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\begin{document}

\title[The leading ideal of a complete intersection]{The leading ideal of a complete
intersection \\ of height two, Part II}


%    Information for first author
\author{Shiro Goto}
\address{Department of Mathematics, School of Science and Technology
Meiji University, 214-8571 Japan}

\email{goto@math.meiji.ac.jp}

\thanks{Shiro Goto is supported by the Grant-in-Aid for Scientific
Researches in Japan (C(2), No.1364044).}

\author{William Heinzer}
\address{Department of Mathematics, Purdue University, West
Lafayette, Indiana 47907}


\email{heinzer@math.purdue.edu}





\author{Mee-Kyoung Kim}
\address{
Department of Mathematics, Sungkyunkwan University
Jangangu Suwon 440-746, Korea}
\email{mkkim@skku.edu}

\thanks{Mee-Kyoung Kim  is supported by the Korea Research Foundation
(R04-2003-000-10113-0)2003.}


\

\date \today

\subjclass{Primary: 13A30, 13C05; Secondary: 13E05, 13H15}

\keywords{ideal of initial forms, associated graded ring,
multiplicity, reduction number,
Gorenstein ring,  Cohen-Macaulay ring}

\maketitle

\begin{abstract}
\noindent Let $(S,\n)$ be a regular local ring and let $I = (f, g)$
be an ideal in $S$  generated by a regular sequence $f, g$ of length
two. Let $R = S/I$ and $\frak{m} = \frak{n}/I$. As in \cite{GHK}, we
examine the leading form ideal $I^*$ of $I$ in the associated graded
ring $G = \gr_{\n}(S)$. If $\mathrm{gr}_{\frak{m}}(R)$ is
Cohen-Macaulay, we describe precisely the Hilbert series
$\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda)$ in terms of the degrees
of homogeneous generators of $I^*$ and of their successive  GCD's.
If $D = \GCD(f^*, g^*)$ is a prime element of $\gr_{\n}(S)$ that is
regular on $\gr_{\n}(S)/(\frac{f^*}{D}, \frac{g^*}{D})$, we prove
that $I^*$ is 3-generated and a perfect ideal. If
$\mathrm{ht}_{\mathrm{gr}_{\frak{n}}(S)}(f^*, g^*, h^*) = 2$, where
$h \in I$ is such that $h^*$ is of minimal degree in $I^* \setminus
(f^*, g^*)\gr_{\n}(S)$, we prove  $I^*$ is 3-generated and a perfect
ideal of $\gr_{\n}(S)$, so $\gr_{\frak{m}}(R) = \gr_{\n}(S)/I^*$ is
a Cohen-Macaulay ring. We give several examples to illustrate our
theorems.
\end{abstract}

\baselineskip 18 pt


\section{Introduction}




 This paper examines generators of the defining ideal of the tangent cone of a complete intersection
 of codimension two.  We fix the following notation.

\begin{setting} \label{1.0}
{\em Let $(S,\frak{n})$ be a regular local ring of dimension $s \geq
2$ and let $I = (f,g)$ be an ideal in $S$ generated by a regular
sequence $f, g$ of length two. For simplicity we assume that the
residue class field $k = S/\fkn$ is infinite. We put $R=S/I$ and
$\fkm = \fkn/I$. Let
$$\mathrm{R}'(\frak{n}) = \sum_{i \in \mathbb Z}\fkn^i t^i \subseteq S[t,t^{-1}]\  \ \ \text{and}
\ \ \mathrm{R}'(\frak{m}) = \sum_{i \in \mathbb Z}\frak{m}^i t^i
\subseteq R[t,t^{-1}]$$ denote the Rees algebras of $\frak{n}$ and
$\frak{m}$ respectively, where $t$ is an indeterminate. We put $$G =
\gr_{\frak{n}}(S) = \mathrm{R}'(\fkn)/t^{-1}\mathrm{R}'(\frak{n}) \
\ \ \text{and} \ \ \gr_{\frak{m}}(R) =
\mathrm{R}'(\frak{m})/t^{-1}\mathrm{R}'(\frak{m}).$$ For each $0 \ne
h \in S$ let $\mathrm{o}(h) = \mathrm{sup}\{i \in \Z \mid h \in
\frak{n}^i \}$ and  put $h^* = \overline{ht^n}$, where $n =
\mathrm{o}(h)$ and $\overline{ht^n}$ denotes the image of $ht^n$ in
$G$.
 The canonical
map $S \to R$ induces the epimorphism $\varphi : G \to
\gr_{\frak{m}}(R)$ of the associated graded rings. We put  $$I^*=
\mathrm{Ker}~(G \overset{\varphi}\to \gr_{\fkm}(R)).$$ Then the
homogeneous components $\{[I^*]_i \}_{i \in \mathbb{Z}}$ of the leading
form ideal $I^*$ of $I$ are given by $$[I^*]_i = \{ \overline{ht^i}
\mid  h \in I \cap \fkn^i\}$$ for each $i \in \mathbb{Z}$. We
throughout assume that $a=\mathrm{o}(f) \leq b=\mathrm{o}(g)$ and
that $f^* \nmid g^*$ in $G$. The latter part of the condition  is
equivalent to saying that $f^*, g^*$ form a part of a minimal
homogeneous system of generators of $I^*$. }
\end{setting}







 The original motivation for our work
 comes from
 a paper of S. C. Kothari \cite{K}. Kothari
answers several questions raised by  Abyhankar concerning
 the local Hilbert function of a pair of plane curves.
Let $\ell_S(*)$  denote  length over $S$. In  the case where $\dim S
= 2$, Kothari   proves that $0 \leq
\dim_k[\mathrm{gr}_{\frak{m}}(R)]_{i} -
\dim_k[\mathrm{gr}_{\frak{m}}(R)]_{i+1} \leq 1$ for all $i \geq a$
and that $\ell_S(R) \geq ab$; moreover,  one has the equality
$\ell_S(R) = ab$ if and only if $f^*,~g^*$ are coprime in $G$, that
is, $f^*,~g^*$ form a $G$-regular sequence.



 We have
 subsequently learned from an informative referee report of other work in this area.
 Indeed, F. Macaulay in a 1904
 paper \cite{M} employs a different method to determine the same
 necessary condition as Kothari on the Hilbert function of a pair of plane
 curves. Using his inverse systems, Macaulay establishes
 the structure of the Hilbert function $H(A)$ of a complete
 intersection quotient $A = k[[x, y]]/(f,g)$ to be of the form
 \begin{equation}\label{11}
 H = (1,2, \ldots , a, t_a, \ldots, t_j, 0),
 \end{equation}
 where $a \ge t_a \ge t_{a+1} \ge \cdots \ge t_j = 1$ and $|t_i -
 t_{i+1}| \le 1$ for all $i$. Thus the Hilbert function $H$ after
an initial rising segment breaks up into platforms and regular
flights of descending stairs, each step of height one.
The structure of $H(A)$ is studied from the point of view of
parametrizations by J. Brian\c{c}on \cite{Br} and by A. Iarrobino
\cite{Ia1} and \cite{Ia2}. These authors prove that every sequence
satisfying  the conditions in Equation \ref{11} is realizable as the
Hilbert function $H(A)$ of some Gorenstein Artin algebra of the form
$A = k[[x, y]]/(f,g)$.




Let
$v(H) = 2 + \#\{ \text{platforms} \}$.
Iarrobino \cite{Ia1}, \cite{Ia2}  proves that $I^*$ needs two initial generators
$f^*, g^*$ and requires a new generator following each platform,
and that $v(H)$ is the minimum possible number of generators of a graded
ideal defining a standard algebra with Hilbert function $H$. In
\cite[Theorem 2.2.A]{Ia1}, Iarrobino characterizes those graded ideals
corresponding to $I^*$ for which $I$ is a complete intersection of
height two. He proves they are exactly the graded ideals with $v(H)$
generators. The referee has pointed out that our results in
Theorem \ref{1.11} and Theorem \ref{1.12} can be deduced from
these results of Iarrobino.  While acknowledging the priority of
these results of Iarrobino,
we hope that our different approach is still of some interest.






\begin{thm} \label{1.11} Let notation be as in Setting \ref{1.0} and
assume that  $\dim S = 2$ and  $n = \mu_G(I^*)$. Then $I^*$ contains
a homogeneous system $\{ \xi_i \}_{1 \leq i \leq n}$ of generators
that satisfy the following three conditions.

\begin{enumerate}
\item[$(1)$] $\xi_1 = f^*$ and $\xi_2 = g^*$.
\item[$(2)$] $\mathrm{deg} \xi_i + 2 \leq \mathrm{deg} \xi_{i+1}$ for all~$2 \leq i \leq n-1$.
\item[$(3)$] $\mathrm{ht}_G(\xi_1, \xi_2, \cdots, \xi_{n-1}) = 1$.
\end{enumerate}
\end{thm}





Let $\{ \xi_i \}_{1 \leq i \leq n}$ be a homogeneous system of
generators of $I^*$ satisfying conditions (1) and (2) in Theorem
\ref{1.11}. We prove that the ideals
$$\{(\xi_j \mid 1
\leq j \leq i )G \}_{1 \leq i \leq n}
$$ of $G$ are independent of the
particular choice of the family $\{ \xi_i \}_{1 \leq i \leq n}$ and
are uniquely determined by $I$. Moreover, if $D_i =
\mathrm{GCD}(\xi_j \mid 1 \leq j \leq i)$ and $d_i =
\mathrm{deg}D_i$, then one has  the strictly descending sequence
$$a=d_1 > d_2
> \cdots > d_{n-1} > d_n =0$$ and
$\displaystyle\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_{1}}{D_{i}},
\frac{\xi_{2}}{D_{i}}, \cdots, \frac{\xi_{i}}{D_{i}})$ for all $1
\leq i \leq n-1$ (Lemma \ref{lem3-2} ). Let $c_i =
\mathrm{deg}~\xi_i$ and let
$$\mathrm{H}(\mathrm{gr}_{\frak{m}}(R),
\lambda)=
\sum_{i=0}^{\infty}\dim_k[\mathrm{gr}_{\frak{m}}(R)]_i\lambda^i
$$
denote the Hilbert series of $\mathrm{gr}_{\frak{m}}(R)$. We
explicitly describe $\mathrm{H}(\mathrm{gr}_{\frak{m}}(R), \lambda)$
and the difference $\ell_S(R) - ab$ in terms of $c_i$ and $d_i$,
sharpening results proved by Kothari in \cite{K}.

\begin{thm}  \label{1.12}
Let notation be as in Setting \ref{1.0} and assume that  $\dim S =
2$ and  $n = \mu_G(I^*)$.  The following assertions hold true.
\begin{enumerate}
\item[$(1)$] $\mathrm{H}(\mathrm{gr}_{\frak{m}}(R), \lambda) =
\frac{\sum_{i=2}^n\lambda^{d_i}(1-\lambda^{d_{i-1}
-d_i})(1-\lambda^{c_i -d_i})}{(1-\lambda)^2}.$
\item[$(2)$] $\ell_S(R) = \sum_{i = 2}^{n} (d_{i-1} - d_i) (c_i - d_i) =
  ab + \sum_{i = 2}^{n-1} d_i{\cdot} [(c_{i+1}- c_i) - (d_{i-1} - d_i)]$.
\item[$(3)$] $c_{i+1} -c_i > d_{i-1} -d_{i} > 0$ for all $2 \leq i \leq n-1$.
\item[$(4)$] {\rm \cite[Corollary1]{K}} $\ell_S(R) = ab$ if
  and only if $n = 2$, i.e., $f^*, g^*$ is a $G$-regular sequence.
\end{enumerate}
\end{thm}



\begin{remark} \label{1.10} {\em
In the case where $\dim S = s > 2$, it is still true that
$\mathrm{ht}_G(f^*, g^*) > 1$ implies  $f^*, g^*$ is a $G$-regular
sequence, and therefore  $I^* =(f^*, g^*)G$ also in this case.
 Thus we assume that  $\mathrm{ht}_G(f^*, g^*)
= 1$ and put $D_2 = \mathrm{GCD}(f^*, g^*)$ and $d_2 = \mathrm{deg}
D_2$. Let $f^* = D_2\xi$ and $g^* = D_2\eta$. Notice that $\xi, \eta$ is
a regular sequence in $G$. We have  $b \geq a > d_2 > 0$, and
$\mu_G(I^*) = n \ge 3$. There exists a minimal homogeneous system
$\{ \xi_1, \xi_2, \ldots, \xi_n \}$ of generators of $I^*$ such that
$\xi_1 = f^*$ and $\xi_2 = g^*$, and  $c_i := \deg \xi_i \le \deg
\xi_{i+1} := c_{i+1}$ for each $i \le n-1$. However, the ideal $I^*$
may fail to be perfect, and it is possible to have $D_3 : =
\GCD(\xi_1, \xi_2, \xi_3) = D_2$ as is illustrated in \cite[Example
1.6]{GHK}. We prove in \cite[Theorem 1.2]{GHK} that $I^*$ is perfect
if $n = 3$. We also prove  in \cite{GHK} that $\xi_3 = h^*$, where
$h$  has the form $h = \alpha f + \beta g \in I$ with
$\mathrm{o}(\alpha) = b-d_2$, and $\mathrm{o}(\beta) = a-d_2$, and
that $c_3 := \mathrm{o}(h)
> a + b -d_2$.  Moreover, if $q = \sigma f + \tau g$ is such that
$q^* \not\in (f^*, g^*)G$ and $\mathrm(o)(\sigma) = b-d_2$, then
$\mathrm{o}(q) = \mathrm{o}(h)$ and $(f^*, g^*, h^*)G = (f^*, g^*,
q^*)G$. Thus the ideal $(\xi_1, \xi_2, \xi_3)G$ is independent of
the choice of $\xi_3$. In the case where $n \ge 4$, we also prove
that $c_4 \ge c_3 + 2$  \cite[Proposition 2.4]{GHK}. However,
examples shown to us by Craig Huneke and Lance Bryant show that it
is possible to have  $c_{i+1} =  c_i$  for $i \ge 4$. This resolves
a question mentioned in  \cite[Discussion 2.5]{GHK}).


}
\end{remark}



If $\mathrm{gr}_{\frak{m}}(R)$ is a Cohen-Macaulay ring, we prove in
Section 4 by passing to the factor ring of $G$ modulo a suitable
linear system of parameters for $\mathrm{gr}_{\frak{m}}(R)$ that it
is possible to  reduce the problems to the case where $\dim S = 2$
and obtain  results corresponding to those proved in Section 3 about
the Hilbert series $\mathrm{H}(\mathrm{gr}_\frak{m}(R), \lambda)$.
In particular, if $I^*$ is perfect, then $c_{i+1} > c_i + 1$ for
each $i$ with $2 \le i \le n-1$.




