x1D6FC}|+1}}$
for
$x_{\unicode[STIX]{x1D6FC}}\in I^{2}$
and
$\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{d})\in \mathbb{Z}^{d}$
with
$\unicode[STIX]{x1D6FC}_{i}\geqslant 0$
$(1\leqslant i\leqslant d)$
, where
$|\unicode[STIX]{x1D6FC}|=\sum _{i=1}^{d}\unicode[STIX]{x1D6FC}_{i}$
, and
$\overline{x_{\unicode[STIX]{x1D6FC}}}$
and
$\overline{x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{|\unicode[STIX]{x1D6FC}|+1}}$
denote the images of
$x_{\unicode[STIX]{x1D6FC}}$
in
$I^{2}/QI$
and
$x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FC}_{d}}t^{|\unicode[STIX]{x1D6FC}|+1}$
in
$L$
.
Then we have the following lemma.
Lemma 2.7. Suppose that
$Q\cap I^{2}=QI$
. Then the map
$\unicode[STIX]{x1D703}:D(-1)\rightarrow L$
is an isomorphism of graded
$T$
-modules.
Proof. We have only to show that
$\ker \unicode[STIX]{x1D703}=(0)$
. Assume that
$\ker \unicode[STIX]{x1D703}\neq (0)$
and let
$n\geqslant 2$
as the least integer so that
$[\ker \unicode[STIX]{x1D703}]_{n}\neq (0)$
(notice that
$[\ker \unicode[STIX]{x1D703}]_{n}=(0)$
for all
$n\leqslant 1$
). Take
$0\neq g\in [\ker \unicode[STIX]{x1D703}]_{n}$
and we set
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E4} & = & \displaystyle \left\{(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{d-1},0)\in \mathbb{Z}^{d}|\unicode[STIX]{x1D6FC}_{i}\geqslant 0\text{ for }1\leqslant i\leqslant d-1,\text{ and }\mathop{\sum }_{i=1}^{d-1}\unicode[STIX]{x1D6FC}_{i}=n-1\right\},\nonumber\\ \displaystyle \unicode[STIX]{x1D6E4}^{\prime } & = & \displaystyle \left\{(\unicode[STIX]{x1D6FD}_{1},\unicode[STIX]{x1D6FD}_{2},\ldots ,\unicode[STIX]{x1D6FD}_{d})\in \mathbb{Z}^{d}|\unicode[STIX]{x1D6FD}_{i}\geqslant 0\text{ for }1\leqslant i\leqslant d-1,\unicode[STIX]{x1D6FD}_{d}\geqslant 1,\text{ and }\!\mathop{\sum }_{i=1}^{d}\unicode[STIX]{x1D6FD}_{i}=n-1\right\}\!.\nonumber\end{eqnarray}$$
Then because
$$\begin{eqnarray}\unicode[STIX]{x1D6E4}\cup \unicode[STIX]{x1D6E4}^{\prime }=\biggl\{(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{d})\in \mathbb{Z}^{d}|\unicode[STIX]{x1D6FC}_{i}\geqslant 0\text{ for }1\leqslant i\leqslant d\text{ and }\mathop{\sum }_{i=1}^{d}\unicode[STIX]{x1D6FC}_{i}=n-1\biggr\}\end{eqnarray}$$
we may write
$$\begin{eqnarray}\displaystyle g & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}\cup \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FC}}}\otimes X_{1}^{\unicode[STIX]{x1D6FC}_{1}}X_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FC}_{d}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}}\overline{x_{\unicode[STIX]{x1D6FC}}}\otimes X_{1}^{\unicode[STIX]{x1D6FC}_{1}}X_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots X_{d-1}^{\unicode[STIX]{x1D6FC}_{d-1}}+\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FD}}}\otimes X_{1}^{\unicode[STIX]{x1D6FD}_{1}}X_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FD}_{d}},\nonumber\end{eqnarray}$$
where
$\overline{x_{\unicode[STIX]{x1D6FC}}},\overline{x_{\unicode[STIX]{x1D6FD}}}$
denote the images of
$x_{\unicode[STIX]{x1D6FC}},x_{\unicode[STIX]{x1D6FD}}\in I^{2}$
in
$I^{2}/QI$
, respectively. We may assume that
$\sum _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FD}}}\otimes X_{1}^{\unicode[STIX]{x1D6FD}_{1}}X_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FD}_{d}}\neq 0$
in
$D$
. Then we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}(g)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}}\overline{x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d-1}^{\unicode[STIX]{x1D6FC}_{d-1}}t^{n}}+\overline{\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}}t^{n}}=0 & & \displaystyle \nonumber\end{eqnarray}$$
so that
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}}x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d-1}^{\unicode[STIX]{x1D6FC}_{d-1}}+\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}}\in Q^{n}I. & & \displaystyle \nonumber\end{eqnarray}$$
Because
$Q^{n}I=(a_{1},a_{2},\ldots ,a_{d-1})^{n}I+a_{d}Q^{n-1}I$
and
$\unicode[STIX]{x1D6FD}_{d}\geqslant 1$
, we may write
$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}}x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d-1}^{\unicode[STIX]{x1D6FC}_{d-1}}+a_{d}\left(\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}-1}\right)=\unicode[STIX]{x1D70F}+a_{d}\unicode[STIX]{x1D70C}\end{eqnarray}$$
with
$\unicode[STIX]{x1D70F}\in (a_{1},a_{2},\ldots ,a_{d-1})^{n}I$
and
$\unicode[STIX]{x1D70C}\in Q^{n-1}I$
. Then because
$$\begin{eqnarray}\displaystyle a_{d}\left(\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}-1}-\unicode[STIX]{x1D70C}\right) & = & \displaystyle \unicode[STIX]{x1D70F}-\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E4}}x_{\unicode[STIX]{x1D6FC}}a_{1}^{\unicode[STIX]{x1D6FC}_{1}}a_{2}^{\unicode[STIX]{x1D6FC}_{2}}\cdots a_{d-1}^{\unicode[STIX]{x1D6FC}_{d-1}}\nonumber\\ \displaystyle & & \displaystyle \in (a_{1},a_{2},\ldots ,a_{d-1})^{n-1}\nonumber\end{eqnarray}$$
we have, by Lemma 2.4,
$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}-1}-\unicode[STIX]{x1D70C}\in (a_{1},a_{2},\ldots ,a_{d-1})^{n-1}\cap Q^{n-2}I^{2}\subseteq Q^{n-1}I.\end{eqnarray}$$
Therefore,
$\sum _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}x_{\unicode[STIX]{x1D6FD}}a_{1}^{\unicode[STIX]{x1D6FD}_{1}}a_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots a_{d}^{\unicode[STIX]{x1D6FD}_{d}-1}\in Q^{n-1}I$
and hence
$\unicode[STIX]{x1D703}(\sum _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FD}}}\otimes X_{1}^{\unicode[STIX]{x1D6FD}_{1}}X_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FD}_{d}-1})=0$
. Then, because
$\sum _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FD}}}\otimes X_{1}^{\unicode[STIX]{x1D6FD}_{1}}X_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FD}_{d}-1}\in [\ker \unicode[STIX]{x1D703}]_{n-1}=(0)$
, we have
$\sum _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6E4}^{\prime }}\overline{x_{\unicode[STIX]{x1D6FD}}}\otimes X_{1}^{\unicode[STIX]{x1D6FD}_{1}}X_{2}^{\unicode[STIX]{x1D6FD}_{2}}\cdots X_{d}^{\unicode[STIX]{x1D6FD}_{d}}=0$
, which is contradiction. Thus
$\ker \unicode[STIX]{x1D703}=(0)$
. Consequently, the map
$\unicode[STIX]{x1D703}:D(-1)\rightarrow L$
is an isomorphism.◻
Thanks to Lemma 2.7, we can prove the following result.
Proposition 2.8. Suppose that
$Q\cap I^{2}=QI$
. Then we have
$$\begin{eqnarray}\displaystyle \ell _{A}(A/I^{n+1}) & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}\!-\{\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)\}\binom{n+d-1}{d-1}\nonumber\\ \displaystyle & & \displaystyle +\,\ell _{A}(I^{2}/QI)\binom{n+d-2}{d-2}-\ell _{A}(C_{n})\nonumber\end{eqnarray}$$
for all
$n\geqslant 0$
.
Proof. We have, for all
$n\geqslant 0$
,
$$\begin{eqnarray}\displaystyle \ell _{A}(S_{n}) & = & \displaystyle \ell _{A}(L_{n})+\ell _{A}(C_{n})\nonumber\\ \displaystyle & = & \displaystyle \ell _{A}(I^{2}/QI)\binom{n+d-2}{d-1}+\ell _{A}(C_{n})\nonumber\\ \displaystyle & = & \displaystyle \ell _{A}(I^{2}/QI)\binom{n+d-1}{d-1}-\ell _{A}(I^{2}/QI)\binom{n+d-2}{d-2}+\ell _{A}(C_{n})\nonumber\end{eqnarray}$$
by the exact sequence
$$\begin{eqnarray}0\rightarrow L\rightarrow S\rightarrow C\rightarrow 0\quad (\dagger )\end{eqnarray}$$
and the isomorphisms
$L\cong D(-1)\cong (I^{2}/QI)\otimes _{A}(A/\text{Ann}_{A}(I^{2}/QI))[X_{1},X_{2},\ldots ,X_{d}]$
of graded
$T$
-modules (see Lemma 2.7). Therefore, we have, for all
$n\geqslant 0$
,
$$\begin{eqnarray}\displaystyle \ell _{A}(A/I^{n+1}) & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}-\{\text{e}_{0}(I)-\ell _{A}(A/I)\}\binom{n+d-1}{d-1}-\ell _{A}(S_{n})\nonumber\\ \displaystyle & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}-\{\text{e}_{0}(I)-\ell _{A}(A/I)\}\binom{n+d-1}{d-1}-\!\{\!\ell _{A}(I^{2}/QI)\nonumber\\ \displaystyle & & \displaystyle \cdot \,\binom{n+d-1}{d-1}-\ell _{A}(I^{2}/QI)\binom{n+d-2}{d-2}+\ell _{A}(C_{n})\!\}\nonumber\\ \displaystyle & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}-\{\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)\}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\binom{n+d-1}{d-1}+\ell _{A}(I^{2}/QI)\binom{n+d-2}{d-2}-\ell _{A}(C_{n})\nonumber\end{eqnarray}$$
by [Reference Goto, Nishida and OzekiGNO, Proposition 2.2 (2)]. ◻
The following result specifies [Reference Goto, Nishida and OzekiGNO, Proposition 2.2(3)] and, by using Propositions 2.2 and 2.8, the proof takes advantage of the same techniques.