With notation as in Setting \ref{1.0}, let  $\mathrm{e}_{\frak{m}}^0
(R)$ denotes the multiplicity of $R$ with respect to $\frak{m}$.
Using Theorem \ref{1.11}, we prove in Section 4:



\begin{thm} \label{1.5}
Assume notation as in Setting \ref{1.0} and Remark \ref{1.10}, and let $D := D_2$, $d :=
d_2$ and $c := c_3$. If $\mathrm{ht}_G(f^*, g^*, h^*)=2$, then the
following assertions hold true.
\begin{enumerate}
\item[$(1)$] $I^* = (f^*, g^*, h^*)$.
\item[$(2)$] $\mathrm{gr}_{\frak{m}}(R)$ is a Cohen-Macaulay ring.
\item[$(3)$] $\mathrm{H}(\mathrm{gr}_\frak{m}(R), \lambda)=\frac{(1-\lambda^{c})(1-\lambda^{d})+
\lambda^{d}(1-\lambda^{a-d})(1-\lambda^{b-d})}{(1-\lambda)^{\dim S}}$.
\item[$(4)$] $\mathrm{e}_{\frak{m}}^0(R) = ab + d{\cdot}[(c+d) -(a+b)].$
\end{enumerate}
\end{thm}





Let $M =[\mathrm{gr}_{\fkm}(R)]_+$ and let
$\mathrm{H}_{M}^{s-2}(\mathrm{gr}_{\fkm}(R))$ denote the
${s-2~}^{\underline{th}}$ local cohomology module of
$\mathrm{gr}_{\fkm}(R)$ with respect to $M$. Recall that
$$\mathrm{a}(\mathrm{gr}_{\fkm}(R)) = \max\{i \in \mathbb{Z}
\mid  [\mathrm{H}_{M}^{s-2}(\mathrm{gr}_{\fkm}(R))]_i \ne (0)\}$$
is the $\mathrm{a}$-invariant of $\mathrm{gr}_{\fkm}(R)$. Using this
notation and setting $Q = (X_1, \ldots, X_{s-2})G$, where $X_1,
\ldots, X_s$ are suitably chosen homogeneous elements of degree one
in $G$ such that $G = k[X_1, \ldots, X_s]$, and using  the formula
$$\mathrm{a}(\mathrm{gr}_{\fkm}(R)/Q \mathrm{gr}_{\fkm}(R)) =
\mathrm{a}(\mathrm{gr}_{\fkm}(R)) + (s-2)$$ of \cite[Remark
(3.1.6)]{GW},  we establish  the following result in Section 4.

\begin{thm} \label{1.6}
Assume notation as in Setting \ref{1.0} and Remark \ref{1.10}. If $\mathrm{gr}_{\fkm}(R)$
is a Cohen-Macaulay ring and  $n = \mu_G(I^*)$, then the following
assertions hold true.
\begin{enumerate}
\item[$(1)$] $\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda)
=\displaystyle\frac{\sum_{i=2}^{n}\lambda^{d_i}(1 -
\lambda^{d_{i-1} -d_i})(1 - \lambda^{c_i -d_i})}{(1 - \lambda)^s}.$
\item[$(2)$] $\mathrm{e}^0_{\fkm}(R) = ab +
\sum_{i = 2}^{n-1}d_i{\cdot}[(c_{i+1} - c_i) - (d_{i-1} - d_i)]$
with $$c_{i+1} - c_i > d_{i-1} - d_i > 0$$ for all $2 \leq i \leq n
-1$.
\item[$(3)$] $\mathrm{e}^0_{\fkm}(R) \leq a{\cdot}[c_n + d_{n-1}-a]$,
where  the equality holds true if and only if $n=2$.
\item[$(4)$] $\mathrm{a}(\mathrm{gr}_{\fkm}(R))=c_n + d_{n-1}-s$.
\end{enumerate}
\end{thm}




Sections 5 is  devoted to some examples, which illustrate our
theorems. Let $H=\left<n_1,n_2,n_3\right>$ be a Gorenstein numerical
semigroup generated by the three integers $n_1, n_2, n_3$, where  $0
< n_1< n_2<n_3$ and $\mathrm{GCD}(n_1,n_2,n_3) =1$. Let $S=k[[X_1,
X_2, X_3]]$ and $T=k[[t]]$ be formal power series rings over a field
$k$. We denote by $\varphi  : S \to T$ the $k$-algebra map defined
by $\varphi(X_i)=t^{n_i}$ for $i = 1, 2, 3$. Let $I=\mathrm{Ker }\,
\varphi, \ R =k[[t^{n_1}, t^{n_2}, t^{n_3}]], \
\frak{n}=(X_1,X_2,X_3)S, \  \text{and} \  \frak{m} =(t^{n_1},
t^{n_2}, t^{n_3})R.$ Then, as was essentially shown in \cite{H2} and
\cite{RV}, $\gr_{\frak{m}}(R)$ is a Cohen-Macaulay ring if and only
if $I^*$ is $3$-generated. We shall recover this result in our
context. In Example \ref{5.5}, we present a family of examples due to Takahumi Shibuta 
that demonstrates that for $I =\mathrm{Ker }\,\varphi$ as above,
there is no bound on the  number of elements needed to generate $I^*$. 


\section{Preliminaries}
Throughout this section, let notation be as in
Setting \ref{1.0}, assume that $\dim S = 2$ and
let $\fkn = (x,y)$.



\begin{Lem}\label{lem2-1}
Let $h \in S$ with $m = \mathrm{o}(h)$ and assume that $x^* \nmid h^*$.
Then $h = \varepsilon y^m + x \varphi$ for some $\varepsilon \in \rmU(S)$ and $\varphi \in \fkn^{m-1}$.
\end{Lem}

\begin{proof}
Let $\overline{S} = S/(x)$ and denote by $\overline{*}$ the image
in $\overline{S}$. Let $\ell = \mathrm{o}(\overline{h})$. Then $\ell
\geq m$ and $\overline{h} = \overline{\varepsilon} \cdot
\overline{y}^\ell$ for some $\varepsilon \in \rmU(S)$. We write $h =
\varepsilon y^\ell + x\varphi$ with $\varphi \in S$. Then $\varphi
\in \fkn^{m-1}$, because $(x) \cap \fkn^m = x \fkn^{m-1}$. Hence
$\ell = m$, as $x^* \nmid h^*$.
\end{proof}

\begin{Lem}\label{lem2-2}
There exist elements $x,y,u,$ and $g_1 \in S$ satisfying the following conditions.
\begin{enumerate}
  \item[$(1)$] $\fkn = (x,y)$ and $x^* \nmid f^*$.
  \item[$(2)$] $u \in \rmU(S)$, $\mathrm{o}(g_1) = b-1$, and $g = uy^{b-a}f + xg_1$
\end{enumerate}
\end{Lem}

\begin{proof}
Let $\fkn = (x,y)$. Then, since $k = S/\fkn$ is infinite, we have
$x^* + cy^* \nmid f^*$ and $x^* + cy^* \nmid g^*$ for some $c \in
k$. Let $c \equiv    \alpha \mod \fkn$ $(\alpha \in S)$ and $z =
x+\alpha y$. Then $\fkn = (z,y)$. Because $z^* \nmid f^*$ and $z^*
\nmid g^*$, by Lemma \ref{lem2-1}, we have
$$
f = \varepsilon y^a + z \xi \ \ \text{and} \ \  g = \tau y^b + z \eta
$$
for some $\varepsilon, \tau \in \rmU(S)$, $\xi \in \fkn^{a-1}$, and $\eta \in \fkn^{b-1}$.
Let $g_1 = \eta - u y^{b-a}\xi$ where $u= \tau \varepsilon^{-1}$.
Then $g = u y^{b-a} f + zg_1$ and $\mathrm{o}(g_1) = b-1$,
because $g_1 \in \fkn^{b-1}$ and $f^* \nmid g^*$.
Replacing $x$ with $z$, we get the required elements $x, y, u$, and $g_1 \in S$ as claimed.
\end{proof}

In what follows let $x,y,u$, and $g_1 \in S$ be elements
which satisfy conditions $(1)$ and $(2)$ in Lemma \ref{lem2-2}.
We put $I_1 = (f,g_1)$, $X = x^*$, and $Y=y^*$.

\begin{Prop} \label{prop2-3}
The following assertions hold true.
\begin{enumerate}
  \item[$(1)$] $I = (f,xg_1)$ and $I:_Sx = I_1$.
  \item[$(2)$] $(f^*, g^*) = (f^*, Xg_1^*)$ whence $f^* \nmid {g_1}^*$.
  \item[$(3)$] $f^*, X$ is a $G$-regular sequence.
  \item[$(4)$] $I=\fkn$, if $b = 1$.
  \item[$(5)$] {\rm (\cite{K})} Suppose that $b > 1$. Then $I_1$
  is a parameter ideal in $S$ and $I^* = (f^*) + X I_1^*$.
  Hence $I^* :_G X = I_1^*$.
\end{enumerate}
\end{Prop}

\begin{proof}
$(1)$ Since $g = uy^{b-a}f + xg_1$, we get $I = (f,xg_1)$, whence $xI_1 \subseteq I$.
Let $\varphi \in I:_S x$ and write $x\varphi = \alpha f + \beta (xg_1)$ ($\alpha, \beta \in S$).
Then $x(\varphi - \beta g_1) \in (f)$ so that $\varphi - \beta g_1 \in (f)$,
because $f,x$ is a regular sequence in $S$ (recall that $x \nmid f$).
Hence $\varphi \in (f,g_1) = I_1$ and thus $I:_Sx = I_1$.

$(2)$ Recall that $g^* = u^* Y^{b-a}f^* + Xg_1^*$.

$(3)$ This is clear, since $X \nmid f^*$.

$(4)$ We have $a=1$, since $a \leq b$. Hence $\mathrm{o}(g_1) = 0$
and $\mathrm{o}(f \mod (x)) = 1$ (cf. Proof of Lemma \ref{lem2-1}),
so that we have $I = (f,xg_1) = (f,x) = \fkn$.

$(5)$ Since $b > 1$, we get $I \subseteq I_1 \subsetneq S$. Hence $I_1$ is a parameter ideal of $S$.
Let $i \geq a-1$ be an integer. Then, thanks to Proof of \cite[Lemma]{K}, we see that for every $k$ -basis
$W_1, W_2, \cdots , W_r$ of $[I_1^*]_i$, the
elements $Y^{i+1-a} f^*, XW_1, XW_2, \cdots, XW_r$ form a $k$-basis of $[I^*]_{i+1}$.
Consequently, $[I^*]_{i+1} \subseteq (f^*) + X I_1^* \subseteq I^*$ (recall that $xI_1 \subseteq I$),
whence $I^* = (f^*) + X I_1^*$, because $[I^*]_i = (0)$ for $i \leq a-1$.
As $f^*, X$ is a $G$-regular sequence, we have the
equality $I^* :_G X = I_1^*$ similarly as in the proof of assertion $(1)$.
\end{proof}


\begin{Cor} \label{cor2-4}
Suppose that $b > 1$. Then
$\rmH(G/I^*, \lambda) = \sum_{i = 0}^{a-1} \lambda^i + \lambda{\cdot} \rmH(G/I_1^*, \lambda)$.
\end{Cor}

\begin{proof}
Notice that $(X, f^*) / I^* = (X, f^*) /[ (f^*) + X I_1^*] \cong (X) / XI_1^* \cong (G/I_1^*)(-1)$,
because $(X) \cap (f^*) = (X f^*)$ and $f^* \in I_1^*$.
Then we get the exact sequence
$$
0 \to (G/I_1^*)(-1) \to G/I^* \to G/(X, f^*) \to 0
$$
of graded $G$-modules, so that
\begin{eqnarray*}
\rmH(G/I^*, \lambda) &=& \rmH(G/(X, f^*), \lambda) + \lambda{\cdot}\rmH(G/I_1^*, \lambda)\\
&=& \sum_{i = 0}^{a-1} \lambda^i + \lambda{\cdot} \rmH(G/I_1^*, \lambda)
\end{eqnarray*}
as claimed.
\end{proof}

The following fact plays a key role in our argument.

\begin{Cor} \label{thm2-5}
Suppose  that $b > 1$. Let $n = \mu_G(I^*)$ and $\ell =
\mu_G(I_1^*)$.
\begin{enumerate}
  \item[$(1)$] Suppose that $a < b$. Then $n = \ell$ and, for every homogeneous
  system $\{\eta_i\}_{1 \leq i \leq n}$ of generators of $I_1^*$ with $\eta_1 = f^*$
  and $\eta_2 = g_1^*$, we have $I^* = (f^*,g^*) + (X\eta_i \mid 3 \leq i \leq n)$.
  \item[$(2)$] Suppose that $a = b$ and $g_1^* \nmid f^*$. Then $n = \ell$ and, for
  every homogeneous system $\{\eta_i\}_{1 \leq i \leq n}$ of generators of $I_1^*$
  with $\eta_1 = g_1^*$ and $\eta_2 = f^*$, we have $I^* = (f^*, g^*) + (X\eta_i \mid 3 \leq i \leq n)$.
  \item[$(3)$] Suppose that $a = b$ but $g_1^* \mid f^*$. Then $n = \ell +1$.
  Choose $f_1 \in S$ so that $\mathrm{o}(f_1) > a$, $I_1 = (g_1, f_1)$, and $g_1^* \nmid f_1^*$.
  Then, for every homogeneous system $\{\eta_i\}_{1 \leq i \leq n-1}$ of generators of $I_1^*$
  with $\eta_1 = g_1^*$ and $\eta_2 = f_1^*$,
  we have $I^* = (f^*, g^*) + (X\eta_i \mid 2 \leq i \leq n - 1)$.
\end{enumerate}
\end{Cor}

\begin{proof}
$(1)$ By Proposition \ref{prop2-3} $(2)$ we have $f^* \nmid g_1^*$.
Let $\{\eta_i\}_{1 \leq i \leq \ell}$ be a homogeneous system of generators of $I_1^*$
with $\eta_1 = f^*$ and $\eta_2 = g_1^*$.
Then, because $I^* = (f^*) + XI_1^*$ and $(f^*, g^*) = (f^*, Xg_1^*)$
(cf. Proposition \ref{prop2-3},  $(2)$ and $(5)$),
we have $I^* = (f^*, X\eta_2) + (X\eta_i \mid 3 \leq i \leq \ell)$.
To see that $n=\ell$, we shall check that $f^*, X\eta_2, X\eta_3, \cdots , X\eta_\ell$
is a minimal system of generators of $I^*$.
Since $f^* \notin (X)$, it suffices to show that
$X\eta_i \notin (f^*) + (X\eta_2, \cdots ,X\eta_{i-1}, X\eta_{i+1},
\cdots ,X\eta_\ell)$ for any $2 \leq i \leq \ell$.
Assume the contrary and write
$X\eta_i = f^* \varphi + \sum_{2\leq j \leq \ell, j \ne i} X\eta_j \varphi_j$
with $\varphi, \varphi_j \in G$.
Then $X[\eta_i - \sum_{2\leq j \leq \ell, j \ne i} \eta_j \varphi_j] \in (f^*)$.
Because $f^*, X$ form a $G$-regular sequence, we get
$\eta_i \in (f^*) + (\eta_2, \cdots ,\eta_{i-1}, \eta_{i+1}, \cdots \eta_\ell)$,
which is impossible (recall that $f^* =\eta_1, \eta_2, \cdots ,\eta_\ell$
is a minimal system of generators of $I_1^*$).
Thus $n = \ell$.