Proposition 2.9. Suppose that
$Q\cap I^{2}=QI$
. Let
$\mathfrak{p}=\mathfrak{m}T$
. Then we have
$$\begin{eqnarray}\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+\ell _{T_{\mathfrak{p}}}(C_{\mathfrak{p}}).\end{eqnarray}$$
Combining Lemma 2.1(3) and Proposition 2.9 we obtain the following result that was proven by Elias and Valla [Reference Elias and VallaEV, Theorem 2.1] in the case where
$I=\mathfrak{m}$
.
Corollary 2.10. Suppose that
$Q\cap I^{2}=QI$
. Then we have
$\text{e}_{1}(I)\geqslant \text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)$
. The equality
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)$
holds true if and only if
$I^{3}=QI^{2}$
. When this is the case,
$\text{e}_{2}(I)=\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)$
if
$d\geqslant 2$
,
$\text{e}_{i}(I)=0$
for all
$3\leqslant i\leqslant d$
, and
$G$
is a Cohen–Macaulay ring.
Let us introduce the relationship between the depth of the module
$C$
and the associated graded ring
$G$
of
$I$
.
Lemma 2.11. Suppose that
$Q\cap I^{2}=QI$
and
$C\neq (0)$
. Let
$s=\text{depth}_{T}C$
. Then we have
$\text{depth}G\geqslant s-1$
. In particular, we have
$\text{depth}G=s-1$
, if
$s\leqslant d-2$
.
Proof. Notice that
$L\cong D(-1)$
by Lemma 2.7 and
$D$
is a
$d$
dimensional Cohen–Macaulay
$T$
-module. Therefore, we have
$s\leqslant \text{depth}_{T}S$
and, if
$s\leqslant d-2$
then
$s=\text{depth}_{T}S$
by the exact sequence
$$\begin{eqnarray}0\rightarrow L\rightarrow S\rightarrow C\rightarrow 0\quad (\dagger ).\end{eqnarray}$$
Because
$\text{depth}G\geqslant \text{depth}_{T}S-1$
and, if
$\text{depth}_{T}S\leqslant d-1$
then
$\text{depth}G=\text{depth}_{T}S-1$
by [Reference Goto, Nishida and OzekiGNO, Proposition 2.2(4)], our assertions follow.◻
3 Proof of Theorem 1.2
The purpose of this section is to prove Theorem 1.2. Throughout this section, let
$I$
be an integrally closed
$\mathfrak{m}$
-primary ideal.
Theorem 3.1. Suppose that
$I$
is integrally closed. Then the following conditions are equivalent:
-
(1)
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
, -
(2)
$\mathfrak{m}C=(0)$
and
$\text{rank}_{B}C=1$
, -
(3) there exists a non-zero graded ideal
$\mathfrak{a}$
of
$B$
such that
$C\cong \mathfrak{a}(-1)$
as graded
$T$
-modules.
To prove Theorem 3.1, we need the following bound on
$\text{e}_{2}(I)$
.
Lemma 3.2. ([Reference ItohI1, Theorem 12], [Reference SallyS2, Corollary 2.5], [Reference Rossi and VallaRV3, Corollary 3.1])
Suppose
$d\geqslant 2$
and let
$I$
be an integrally closed ideal. Then
$\text{e}_{2}(I)\geqslant \text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I).$
Proof of Theorem 3.1.
Let
$\mathfrak{p}=\mathfrak{m}T$
; then we see that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+\ell _{T_{\mathfrak{p}}}(C_{\mathfrak{p}})$
by Lemma 2.9 and
$\text{Ass}_{T}C=\{\mathfrak{p}\}$
byProposition 2.2.
$(1)\Rightarrow (2)$
Since
$\ell _{T_{\mathfrak{p}}}(C_{\mathfrak{p}})=1$
and
$\text{Ass}_{T}C=\{\mathfrak{p}\}$
, we have
$\mathfrak{m}C=(0)$
and
$\text{rank}_{B}C=1$
.
$(2)\Rightarrow (1)$
This is clear, because assertion
$(1)$
is equivalent to saying that
$\ell _{T_{\mathfrak{p}}}(C_{\mathfrak{p}})=1$
.
$(3)\Rightarrow (2)$
This is obvious.
$(2)\Rightarrow (3)$
Because
$\text{Ass}_{T}C=\{\mathfrak{p}\}$
,
$C$
is a torsion free
$B$
-module. If
$C$
is
$B$
-free, then we have
$C\cong B(-2)$
as graded
$B$
-modules, because
$C_{2}\neq (0)$
and
$C_{n}=(0)$
for
$n\leqslant 1$
. Hence
$C\cong X_{1}B(-1)$
as graded
$B$
-modules with
$0\neq X_{1}\in B_{1}$
.
Suppose that
$C$
is not
$B$
-free. Then we have
$d=\dim A\geqslant 2$
. Because
$\text{rank}_{B}C=1$
, there exists a graded ideal
$\mathfrak{a}$
of
$B$
such that
$$\begin{eqnarray}C\cong \mathfrak{a}(r)\end{eqnarray}$$
as graded
$B$
-modules for some
$r\in \mathbb{Z}$
. Since every height one prime in the polynomial ring
$B$
is principal, we may choose
$\mathfrak{a}$
with
$\text{ht}_{B}\mathfrak{a}\geqslant 2$
. Then since
$\mathfrak{a}_{r+2}\cong \mathfrak{a}(r)_{2}\cong C_{2}\neq (0)$
and
$\mathfrak{a}_{n}=(0)$
for all
$n\leqslant 0$
, we have
$r+2\geqslant 1$
so that
$r\geqslant -1$
. It is now enough to show that
$r=-1$
. Applying the exact sequence
$$\begin{eqnarray}0\rightarrow C\rightarrow B(r)\rightarrow (B/\mathfrak{a})(r)\rightarrow 0\end{eqnarray}$$
of graded
$B$
-modules, we have
$$\begin{eqnarray}\displaystyle \ell _{A}(C_{n}) & = & \displaystyle \ell _{A}(B_{r+n})-\ell _{A}([B/\mathfrak{a}]_{r+n})\nonumber\\ \displaystyle & = & \displaystyle \binom{n+r+d-1}{d-1}-\ell _{A}([B/\mathfrak{a}]_{r+n})\nonumber\\ \displaystyle & = & \displaystyle \binom{n+d-1}{d-1}+r\binom{n+d-2}{d-2}+\text{(lower terms)}\nonumber\end{eqnarray}$$
for all
$n\gg 0$
, because
$\text{ht}_{B}\mathfrak{a}\geqslant 2$
. Therefore, thanks to Proposition 2.8, we have
$$\begin{eqnarray}\displaystyle \ell _{A}(A/I^{n+1}) & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}-\{\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)\}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\binom{n+d-1}{d-1}+\ell _{A}(I^{2}/QI)\binom{n+d-2}{d-2}-\ell _{A}(C_{n})\nonumber\\ \displaystyle & = & \displaystyle \text{e}_{0}(I)\binom{n+d}{d}-\{\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1\}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\binom{n+d-1}{d-1}+\{\ell _{A}(I^{2}/QI)\,-r\}\binom{n+d-2}{d-2}+\text{(lower terms)}\nonumber\end{eqnarray}$$
for all
$n\gg 0$
. Therefore,
$\text{e}_{2}(I)=\ell _{A}(I^{2}/QI)-r$
. Then, because
$\text{e}_{2}(I)\geqslant \text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)=\ell _{A}(I^{2}/QI)+1$
by Lemma 3.2 we have
$r\leqslant -1$
. Thus
$r=-1$
and so
$C\cong \mathfrak{a}(-1)$
as graded
$B$
-modules.◻
As a direct consequence of Theorem 3.1 the following result holds true.
Proposition 3.3. Assume that
$I$
is integrally closed. Suppose that
$\text{e}_{1}(I)\,=\,\text{e}_{0}(I)\,-\,\ell _{A}(\!A/I)\,+\,\ell _{A}(I^{2}/QI)\,+\,1\!$
and
$I^{4}\,=\,QI^{3}$
. Let
$c\,=\,\ell _{A}(I^{3}/QI^{2})$
. Then
-
(1)
$1\leqslant c\leqslant d$
and
$\unicode[STIX]{x1D707}_{B}(C)=c$
. -
(2)
$\text{depth}~G\geqslant d-c$
and
$\text{depth}_{T}C=d-c+1$
, -
(3)
$\text{depth}~G=d-c$
, if
$c\geqslant 2$
. -
(4) Suppose
$c=1<d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 0$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)+1 & \quad \text{if }i=2,\\ 1 & \quad \text{if }i=3\text{ and }d\geqslant 3,\\ 0 & \quad \text{if }4\leqslant i\leqslant d.\end{array}\right.\end{eqnarray}$$
-
(5) Suppose
$2\leqslant c<d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 0$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I) & \quad \text{if }i=2,\\ 0 & \quad \text{if }i\neq c+1,c+2,3\leqslant i\leqslant d,\\ (-1)^{c+1} & \quad \text{if }i=c+1,c+2,3\leqslant i\leqslant d,\end{array}\right.\end{eqnarray}$$
-
(6) Suppose
$c=d$
. Then
$HP_{I}(n)=\ell _{A}(A/I^{n+1})$
for all
$n\geqslant 2$
and
$$\begin{eqnarray}\text{e}_{i}(I)=\left\{\begin{array}{@{}ll@{}}\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I) & \quad \text{if }i=2\text{ and }d\geqslant 2,\\ 0 & \quad \text{if }3\leqslant i\leqslant d,\end{array}\right.\end{eqnarray}$$
-
(7) The Hilbert series
$HS_{I}(z)$
is given by
$$\begin{eqnarray}HS_{I}(z)=\frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-\ell _{A}(I^{2}/QI)-1\}z+\{\ell _{A}(I^{2}/QI)+1\}z^{2}+(1-z)^{c+1}z}{(1-z)^{d}}.\end{eqnarray}$$
Proof.
$(1)$
We have
$C=TC_{2}$
, since
$I^{4}=QI^{3}$
(cf. Lemma 2.1(5)). Therefore, thanks to Theorem 3.1,
$C\cong \mathfrak{a}(-1)$
as graded
$T$
-modules, where
$\mathfrak{a}=(X_{1},X_{2},\ldots ,X_{c})B$
is an ideal generated by linear forms
$\{X_{i}\}_{1\leqslant i\leqslant c}$
of
$B$
. Hence, we get
$1\leqslant c\leqslant d$
and
$\unicode[STIX]{x1D707}_{B}(C)=c$
.
$(4)$
,
$(5)$
,
$(6)$
Let us consider the exact sequence
$$\begin{eqnarray}0\rightarrow C\rightarrow B(-1)\rightarrow (B/\mathfrak{a})(-1)\rightarrow 0\quad (\ast _{5})\end{eqnarray}$$
of graded
$B$
-modules. Then, we have
$$\begin{eqnarray}\displaystyle \ell _{A}(C_{n}) & = & \displaystyle \ell _{A}(B_{n-1})-\ell _{A}([B/\mathfrak{a}]_{n-1})\nonumber\\ \displaystyle & = & \displaystyle \binom{n-1+d-1}{d-1}-\binom{n-1+d-c-1}{d-c-1}\nonumber\\ \displaystyle & = & \displaystyle \binom{n+d-1}{d-1}-\binom{n+d-2}{d-2}-\binom{n+d-c-1}{d-c-1}\nonumber\\ \displaystyle & & \displaystyle +\,\binom{n+d-c-2}{d-c-2}\nonumber\end{eqnarray}$$
for all
$n\geqslant 0$
(resp.
$n\geqslant 2$
) if
$1\leqslant c\leqslant d-1$
(resp.
$c=d$
). Therefore, our assertions
$(4)$
,
$(5)$
, and
$(6)$
follow by Proposition 2.8.
(7) We have
$$\begin{eqnarray}HS_{C}(z)=HS_{B}(z)z-HS_{B/\mathfrak{a}}(z)z=\frac{z-(1-z)^{c}z}{(1-z)^{d}}\end{eqnarray}$$
by the above exact sequence
$(\ast _{5})$
, where
$HS_{\ast }(z)$
denotes the Hilbert series of the graded modules. We also have
$$\begin{eqnarray}HS_{S}(z)=HS_{L}(z)+HS_{C}(z)=\frac{\{\ell _{A}(I^{2}/QI)+1\}z-(1-z)^{c}z}{(1-z)^{d}}\end{eqnarray}$$
by the exact sequence
$$\begin{eqnarray}0\rightarrow L\rightarrow S\rightarrow C\rightarrow 0\quad (\dagger )\end{eqnarray}$$
and the isomorphism
$L\cong D(-1)$
of graded
$T$
-modules (Lemma 2.7). Then, because
$$\begin{eqnarray}HS_{I}(z)=\frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)\}z}{(1-z)^{d}}-(1-z)HS_{S}(z)\end{eqnarray}$$
([Reference Vaz PintoVP], [Reference Rossi and VallaRV3, Proposition 6.3]), we can get the required result.
We prove now
$(2)$
,
$(3).$
We have
$\text{depth}_{T}C=d-c+1$
by the exact sequence
$(\ast _{5})$
so that
$\text{depth}G\geqslant d-c$
and, if
$c\geqslant 3,$
then
$\text{depth}G=d-c$
by Lemma 2.11. Let us consider the case where
$c=2$
and we need to show
$\text{depth}G=d-2$
. Assume
$\text{depth}G\geqslant d-1$
; then
$S$
is a Cohen–Macaulay
$T$
-module by [Reference Goto, Nishida and OzekiGNO, Proposition 2.2]. Taking the local cohomology functors
$\text{H}_{M}^{i}(\ast )$
of
$T$
with respect to the graded maximal ideal
$M=\mathfrak{m}T+T_{+}$
to the above exact sequence
$(\dagger )$
of graded
$T$
-modules, we get a monomorphism
$$\begin{eqnarray}\text{H}_{M}^{d-1}(C){\hookrightarrow}\text{H}_{M}^{d}(L)\end{eqnarray}$$
of graded
$T$
-modules. Because
$C\cong (X_{1},X_{2})B(-1)$
, we have
$\text{H}_{M}^{d-1}(C)\cong \text{H}_{M}^{d-2}(B/(X_{1},X_{2})B)(-1)$
as graded
$T$
-modules so that
$[\text{H}_{M}^{d-1}(C)]_{-d+3}\neq (0)$
(notice that
$B/(X_{1},X_{2})B\cong (A/\mathfrak{m})[X_{3},X_{4},\ldots ,X_{d}]$
). On the other hand, we have
$[\text{H}_{M}^{d}(L)]_{n}=(0)$
for all
$n\geqslant -d+2$
, because
$L\cong D(-1)\cong (I^{2}/QI)\otimes _{A}(A/\text{Ann}_{A}(I^{2}/QI))[X_{1},X_{2},\ldots ,X_{d}](-1)$
by Lemma 2.7. However, this is impossible. Therefore,
$\text{depth}~G=d-2$
if
$c=2$
.◻
We prove now Theorem 1.2. Assume assertion
$(1)$
in Theorem 1.2. Then we have an isomorphism
$C\cong \mathfrak{a}(-1)$
as graded
$B$
-modules for a graded ideal
$\mathfrak{a}$
in
$B$
by Theorem 3.1. Once we are able to show
$I^{4}=QI^{3}$
, then, because
$C=TC_{2}$
by Lemma 2.1(5), the ideal
$\mathfrak{a}$
is generated by linearly independent linear forms
$\{X_{i}\}_{1\leqslant i\leqslant c}$
of
$B$
with
$c=\ell _{A}(I^{3}/QI^{2})$
(recall that
$\mathfrak{a}_{1}\cong C_{2}\cong I^{3}/QI^{2}$
by Lemma 2.1
$(2)$
). Therefore, the implication
$(1)\Rightarrow (3)$
in Theorem 1.2 follows. We also notice that, the last assertions of Theorem 1.2 follow by Proposition 3.3.
Thus our Theorem 1.2 has been proven modulo the following theorem.
Theorem 3.4. Assume that
$I$
is integrally closed. Suppose that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
. Then
$I^{4}=QI^{3}$
.
Proof. We proceed by induction on
$d$
. Suppose that
$d=1$
. Then the result follows by [Reference Huckaba and MarleyHM, Proposition 4.6] since
$\text{e}_{1}(I)=\sum _{i\geqslant 0}\ell _{A}(I^{i+1}/QI^{i}).$
Assume that
$d\geqslant 2$
and that our assertion holds true for
$d-1$
. Since the residue class field
$A/\mathfrak{m}$
of
$A$
is infinite, without loss of generality, we may assume that
$a_{1}$
is a superficial element of
$I$
and
$I/(a_{1})$
is integrally closed (cf. [Reference ItohI1, p. 648], [Reference Rossi and VallaRV3, Proposition 1.1]). We set
$A^{\prime }=A/(a_{1}),I^{\prime }=I/(a_{1}),Q^{\prime }=Q/(a_{1})$
. We then have
$\text{e}_{1}(I^{\prime })=\text{e}_{0}(I^{\prime })-\ell _{A^{\prime }}(A^{\prime }/I^{\prime })+\ell _{A^{\prime }}({I^{\prime }}^{2}/Q^{\prime }I^{\prime })+1$
because
$\text{e}_{i}(I^{\prime })=\text{e}_{i}(I)$
for
$0\leqslant i\leqslant d-1$
,
$\ell _{A^{\prime }}(A^{\prime }/I^{\prime })=\ell _{A}(A/I)$
, and since
$(a_{1})\cap I^{2}=a_{1}I$
,
$\ell _{A^{\prime }}({I^{\prime }}^{2}/Q^{\prime }I^{\prime })=\ell _{A}(I^{2}/QI)$
. Then the inductive assumption on
$d$
says that
${I^{\prime }}^{4}=Q^{\prime }{I^{\prime }}^{3}$
holds true. If
$\text{depth}\text{G}(I^{\prime })>0$
then, thanks to Sally’s technique (cf. [Reference SallyS1], [Reference Huckaba and MarleyHM, Lemma 2.2]),
$a_{1}t$
is a non-zero divisor on
$G$
. Then we have
$I^{4}=QI^{3}$
.
Assume that
$\text{depth}\text{G}(I^{\prime })=0$
. Then because
$\text{e}_{1}(I^{\prime })=\text{e}_{0}(I^{\prime })-\ell _{A^{\prime }}(A^{\prime }/I^{\prime })+\ell _{A^{\prime }}({I^{\prime }}^{2}/Q^{\prime }I^{\prime })+1$
and
${I^{\prime }}^{4}=Q^{\prime }{I^{\prime }}^{3}$
, we have
$\ell _{A^{\prime }}({I^{\prime }}^{3}/Q^{\prime }{I^{\prime }}^{2})=d-1$
by Proposition 3.3(2). Since
${I^{\prime }}^{3}/Q^{\prime }{I^{\prime }}^{2}$
is a homomorphic image of
$I^{3}/QI^{2}$
, we have
$\ell _{A}(C_{2})=\ell _{A}(I^{3}/QI^{2})\geqslant d-1$
. Let us now take an isomorphism
$$\begin{eqnarray}\unicode[STIX]{x1D711}:C\rightarrow \mathfrak{a}(-1)\end{eqnarray}$$
of graded
$B$
-modules, where
$\mathfrak{a}$
is a graded ideal of
$B$
(cf. Theorem 3.1). Then since
$\ell _{A}(\mathfrak{a}_{1})=\ell _{A}(C_{2})\geqslant d-1$
, we have
$Y_{1},Y_{2},\ldots ,Y_{d-1}\in \mathfrak{a}$
where
$Y_{1},Y_{2},\ldots ,Y_{d-1}$
denote linearly independent linear forms of
$B$
, which we enlarge to a basis
$Y_{1},Y_{2},\ldots ,Y_{d-1},Y_{d}$
of
$B_{1}$
. If
$\mathfrak{a}=(Y_{1},Y_{2},\ldots ,Y_{d-1})B$
then, because
$C=TC_{2}$
, we have
$I^{4}=QI^{3}$
by Lemma 2.1(5). Assume that
$\mathfrak{a}\neq (Y_{1},Y_{2},\ldots ,Y_{d-1})B$
. Then, since
$B=k[Y_{1},Y_{2},\ldots ,Y_{d}]$
, the ideal
$\mathfrak{a}/(Y_{1},Y_{2},\ldots ,Y_{d-1})$
is principal so that
$\mathfrak{a}=(Y_{1},Y_{2},\ldots ,Y_{d-1},Y_{d}^{\ell })B$
for some
$\ell \geqslant 1$
.