$(2)$ Let $\{\eta_i\}_{1 \leq i \leq \ell}$ be a homogeneous system of
generators of $I_1^*$ with $\eta_1 = g_1^*$ and $\eta_2 = f^*$.
Then $I^* = (f^*, X\eta_1) + (X\eta_i \mid 3 \leq i \leq \ell)$.
For the same reason as in the proof of assertion $(1)$, $f^*, X\eta_1, X\eta_3, \cdots ,X\eta_\ell$
is a minimal system of generators of $I^*$ and we get $n = \ell$.

$(3)$ Let $\{\eta_i\}_{1 \leq i \leq \ell}$ be a homogeneous system
of generators of $I_1^*$ such that $\eta_1 = g_1^*$ and $\eta_2 = f_1^*$.
Then $I^* = (f^*, X\eta_1) + (X\eta_i \mid 2 \leq i \leq \ell)$.
We want to show that $f^*, X\eta_1, X\eta_2, \cdots ,X\eta_\ell$ is
a minimal system of generators of $I^*$.
Let $1 \leq i \leq \ell$ and assume that $X\eta_i \in (f^*) +
(X\eta_1, \cdots ,X\eta_{i-1}, X\eta_{i+1}, \cdots X\eta_\ell)$.
Then $X[\eta_i - \sum_{1 \leq j \leq \ell, j \ne i} \eta_j \varphi_j] \in (f^*)$
for some $\varphi_j \in G$, so that we have $\eta_i \in (f^*) +
(\eta_1, \cdots ,\eta_{i-1}, \eta_{i+1}, \eta_\ell)$. If $i=1$,
then $\eta_1 = g_1^* \in (f^*) + (\eta_2, \eta_3, \cdots, \eta_\ell)$.
Since $\deg f^* = a > \deg g_1^* = a-1$, this forces
$\eta_1 \in (\eta_2, \eta_3, \cdots ,\eta_\ell)$, which is impossible.
Hence $i > 1$. Then, because $\eta_1 \mid f^*$, we have
$\eta_i \in (\eta_1, \cdots ,\eta_{i-1}, \eta_{i+1}, \cdots ,\eta_\ell)$,
which is absurd. Thus $f^*, X\eta_1, X\eta_2, \cdots ,X\eta_\ell$ constitute a minimal
system of generators of $I^*$ and so $n = \ell+1$.
\end{proof}

We close this section with the following.

\begin{Prop} \label{prop2-6}
Let $P = k[X,Y]$ be the polynomial ring in two variables $X, Y$ over a field $k$.
Let $J$ be a graded ideal of $P$ with $\mu_P(J) = n$ and $\sqrt{J} = (X,Y)$.
Let $\{\xi_i\}_{1 \leq i \leq n}$be a homogeneous system of generators of $J$ and
set $D_i = \mathrm{GCD}(\xi_1, \xi_2, \cdots ,\xi_i)$ for $1 \leq i \leq n$.
If $\deg D_i > \deg D_{i+1}$ and
$\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i},
\frac{\xi_2}{D_i}, \cdots ,\frac{\xi_i}{D_i})$ for all $1 \leq i \leq n-1$,
then the Hilbert series $\rmH(P/J, \lambda) =
\sum_{i=0}^{\infty} \dim_k[P/J]_i \lambda^i$ of $P/J$ is given by the formula
$$
\rmH(P/J,\lambda) =
\frac{\sum_{i = 2}^{n} \lambda^{\deg D_i}(1-\lambda^{\deg D_{i-1} -
\deg D_i})(1-\lambda^{\deg \xi_i - \deg D_i})}{(1-\lambda)^2}.
$$
In particular,
$$
\mathrm{H}(G/(X^{3i}Y^{m-i-1} \mid 0 \leq i \leq m-1), \lambda)
=\displaystyle\frac{\sum_{i=2}^{m}\lambda^{m-i}(1 - \lambda^{3(i-1)})}{1-\lambda}$$
for all $2 \leq m \in {\mathbb Z}$.
\end{Prop}

\begin{proof}
If $n = 2$, then $\xi_1, \xi_2$ is a $P$-regular sequence and we get
$\rmH(P/J, \lambda) = \frac{(1-\lambda^{\deg \xi_1})(1-\lambda^{\deg \xi_2})}{(1-\lambda)^2}.$
Suppose that $n > 2$ and that our assertion holds true for $n-1$. Let $D=D_{n-1}$.
Then $J \subseteq (D, \xi_n)$ and $D, \xi_n$ form a $P$-regular
sequence (recall that $\mathrm{GCD}(D,\xi_n) = 1)$.
We write $\xi_i = D\eta_i$ ($1 \leq i \leq n-1$) and put $K = (\eta_i \mid 1 \leq i \leq n - 1)$.
Then $\xi_n = \frac{\xi_n}{D_n} \in
(\frac{\xi_1}{D},\frac{\xi_2}{D},\cdots ,\frac{\xi_{n-1}}{D}) = K$ and so $\mu_P(K) = n-1$,
since $J = DK + (\xi_n)$.
Let $E_i = \mathrm{GCD}(\eta_1, \eta_2, \cdots ,\eta_i)$.
Then $D_i = D E_i$ so that we have $\deg E_i > \deg E_{i+1}$ and
$\frac{\eta_{i+1}}{E_{i+1}} = \frac{\xi_{i+1}}{D_{i+1}}
\in (\frac{\xi_1}{D_i}, \frac{\xi_2}{D_i}, \cdots ,\frac{\xi_i}{D_i}) =
(\frac{\eta_1}{E_i}, \frac{\eta_2}{E_i}, \cdots ,\frac{\eta_i}{E_i})$ for all $1 \leq i \leq n-2$.
Therefore, thanks to the exact sequence
$$
0 \to (P/K)(-\deg D) \to P/J \to P/(D,\xi_n) \to 0
$$
of graded $P$-modules (recall that
$(D,\xi_n)/J = (D,\xi_n)/[DK+(\xi_n)] \cong (D)/[DK+(D)\cap(\xi_n)] =
(D) / DK \cong (P/K)(-\deg D)$, since $(D) \cap (\xi_n) = (D\xi_n)$
and $\xi_n \in (K)$) and the hypothesis of induction on $n$, we get
\begin{eqnarray*}
\rmH(P/J,\lambda) &=& \rmH(P/(D,\xi_n) ,\lambda) + \lambda^{\deg D} {\cdot}\rmH(P/K,\lambda)\\
&=& \frac{(1-\lambda^{\deg D})(1-\lambda^{\deg \xi_n})}{(1-\lambda)^2}\\
&+& \frac{\lambda^{\deg D}{\cdot}\sum_{i=2}^{n-1}
\lambda^{\deg E_i}(1-\lambda^{\deg E_{i-1} -
\deg E_i})(1-\lambda^{\deg \eta_i - \deg E_i})}{(1-\lambda )^2}\\
&=& \frac{\sum_{i = 2}^{n} \lambda^{\deg D_i}(1-\lambda^{\deg D_{i-1} -
\deg D_i})(1-\lambda^{\deg \xi_i - \deg D_i})}{(1-\lambda )^2}
\end{eqnarray*}
as claimed.


For the last assertion, let $\xi_i = X^{3(i-1)}Y^{m-i}$ for $1 \leq i \leq m$.
Then $D_i = Y^{m-i}$ and $\displaystyle\frac{\xi_i}{D_i} = X^{3(i-1)}$ for all $1 \leq i \leq m$. Hence
\begin{eqnarray*}
\mathrm{H}(G/(X^{3i}Y^{m-i-1} \mid 0 \leq i \leq m-1), \lambda)
&=&\displaystyle\frac{\sum_{i=2}^{m}\lambda^{m-i}(1-\lambda)(1 - \lambda^{3(i-1)})}{(1-\lambda)^2}\\
&=&\displaystyle\frac{\sum_{i=2}^{m}\lambda^{m-i}(1 - \lambda^{3(i-1)})}{1-\lambda}.
\end{eqnarray*}
\end{proof}






\section{Proof of Theorems \ref{1.11} and \ref{1.12} }

\noindent {\bf Proof of Theorem \ref{1.11}.}
Assume  that  Theorem \ref{1.11} fails to hold and choose the ideal $I$ so that
$a =\mathrm{o}(I) := \sup \{i \in \mathbb{Z} \mid  I \subseteq \fkn^i\}$ is as small
as possible among the counterexamples.
We furthermore choose our ideal $I$ so that $b = \mathrm{o}(g)$ is the
smallest among the counterexamples $I$ with $\mathrm{o}(I) = a$.
Then $n > 2$, whence $b > 1$ (Proposition \ref{prop2-3} $(4)$).
Choose elements $x,y,u, $ and $g_1 \in S$ so that conditions $(1)$
and $(2)$ in Lemma \ref{lem2-2} are satisfied and put $I_1 = (f, g_1)$.
We then have the following three cases: (i) $a < b$, (ii) $a = b$
and $g_1^* \nmid f^*$, and (iii) $a = b$ but $g_1^* \mid f^*$.

Suppose that case (i) occurs. Then $\mu_G(I_1^*) = n$ (cf. Corollary \ref{thm2-5}).
Since $\mathrm{o}(I_1) =  a$ but $\mathrm{o}(g_1) = b-1$, we may choose a minimal
homogeneous system $\{\eta_i\}_{1 \leq i \leq n}$ of generators of $I_1^*$ so that

\begin{enumerate}
  \item[$(1)$] $\eta_1 = f^*$ and $\eta_2 = g_1^*$,
  \item[$(2)$] $\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-1$, and
  \item[$(3)$] $\mathrm{ht}_G(\eta_1, \eta_2, \cdots ,\eta_{n-1}) = 1$.
\end{enumerate}
Then, thanks to Corollary  \ref{thm2-5} (1), we get $I^* = (f^*, g^*) + (X\eta_3, \cdots, X\eta_n)$.
Letting $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_i$ ($3 \leq i \leq n$),
we certainly have conditions (1) and (2) in Theorem \ref{1.11},
because $\deg g_1^* = b-1 \leq \deg \eta_3 - 2$.
Since $(\xi_1, \xi_2, \cdots, \xi_{n-1}) =
(f^*,g^*) + (\xi_3, \cdots ,\xi_{n-1}) = (f^*, Xg_1^*) +
(X\eta_3, \cdots ,X\eta_{n-1}) \subseteq (\eta_1, \eta_2) + (\eta_3, \cdots ,\eta_{n-1})$,
we get $\mathrm{ht}_G(\xi_1, \xi_2, \cdots ,\xi_{n-1}) = 1$.
Thus case (i) cannot occur.

Suppose case (ii) occurs.
Then $\mu_G(I_1^*) = n$. Since $\mathrm{o}(I_1) = a-1$, we may choose a minimal
homogeneous system $\{\eta_i\}_{1 \leq i \leq n}$ of generators of $I_1^*$ so that
\begin{enumerate}
  \item[$(1)$] $\eta_1 = g_1^*$ and $\eta_2 = f^*$,
  \item[$(2)$] $\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-1$, and
  \item[$(3)$] $\mathrm{ht}_G(\eta_1, \eta_2, \cdots ,\eta_{n-1}) = 1$.
\end{enumerate}
Then $I^* = (f^*, g^*) + (X\eta_3, \cdots ,X\eta_n)$ by Corollary  \ref{thm2-5} (2).
Let $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_i$ ($3 \leq i \leq n$).
Then $\deg \xi_2 = b=a$ and $\deg \xi_3 = \deg \eta_3 + 1 \geq \deg \eta_2 + 3 = a +3$,
so that conditions (1) and (2) in Theorem \ref{1.11} are
safely satisfied for the family $\{\xi_i\}_{1 \leq i \leq n}$.
Since $(\xi_1, \xi_2, \cdots ,\xi_{n-1}) = (f^*, g^*) + (\xi_3, \cdots, \xi_{n-1}) =
(f^*, Xg_1^*)+ (\xi_3, \cdots, \xi_{n-1}) \subseteq (\eta_1, \eta_2, \cdots ,\eta_{n-1})$,
we also have condition (3) in Theorem \ref{1.11} to be satisfied.
Hence case (ii) cannot occur.