We need the following.
Claim 1. We have
$\ell =1$
or
$\ell =2$
.
Proof of Claim 1.
Assume that
$\ell \geqslant 3$
. Then
$I^{4}/[QI^{3}+\mathfrak{m}I^{4}]\cong [C/MC]_{3}=(0)$
, where
$M=\mathfrak{m}T+T_{+}$
denotes the graded maximal ideal of
$T$
. Therefore,
$I^{4}=QI^{3}$
by Nakayama’s lemma so that
$C=TC_{2}$
, which is impossible. Thus
$\ell =1$
or
$\ell =2$
.◻
We have to show that
$\ell =1$
. Assume that
$\ell =2$
. Let us write, for each
$1\leqslant i\leqslant d$
,
$Y_{i}=\overline{b_{i}t}$
with
$b_{i}\in Q$
, where
$\overline{b_{i}t}$
denotes the image of
$b_{i}t\in T$
in
$B$
. We notice here that
$Q=(b_{1},b_{2},\ldots ,b_{d})$
, because
$Y_{1},Y_{2},\ldots ,Y_{d}$
forms a
$k$
-basis of
$B_{1}$
.
Let us choose elements
$f_{i}\in C_{2}$
for
$1\leqslant i\leqslant d-1$
and
$f_{d}\in C_{3}$
so that
$\unicode[STIX]{x1D711}(f_{i})=Y_{i}$
for
$1\leqslant i\leqslant d-1$
and
$\unicode[STIX]{x1D711}(f_{d})=Y_{d}^{2}$
. Let
$z_{i}\in I^{3}$
for
$1\leqslant i\leqslant d-1$
and
$z_{d}\in I^{4}$
so that
$\{f_{i}\}_{1\leqslant i\leqslant d-1}$
and
$f_{d}$
are, respectively, the images of
$\{z_{i}t^{2}\}_{1\leqslant i\leqslant d-1}$
and
$z_{d}t^{3}$
in
$C$
. Let us now consider the relations
$Y_{d}^{2}f_{i}=Y_{i}f_{d}$
in
$C$
for
$1\leqslant i\leqslant d-1$
, that is
$$\begin{eqnarray}b_{d}^{2}z_{i}-b_{i}z_{d}\in Q^{3}I^{2}\end{eqnarray}$$
for
$1\leqslant i\leqslant d-1$
. Notice that
$$\begin{eqnarray}(b_{1},b_{d})\cap Q^{3}I^{2}=(b_{1},b_{d})Q^{2}I^{2}\end{eqnarray}$$
by Lemma 2.3 and write
$$\begin{eqnarray}b_{d}^{2}z_{1}-b_{1}z_{d}=b_{1}\unicode[STIX]{x1D70F}_{1}+b_{d}\unicode[STIX]{x1D70F}_{d}\end{eqnarray}$$
with
$\unicode[STIX]{x1D70F}_{1},\unicode[STIX]{x1D70F}_{d}\in Q^{2}I^{2}$
. Then we have
$$\begin{eqnarray}b_{d}(b_{d}z_{1}-\unicode[STIX]{x1D70F}_{d})=b_{1}(z_{d}+\unicode[STIX]{x1D70F}_{1})\end{eqnarray}$$
so that
$b_{d}z_{1}-\unicode[STIX]{x1D70F}_{d}\in (b_{1})$
because
$b_{1},b_{d}$
forms a regular sequence on
$A$
. Since
$\unicode[STIX]{x1D70F}_{d}\in (b_{1},b_{d})\cap Q^{2}I^{2}=(b_{1},b_{d})QI^{2}$
by Lemma 2.3, there exist elements
$\unicode[STIX]{x1D70F}_{1}^{\prime },\unicode[STIX]{x1D70F}_{d}^{\prime }\in QI^{2}$
such that
$\unicode[STIX]{x1D70F}_{d}=b_{1}\unicode[STIX]{x1D70F}_{1}^{\prime }+b_{d}\unicode[STIX]{x1D70F}_{d}^{\prime }$
. Then by the equality
$(1)$
we have
$$\begin{eqnarray}b_{d}^{2}(z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime })=b_{1}(z_{d}+\unicode[STIX]{x1D70F}_{1}+b_{d}\unicode[STIX]{x1D70F}_{1}^{\prime })\end{eqnarray}$$
so that we have
$z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime }\in (b_{1})$
. Hence
$z_{1}\in QI^{2}+(b_{1})$
. The same argument works for each
$1\leqslant i\leqslant d-1$
to see that
$z_{i}\in QI^{2}+(b_{i})$
. Therefore, because
$I^{3}=QI^{2}+(z_{1},z_{2},\ldots ,z_{d-1})$
, we have
$I^{3}\subseteq b_{d}I^{2}+(b_{1},b_{2},\ldots ,b_{d-1})$
and hence
$$\begin{eqnarray}I^{4}\subseteq b_{d}^{2}I^{2}+(b_{1},b_{2},\ldots ,b_{d-1}).\end{eqnarray}$$
Then, because
$z_{d}+\unicode[STIX]{x1D70F}_{1}+b_{d}\unicode[STIX]{x1D70F}_{1}^{\prime }\in I^{4}$
, there exist elements
$h\in I^{2}$
and
$\unicode[STIX]{x1D702}\in (b_{1},b_{2},\ldots ,b_{d-1})$
such that
$z_{d}+\unicode[STIX]{x1D70F}_{1}+b_{d}\unicode[STIX]{x1D70F}_{1}^{\prime }=b_{d}^{2}h+\unicode[STIX]{x1D702}$
. Then we have
$$\begin{eqnarray}b_{d}^{2}(z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime }-b_{1}h)=b_{1}\unicode[STIX]{x1D702}\end{eqnarray}$$
by the equality (2). Since
$b_{1},b_{2},\ldots ,b_{d}$
is a regular sequence on
$A$
,
$\unicode[STIX]{x1D702}\in (b_{d}^{2})\cap (b_{1},b_{2},\ldots ,b_{d-1})=b_{d}^{2}(b_{1},b_{2},\ldots ,b_{d-1})$
. Write
$\unicode[STIX]{x1D702}=b_{d}^{2}\unicode[STIX]{x1D702}^{\prime }$
with
$\unicode[STIX]{x1D702}^{\prime }\in (b_{1},b_{2},\ldots ,b_{d-1})$
, then we have
$$\begin{eqnarray}b_{d}^{2}(z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime }-b_{1}h)=b_{1}b_{d}^{2}\unicode[STIX]{x1D702}^{\prime }\end{eqnarray}$$
by the equality (3), so that
$z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime }-b_{1}h=b_{1}\unicode[STIX]{x1D702}^{\prime }$
. Then we have
$b_{1}\unicode[STIX]{x1D702}^{\prime }=z_{1}-\unicode[STIX]{x1D70F}_{d}^{\prime }-b_{1}h\in Q^{2}\cap I^{3}=Q^{2}I$
, since
$Q^{2}\cap I^{3}\subseteq Q^{2}\cap \overline{Q^{3}}=Q^{2}\overline{Q}=Q^{2}I$
(cf. [Reference HunekeH, Reference ItohI2]), where
$\overline{J}$
denotes the integral closure of an ideal
$J$
. Hence
$z_{1}\in QI^{2}$
, because
$\unicode[STIX]{x1D70F}_{d}^{\prime },b_{1}h,b_{1}\unicode[STIX]{x1D702}^{\prime }\in QI^{2}$
. Therefore,
$f_{1}=0$
in
$C$
, which is impossible. Thus
$\ell =1$
so that we have
$\mathfrak{a}=(X_{1},X_{2},\ldots ,X_{d})B$
. Therefore,
$C=TC_{2}$
, that is
$I^{4}=QI^{3}$
. This completes the proof of Theorem 3.4 and that of Theorem 1.2 as well.◻
4 Consequences
The purpose of this section is to present some consequences of Theorem 1.2. Let us begin with the following which is exactly the case where
$c=1$
in Theorem 1.2.
Theorem 4.1. Assume that
$I$
is integrally closed. Then the following conditions are equivalent.
-
(1)
$C\cong B(-2)$
as graded
$T$
-modules. -
(2)
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/IQ)+1$
, and if
$d\geqslant 2$
then
$\text{e}_{2}(I)\neq \text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)$
. -
(3)
$\ell _{A}(I^{3}/QI^{2})=1$
and
$I^{4}=QI^{3}$
.
When this is the case, the following assertions follow.
-
(i)
$\text{depth}G\geqslant d-1$
. -
(ii)
$\text{e}_{2}(I)=\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)+1$
if
$d\geqslant 2$
. -
(iii)
$\text{e}_{3}(I)=1$
if
$d\geqslant 3$
, and
$\text{e}_{i}(I)=0$
for
$4\leqslant i\leqslant d$
. -
(iv) The Hilbert series
$HS_{I}(z)$
is given by
$$\begin{eqnarray}HS_{I}(z)=\frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-\ell _{A}(I^{2}/QI)\}z+\{\ell _{A}(I^{2}/QI)-1\}z^{2}+z^{3}}{(1-z)^{d}}.\end{eqnarray}$$
Proof. For
$(1)\Leftrightarrow (2)$
,
$(1)\Rightarrow (3)$
and the last assertions see Theorem 1.2 with
$c=1$
.
$(3)\Rightarrow (1)$
: Since
$\mathfrak{m}I^{n+1}=Q^{n-1}I^{2}$
for all
$n\geqslant 2$
, we have
$\mathfrak{m}C=(0)$
by Lemma 2.1(4). Then we have an epimorphism
$B(-2)\rightarrow C\rightarrow 0$
of graded
$T$
-modules, which must be an isomorphism because
$\dim _{T}C=d$
(Proposition 2.2).◻
Let
$\widetilde{J}=\bigcup _{n\geqslant 1}[J^{n+1}:_{A}J^{n}]=\bigcup _{n\geqslant 1}J^{n+1}:_{A}(a_{1}^{n},a_{2}^{n},\ldots ,a_{d}^{n})$
denote the Ratliff–Rush closure of an
$\mathfrak{m}$
-primary ideal
$J$
in
$A$
, which is the largest
$\mathfrak{m}$
-primary ideal in
$A$
such that
$J\subseteq \widetilde{J}$
and
$\text{e}_{i}(\widetilde{J})=\text{e}_{i}(J)$
for all
$0\leqslant i\leqslant d$
(cf. [Reference Ratliff and RushRR]).
Let us note the following remark.
Remark 4.2. Assume that
$I$
is integrally closed. Then, by [Reference HunekeH, Reference ItohI1],
$Q^{n}\cap \overline{I^{n+1}}=Q^{n}\cap \overline{Q^{n+1}}=Q^{n}\overline{Q}=Q^{n}I$
holds true for
$n\geqslant 1$
, where
$\overline{J}$
denotes the integral closure of an ideal
$J$
. Thus, we have
$Q^{n}\cap \widetilde{I^{n+1}}=Q^{n}I$
for all
$n\geqslant 1$
.
The following result corresponds to the case where
$c=d$
in Theorem 1.2. In Section 5 we give an example of the maximal ideal which satisfies assertion
$(1)$
in Theorem 4.3.
Theorem 4.3. Suppose that
$d\geqslant 2$
and assume that
$I$
is integrally closed. Then the following conditions are equivalent:
-
(1)
$C\cong B_{+}(-1)$
as graded
$T$
-modules. -
(2)
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
,
$\text{e}_{2}(I)=\ell _{A}(I^{2}/QI)+1$
, and
$\text{e}_{i}(I)=0$
for all
$3\leqslant i\leqslant d$
. -
(3)
$\ell _{A}(\widetilde{I^{2}}/I^{2})=1$
and
$\widetilde{I^{n+1}}=Q\widetilde{I^{n}}$
for all
$n\geqslant 2$
.
When this is the case, the associated graded ring
$G$
of
$I$
is a Buchsbaum ring with
$\text{depth}~G=0$
and the Buchsbaum invariant
$\mathbb{I}(G)=d$
.
Proof. We set
$c=\ell _{A}(I^{3}/QI^{2})$
and
${\mathcal{F}}=\{\widetilde{I^{n}}\}_{n\geqslant 0}$
. Let
$\text{R}^{\prime }({\mathcal{F}})=\sum _{n\in \mathbb{Z}}\widetilde{I^{n}}t\subseteq A[t,t^{-1}]$
and
$\text{G}({\mathcal{F}})=\text{R}^{\prime }({\mathcal{F}})/t^{-1}\text{R}^{\prime }({\mathcal{F}})$
. Let
$\text{e}_{i}({\mathcal{F}})$
denote the
$i$
th Hilbert coefficients of the filtration
${\mathcal{F}}$
for
$0\leqslant i\leqslant d$
.
$(1)\Rightarrow (2)$
follows from Theorem 1.2, because
$c=\ell _{A}(C_{2})=d$
.
$(2)\Rightarrow (1)$
Because
$\text{e}_{2}(I)=\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)$
and
$\text{e}_{i}(I)=0$
for all
$3\leqslant i\leqslant d$
, we have
$c=d$
by Theorem 1.2. Therefore
$C\cong B_{+}(-1)$
as graded
$T$
-modules.
$(1)\Rightarrow (3)$
Since
$c=d$
, we have
$\text{depth}G=0$
by Theorem 1.2
$(ii)$
. We apply local cohomology functors
$\text{H}_{M}^{i}(\ast )$
of
$T$
with respect to the graded maximal ideal
$M=\mathfrak{m}T+T_{+}$
of
$T$
to the exact sequences
$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow I^{2}R(-1)\rightarrow IR(-1)\rightarrow G_{+}\rightarrow 0\qquad \text{and}\nonumber\\ \displaystyle & & \displaystyle \qquad 0\rightarrow I^{2}T(-1)\rightarrow I^{2}R(-1)\rightarrow C\rightarrow 0\nonumber\end{eqnarray}$$
of graded
$T$
-modules and get derived monomorphisms
$$\begin{eqnarray}\text{H}_{M}^{0}(G_{+}){\hookrightarrow}\text{H}_{M}^{1}(I^{2}R)(-1)\qquad \text{and}\qquad \text{H}_{M}^{1}(I^{2}R)(-1){\hookrightarrow}\text{H}_{M}^{1}(C)\end{eqnarray}$$
because
$\text{depth}_{T}IR>0$
and
$\text{depth}_{T}I^{2}T\geqslant 2$
(recall that
$T$
is a Cohen–Macaulay ring with
$\dim T=d+1$
and
$\text{depth}_{T}T/I^{2}T\geqslant 1$
by Lemma 2.3). We furthermore have
$\text{H}_{M}^{1}(C)\cong (B/B_{+})(-1)$
since
$C\cong B_{+}(-1)$
. Since
$I$
is integrally closed, we have
$[\text{H}_{M}^{0}(G)]_{0}=(0)$
so that
$\text{H}_{M}^{0}(G)\cong \text{H}_{M}^{0}(G_{+})\neq (0)$
. Then because
$\ell _{A}(B/B_{+})=1$
, we have isomorphisms
$$\begin{eqnarray}\text{H}_{M}^{0}(G)\cong \text{H}_{M}^{1}(I^{2}R)(-1)\cong \text{H}_{M}^{1}(C)\cong B/B_{+}(-1)\end{eqnarray}$$
of graded
$B$
-modules and hence
$\text{H}_{M}^{0}(G)=[\text{H}_{M}^{0}(G)]_{1}\cong B/B_{+}$
. Then since
$[\text{H}_{M}^{0}(G)]_{1}\cong \widetilde{I^{2}}/I^{2}$
we have
$\ell _{A}(\widetilde{I^{2}}/I^{2})=1$
. Hence we have
$$\begin{eqnarray}\displaystyle \text{e}_{1}({\mathcal{F}}) & = & \displaystyle \text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1\nonumber\\ \displaystyle & = & \displaystyle \text{e}_{0}({\mathcal{F}})-\ell _{A}(A/I)+\ell _{A}(\widetilde{I^{2}}/QI)+1-\ell _{A}(\widetilde{I^{2}}/I^{2})\nonumber\\ \displaystyle & = & \displaystyle \text{e}_{0}({\mathcal{F}})-\ell _{A}(A/I)+\ell _{A}(\widetilde{I^{2}}/Q\cap \widetilde{I^{2}})\nonumber\end{eqnarray}$$
because
$\widetilde{I}=I$
,
$Q\cap \widetilde{I^{2}}=QI$
by Remark 4.2, and
$\text{e}_{i}({\mathcal{F}})=\text{e}_{i}(I)$
for
$i=0,1$
. Therefore,
$\widetilde{I^{n+1}}=Q\widetilde{I^{n}}$
for all
$n\geqslant 2$
by [Reference Guerrieri and RossiGR, Theorem 2.2].
$(3)\Rightarrow (2)$
Because
$Q\cap \widetilde{I^{2}}=QI$
by Remark 4.2 and
$\widetilde{I^{n+1}}=Q\widetilde{I^{n}}$
for all
$n\geqslant 2$
, we have
$\text{e}_{1}({\mathcal{F}})=\text{e}_{0}({\mathcal{F}})-\ell _{A}(A/I)+\ell _{A}(\widetilde{I^{2}}/QI)$
and
$\text{G}({\mathcal{F}})$
is a Cohen–Macaulay ring by [Reference Guerrieri and RossiGR, Theorem 2.2]. Then, since
$\ell _{A}(\widetilde{I^{2}}/I^{2})=1$
, we have
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
. We furthermore have
$\text{e}_{2}(I)=\text{e}_{2}({\mathcal{F}})=\ell _{A}(\widetilde{I^{2}}/QI)=\ell _{A}(I^{2}/QI)+1=\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)$
, and
$\text{e}_{i}(I)=\text{e}_{i}({\mathcal{F}})=0$
for
$3\leqslant i\leqslant d$
, because
$\text{G}({\mathcal{F}})$
is a Cohen–Macaulay ring (cf. [Reference Huckaba and MarleyHM, Proposition 4.6]).
Assume one of the equivalent conditions. We have
$\text{H}_{M}^{0}(G)=[\text{H}_{M}^{0}(G)]_{1}$
by the proof of the implication
$(1)\Rightarrow (3)$
. Let
$n\geqslant 3$
be an integer. We then have
$$\begin{eqnarray}\widetilde{I^{n}}/I^{n}=\widetilde{I^{n}}\cap I^{n-1}/I^{n}\cong [\text{H}_{M}^{0}(G)]_{n-1}=(0)\end{eqnarray}$$
because
$\widetilde{I^{n}}=Q^{n-2}\widetilde{I^{2}}\subseteq I^{n-1}$
. Therefore, we have
$\widetilde{I^{n}}=I^{n}$
for all
$n\geqslant 3$
.