Thus we have case (iii). Hence  $\mu_G(I_1^*) = n-1$.
We choose $f_1 \in S$ so that $\mathrm{o}(f_1)=a_1 > a$, $I_1 = (g_1, f_1)$, and $g_1^* \nmid f_1^*$.
Because $\mathrm{o}(I_1) = a-1 < a = \mathrm{o}(I)$, we may choose a minimal homogeneous
system $\{\eta_i\}_{1 \leq i \leq n-1}$ of generators of $I_1^*$ so that
\begin{enumerate}
  \item[$(1)$] $\eta_1 = g_1^*$ and $\eta_2 = f_1^*$,
  \item[$(2)$] $\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-2$, and
  \item[$(3)$] $\mathrm{ht}_G(\eta_1, \eta_2, \cdots ,\eta_{n-2}) = 1$.
\end{enumerate}
Then $I^* = (f^*,g^*) + (X\eta_2, X\eta_3, \cdots ,X\eta_{n-1})$.
Let $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_{i-1}$ for $3 \leq i \leq n$.
Because $\deg \eta_2 = a_1 > a$, we have $\deg \xi_3 \geq a+2$, so that conditions (1) and (2)
in Theorem \ref{1.11} are satisfied for the family $\{\xi_i\}_{1 \leq i \leq n}$.
Since $(f^*, g^*) + (\xi_3, \cdots ,\xi_{n-1}) = (f^*, Xg_1^*) + (\xi_3, \cdots ,\xi_{n-1})
\subseteq (g_1^*) + (\eta_2, \cdots ,\eta_{n-2}) =
(\eta_1, \eta_2, \cdots , \eta_{n-2})$ (recall that $g_1^* \mid f^*$),
we also have condition (3).
This is absurd and thus Theorem \ref{1.11} holds true.



\begin{dis} \label{3.11} {\em
Let $\xi_1, \xi_2, \cdots ,\xi_n$  be a homogeneous system of
generators for $I^*$ which  satisfies  conditions (1) and (2) in
Theorem \ref{1.11}. Let $c_i = \deg \xi_i$ for $1 \leq i \leq  n$,
and let $G_+ =\sum_{i > 0}G_i$. We then have $\{c_1, c_2, \cdots ,
c_n\} = \{i \in \mathbb{Z} \mid [I^*/ G_+{\cdot} I^*]_i \ne (0)\}$,
whence the $degree$ sequence $(c_1,c_2, \cdots ,c_n)$ is independent
of the choice of $\{\xi_i \}_{1 \leq i \leq n}$. Because $\xi_1=
f^*$, $\xi_2 = g^*$, and $c_1 = a \leq c_2 = b < c_3 < \cdots <
c_n$, the ideals $(\xi_1, \xi_2, \cdots ,\xi_i)$ ($1 \leq i \leq
n)G$  also do not depend on the choice of $\{\xi_i\}_{1\leq i \leq
n}$. We put $D_i = \mathrm{GCD}(\xi_1, \xi_2, \cdots ,\xi_i)$ and
$d_i = \deg D_i$ ($1 \leq i \leq n$). (Hence $D_1 = \xi_1$ and $D_n
= 1$.) Since the ideal $(\xi_1, \xi_2, \cdots ,\xi_{n-1})$ is
independent of the choice of $\{\xi_i\}_{1\leq i \leq n}$, we have
condition (3) in Theorem \ref{1.11} that $\mathrm{ht}_G(\xi_1,
\xi_2, \cdots ,\xi_{n-1}) = 1$ is always satisfied for every
homogeneous system of generators $\{\xi_i\}_{1\leq i \leq n}$ of
$I^*$ which satisfies conditions (1) and (2) of Theorem \ref{1.11}.
Similarly, the fact whether $\frac{\xi_{i+1}}{D_{i+1}} \in
(\frac{\xi_1}{D_i}, \frac{\xi_2}{D_i},\cdots ,\frac{\xi_i}{D_i})$ or
not does not depend on the particular choice of a homogeneous system
$\{\xi_i\}_{1 \leq i \leq n}$ of generators of $I^*$ which satisfies
conditions (1) and (2) in Theorem \ref{1.11}.

}
\end{dis}



\begin{Lem} \label{lem3-2}
With notation as in Discussion \ref{3.11}, the following assertions
hold true.
\begin{enumerate}
  \item[$(1)$] $d_i > d_{i+1}$ and $\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i},
  \frac{\xi_2}{D_i}, \cdots ,\frac{\xi_i}{D_i})$ for all $1 \leq i \leq n-1$.
  \item[$(2)$] $c_{i+1} - c_i > d_{i-1} - d_i > 0$ for all $2 \leq i \leq n-1$.
  \item[$(3)$] $c_n + d_{n-1} \geq d_{i-1} + c_i - d_i$ for all $2 \leq i \leq n$.
  \item[$(4)$] $c_n + d_{n-1} \geq a+b$.
\end{enumerate}
\end{Lem}

\begin{proof}
Assume that  Lemma \ref{lem3-2} is false and choose an  ideal $I$ so
that $a =\mathrm{o}(I) = \sup \{i \in \mathbb{Z} \mid I \subseteq
\fkn^i\}$ is as small as possible among the counterexamples. We
furthermore choose the ideal $I$ so that $b = \mathrm{o}(g)$ is the
smallest among the counterexamples $I$ with $\mathrm{o}(I) = a$.
Then $b > 1$, since $n > 2$. Let $x,y,u,$ and $g_1 \in S$ be
elements which satisfy conditions (1) and (2) in Lemma \ref{lem2-2}.
We put $I_1 = (f, g_1)$. Then we have the following three cases: (i)
$a<b$, (ii) $a = b$ and $g_1^* \nmid f^*$, and (iii) $a = b$ but
$g_1^* \mid f^*$. For case (i) we have $f^* \nmid g_1^*$ and for
case (iii) we have some $f_1 \in S$ with $\mathrm{o}(f_1)=a_1 > a$
such that $I_1 = (g_1, f_1)$ and $g_1^* \nmid f_1^*$. In any case,
because the value $a$ or the value $b$ for $I_1$ is less than that
for $I$,  Lemma \ref{lem3-2} holds true for the ideal $I_1$. In what
follows, we shall establish a contradiction by showing  (i),(ii), and (iii)
cannot occur.

Suppose that case (i) occurs. Then $\mu_G(I_1^*) = n$.
Let $\{\eta_i\}_{1 \leq i \leq n}$ be a homogeneous system of generators of $I_1^*$
such that $\eta_1 = f^*$, $\eta_2 = g_1^*$, and
$\deg \eta_i +2 \leq \deg \eta_{i+1}$ for $2 \leq i \leq n-1$.
Then Lemma \ref{lem3-2} holds true for the family $\{\eta_i\}_{1 \leq i \leq n}$ and
by Corollary \ref{thm2-5} we have
$$
I^* = (f^*,g^*) + (X\eta_3, \cdots, X\eta_n).
$$
Let $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_i$ ($3\leq i \leq n$).
Then the homogeneous system $\{\xi_i\}_{1 \leq i \leq n}$ of
generators of $I^*$ satisfies conditions (1) and (2) in Theorem \ref{1.11}.
We put $c_i ' = \deg \eta_i$, $D_i' =
\mathrm{GCD}(\eta_1, \eta_2, \cdots ,\eta_i)$, and $d_i' = \deg D_i'$.
Then, because $(\xi_1, \xi_2) = (f^*,g^*) = (f^*,Xg_1^*) =
(\eta_1, X\eta_2)$ and $X \nmid f^*$, we have $D_i = D_i'$ for all $1 \leq i \leq n$,
while $c_1' =a = c_1$ and $c_i' = c_i -1$ for all $2 \leq i \leq n$.
Consequently, assertions (2),  (3),  (4), and the former part of assertion (1) in Lemma \ref{lem3-2}
are safely deduced from those on the ideal $I_1$.
Let us check that $\frac{\xi_{i+1}}{D_{i+1}} \in
(\frac{\xi_1}{D_i} , \frac{\xi_2}{D_i} , \cdots ,\frac{\xi_i}{D_i})$.
Since $D_1 = \xi_1$, we may assume $i \geq 2$.
First of all, recall that $\frac{\eta_{i+1}}{D_{i+1}} \in (\frac{\eta_1}{D_i} ,
\frac{\eta_2}{D_i} , \cdots ,\frac{\eta_i}{D_i})$ and we have
$\frac{X\eta_{i+1}}{D_{i+1}} \in (\frac{\eta_1}{D_i} , \frac{X\eta_2}{D_i} ,
\cdots ,\frac{X\eta_i}{D_i}) = (\frac{\xi_1}{D_i} , \frac{\xi_2}{D_i} , \cdots ,\frac{\xi_i}{D_i})$,
because $(\eta_1, X\eta_2, \cdots ,X\eta_i) = (\xi_1, \xi_2, \cdots ,\xi_i)$.
Hence $\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i} ,
\frac{\xi_2}{D_i} , \cdots ,\frac{\xi_i}{D_i})$ as $\xi_{i+1} = X\eta_{i+1}$.
Thus case (i) does not occur.

Suppose case (ii).
Then $\mu_G(I_1^*) = n$.
Let $\{\eta_i\}_{1 \leq i \leq n}$ be a homogeneous system of generators of $I_1^*$ such
that $\eta_1 = g_1^*$, $\eta_2 = f^*$, and
$\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-1$.
Then $I^* = (f^*,g^*) + (X\eta_3, \cdots ,X\eta_n)$ by Corollary  \ref{thm2-5}.
Let $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_i$ for $3 \leq i \leq n$.
Then $\{\xi_i\}_{1 \leq i \leq n}$ is a homogeneous system of generators of $I^*$
which satisfies conditions (1) and (2) in Theorem \ref{1.11}.
We put $c_i' = \deg \eta_i$, $D_i' = \mathrm{GCD}(\eta_1, \eta_2, \cdots ,\eta_i)$,
and $d_i' = \deg D_i'$ for each $1 \leq i \leq n$.
Then $c_1' = a-1 = c_1 - 1$, $c_2' = a = c_2$, and $c_i' = c_i -1$ for $3 \leq i \leq n$.
Because $X \nmid f^*$ and $(\xi_1, \xi_2) = (f^*, g^*) = (\eta_2, X\eta_1)$,
we get $D_1' = \eta_1 = g_1^*$ and $D_i' = D_i$ for $2 \leq i \leq n$.
Hence $d_1' = a-1 = d_1-1$ and $d_i' = d_i$ for all $2 \leq i \leq n$.
Consequently, it is direct to check that assertions (2), (3), (4),
and the former part of assertion (1) hold true for the ideal $I$.
Let us show $\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i},
\frac{\xi_2}{D_i}, \cdots ,\frac{\xi_i}{D_i})$ for all $1 \leq i \leq n-1$.
We may assume $i \geq 2$. Because
$\frac{\eta_{i+1}}{D_{i+1}} \in (\frac{\eta_1}{D_i}, \frac{\eta_2}{D_i}, \cdots ,\frac{\eta_i}{D_i})$,
we have $\frac{X\eta_{i+1}}{D_{i+1}} \in (\frac{X\eta_1}{D_i}, \frac{X\eta_2}{D_i},
\cdots ,\frac{X\eta_i}{D_i})
\subseteq (\frac{\xi_1}{D_i}) + (\frac{X\eta_j}{D_i} | 1 \leq j \leq i, j \ne 2) =
(\frac{\xi_1}{D_i}, \frac{\xi_2}{D_i}, \cdots , \frac{\xi_i}{D_i})$
(use the fact $(\xi_1, \xi_2) = (\xi_1, X\eta_1)$).
Hence $\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i},
\frac{\xi_2}{D_i}, \cdots ,\frac{\xi_i}{D_i})$ as $\xi_{i+1} = X\eta_{i+1}$.
Thus case (ii) does not occur.


Now we consider case (iii).
We have $\mu_G(I_1^*) = n-1$. Let $f_1 \in S$ such that
$\mathrm{o}(f_1) = a_1 > a$, $I_1 = (g_1, f_1)$, and $g_1^* \nmid f_1^*$.
Choose a homogeneous system $\{\eta_i\}_{1 \leq i \leq n - 1}$ of generators for $I_1^*$
so that $\eta_1 = g_1^*$, $\eta_2 = f_1^*$, and
$\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-2$.
Then $I^* = (f^*, g^*) + (X\eta_2, \cdots , X\eta_{n-1})$.
We put $\xi_1 = f^*$, $\xi_2 = g^*$, and $\xi_i = X\eta_{i-1}$ for $3 \leq i \leq n-1$.
Then the homogeneous system $\{ \xi_i \}_{1 \leq i \leq n}$ of generators of
$I^*$ satisfies conditions (1) and (2) in Theorem \ref{1.11}.
Let $D_i' = \mathrm{GCD}(\eta_1, \eta_2, \cdots , \eta_i)$, $d_i' =
\deg D_i'$, and $c_i' = \deg \eta_i$ for each $1 \leq i \leq n-1$.
Then $D_i' = D_{i+1}$ for $1 \leq i \leq n-1$ (recall that $g_1^* \mid f^*$ and
$(\xi_1, \xi_2) = (f^*, g^*) = (f^*, Xg_1^*) = (f^*,X\eta_1))$.
Hence $c_1' = a-1 = c_1 - 1$, $c_i' = c_{i+1} - 1$
for $2 \leq i \leq n-1$, and $d_i' = d_{i+1}$ for $1 \leq i \leq n-1$.
Consequently, assertions (2), (3), (4), and the former part of assertion (1) hold true
(use the fact that $c_3 = a_1+1 \geq a+2$, $d_1' = a-1$, and $c_n \geq a+2$).
Let us check the latter part of assertion (1).
We may assume $i \geq 2$.
Then, since $\frac{\eta_i}{D_i'} \in (\frac{\eta_1}{D_{i-1}'},
\frac{\eta_2}{D_{i-1}'}, \cdots ,\frac{\eta_{i-1}}{D_{i-1}'})$,
we get $\frac{X\eta_i}{D_{i+1}} \in (\frac{f^*}{D_i}, \frac{X\eta_1}{D_i},
\cdots ,\frac{X\eta_{i-1}}{D_i})$.
Hence $\frac{\xi_{i+1}}{D_{i+1}} \in (\frac{\xi_1}{D_i}, \frac{\xi_2}{D_i}, \cdots ,
\frac{\xi_i}{D_i})$, because $\xi_{i+1} = X\eta_i$ and $(f^*,X\eta_1) = (\xi_1, \xi_2)$.
Thus even case (iii) cannot occur. We  conclude  that Lemma \ref{lem3-2} holds true.
\end{proof}


\noindent {\bf Proof of Theorem \ref{1.12}.}
 Items (1) and (3)  follow from
Proposition \ref{prop2-6} and Lemma \ref{lem3-2}.