We set
$W=\text{R}^{\prime }({\mathcal{F}})/R^{\prime }$
and look at the exact sequence
$$\begin{eqnarray}0\rightarrow R^{\prime }\rightarrow \text{R}^{\prime }({\mathcal{F}})\rightarrow W\rightarrow 0\quad (\ast ^{\prime })\end{eqnarray}$$
of graded
$R^{\prime }$
-modules. Since
$\widetilde{I^{n}}=I^{n}$
for all
$n\neq 2$
, we have
$W=W_{2}=\widetilde{I^{2}}/I^{2}$
so that
$\ell _{A}(W)=1$
. Then, because
$\text{G}({\mathcal{F}})=\text{R}^{\prime }({\mathcal{F}})/t^{-1}\text{R}^{\prime }({\mathcal{F}})$
is a Cohen–Macaulay ring, so is
$\text{R}^{\prime }({\mathcal{F}})$
. Let
$M^{\prime }=(\mathfrak{m},R_{+},t^{-1})R^{\prime }$
be the unique graded maximal ideal in
$R^{\prime }$
. Then applying local cohomology functors
$\text{H}_{M^{\prime }}^{i}(\ast )$
to the exact sequence
$(\ast ^{\prime })$
yields
$\text{H}_{M^{\prime }}^{i}(R^{\prime })=(0)$
for all
$i\neq 1,d+1$
and
$\text{H}_{M^{\prime }}^{1}(R^{\prime })=W$
. Since
$\mathfrak{m}W=(0)$
, we have
$\mathfrak{m}\text{H}_{M^{\prime }}^{1}(R^{\prime })=(0)$
. Thus,
$R^{\prime }$
is a Buchsbaum ring with the Buchsbaum invariant
$$\begin{eqnarray}\mathbb{I}(R^{\prime })=\mathop{\sum }_{i=0}^{d}\binom{d}{i}\ell _{A}(\text{H}_{M^{\prime }}^{i}(R^{\prime }))=d\end{eqnarray}$$
and hence so is the graded ring
$G=R^{\prime }/t^{-1}R^{\prime }$
. This completes the proof of Theorem 4.3.◻
In the rest of this section, we explore the relationship between the inequality of Northcott [Reference NorthcottN] and the structure of the graded module
$C$
of an integrally closed ideal.
It is well known that the inequality
$\text{e}_{1}(I)\geqslant \text{e}_{0}(I)-\ell _{A}(A/I)$
holds true [Reference NorthcottN] and the equality holds if and only if
$I^{2}=QI$
[Reference HunekeH, Theorem 2.1]. When this is the case, the associated graded ring
$G$
of
$I$
is Cohen–Macaulay.
Suppose that
$I$
is integrally closed and
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+1$
then, thanks to [Reference ItohI1, Corollary 14], we have
$I^{3}=QI^{2}$
and the associated graded ring
$G$
of
$I$
is Cohen–Macaulay. Thus the integrally closed ideal
$I$
with
$\text{e}_{1}(I)\leqslant \text{e}_{0}(I)-\ell _{A}(A/I)+1$
seems understood in a satisfactory way. In this section, we briefly study the integrally closed ideals
$I$
with
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+2$
, and
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
.
Let us begin with the following.
Theorem 4.4. Assume that
$I$
is integrally closed. Suppose that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+2$
and
$I^{3}\neq QI^{2}$
. Then the following assertions hold true.
-
(1)
$\ell _{A}(I^{2}/QI)=\ell _{A}(I^{3}/QI^{2})=1$
, and
$I^{4}=QI^{3}$
. -
(2)
$C\cong B(-2)$
as graded
$T$
-modules. -
(3)
$\text{depth}~G=d-1$
. -
(4)
$\text{e}_{2}(I)=3$
if
$d\geqslant 2$
,
$\text{e}_{3}(I)=1$
if
$d\geqslant 3$
, and
$\text{e}_{i}(I)=0$
for
$4\leqslant i\leqslant d$
. -
(5) The Hilbert series
$HS_{I}(z)$
is given by
$$\begin{eqnarray}HS_{I}(z)=\frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-1\}z+z^{3}}{(1-z)^{d}}.\end{eqnarray}$$
Proof. Because
$I^{3}\neq QI^{2}$
, it follows from Corollary 2.10 that
$$\begin{eqnarray}0<\ell _{A}(I^{2}/QI)<\text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)=2.\end{eqnarray}$$
Thus,
$\ell _{A}(I^{2}/QI)=1$
and
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
. Let
$I^{2}=QI+(xy)$
with
$x$
,
$y\in I\backslash Q$
. Then
$I^{3}=QI^{2}+(x^{2}y)$
, so that
$\ell _{A}(I^{3}/QI^{2})=1$
since
$I^{3}\neq QI^{2}$
and
$\mathfrak{m}I^{2}\subseteq QI$
. Thanks to Theorem 1.2,
$C\cong B(-2)$
as graded
$T$
-modules, so that assertions
$(1)$
,
$(2)$
,
$(4)$
, and
$(5)$
follow, and
$\text{depth}~G\geqslant d-1$
by Theorem 4.1. Since
$I^{3}\subseteq QI$
,
$G$
is not a Cohen–Macaulay ring, for otherwise
$I^{3}=Q\cap I^{3}=QI^{2}$
, so that
$\text{depth}G=d-1$
. This completes the proof of Theorem 4.4.◻
Notice that the following result also follows by [Reference Rossi and VallaRV3, Theorem 4.6].
Corollary 4.5. Assume that
$I$
is integrally closed and suppose that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+2$
. Then
$\text{depth}~G\geqslant d-1$
and
$I^{4}=QI^{3}$
, and the graded ring
$G$
is Cohen–Macaulay if and only if
$I^{3}=QI^{2}$
.
Before closing this section, we briefly study the integrally closed ideal
$I$
with
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
. Suppose that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
then we have
$$\begin{eqnarray}0<\ell _{A}(I^{2}/QI)\leqslant \text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)=3\end{eqnarray}$$
by Corollary 2.10. If
$\ell _{A}(I^{2}/QI)=1$
then we have
$\text{depth}G\geqslant d-1$
by [Reference Rossi and VallaRV1, Reference WangW]. If
$\ell _{A}(I^{2}/QI)=3$
then the equality
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)$
holds true, so that
$I^{3}=QI^{2}$
and the associated graded ring
$G$
of
$I$
is Cohen–Macaulay by Corollary 2.10. Thus we need to consider the following.
Theorem 4.6. Suppose that
$d\geqslant 2$
. Assume that
$I$
is integrally closed and
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
and
$\ell _{A}(I^{2}/QI)=2$
. Let
$c=\ell _{A}(I^{3}/QI^{2})$
. Then the following assertions hold true.
-
(1) Either
$C\cong B(-2)$
as graded
$T$
-modules or there exists an exact sequence of graded
$$\begin{eqnarray}0\rightarrow B(-3)\rightarrow B(-2)\oplus B(-2)\rightarrow C\rightarrow 0\end{eqnarray}$$
$T$
-modules.
-
(2)
$1\leqslant c\leqslant 2$
and
$I^{4}=QI^{3}$
. -
(3) Suppose
$c=1$
; then
$\text{depth}G\geqslant d-1$
and
$\text{e}_{2}(I)=4$
,
$\text{e}_{3}(I)=1$
if
$d\geqslant 3$
, and
$\text{e}_{i}(I)=0$
for
$4\leqslant i\leqslant d$
. -
(4) Suppose
$c=2$
; then
$\text{depth}G=d-2$
and
$\text{e}_{2}(I)=3$
,
$\text{e}_{3}(I)=-1$
if
$d\geqslant 3$
,
$\text{e}_{4}(I)=-1$
if
$d\geqslant 4$
, and
$\text{e}_{i}(I)=0$
for
$5\leqslant i\leqslant d$
. -
(5) The Hilbert series
$HS_{I}(z)$
is given by
$$\begin{eqnarray}HS_{I}(z)=\left\{\begin{array}{@{}rl@{}}\displaystyle \frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-2\}z+z^{2}+z^{3}}{(1-z)^{d}} & \quad \text{if }c=1,\\ \displaystyle \frac{\ell _{A}(A/I)+\{\text{e}_{0}(I)-\ell _{A}(A/I)-2\}z+3z^{3}-z^{4}}{(1-z)^{d}} & \quad \text{if }c=2.\end{array}\right.\end{eqnarray}$$
Proof. Since
$\ell _{A}(I^{2}/QI)=2$
and
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
, we have
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+\ell _{A}(I^{2}/QI)+1$
. We also have
$1\leqslant \ell _{A}(I^{3}/QI^{2})\leqslant 2$
(see the proof of [Reference Rossi and VallaRV2, Propositions 2.1 and 2.2]). Then, thanks to Theorem 1.2,
$C\cong X_{1}B(-1)$
or
$C\cong (X_{1},X_{2})B(-1)$
as graded
$T$
-modules, where
$X_{1}$
,
$X_{2}$
denote the linearly independent linear forms of
$B$
. Thus all assertions follow by Theorem 1.2.◻
We remark that
$\ell _{A}(I^{2}/QI)$
measures how far is the multiplicity of
$I$
from the minimal value (see [Reference Rossi and VallaRV3, Corollary 2.1]). If
$\ell _{A}(I^{2}/QI)\leqslant 1,$
then
$\text{depth}~G\geqslant d-1,$
but it is still open the problem whether
$\text{depth}G\geqslant d-2,$
assuming
$\ell _{A}(I^{2}/QI)=2$
. Theorem 4.6 confirms the conjectured bound.
Corollary 4.7. Assume that
$I$
is integrally closed. Suppose that
$\text{e}_{1}(I)=\text{e}_{0}(I)-\ell _{A}(A/I)+3$
. Then
$\text{depth}~G\geqslant d-2$
.
Proof. We have
$0<\ell _{A}(I^{2}/QI)\leqslant \text{e}_{1}(I)-\text{e}_{0}(I)+\ell _{A}(A/I)=3$
by Corollary 2.10. If
$\ell _{A}(I^{2}/QI)=1$
or
$\ell _{A}(I^{2}/QI)=3$
then we have
$\text{depth}G\geqslant d-1$
as above. Suppose that
$\ell _{A}(I^{2}/QI)=2$
then we have
$\text{depth}G\geqslant d-2$
by Theorem 4.6(3), (4). This completes a proof of Corollary 4.7.◻
5 An example
The goal of this section is to construct an example of a Cohen–Macaulay local ring with the maximal ideal
$\mathfrak{m}$
satisfying the equality in Theorem 1.2(1). The class of examples we exhibit includes an interesting example given by Wang, see [Reference Rossi and VallaRV3, Example 3.2].