For items (2) and (4), since $$ \rmH(\mathrm{gr}_{\fkm}(R), \lambda)
= \sum_{i = 2}^n \lambda^{d_i} (\sum_{j = 0}^{d_{i-1} - d_i -
1}\lambda^j) (\sum_{j=0}^{c_i - d_i - 1}\lambda^j)$$ and $\ell_S(R)
= \dim_k \mathrm{gr}_{\fkm}(R) $, we readily get $\ell_S(R) =
\sum_{i=2}^n (d_{i-1} -d_i) (c_i - d_i) = c_1 c_2 + \sum_{i=2}^{n-1}
d_i{\cdot}[(c_{i+1} - c_i)- (d_{i-1} - d_i)] = ab + \sum_{i=2}^{n-1}
d_i{\cdot}[(c_{i+1} - c_i)- (d_{i-1} - d_i)]$. We have $\ell_S(R) =
ab$ if and only if $n = 2$, because $(c_{i+1} - c_i) - (d_{i-1} -
d_i) > 0$ for all $2 \leq i \leq n-1$ by Lemma \ref{lem3-2} (2).
Since $I^* = (f^*, g^*)$ if and only if $n = 2$, we have $\ell_S(R)
= ab$ if and only if $f^*, g^*$ form a regular sequence in $G$.



\begin{Cor} \label{cor3-4}
Assume notation as in Theorem \ref{1.12}, and let \newline
$a(\mathrm{gr}_{\fkm}(R)) = \max\{i \in \mathbb{Z} \mid
[\mathrm{gr}_{\fkm}(R)]_i \ne (0)\}$. The following assertions hold
true.
\begin{enumerate}
  \item[$(1)$] $a(\mathrm{gr}_{\fkm}(R)) = c_n + d_{n-1} - 2$.
  \item[$(2)$] $\ell_S(R) \leq a{\cdot}[c_n + d_{n-1} - a]$.
  \item[$(3)$] $\ell_S(R) = a{\cdot}[c_n + d_{n-1} -a]$ if and only if $n = 2$.
\end{enumerate}
\end{Cor}

\begin{proof}
Since $a(\mathrm{gr}_{\fkm}(R)) = \deg \rmH(\mathrm{gr}_{\fkm}(R), \lambda)$, thanks to
Theorem \ref{1.12} (1), we have
$a(\mathrm{gr}_{\fkm}(R)) = \max\{d_i + (d_{i-1} - d_i - 1) + (c_i - d_i - 1) \mid 2 \leq i \leq n\}$.
Hence $ a(\mathrm{gr}_{\fkm}(R)) = c_n + d_{n-1} - 2$ by Lemma \ref{lem3-2} (3).
Because $d_1 - 1 = a-1 \geq d_i$ for all $2 \leq i \leq n-1$ and $c_n + d_{n-1} \geq a+b$
by Lemma \ref{lem3-2} (1), (4), we get by Theorem \ref{1.12} (2) that
\begin{eqnarray*}
\ell_S(R) &\leq& ab + (a-1){\cdot}\sum_{i=2}^{n-1} [(c_{i+1} - c_i) - (d_{i-1} - d_i)]\\
&=& ab + (a-1){\cdot} [c_n + d_{n-1} - (a + b)]\\
&=& a{\cdot}[c_n + d_{n-1} - a] - [c_n + d_{n-1} - (a+b)]\\
&\leq& a{\cdot}[c_n + d_{n-1} - a].
\end{eqnarray*}
If the equality $\ell_S(R) = a{\cdot}[c_n + d_{n-1} - a]$ holds true,
then $c_n + d_{n-1} - (a+b) = 0$, so that $\ell_S(S/I) = a{\cdot}[c_n + d_{n-1} - a]= ab$, whence $n = 2$.
Since $c_2 = b$ and $d_1 = a$, we certainly have $\ell_S(R) = a(c_n + d_{n-1} - a)$ if $n = 2$.
This completes the proof of Corollary \ref{cor3-4}.
\end{proof}

Suppose that $\mathrm{ht}_G(f^*, g^*) = 1$. Let $D = \mathrm{GCD}(f^*, g^*)$ and $ d = \deg D$.
We write $f^* = D\xi$ and $g^* = D \eta$ with $\xi, \eta \in G$.
Then $b \geq a > d > 0$ and by \cite[Proposition 2.2]{GHK} we may choose
$h = \alpha f + \beta g$ with $\alpha, \beta \in S$ so that
$\mathrm{o}(\alpha) = b-d$, $\mathrm{o}(\beta) = a-d$, and $h^* \notin (f^*, g^*)$.
We call such an element $h^*$ the third generator of $I^*$.
We put $c = \mathrm{o}(h)$. With this notation we have the following.




\begin{Cor} \label{cor3-5}
Suppose that $\mathrm{ht}_{G}(f^*, g^*) = 1$ and $\mathrm{ht}_G(f^*, g^*, h^*)=2$. Then  the following assertions hold true.
\begin{enumerate}
\item[$(1)$] $I^* = (f^*, g^*, h^*)$.
\item[$(2)$] $\mathrm{H}(\mathrm{gr}_\frak{m}(R), \lambda)=\frac{(1-\lambda^{c})(1-\lambda^{d})+
\lambda^{d}(1-\lambda^{a-d})(1-\lambda^{b-d})}{(1-\lambda)^{2}}$.
\item[$(3)$] $\mathrm{e}_{\frak{m}}^0(R) = ab + d{\cdot}[(c+d) -(a+b)].$
\end{enumerate}
\end{Cor}

\begin{proof}
For each $3 \leq i \leq n$, let $\xi_i = \overline{h_i t^{c_i}}$ where $h_i \in I$ with $\mathrm{o}(h_i) = c_i$.
We write $h_i = \alpha_i f + \beta_i g$ with $\alpha_i, \beta_i \in S$. Then
$\mathrm{o}(\alpha_i) \geq b-d$ and
$\mathrm{o}(h_i) \geq \mathrm{o}(h) + [\mathrm{o}(\alpha_i) - (b-d)] \geq \mathrm{o}(h)$ (cf. \cite[Proposition 2.4 (1)]{GHK}).
Let $h^* = \sum_{i=1}^n \xi_i \varphi_i$ with $\varphi_i \in G_{c-c_i}$.
Then, since $h^* \notin (f^*, g^*) = (\xi_1, \xi_2)$, we have $c - c_i \geq 0$ for
some $3 \leq i \leq n$. Hence $c \geq c_i \geq c_3$, so that $c = c_3$,
because $c_3 \geq c + [\mathrm{o}(\alpha_3) - (b-d)] \geq c$.
We furthermore have $\mathrm{o}(\alpha_3) = b-d$, whence, thanks to \cite[Proposition 2.4 (3)]{GHK},
we get $(f^*, g^*, \xi_3) = (f^*, g^*, h^*)$.
Thus $n = 3$ by Theorem \ref{1.11} (3), because $\mathrm{ht}_G(f^*, g^*,h^*) = 2$ by our assumption,
so that $I^* = (\xi_1, \xi_2, \xi_3) = (f^*, g^*, h^*)$ as claimed.
Assertions (2) and (3) now readily follow from Theorem \ref{1.12} (1) and (2).
\end{proof}


\begin{remark} \label{3.6} {\em  With notation as in Setting
\ref{1.0},  it follows from Part (1) of  Lemma \ref{lem3-2}  that
there exists a strictly descending chain $$ (\frac{\xi_1}{D_2},
\frac{\xi_2}{D_2})G \supset (\frac{\xi_1}{D_3}, \frac{\xi_2}{D_3},
\frac{\xi_3}{D_3})G \supset \cdots \supset (\frac{\xi_1}{D_{n-1}},
\frac{\xi_2}{D_{n-1}}, \cdots  \frac{\xi_{n-1}}{D_{n-1}})G \supset
I^*
$$
of height-two ideals of $G$. In particular,  $I^*$ is contained in
the ideal $(\frac{\xi_1}{D_2}, \frac{\xi_2}{D_2})G$. This behavior
fails to hold in general in the higher dimensional case. The
leading ideal of a complete intersection of height two in a
three-dimensional regular local ring may fail to have this 
property as is demonstrated by Example
1.6 of \cite{GHK}.


}
\end{remark}



\section{Proof of Theorems \ref{1.5} and \ref{1.6}}

The goal of this section is to prove Theorems \ref{1.5},  and
\ref{1.6} and deduce several consequences of these theorems. We use
the following lemma.


\begin{Lem} \label{lem4-1} Assume notation as in Setting \ref{1.0}.
Let $0 \ne h \in \fkn$ and $m = \mathrm{o}(h)$.
Let $X_1, X_2, \cdots, X_{s-1} \in G$ be a linear system of parameters for
the graded ring $G/(h^*)$ and write $X_i = x_i^*$ with $x_i \in \fkn$.
Then $x_1 ,x_2, \cdots ,x_{s-1}$ is a part of a regular system of parameters of $S$
and for all $1 \leq \ell \leq s-1$, we have $\mathrm{o}(\overline{h}) = m$,
where $\overline{h}$ denotes the image of $h$ in $\overline{S} = S/(x_1, x_2, \cdots ,x_\ell)$.
\end{Lem}

\demo
Since $X_1, X_2, \cdots ,X_{s-1}$ are algebraically independent over $k$,
the elements $x_1, x_2, \cdots ,x_{s-1}$ form a part of a regular system of parameters in $S$.
If $$h \in \fkn^{m+1} + (x_1, x_2, \cdots ,x_\ell),$$ then
since $(x_1, x_2, \cdots ,x_\ell) \cap \fkn^m = (x_1, x_2, \cdots ,x_\ell)\fkn^{m-1}$,
we get $$h \in \fkn^{m+1} + (x_1, x_2, \cdots ,x_\ell) \fkn^{m-1}.$$
Thus $h^* \in (X_1, X_2, \cdots ,X_\ell)$, which is impossible,
because $X_1, X_2, \cdots , X_\ell , h^*$ forms a regular sequence in $G$.
Hence $\mathrm{o}(\overline{h}) = m$ as claimed. \QED


\medskip

\noindent {\bf Proof of Theorem \ref{1.5}.} By  Corollary \ref{cor3-5},  
we may assume that  $\dim S = s > 2$. Choose $X_1, X_2,
\cdots ,X_{s-1} \in G_1$ so that $X_1, X_2, \cdots ,X_{s-1}$ is a
homogeneous system of parameters for the graded rings $G/(f^*)$,
$G/(g^*)$, $G/(h^*)$, $G/(\alpha^*)$, $G/(\beta^*)$, and $G/(D)$ and
$X_1, X_2, \cdots ,X_{s-2}$ is a homogeneous system of parameters
for the graded rings $G/(f^*,g^*,h^*)$, $G/(\xi, \eta)$, and
$\mathrm{gr}_{\fkm}(R)$. For each $i$ with $1 \le i \le s-1$, choose $x_i \in
\fkn$ such that $x_i^* = X_i$. Then $x_1, x_2, \cdots ,x_{s-1}$ form
a part of a regular system of parameters for $S$. Let $\frak{q} =
(x_1, x_2, \cdots ,x_{s-2})S$. We put $\overline{S} = S/\frak{q}$,
$\overline{\fkn} = \fkn/\frak{q}$, and $\overline{I} =
(\overline{f}, \overline{g})$, where overline  denotes image in
$\overline{S}$. Notice that   $\frak{q}R$ is a minimal reduction of
$\fkm$. Thus $I + \frak{q}$ is a parameter ideal for $S$ and
$\overline{I} = (\overline{f}, \overline{g})\overline{S}$ is a
parameter ideal in the regular local ring $\overline{S}$ of
dimension $2$. Lemma \ref{lem4-1} implies that
$\mathrm{o}(\overline{f}) = a$,  $\mathrm{o}(\overline{g}) = b$ and
$\mathrm{o}(\overline{h}) = c$.

Let $Q = (X_1, X_2, \ldots , X_{s-2})G$. We prove that the following
diagram is commutative:
$$
\CD
 S  @>>>   S/\frak{q} := \overline{S}  \\
 @V\varphi_1VV      @V\varphi_2VV                   \\
 G := \gr_{\n}(S)  @>>>  G/Q := \widetilde{G} \cong
\gr_{\overline \fkn}(\overline S).
\endCD
$$
\vskip 5 pt

\noindent Here $\varphi_1$ and $\varphi_2$ denote the canonical maps
associating an element with its leading form in the associated
graded ring, and the identification $\widetilde{G} \cong
\gr_{\overline \n}(\overline S)$ is because $Q$ is the leading ideal
in $G$ of the ideal $\frak{q}$ of $S$. We denote with a tilde the
image in $G/Q$ of elements and ideals of $G$. Since $X_1, X_2,
\cdots ,X_{s-2}, \xi, \eta$ is a homogeneous system of parameters in
$G$, $\widetilde \xi$, $\widetilde \eta$ is a homogeneous system of
parameters in  $G/Q$. Thus $\GCD(\widetilde \xi, \widetilde \eta) =
1$, and $\widetilde D = \GCD(\widetilde{f^*}, \widetilde{g^*})$.
Since $\mathrm{o}(f) = \mathrm{o}(\overline{f})$, we have
$\widetilde{f^*} = \overline{f}^*$. Similarly, $\widetilde{g^*} =
\overline{g}^*$ and $\widetilde{h^*} = \overline{h}^*$.  We have
$\widetilde{I^*} \subseteq \overline{I}^*$. Moreover,
$\widetilde{I^*} = \overline{I}^*$ if and only if $X_1, \ldots,
X_{s-2}$ is  a regular sequence on $G/I^*$. Thus $\widetilde{I^*} =
\overline{I}^*$ if and only if $I^*$ is a perfect ideal of $G$.



 We furthermore have the following.

\begin{Claim} \label{claim4-3}
The following assertions hold true.
\begin{enumerate}
  \item[$(1)$] $\overline{f}^* \nmid \overline{g}^*$ in
  $\mathrm{gr}_{\overline{\fkn}}(\overline{S})$.
  \item[$(2)$] $\mathrm{o}(\overline{\alpha}) = b-d$,
  $\mathrm{o}(\overline{\beta}) = a-d$, and $\mathrm{o}(\overline{h}) = c$.
  \item[$(3)$] $\overline{h}^* \notin (\overline{f}^* , \overline{g}^*)$.
\end{enumerate}
Thus $\overline{h}^*$ is the third generator of
$\overline{I}^*$ in $\mathrm{gr}_{\overline{\fkn}}(\overline{S})$.
\end{Claim}

\begin{proof}[Proof of Claim $\ref{claim4-3}$]
(1) Suppose that $\overline{f}^* \mid \overline{g}^*$. Then, via the
identification $G/Q = \mathrm{gr}_{\overline{\fkn}}(\overline{S}),$ we have $g^* \in (f^*) + Q$.
Let us write $g^* = f^*\varphi + \sum_{i=1}^{s-2} X_i \varphi_i$ with $\varphi, \varphi_i \in G$.
Then, since $f^* = D\xi$ and $g^* = D\eta$, we have $D(\eta - \xi \varphi) \in Q$.
Hence $\eta - \xi\varphi \in Q$,
because $X_1, X_2, \cdots ,X_{s-2}, D$ is a regular sequence in $G$.
Thus  $\eta \in Q + (\xi)$, which is impossible,
because $X_1, X_2, \cdots , X_{s-2} , \xi , \eta$ is a $G$-regular sequence.
Hence $\overline{f}^* \nmid \overline{g}^*$.

(2) See Lemma \ref{lem4-1}

(3) We have $h^* \in (\xi, \eta)$ (\cite[Remark 2.3]{GHK}; recall
that $h \in (\alpha, \beta)$). Write $h^* = \xi \varphi + \eta \psi$ with $\varphi, \psi \in G$.
Then
$$
(f^*, g^*, h^*) = \rm\mathrm{I}_2\left(
\begin{array}{c|c|c} \varphi & -\psi & D \\\hline \xi & \eta & 0 \end{array} \right),
$$
so that $(f^*, g^*, h^*)$ is a perfect ideal with $\mu_G(f^*, g^*, h^*) =3$,
since $\mathrm{ht}_G(f^*, g^*, h^*) = 2$. Therefore $G/(f^*, g^*, h^*)$ is
a Cohen-Macaulay ring, whence $X_1, X_2, \cdots , X_{s-2}$ form a regular sequence in $G/(f^*, g^*, h^*)$.
Thus $\overline{h}^* \notin (\overline{f}^*, \overline{g}^*)$,
because $\mu_{\mathrm{gr}_{\overline{\fkn}}(\overline{S})} (\overline{f}^*,
\overline{g}^*, \overline{h}^*)=3$.
\end{proof}

Therefore $\overline{I}^* = (\overline{f}^*, \overline{g}^*,
\overline{h}^*)$ by Corollary \ref{cor3-5}, because
$\overline{h}^*$ is the third generator of $\overline{I}^*$ in
$\mathrm{gr}_{\overline{\fkn}}(\overline{S})$ with
$\mathrm{ht}_{\mathrm{gr}_{\overline{\fkn}}(\overline{S})}(\overline{f}^*,
\overline{g}^*, \overline{h}^*) = 2$. We now look at the estimation
$(*)$:
\begin{eqnarray*}
\ell_R(R/\frak{q}R) = \ell_{\overline{S}}(\overline{S} / \overline{I})
&=& \dim_k \mathrm{gr}_{\overline{\fkn}}(\overline{S})/ (\overline{f}^*, \overline{g}^*, \overline{h}^*)\\
&=& \dim_k G/[Q + (f^*, g^*, h^*)]\\
&\geq & \dim_k G/[Q + I^*]\\
&=& \dim_k \mathrm{gr}_{\fkm}(R) / Q\mathrm{gr}_{\fkm}(R)\\
&\geq & {\mathrm{e}^0}_{Q \mathrm{gr}_{\fkm}(R)} (\mathrm{gr}_{\fkm}(R))\\
&=& \mathrm{e}^0_{\fkm} (R)\\
&=& \ell_R(R/\frak{q}R),
\end{eqnarray*}
since $\frak{q}R$ is a minimal reduction of $\fkm$.
Thus $\mathrm{gr}_{\fkm}(R)=G/I^*$ is Cohen-Macaulay, since
$\dim_k \mathrm{gr}_{\fkm}(R)/ Q \mathrm{gr}_{\fkm}(R) =
\mathrm{e}^0_{Q\mathrm{gr}_{\fkm}(R)} (\mathrm{gr}_{\fkm}(R))$
(cf. estimation $(*)$),
and so the sequence $X_1, X_2, \cdots ,X_{s-2}$ is $\mathrm{gr}_{\fkm}(R)$-regular.
Hence $I^* = (f^*, g^*, h^*)$, because $Q + (f^*, g^*, h^*) = Q + I^*$ and $Q \cap I^* = QI^*$.
We furthermore have that
$$
\rmH(\mathrm{gr}_{\fkm}(R), \lambda) =
\frac{\rmH(\mathrm{gr}_{\overline{\fkn}}(\overline{S}) / (\overline{f}^*,
\overline{g}^*, \overline{h}^*) , \lambda)}{(1-\lambda)^{s-2}},
$$
whence
$$
\rmH(\mathrm{gr}_{\fkm}(R) , \lambda) =
\frac{(1-\lambda^c)(1-\lambda^d) + \lambda^d(1-\lambda^{a-d})(1-\lambda^{b-d})}{(1-\lambda)^s}
$$
by Corollary \ref{cor3-5}. Thus $\mathrm{e}_{\fkm}^0(R) = ab +
d{\cdot}[(c+d) -(a+b)]$ as claimed. This completes the proof of
Theorem \ref{1.5}.  \QED

\begin{remark} {\em
Without the assumption in Theorem \ref{1.5} that $\hgt(f^*, g^*,
h^*) = 2$, it is still possible to specialize via $\frak{q}$ and $Q$
to obtain $\widetilde{f^*} = \overline{f}^*$, $\widetilde{g^*} =
\overline{g}^*$ and $\widetilde D = \GCD(\widetilde{f^*},
\widetilde{g^*})$. However, $\widetilde{h^*} = \overline{h}^*$ may
fail to be a minimal generator of $\overline{I}^*$  as we
demonstrate in Example \ref{2.3}.

}
\end{remark}



\begin{exam} \label{2.3} {\em  Let $S = k[[x, y, z]]$ be the formal power series
ring in the three variables $x, y, z$ over a field $k$, and let $X,
Y, Z$ denote the leading forms of $x, y, z$ in $G = \gr_{\n}(S) =
k[X, Y, Z]$. As in \cite[Example 1.6]{GHK}, let $I = (f, g)$, where
$f = z^2 - x^5$ and $g = zx - y^3$. Thus $R = S/I$ is a complete
intersection of dimension one.  We have  $I^* = (Z^2, ZX, ZY^3,
Y^6)G$. We consider several choices for an element  $w \in \n
\setminus \n^2$ and behavior of the specialization $S \to S/wS =
\overline{S}$.  Since $\dim G/I^* = 1$ and $I^*$ is not a perfect
ideal, one always has the strict inequality $I^* \widetilde{G}
\subsetneq (I\overline S)^*$.
\begin{enumerate}
\item
Let  $w = x$.  Then $\overline S = k[[y, z]]$, $\overline f = z^2$
and $\overline g = -y^3$. We have
$$
I^* \widetilde{G} = (Z^2, ZY^3, Y^6) \widetilde{G} \subsetneq
(I\overline S)^* = (Z^2, Y^3)k[Y, Z].
$$
The multiplicity of $G/I^*$ is $6$ as is the multiplicity of
$\gr_{\overline \n}(\overline S)/(I\overline S)^*$.  The  Hilbert
series for $G/I^*$ is
$$
H(G/I^*, \lambda) = \frac{1 + 2\lambda + \lambda^2 + \lambda^3 +
\lambda^5}{1 - \lambda},
$$
while the Hilbert series  for  $\gr_{\overline \n}(\overline
S)/(I\overline S)^*$ is
$$
H(\gr_{\overline \n}(\overline S)/(I\overline S)^*, \lambda) =
\frac{ 1 + 2\lambda  + 2\lambda^2 + \lambda^3}{1 - \lambda}.
$$ The multiplicity of $\widetilde{G}/I^* \widetilde{G}$ is 9,
and the Hilbert series for $\widetilde{G}/I^* \widetilde{G}$ is
$$ H(\widetilde{G}/I^* \widetilde{G}, \lambda) = \frac{
1 + 2\lambda + 2\lambda^2 + 2\lambda^3 + \lambda^4 +
\lambda^5}{1-\lambda}.
$$
\item
Let $w = x-y$ and use this to eliminate $x$. Then $\overline S =
k[[y, z]]$, $\overline f = z^2 - y^5$ and $\overline g = zy - y^3$.
We have
$$
I^* \widetilde{G} = (Z^2, ZY, Y^6) \widetilde{G} \subsetneq
(I\overline S)^* = (Z^2, ZY, Y^5) k[Y, Z].
$$
The multiplicity  and Hilbert series of $G/I^*$ are as given in part
(1). The multiplicity of $\gr_{\overline \n}(\overline
S)/(I\overline S)^*$ is 6, while the multiplicity of
$\widetilde{G}/I^* \widetilde{G}$ is 7. The Hilbert series of
$\gr_{\overline \n}(\overline S) /(I\overline S)^*$ is
$$
H(\gr_{\overline \n}(\overline S) /(I\overline S)^*, \lambda) =
\frac{1 + 2\lambda + \lambda^2 + \lambda^3 + \lambda^4}{1 -
\lambda}, $$
 while the Hilbert series of
$\widetilde{G}/I^* \widetilde{G}$ is
$$
H(\widetilde{G}/I^* \widetilde{G}, \lambda) = \frac{1 + 2\lambda +
\lambda^2 + \lambda^3 + \lambda^4 + \lambda^5}{1 - \lambda}.
$$
\end{enumerate}
}
\end{exam}

\begin{exam} {\em  Let $S = k[[x, y, z, u]]$ be the formal power series
ring in the four  variables $x, y, z, u$ over a field $k$, and let
$X, Y, Z, U$ denote the leading forms of $x, y, z, u$ in $G =
\gr_{\n}(S) = k[X, Y, Z, U]$. Let $I = (f, g)S$, where $f = xy$ and
$g = xz + u^3$. Thus $R = S/I$ is a complete intersection of
dimension two.  It can be seen directly, and also is a consequence
of Theorem \ref{1.5}, that $I^* = (XY, XZ, YU^3)G$. Since $I^*$ is a
perfect ideal and $\dim G/I^* = 2$, it is possible to choose $Q =
(X_1, X_2)G$, the leading form ideal of $\frak{q} = (x_1, x_2)S$
such that $\widetilde{I^*} = \overline{I}^*$. We illustrate how to
successively choose $x_1$ and $x_2$.

\begin{enumerate}
\item
Let $x_1 = y-u$ and use this to eliminate $u$. Thus $\overline S =
k[[x, y, z]]$, $\overline f = xy$ and $\overline g = xz + y^3$. We
have
$$
I^* \widetilde{G} = (XY, XZ, Y^4) \widetilde{G} = (I\overline S)^* =
(XY, XZ, Y^4) k[X, Y, Z].
$$
We now apply the process again:
\item
Let $x_2 = z-x$ and use this to eliminate $z$. Thus $\overline S =
k[[x, y]]$, $\overline f = xy$ and $\overline g = x^2 + y^3$. We
have
$$
I^* \widetilde{G} = (XY, X^2, Y^4) \widetilde{G} = (I\overline S)^*
= (XY, X^2, Y^4)k[X, Y].
$$
The numerator polynomial of the Hilbert series in each case is $1 +
2t + t^2 + t^3$.
\end{enumerate}
}
\end{exam}




We record the following corollary to Theorem \ref{1.5}.



\begin{cor} \label{2.5}
Assume  notation as in Setting \ref{1.0} and Remark \ref{1.10}. If $D :=
D_2$ is a prime element of $G$ that is regular on $G/(\xi, \eta)$,
then $\mu(I^*) = 3$ and $I^*$ is perfect.
\end{cor}

\begin{proof} It suffices to show that $\GCD(f^*, g^*, h^*) = 1$. If
this fails, then $$ h^* \in (D) \cap (\xi, \eta) = (D\xi, D\eta) =
(f^*, g^*),
$$
a contradiction to the assumption that $h^*$ is the third generator
of $I^*$.
\end{proof}


\begin{exam} {\em  Let $S = k[[x, y, z]]$ be the formal power series
ring in the three variables $x, y, z$ over a field $k$. Let $f =
xy^i + z^s$ and $g = xz^j$, where $s > i + 1$ and $i$ and $j$ are
positive. By Corollary \ref{2.5},  $\mu(I^*) = 3 $ and $I^*$ is
perfect.


}
\end{exam}


We use Lemma \ref{lem4-1} and  Theorem \ref{1.11}  to establish
in Theorem \ref{prop4-4} conditions on the degrees of a minimal
homogeneous system of generators for $I^*$ in the case where $I^*$
is perfect. 





\begin{Thm} \label{prop4-4}
Assume  notation as in Setting \ref{1.0} and Remark \ref{1.10}. If
$\mathrm{gr}_{\fkm}(R)$ is a Cohen-Macaulay ring and $n = \mu_G(I^*)$,
then there exist homogeneous elements $\{\xi_i\}_{1 \leq i \leq n}$ of $G$ such that
\begin{enumerate}
  \item[$(1)$] $I^* = (\xi_1, \xi_2, \cdots ,\xi_n)$,
  \item[$(2)$] $\xi_1 = f^*$ and $\xi_2 = g^*$,
  \item[$(3)$] $\deg \xi_i + 2 \leq \deg \xi_{i+1}$ for all $2 \leq i \leq n-1$, and
  \item[$(4)$] $\mathrm{ht}_G(\xi_1, \xi_2, \cdots ,\xi_{n-1}) = 1$.
\end{enumerate}
\end{Thm}

\begin{proof}
By Theorem \ref{1.11}, we may assume $s > 2$. If $n = 2$, then
$I^* = (f^*, g^*)$ and there is nothing to prove. Assume $n > 2$ and
let $D = \mathrm{GCD}(f^*, g^*)$. We write $f^* =D\xi$ and $g^*
=D\eta$; hence $\xi, \eta$ is a $G$-regular sequence. We choose,
similarly as in the proof of Theorem \ref{1.5}, the elements $X_1,
X_2, \cdots, X_{s-1} \in G_1$ so that $\{X_i\}_{1 \leq i \leq s-1}$
is a homogeneous system of parameters for the rings $G/(f^*),
G/(g^*)$, and $G/(D)$ and $\{X_i\}_{1 \leq i \leq s-2}$ is a
homogeneous system of parameters for the rings $G/(\xi, \eta)$ and
$\mathrm{gr}_{\fkm}(R)$. Let $x_i \in \fkn$ with $X_i = x_i^*$. We
put $\frak{q}=(x_i \mid 1 \leq i \leq s-2)$, $\overline{S} =
S/\frak{q}$, $\overline{\fkn}=\fkn/\frak{q}$, and $\overline{I} =
(\overline{f}, \overline{g})$, where $\overline{f}$ and
$\overline{g}$ respectively  denote the images of $f$ and $g$ in
$\overline{S}$.  Then $\overline{f}^* \nmid \overline{g}^*$ (cf.
Proof of Claim \ref{claim4-3} (1)). The sequence $X_1, X_2, \cdots,
X_{s-2}$ is regular in the ring $\mathrm{gr}_{\fkm}(R)$, because
$\mathrm{gr}_{\fkm}(R)$ is Cohen-Macaulay. We identify
$$\mathrm{gr}_{\overline{\fkn}}(\overline{S}) = G/Q \ \ \text{and} \
\ \overline{I}^*= [I^* + Q]/Q,$$ where $Q = (X_i \mid 1 \leq i \leq
s-2)$. Therefore, since
$\mu_{\mathrm{gr}_{\overline{\fkn}}\overline{S}}(\overline{I}^*)
=\mu_G(I^*) = n$, thanks to Theorem \ref{1.11}, the ideal
$\overline{I}^*$ contains a homogeneous system $\{\eta_i\}_{1 \leq i
\leq n}$ of generators which satisfies the conditions
\begin{enumerate}
  \item[$(1)$] $\eta_1 = \overline{f}^*$ and $\eta_2 = \overline{g}^*$,
  \item[$(2)$] $\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-1$, and
  \item[$(3)$]
  $\mathrm{ht}_{\mathrm{gr}_{\overline{\fkn}}(\overline{S})}(\eta_i \mid 1 \leq i \leq n - 1) = 1$.
\end{enumerate}
Thus, taking $\xi_{i} \in I^*$ to be a preimage of $\eta_i$, we readily get a
homogeneous system $\{\xi_i\}_{1 \leq i \leq n}$ of generators of $I^*$ which
satisfies conditions (2) and (3) in Theorem \ref{prop4-4}.