Theorem 5.1. Let
$d\geqslant c\geqslant 1$
be integers. Then there exists a Cohen–Macaulay local ring
$(A,\mathfrak{m})$
such that
$$\begin{eqnarray}d=\dim A,\quad \text{e}_{1}(\mathfrak{m})=\text{e}_{0}(\mathfrak{m})+\ell _{A}(\mathfrak{m}^{2}/Q\mathfrak{m}),\quad \text{and}\quad c=\ell _{A}(\mathfrak{m}^{3}/Q\mathfrak{m}^{2})\end{eqnarray}$$
for some minimal reduction
$Q=(a_{1},a_{2},\ldots ,a_{d})$
of
$\mathfrak{m}$
.
To construct necessary examples we may assume that
$c=d$
. In fact, suppose that
$0<c<d$
and assume that we have already chosen a certain Cohen–Macaulay local ring
$(A_{0},\mathfrak{m}_{0})$
such that
$c=\operatorname{dim}A_{0}$
,
$\text{e}_{1}(\mathfrak{m}_{0})=\text{e}_{0}(\mathfrak{m}_{0})+\ell _{A_{0}}(\mathfrak{m}_{0}^{2}/Q_{0}\mathfrak{m}_{0})$
, and
$c=\ell _{A_{0}}(\mathfrak{m}_{0}^{3}/Q_{0}\mathfrak{m}_{0}^{2})$
with
$Q_{0}=(a_{1},a_{2},\ldots ,a_{c})A_{0}$
a minimal reduction of
$\mathfrak{m}_{0}$
. Let
$n=d-c$
and let
$A=A_{0}[[X_{1},X_{2},\ldots ,X_{n}]]$
be the formal power series ring over the ring
$A_{0}$
. We set
$\mathfrak{m}=\mathfrak{m}_{0}A+(X_{1},X_{2},\ldots ,X_{n})A$
and
$Q=Q_{0}A+(X_{1},X_{2},\ldots ,X_{n})A$
. Then
$A$
is a Cohen–Macaulay local ring with
$\operatorname{dim}A=d$
and maximal ideal
$\mathfrak{m}=\mathfrak{m}_{0}A+(X_{1},X_{2},\ldots ,X_{n})A$
. The ideal
$Q$
is a reduction of
$\mathfrak{m}$
and because
$X_{1},X_{2},\ldots ,X_{n}$
forms a super regular sequence in
$A$
with respect to
$\mathfrak{m}$
(recall that
$\text{G}(\mathfrak{m})=\text{G}(\mathfrak{m}_{0})[Y_{1},Y_{2},\ldots ,Y_{n}]$
is the polynomial ring, where
$Y_{i}$
’s are the initial forms of
$X_{i}$
’s), we have
$\text{e}_{i}(\mathfrak{m})=\text{e}_{i}(\mathfrak{m}_{0})$
for
$i=0,1$
,
$\mathfrak{m}^{2}/Q\mathfrak{m}\cong \mathfrak{m}_{0}^{2}/Q_{0}\mathfrak{m}_{0}$
, and
$\mathfrak{m}^{3}/Q\mathfrak{m}^{2}\cong \mathfrak{m}_{0}^{3}/Q_{0}\mathfrak{m}_{0}^{2}$
. Thus we have
$\text{e}_{1}(\mathfrak{m})=\text{e}_{0}(\mathfrak{m})+\ell _{A}(\mathfrak{m}^{2}/Q\mathfrak{m})$
and
$\ell _{A}(\mathfrak{m}^{3}/Q\mathfrak{m}^{2})=c$
. This observation allows us to concentrate our attention on the case where
$c=d$
.
Let
$m\geqslant 0$
and
$d\geqslant 1$
be integers. Let
$$\begin{eqnarray}D=k[[\{X_{j}\}_{1\leqslant j\leqslant m},Y,\{V_{i}\}_{1\leqslant i\leqslant d},\{Z_{i}\}_{1\leqslant i\leqslant d}]]\end{eqnarray}$$
be the formal power series ring with
$m+2d+1$
indeterminates over an infinite field
$k$
, and let
$$\begin{eqnarray}\displaystyle \mathfrak{a} & = & \displaystyle [(X_{j}|1\leqslant j\leqslant m)+(Y)]\cdot [(X_{j}|1\leqslant j\leqslant m)+(Y)+(V_{i}|1\leqslant i\leqslant d)]\nonumber\\ \displaystyle & & \displaystyle +\,(V_{i}V_{j}|1\leqslant i,j\leqslant d,i\neq j)+(V_{i}^{3}-Z_{i}Y|1\leqslant i\leqslant d).\nonumber\end{eqnarray}$$
We set
$A=D/\mathfrak{a}$
and denote the images of
$X_{j}$
,
$Y$
,
$V_{i}$
, and
$Z_{i}$
in
$A$
by
$x_{j}$
,
$y$
,
$v_{i}$
, and
$a_{i}$
, respectively. Then, since
$\sqrt{\mathfrak{a}}=(X_{j}|1\leqslant j\leqslant m)+(Y)+(V_{i}|1\leqslant i\leqslant d)$
, we have
$\dim A=d$
. Let
$\mathfrak{m}=(x_{j}|1\leqslant j\leqslant m)+(y)+(v_{i}|1\leqslant i\leqslant d)+(a_{i}|1\leqslant i\leqslant d)$
be the maximal ideal in
$A$
and we set
$Q=(a_{i}|1\leqslant i\leqslant d)$
. Then,
$\mathfrak{m}^{2}=Q\mathfrak{m}+(v_{i}^{2}|1\leqslant i\leqslant d)$
,
$\mathfrak{m}^{3}=Q\mathfrak{m}^{2}+Qy$
, and
$\mathfrak{m}^{4}=Q\mathfrak{m}^{3}$
. Therefore
$Q$
is a minimal reduction of
$\mathfrak{m}$
, and
$a_{1},a_{2},\ldots ,a_{d}$
is a system of parameters for
$A$
.
We are now interested in the Hilbert coefficients
$\text{e}_{i}(\mathfrak{m})$
of the maximal ideal
$\mathfrak{m}$
as well as the structure of the associated graded ring
$\text{G}(\mathfrak{m})$
and the module
$\text{C}_{Q}(\mathfrak{m})$
of
$\mathfrak{m}$
.
Theorem 5.2. The following assertions hold true.
-
(1)
$A$
is a Cohen–Macaulay local ring with
$\dim A=d$
. -
(2)
$\text{C}_{Q}(\mathfrak{m})\cong B_{+}(-1)$
as graded
$T$
-modules. Therefore,
$\ell _{A}(\mathfrak{m}^{3}/Q\mathfrak{m}^{2})=d$
. -
(3)
$\text{e}_{0}(\mathfrak{m})=m+2d+2$
,
$\text{e}_{1}(\mathfrak{m})=m+3d+2$
. -
(4)
$\text{e}_{2}(\mathfrak{m})=d+1$
if
$d\geqslant 2$
, and
$\text{e}_{i}(\mathfrak{m})=0$
for all
$3\leqslant i\leqslant d$
. -
(5)
$\text{G}(\mathfrak{m})$
is a Buchsbaum ring with
$\text{depth}\text{G}(\mathfrak{m})=0$
and
$\mathbb{I}(\text{G}(\mathfrak{m}))=d$
. -
(6) The Hilbert series
$HS_{\mathfrak{m}}(z)$
of
$A$
is given by
$$\begin{eqnarray}HS_{\mathfrak{m}}(z)=\frac{1+\{m+d+1\}z+\mathop{\sum }_{j=3}^{d+2}(-1)^{j-1}\binom{d+1}{j-1}z^{j}}{(1-z)^{d}}.\end{eqnarray}$$
Notice that Wang’s example before quoted corresponds to the particular case
$m=0$
and
$d=2.$
Let us divide the proof of Theorem 5.2 into two steps. Let us begin with the following.
Proposition 5.3. Let
$\mathfrak{p}=(X_{j}|1\leqslant j\leqslant m)+(Y)+(V_{i}|1\leqslant i\leqslant d)$
in
$D$
. Then
$\ell _{D_{\mathfrak{p}}}(A_{\mathfrak{p}})=m+2d+2$
.