Let us check condition (4) is also satisfied. Assume the contrary and rechoose
the system $\{X_i\}_{1 \leq i \leq s-1}$ so that $\{X_i\}_{1 \leq i \leq s-2}$ is also a
homogeneous system of parameters for the ring $G/(\xi_i \mid 1 \leq i \leq n-1)$ of
dimension $s-2$. Let $\overline{\xi_i}$ denote the image of $\xi_i$ in $G/Q$.
Then $\{\overline{\xi_i}\}_{1 \leq i \leq n}$ constitutes a minimal homogeneous
system of generators of $\overline{I}^* = [I^* + Q]/Q$ with
$\deg \overline{\xi_i} \leq \deg \overline{{\xi}_{i+1}}$ for all $2 \leq i \leq n - 1$.
Consequently, even though we do not necessarily have $\eta_i = \overline{\xi_i}$~$(1 \leq i \leq n)$
for the second choice of $\{X_i\}_{1 \leq i \leq s -1}$, we still
have $(\eta_i \mid 1 \leq i \leq n-1)= (\overline{\xi_i} \mid 1 \leq i \leq n-1)$,
because the ideals $\{(\eta_j \mid 1 \leq j \leq i)\}_{1 \leq i \leq n}$ of
$\mathrm{gr}_{\overline{\fkn}}(\overline{S})$ are independent of the choice of
minimal homogeneous systems $\{\eta_i\}_{1 \leq i \leq n}$ of generators of $\overline{I}^*$
which satisfy the condition that $\eta_1 = \overline{f}^*$, $\eta_2 = \overline{g}^*$, and
$\deg \eta_i + 2 \leq \deg \eta_{i+1}$ for all $2 \leq i \leq n-1$. This is however
impossible, since
$\mathrm{ht}_{\mathrm{gr}_{\overline{\fkn}}(\overline{S})}(\eta_i\mid 1 \leq i \leq n-1) = 1$
while $\dim G/[Q + (\xi_i \mid 1 \leq  i \leq n-1)]=0$.
Thus $\mathrm{ht}_G(\xi_i \mid 1 \leq i \leq n - 1)=1$ as claimed.
\end{proof}


\noindent {\bf Proof of Theorem \ref{1.6}.}  Assume that
$\mathrm{gr}_{\fkm}(R)$ is a Cohen-Macaulay ring and let
$\{\xi_i\}_{1 \leq i \leq n}$ be a homogeneous system of generators
of $I^*$ which satisfies conditions (2) and (3) in Theorem
\ref{prop4-4}. Let $X_1,
X_2, \cdots , X_{s-2} \in G_1$ and write $X_i = x_i^*$ with $x_i \in
\fkn$. We put $\frak{q} = (x_i \mid 1 \leq i \leq s - 2)$,
$\overline{S} = S/\frak{q}$, $\overline{\fkn} = \fkn/\frak{q}$, and
$\overline{I} = (\overline{f}, \overline{g})$, where $\overline{f}$
and $\overline{g}$ respectively denote the images of $f$ and $g$ in
$\overline{S}$. We put $Q = (X_i \mid 1 \leq i \leq s - 2)$. Then,
choosing $\{X_i\}_{1 \leq i \leq s - 2}$ to be sufficiently general,
we may assume that
\begin{enumerate}
\item[$(1)$] $\{X_i\}_{1 \leq i \leq s - 2}$ is a homogeneous system of
parameters for $\mathrm{gr}_{\fkm}(R)$, so that $\overline{S}$ is a regular local ring
of dimension $2$ with the parameter ideal $\overline{I}$, and
\item[$(2)$] $\widetilde{D_i} = \mathrm{GCD}(\widetilde{\xi_1},
\widetilde{\xi_2}, \cdots, \widetilde{\xi_i})$ for all $1 \leq i \leq n$,
\end{enumerate}
where $\widetilde{D_i}$ and $\widetilde{\xi_i}$ respectively denote
the image of $D_i$ and $\xi_i$ in $G/Q=
\mathrm{gr}_{\overline{\fkn}}(\overline{S})$.
 Then the minimal homogeneous system $\{\widetilde{\xi_i}\}_{1 \leq i \leq n}$ of generators
 of the ideal $\widetilde{I}^* = \overline{I}^*$ in $G/Q= \mathrm{gr}_{\overline{\fkn}}(\overline{S})$
 satisfies conditions (1) and (2) in Theorem \ref{1.11}. We have
$$
\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda) =
\displaystyle\frac{\mathrm{H}(\mathrm{gr}_{\overline{\fkn}}(\overline{S})/\overline{I}^*,
\lambda)}{(1 - \lambda)^{s-2}},$$ because $X_1, X_2, \cdots,
X_{s-2}$ form a regular sequence in $\mathrm{gr}_{\fkm}(R)$. The
assertions in Theorem \ref{1.6} readily follow from this. \QED


\begin{ques}  {\em
With notation as in Setting \ref{1.0} and Remark \ref{1.10}, if $I^*$ is perfect,
does it follow that
$I^* \subseteq (\xi_1/D_2, \xi_2/D_2)G$?
}
\end{ques}




\section{Examples with $\mu_G(I^*) = 3$ and with given $\mu_G(I^*)$}
Let $0 < n_1 < n_2 < n_3$ be integers such that $\mathrm{GCD}(n_1,n_2,n_3) = 1$ and
let $S=k[[X_1, X_2, X_3]]$ and $T=k[[t]]$ be the formal power series rings over a
field $k$. We denote by $\varphi  : S \to T$ the $k$-algebra map defined
by $\varphi(X_i)=t^{n_i}$ for $i = 1, 2, 3$. Let
$$I=\mathrm{Ker} \varphi, \ R =k[[t^{n_1}, t^{n_2}, t^{n_3}]], \ \fkn=(X_1,X_2,X_3)S, \
\text{and} \  \frak{m} =(t^{n_1}, t^{n_2}, t^{n_3})R.$$ We then have the following, which
is essentially due to J. Herzog \cite{H2} (see p.191--192) and  L. Robbiano and G. Valla \cite{RV}.
Let us include a brief proof in our context for the sake of completeness.

\begin{thm}
Suppose that $\mu_S(I)=2$, namely, $R$ is a Gorenstein ring. Then $\gr_{\fkm}(R)$ is a
Cohen-Macaulay ring if and only if the leading form ideal $I^*$ of $I$ is $3$-generated.
\end{thm}

\begin{proof}
See \cite[Theorem 1.2]{GHK} for the proof of the $if$ part. Suppose now that $\gr_{\fkm}(R)$ is a
Cohen-Macaulay ring. Let $G=\gr_{\fkn}(S)$, which we shall identify with the polynomial
ring $k[X_1, X_2, X_3]$ over $k$. We will show that $\mu_G(I^*) \leq 3$.
Since $\mu_S(I)=2$, as for the system of generators of $I$
we distinguish the following four cases (\cite{H1}):
\begin{enumerate}
\item $I=(X_1^{c_1} - X_2^{c_2}, X_1^{c_1} - X_3^{c_3})$,
\item $I=(X_2^{c_2} - X_3^{c_3}, X_1^{c_1} - X_2^{s_{12}}X_3^{s_{13}})$~$(s_{12} > 0, s_{13} > 0)$,
\item  $I=(X_1^{c_1} - X_3^{c_3},  X_2^{c_2} - X_1^{s_{21}}X_3^{s_{23}})$~$(s_{21} > 0, s_{23} > 0)$, and
\item $I=(X_1^{c_1} - X_2^{c_2}, X_3^{c_3} - X_1^{s_{31}}X_2^{s_{32}})$~$(s_{31} > 0, s_{32} > 0)$
\end{enumerate}
\noindent
where $c_i = \min\{0 < c \in {\mathbb Z} \mid 0 \ne X_i^{c} -
X_1^{\alpha_1}X_2^{\alpha_2}X_3^{\alpha_3} \in I \ \text{for~some}\
0 \leq \alpha_1, \alpha_2, \alpha_3 \in {\mathbb Z} \}$.
For cases (1), (3), and (4), the ideal $I+(X_1)$ is generated by monomials in
$X_1, X_2, X_3$ and so, thanks to \cite[Theorem  1]{H2}, we have $\mu_G(I^*)=\mu_S(I)= 2$,
once $\gr_{\fkm}(R)$ is a Cohen-Macaulay ring. We are now concentrated in case (2),
where
$$I=(X_2^{c_2} - X_3^{c_3}, X_1^{c_1} - X_2^{s_{12}}X_3^{s_{13}})$$ for some integers
$s_{12} > 0$ and $s_{13} > 0$. Then $c_1 =(n_2,n_3)$, $n_2 = c_1c_3$, and $n_3 =c_1 c_2$ (\cite{H1});
hence $c_3 < c_2$. We write $s_{13} = c_3q + s_{13}'$ with integers $q, s_{13}'$ such that
$0 \leq q, 0 \leq s_{13}' < c_3$ and put $s_{12}' = c_2 q + s_{12}$.
Then $$s_{13}'= 0 \ \ \text{or} \ \ c_1 + c_3 - s_{13}' \geq c_2 + s_{12}',$$
because $\gr_{\fkm}(R)$ is a Cohen-Macaulay ring (see \cite[p.192]{H2}). Let $f = X_2^{c_2}-X_3^{c_3}$
and $g =X_1^{c_1} - X_2^{s_{12}'}X_3^{s_{13}'}$. Then $I =(f,g)$,
since $g \equiv X_1^{c_1} - X_2^{s_{12}}X_3^{s_{13}} \mod (f)$. If $s_{13}' = 0$,
then $g^*=X_1^{c_1}$ if $c_1 < s_{12}'$, $g^* = X_1^{c_1} -X_2^{s_{12}'}$ if
$c_1 = s_{12}'$, and $g^* =  -X_2^{s_{12}'}$ if $c_1 > s_{12}'$.
Since $f^*=-X_3^{c_3}$ (recall that $c_3 < c_2$), in any case the forms $f^*, g^*$
constitute a regular sequence in $G$, so that we have $I^* =(f^*,g^*)$.

Assume that $s_{13}' > 0$. Then $g^* = - X_2^{s_{12}'}X_3^{s_{13}'}$,
since $c_1 -(s_{12}' + s_{13}') \geq c_2 - c_3 > 0$.
We put  $h := X_2^{s_{12}'}f +X_3^{c_3 -s_{13}'}g = X_1^{c_1}X_3^{c_3- s_{13}'} -X_2^{c_2 + s_{12}'}$.
Let $J= (f^*, g^*, h^*) \subseteq I^*$. Then
$$J=(X_2^{c_2+s_{12}'}, X_2^{s_{12}'}X_3^{s_{13}'}, X_3^{c_3})=\mathrm{I}_2\begin{pmatrix}
0&X_2^{c_2}&X_3^{s_{13}'}\\
X_2^{s_{12}'}&X_3^{c_3-s_{13}'}&0\\
\end{pmatrix}$$
$$(\text{resp.}~J=(X_1^{c_1}X_3^{c_3- s_{13}'} -X_2^{c_2 + s_{12}'},
X_2^{s_{12}'}X_3^{s_{13}'}, X_3^{c_3})=\mathrm{I}_2\begin{pmatrix}
X_1^{c_1}&X_2^{c_2}&X_3^{s_{13}'}\\
X_2^{s_{12}'}&X_3^{c_3-s_{13}'}&0\\
\end{pmatrix})$$  if $c_1 + c_3 - s_{13}' > c_2 + s_{12}'$ (resp. $c_1 + c_3 - s_{13}'= c_2 + s_{12}')$.
We now want to show $I^* = J$. For this purpose we firstly look at the exact sequence
$$0 \to I^*/J \to G/J \to \gr_{\fkm}(R) \to 0.$$
Then, since $a =t^{n_1}$ is a minimal reduction of the ideal $\fkm$, the
element $X_1 \in G$ acts on the Cohen-Macaulay ring $\gr_{\fkm}(R)$ as a non-zerodivisor,
whence we have the exact sequence
$$0 \to (I^*/J)/X_1(I^*/J)  \to G/[(X_1)+ J] \overset{\varepsilon}{\to} \gr_{\fkm /(a)}(R/(a)) \to 0.$$
Therefore, to show $I^* = J$, by Nakayama's lemma it is
enough to check that $\varepsilon$ is an isomorphism, or equivalently, to check that
$$\dim_kG/[(X_1)+ J] \leq \dim_k\gr_{\fkm /(a)}(R/(a)).$$
We have $$\dim_k\gr_{\fkm /(a)}(R/(a))=\ell_R(R/(a))=\mathrm{e}^0_{\fkm}(R)=n_1$$ and
$n_1 =c_3s_{12}' + c_2s_{13}'$~(recall that $n_1c_1=n_2s_{12}' + n_3s_{13}'$, $n_2 =c_1c_3$,
and $n_3 =c_1c_2$). On the other hand, since
$$G/[(X_1)+J]\cong k[X_2,X_3]/(X_2^{c_2 + s_{12}'}, X_2^{s_{12}'}X_3^{s_{13}'}, X_3^{c_3}),$$
we readily get $\dim_kG/[(X_1)+J] \leq c_3(c_2 + s_{12}')
-c_2(c_3 - s_{13}') =c_3s_{12}' + c_2s_{13}' = n_1$.
Hence $I^* = J$ so that we have $\mu_G(I^*) = 3$ as claimed.
\end{proof}