Proof. Let
$\widetilde{k}=k[\{Z_{i}\}_{1\leqslant i\leqslant d},\{\frac{1}{Z_{i}}\}_{1\leqslant i\leqslant d}]$
and
$\widetilde{D}=D[\{\frac{1}{Z_{i}}\}_{1\leqslant i\leqslant d}]$
. We set
$X_{j}^{\prime }=\frac{X_{j}}{Z_{1}}$
for
$1\leqslant j\leqslant m$
,
$V_{i}^{\prime }=\frac{V_{i}}{Z_{1}}$
for
$1\leqslant i\leqslant d$
, and
$Y^{\prime }=\frac{Y}{Z_{1}}$
. Then we have
$$\begin{eqnarray}\displaystyle \widetilde{D} & = & \displaystyle \widetilde{k}[\{X_{j}^{\prime }|1\leqslant j\leqslant m\},Y^{\prime },\{V_{i}^{\prime }|1\leqslant i\leqslant d\}],\nonumber\\ \displaystyle \mathfrak{a}\widetilde{D} & = & \displaystyle [(X_{j}^{\prime }|1\leqslant j\leqslant m)+(Y^{\prime })]\cdot [(X_{j}^{\prime }|1\leqslant j\leqslant m)+(Y^{\prime })+(V_{i}^{\prime }|1\leqslant i\leqslant d)]\nonumber\\ \displaystyle & & \displaystyle +\,(V_{i}^{\prime }V_{j}^{\prime }|1\leqslant i,j\leqslant d,~i\neq j)+\biggl(\frac{Z_{1}^{2}}{Z_{i}}{V_{i}^{\prime }}^{3}-Y^{\prime }|1\leqslant i\leqslant d\biggr),\nonumber\end{eqnarray}$$
and
$\{X_{j}^{\prime }\}_{1\leqslant j\leqslant m}$
,
$Y^{\prime }$
, and
$\{V_{i}^{\prime }\}_{1\leqslant i\leqslant d}$
are algebraically independent over
$\widetilde{k}$
. Let
$$\begin{eqnarray}W=\widetilde{k}[\{X_{j}^{\prime }|1\leqslant j\leqslant m\},\{V_{i}^{\prime }|1\leqslant i\leqslant d\}]\end{eqnarray}$$
in
$\widetilde{D}$
and
$$\begin{eqnarray}\displaystyle \mathfrak{b} & = & \displaystyle [(X_{j}^{\prime }|1\leqslant j\leqslant m)+({V_{1}^{\prime }}^{3})]\cdot [(X_{j}^{\prime }|1\leqslant j\leqslant m)+(V_{i}^{\prime }|1\leqslant i\leqslant d)]\nonumber\\ \displaystyle & & \displaystyle +\,(V_{i}^{\prime }V_{j}^{\prime }|1\leqslant i,j\leqslant d,~i\neq j)+\biggl(\frac{Z_{1}^{2}}{Z_{i}}{V_{i}^{\prime }}^{3}-Z_{1}{V_{1}^{\prime }}^{3}|2\leqslant i\leqslant d\biggr)\nonumber\end{eqnarray}$$
in
$W$
. Then substituting
$Y^{\prime }$
with
$Z_{1}{V_{1}^{\prime }}^{3}$
in
$\widetilde{D}$
, we get the isomorphism
$$\begin{eqnarray}\widetilde{D}/\mathfrak{a}\widetilde{D}\cong W/\mathfrak{b}\end{eqnarray}$$
of
$\widetilde{k}$
algebras. Then the prime ideal
$\mathfrak{p}\widetilde{D}/\mathfrak{a}\widetilde{D}$
corresponds to the prime ideal
$P/\mathfrak{b}$
of
$W/\mathfrak{b}$
, where
$P=W_{+}=(X_{j}^{\prime }|1\leqslant j\leqslant m)+(V_{i}^{\prime }|1\leqslant i\leqslant d)$
. Then because
$$\begin{eqnarray}\displaystyle \mathfrak{b}+({V_{1}^{\prime }}^{3}) & = & \displaystyle (X_{j}^{\prime }|1\leqslant j\leqslant m)\cdot [(X_{j}^{\prime }|1\leqslant j\leqslant m)+(V_{i}^{\prime }|1\leqslant i\leqslant d)]\nonumber\\ \displaystyle & & \displaystyle +\,(V_{i}^{\prime }V_{j}^{\prime }|1\leqslant i,j\leqslant d,~i\neq j)+({V_{i}^{\prime }}^{3}|1\leqslant i\leqslant d)\nonumber\end{eqnarray}$$
and
$\ell _{W_{P}}([\mathfrak{b}+({V_{1}^{\prime }}^{3})]W_{P}/\mathfrak{b}W_{P})=1$
, we get
$$\begin{eqnarray}\displaystyle \ell _{W_{P}}(W_{P}/\mathfrak{b}W_{P}) & = & \displaystyle \ell _{W_{P}}(W_{P}/[\mathfrak{b}+({V_{1}^{\prime }}^{3})]W_{P})+\ell _{W_{P}}([\mathfrak{b}+({V_{1}^{\prime }}^{3})]W_{P}/\mathfrak{b}W_{P})\nonumber\\ \displaystyle & = & \displaystyle (m+2d+1)+1=m+2d+2.\nonumber\end{eqnarray}$$
Thus
$\ell _{A_{\mathfrak{p}}}(A_{\mathfrak{p}})=\ell _{W_{P}/\mathfrak{b}W_{P}}(W_{P}/\mathfrak{b}W_{P})=m+2d+2$
.◻
Thanks to the associative formula of multiplicity, we have
$$\begin{eqnarray}\text{e}_{0}(Q)=\ell _{A_{\mathfrak{p}}}(A_{\mathfrak{p}})\cdot \text{e}_{0}^{A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}}([Q+\mathfrak{p}A]/\mathfrak{p}A)=m+2d+2,\end{eqnarray}$$
because
$\mathfrak{p}=\sqrt{\mathfrak{a}}$
and
$A/\mathfrak{p}A=D/\mathfrak{p}\cong k[[Z_{i}|1\leqslant i\leqslant d]]$
. On the other hand, we have
$$\begin{eqnarray}A/Q\cong k[[\{X_{j}\}_{1\leqslant j\leqslant m},Y,\{V_{i}\}_{1\leqslant i\leqslant d}]]/\mathfrak{c}\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle \mathfrak{c} & = & \displaystyle (\,[(X_{j}|1\leqslant j\leqslant m)+(Y)]\cdot [(X_{j}|1\leqslant j\leqslant m)+(Y)+(V_{i}|1\leqslant i\leqslant d)]\nonumber\\ \displaystyle & & \displaystyle +\,(V_{i}V_{j}|1\leqslant i,j\leqslant d,~i\neq j)+({V_{i}}^{3}|1\leqslant i\leqslant d).\nonumber\end{eqnarray}$$
Therefore,
$\ell _{A}(A/Q)=m+2d+2$
. Thus
$\text{e}_{0}(Q)=\ell _{A}(A/Q)$
so that
$A$
is a Cohen–Macaulay local ring with
$\text{e}_{0}(Q)=m+2d+2$
.
Let
$K_{0}=A$
,
$K_{1}=\mathfrak{m}$
, and
$K_{n}=\mathfrak{m}^{n}+y\mathfrak{m}^{n-2}$
for
$n\geqslant 2$
, and we set
${\mathcal{K}}=\{K_{n}\}_{n\geqslant 0}$
. Let
$\text{e}_{i}({\mathcal{K}})$
denote the
$i$
th Hilbert coefficients of the filtration
${\mathcal{K}}$
for
$0\leqslant i\leqslant d$
.
Lemma 5.4. The following assertions hold true.
-
(1)
$\ell _{A}(K_{2}/\mathfrak{m}^{2})=1$
and
$K_{n}=\mathfrak{m}^{n}$
for all
$n\geqslant 3$
. -
(2)
$Q\cap K_{2}=QK_{1}$
and
$K_{n+1}=QK_{n}$
for all
$n\geqslant 2$
. Therefore,
$\text{e}_{1}({\mathcal{K}})=\text{e}_{0}({\mathcal{K}})-\ell _{A}(A/K_{1})+\ell _{A}(K_{2}/QK_{1})$
,
$\text{e}_{2}({\mathcal{K}})=\ell _{A}(K_{2}/QK_{1})$
if
$d\geqslant 2$
, and
$\text{e}_{i}({\mathcal{K}})=0$
for
$3\leqslant i\leqslant d$
.
Proof. (1) Since
$K_{2}=\mathfrak{m}^{2}+(y)$
, we have
$\ell _{A}(K_{2}/\mathfrak{m}^{2})=1$
. We have
$K_{n}=\mathfrak{m}^{n}+y\mathfrak{m}^{n-2}=\mathfrak{m}^{n}$
for all
$n\geqslant 3$
, because
$y\mathfrak{m}=(yv_{i}|1\leqslant i\leqslant d)=({v_{i}}^{3}|1\leqslant i\leqslant d)\subseteq \mathfrak{m}^{3}$
.
$(2)$
Since
$K_{n}=\mathfrak{m}^{n}$
for all
$n\geqslant 3$
by assertion
$(1)$
, we have
$K_{2}\subseteq \widetilde{\mathfrak{m}^{2}}$
. Therefore,
$Q\cap K_{2}\subseteq Q\cap \widetilde{\mathfrak{m}^{2}}=Q\mathfrak{m}=QK_{1}$
by Remark 4.2. It is routine to check that
$K_{n+1}=QK_{n}$
for all
$n\geqslant 2$
. Thus
$\text{e}_{1}({\mathcal{K}})=\text{e}_{0}({\mathcal{K}})-\ell _{A}(A/K_{1})+\ell _{A}(K_{2}/QK_{1})$
by [Reference Guerrieri and RossiGR, Theorem 2.2]. We also have
$\text{e}_{2}({\mathcal{K}})=\ell _{A}(K_{2}/QK_{1})$
if
$d\geqslant 2$
, and
$\text{e}_{i}({\mathcal{K}})=0$
for
$3\leqslant i\leqslant d$
by [Reference Huckaba and MarleyHM, Proposition 4.6].◻
We prove now Theorem 5.2.
Proof of Theorem 5.2.
Since
$K_{n}=\mathfrak{m}^{n}$
for all
$n\geqslant 3$
by Lemma 5.4(1), we have
$\text{e}_{i}({\mathcal{K}})=\text{e}_{i}(\mathfrak{m})$
for
$0\leqslant i\leqslant d$
. Therefore,
$\text{e}_{1}(\mathfrak{m})=\text{e}_{0}(\mathfrak{m})+\ell _{A}(\mathfrak{m}^{2}/Q\mathfrak{m})$
,
$\text{e}_{2}(\mathfrak{m})=\ell _{A}(\mathfrak{m}^{2}/Q\mathfrak{m})+1$
if
$d\geqslant 2$
, and
$\text{e}_{i}(\mathfrak{m})=0$
for all
$3\leqslant i\leqslant d$
, because
$\ell _{A}(K_{2}/\mathfrak{m}^{2})=1$
,
$\text{e}_{1}({\mathcal{K}})=\text{e}_{0}({\mathcal{K}})+\ell _{A}(K_{2}/QK_{1})-1$
,
$\text{e}_{2}({\mathcal{K}})=\ell _{A}(K_{2}/QK_{1})$
if
$d\geqslant 2$
, and
$\text{e}_{i}({\mathcal{K}})=0$
for
$3\leqslant i\leqslant d$
by Lemma 5.4. Then we have
$\text{e}_{1}(\mathfrak{m})=m+3d+2$
and
$\text{e}_{2}(\mathfrak{m})=d+1$
because
$\text{e}_{0}(\mathfrak{m})=m+2d+2$
and
$\ell _{A}(\mathfrak{m}^{2}/Q\mathfrak{m})=d$
. The ring
$\text{G}(\mathfrak{m})$
is Buchsbaum ring with
$\text{depth}\text{G}(\mathfrak{m})=0$
and
$\mathbb{I}(\text{G}(\mathfrak{m}))=d$
by Theorem 4.3. This completes the proof of Theorem 5.2.◻