\begin{cor}[to the proof] \label{1.1}
Assume that $\mu_S(I)=2$. Then $\mu_G(I^*)=3$ if and only if there exist integers
$\alpha, \beta \in {\mathbb Z}$ such that 
$0 < \alpha, 0 < \beta < c_3, c_1 + c_3 \geq c_2 +(\alpha + \beta)$,
and $I = (X_2^{c_2} - X_3^{c_3}, X_1^{c_1} - X_2^{\alpha}X_3^{\beta})$. When this is the case,
we have $c_1 = \mathrm{GCD}(n_2, n_3), n_2 = c_1c_3, n_3 = c_1c_2, n_1 =c_3\alpha + c_2 \beta$, and
the leading form ideal $I^*$ of $I$ is given by
$$I^* = \mathrm{I}_2\begin{pmatrix}
0 & X_2^{c_2}&X_3^{\beta} \\
X_2^{\alpha} & X_3^{c_3 -\beta} & 0 \\
\end{pmatrix}~(\text{resp.}~ I^* = \mathrm{I}_2\begin{pmatrix}
X_1^{c_1} & X_2^{c_2}&X_3^{\beta} \\
X_2^{\alpha} & X_3^{c_3 -\beta} & 0 \\
\end{pmatrix})$$
if $c_1 + c_3 > c_2 + (\alpha + \beta)$~$($resp. $c_1 + c_3 = c_2 + (\alpha + \beta)$ $)$.
\end{cor}

\begin{remark}\label{1.2}
{\em
This result classifies Gorenstein numerical semigroups $H=\left<n_1,n_2,n_3\right>$ generated by 3
integers $n_i's$ with $0 < n_1< n_2<n_3$ and $\mathrm{GCD}(n_1,n_2,n_3) =1$, for which the
associated graded rings $\gr_{\fkm}(R)$~$(R=k[[t^{n_1}, t^{n_2}, t^{n_3}]], k~\text{a~field})$
are non-Gorenstein Cohen-Macaulay rings. In fact, firstly we choose integers
$c_2, c_3$ so that $2 \leq  c_3 < c_2$ and $\mathrm{GCD}(c_2,c_3)=1$. Let $\alpha, \beta$
be integers such that $0 < \alpha, 0 < \beta < c_3$ and put $n_1=c_3\alpha + c_2\beta$.
We choose an integer $c_1$ so that $c_1 > \frac{n_1}{c_3}$, $\mathrm{GCD}(n_1,c_1)=1$,
and $c_1 + c_3 \geq c_2 + (\alpha + \beta)$. Lastly let $n_2 = c_1c_3$ and $n_3 = c_1c_2$.
Then for $H=\left<n_1,n_2,n_3\right>$ we easily get the equality
$$I= (X_2^{c_2} - X_3^{c_3}, X_1^{c_1} - X_2^{\alpha}X_3^{\beta})$$ and
$c_i = \min\{0 < c \in {\mathbb Z} \mid 0 \ne X_i^{c} - 
X_1^{\alpha_1}X_2^{\alpha_2}X_3^{\alpha_3} \in I \
\text{for~some}\  0 \leq \alpha_1, \alpha_2, \alpha_3 \in {\mathbb Z} \}$ as well for
each $i=1,2,3$. Hence by Corollary \ref{1.1} the ring $\gr_{\fkm}(R)$ is a
non-Gorenstein Cohen-Macaulay ring. Let $f=X_2^{c_2} - X_3^{c_3}$, $g= X_1^{c_1} - X_2^{\alpha}X_3^{\beta}$,
and  $h = X_1^{c_1}X_3^{c_3- \beta} -X_2^{c_2 + \alpha}~(= X_2^{\alpha}f +X_3^{c_3 -\beta}g)$.
Then, since $c_2 > c_3$ and $c_1-(\alpha + \beta) \geq c_2 - c_3 > 0$, we have
$f^* = - X_3^{c_3}$ and  $g^* = - X_2^{\alpha}X_3^{\beta}$ whence $\mathrm{GCD}(f^*, g^*) = X_3^{\beta}$,
while $h^*=-X_2^{c_2 + \alpha}$~(resp. $h^* = X_1^{c_1}X_3^{c_3- \beta} -X_2^{c_2 + \alpha}$)
if $c_1 + c_3 > c_2 + (\alpha + \beta)$~$($resp. $c_1 + c_3 = c_2 + (\alpha + \beta))$,
which is the third generator of $I^*$. Hence
$$\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda) =
\displaystyle\frac{(1 - \lambda^{c_2 + \alpha})(1 - \lambda^{\beta})
+ \lambda^{\beta}(1 - \lambda^{c_3 - \beta})(1 - \lambda^{\alpha})}{(1 - \lambda)^3}$$  and
$\mathrm{e}^0_{\fkm}(R) = c_3(\alpha + \beta) + \beta [\{(c_2 + \alpha)
+ \beta\} -\{c_3 + (\alpha + \beta) \}]
= c_3 \alpha + c_2 \beta = n_1$  by Theorem  \ref{1.5}.


}
\end{remark}


Let us note more concrete examples.


\begin{exam} {\em $(1)$ Let $q \geq 0$ be an integer and put $n_1 = 6q+5$, $n_2 =2(3q + 4)$,
and $n_3 = 3(3q+4)$. Then, letting $c_2 = 3, c_3=2, \alpha = 3q+1$,  $\beta = 1$, and $c_1 = 3q + 4$,
by Corollary \ref{1.1} and Remark \ref{1.2} we get
$I^* = I_2\begin{pmatrix}
0&X_2^3 & X_3\\
X_2^{3q+1}&X_3&0\\
\end{pmatrix}$. If we take $q = 0$, then $n_1 = 5, n_2=8, n_3 = 12$.

$(2)$ Similarly,
let $q \geq 0$ be an integer and put $n_1 = 6q+5$, $n_2 =2(3q + 3)$, and $n_3 = 3(3q+3)$.
Then, letting $c_2 = 3, c_3=2, \alpha = 3q+1$,  $\beta = 1$, and $c_1 = 3q + 3$,
by Corollary \ref{1.1}  and Remark \ref{1.2} we get
$I^* = I_2\begin{pmatrix}
X_1^{3q+3}&X_2^3 & X_3\\
X_2^{3q+1}&X_3&0\\
\end{pmatrix}$. If we take $q = 0$, then $n_1 = 5, n_2=6, n_3 = 9$, which is \cite[Example 1.5]{GHK}.}
\end{exam}


We close this section with an example due to Takahumi Shibuta (Kyusyu University). His example
shows that, unless $\mathrm{gr}_{\fkm}(R)$ is Cohen-Macaulay,
we do not necessarily have the descending sequence
$$a = d_1 > d_2> \cdots > d_{n-1} > d_n = 0$$
of degrees of GCD's of $\xi_i's$ even for a minimal homogeneous system $\{\xi_i\}_{1 \leq i \leq n}$
of generators of $I^*$ which satisfies the conditions in Theorem \ref{prop4-4}.


\begin{exam} \label{5.5}
{\em
Let $2 \leq m \in {\mathbb Z}$ and put $n_1 = 3m, n_2= 3m + 1$, and $n_3 = 6m + 3$.
Then $I = (X_1^{2m+1} - X_3^m, X_2^3 - X_1X_3)$ in $S$  and
$I^* =(X_1X_3)+(X_2^{3i}X_3^{m-i} \mid 0 \leq i \leq m)$
in $G = k[X_1, X_2, X_3]$ with $\mu_G(I^*) = m + 2$. Letting $\xi_1 = X_1X_3$
and $\xi_i =X_2^{3(i-2)}X_3^{m - i + 2}$ for $2 \leq i \leq m+2$, we see that the minimal
homogeneous system $\{\xi_i\}_{1 \leq i \leq m+2}$ of generators of $I^*$
satisfies the conditions in Theorem \ref{prop4-4}, while $\mathrm{GCD}(\xi_1, \xi_2, \cdots, \xi_i) = X_3$
for $2 \leq i \leq m+1$. We have
$$\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda) =\displaystyle\frac{\sum_{i=2}^{m}\lambda^{m - i +1}
- \sum_{i=2}^{m}\lambda^{m + 2i - 2} + \sum_{i=0}^{3m-1}\lambda^i}{1 - \lambda}.$$


}
\end{exam}

\begin{proof}
It is routine to check that $I = (X_1^{2m+1} - X_3^m, X_2^3 - X_1X_3)S$.
Hence we have $X_1X_3, X_3^{m} \in I^*$. Let $h_i =X_1^{2m + i + 1} - X_2^{3i}X_3^{m-i}$
for $1  \leq i \leq m$. We put $J =(X_1X_3)+(X_2^{3i}X_3^{m-i} \mid 0 \leq i \leq m)$ in $G$.
Then $h_i \in I$ for all $1 \leq i \leq m$, whence
$J \subseteq I^*$. Let $K = (X_1)+(X_2^{3i}X_3^{m-i-1} \mid 0 \leq i \leq m-1)$ in $G$.
Then $\sqrt{K} = G_+=(X_1, X_2, X_3)$, $J :_GX_3 = K$, and $(X_3) + J = (X_2^{3m}, X_3)$.
Consequently, $\mathrm{H}_{N}^0(G/J) = (\overline{X_3})$, where $\overline{X_3}$ is the image
of $X_3$ in $G/J$ and $\mathrm{H}_{N}^0(G/J) $ denotes the $0^{\underline{th}}$ local cohomology
module of $G/J$ with respect to $N = G_+$. Hence $$(0) :_{G/J} N =
\sum_{i=1}^{m-1}k\overline{X_2^{3i-1}X_3^{m-i}},$$
because $(\overline{X_3}) \cong  [G/K](-1)$ and the $k$-vector space $(0) :_{G/K}N$ is spanned  by the
images of $\{X_2^{3i-1}X_3^{m-i-1}\}_{1 \leq i \leq m-1}$.


Let $\overline{\theta} : G/[J + (X_3)] \to \mathrm{gr}_{\fkm}(R)/\mathrm{H}^0_N(\mathrm{gr}_{\fkm}(R))$
be the epimorphism induced from the canonical epimorphism $G \to \mathrm{gr}_{\fkm}(R)$.
Recall that  $X_1$ is a parameter for the ring $\mathrm{gr}_{\fkm}(R)$, since $t^{3m}$ is a minimal
reduction of $\fkm$, so that $X_1$ is a non-zerodivisor in the Cohen-Macaulay ring
$\overline{\mathrm{gr}_{\fkm}(R)} =\mathrm{gr}_{\fkm}(R)/\mathrm{H}^0_N(\mathrm{gr}_{\fkm}(R))$.
Hence $\overline{\theta}$ is an isomorphism, because $\dim_k G/[J + (X_1, X_3)] = 3m$ and
\begin{eqnarray*}
\dim_k \mathrm{gr}_{\fkm}(R)/[\mathrm{H}^0_N(\mathrm{gr}_{\fkm}(R)) + X_1\mathrm{gr}_{\fkm}(R)]
&=& \mathrm{e}^0_{X_1\mathrm{gr}_{\fkm}(R)}(\overline{\mathrm{gr}_{\fkm}(R)}) \\
&=& \mathrm{e}^0_{X_1\mathrm{gr}_{\fkm}(R)}(\mathrm{gr}_{\fkm}(R)) \\
&=& \mathrm{e}^0_{\fkm}(R) \\
&=& 3m.
\end{eqnarray*}
Therefore, the kernel of the epimorphism $\theta : G/J \to \mathrm{gr}_{\fkm}(R)$ induced from
the canonical epimorphism $G \to \mathrm{gr}_{\fkm}(R)$ is contained in
$(\overline{X_3}) = \mathrm{H}^0_N(G/J)$
and so, to see that $\theta$ is an isomorphism, it suffices to show that $\theta$ is injective
on the socle $$(0) :_{G/J}N = \sum_{i=1}^{m-1}k\overline{X_2^{3i-1}X_3^{m-i}}$$ of $G/J$, that is, it is
enough to show $\theta(\overline{X_2^{3i-1}X_3^{m-i}}) \ne 0$ in $\mathrm{gr}_{\fkm}(R)$
for any $1 \leq i \leq m-1$,
because the degrees of $\overline{X_2^{3i-1}X_3^{m-i}}$ are distinct.


Let $x = t^{n_1}, y= t^{n_2}$, and $z = t^{n_3}$. We put $U = k[x, y, z]$ in $R$. Hence $U$ is a
graded ring with $\deg x = n_1, \deg y = n_2,$ and $\deg z = n_3$. Let $M = U_+ = (x, y, z)U$.
We denote by $U_i$ the homogeneous component of $U$ of degree $i$. In what follows we will show
that $y^{3i-1}z^{m-i} \notin \fkm^{m + 2i}$ for any $1 \leq i \leq m -1$. Assume that
$y^{3i-1}z^{m-i} \in \fkm^{m + 2i}$, or equivalently, assume that $y^{3i-1}z^{m-i} \in M^{m + 2i}$
for some $1 \leq i \leq m -1$. Then we have the following.

\begin{claim}\label{5.6}
$y^{3i-1}z^{m-i} \in M^{m + 2i + \ell}$ for all $0 \leq \ell \leq m - i$.
\end{claim}

\begin{proof}[$Proof~of~Claim~\ref{5.6}$]
When $\ell = 0$, we have nothing to prove. Assume that $0 \leq \ell < m-i$ and that our
assertion holds true for $\ell$. We put $\delta = (3i-1)n_2 + (m-i)n_3 = 6m^2 + 3mi -1$.
Then $t^\delta = y^{3i-1}z^{m-i}   \in M^{m + 2i + \ell} =
\sum_{\alpha = 0}^{m + 2i + \ell}(x, y)^{m + 2i + \ell -\alpha}{\cdot}z^\alpha$ in $U$.
Take $0 \leq \alpha \in {\mathbb Z}$ and assume that $m - i -\ell \leq \alpha \leq m + 2i + \ell$.
Then $$(x, y)^{m + 2i + \ell -\alpha}{\cdot}z^\alpha =
(x^\beta y^\gamma z^\alpha \mid 0 \leq \beta, \gamma \in {\mathbb Z}~\text{such~that}~ \beta +
\gamma = m + 2i + \ell - \alpha).$$ We now choose $0 \leq \beta, \gamma \in {\mathbb Z}$
so that $\beta + \gamma = m + 2i + \ell - \alpha$ and put $\eta = \beta n_1 + \gamma n_2 + \alpha n_3$.
Then \begin{eqnarray*}
\eta &\geq&(\beta + \gamma)n_1 + \alpha n_3\\
&=& 3(m + 2i + \ell - \alpha)m + \alpha (6m+3)\\
&=& 3m^2 + 6mi + 3m\ell + 3m\alpha + 3\alpha\\
&\geq& 6m^2 + 3mi + 3(m-i-\ell )\ \ \ (\text{since}~\alpha \geq m - i - \ell )\\
&\geq&6m^2+3mi\\
&>& \delta = 6m^2 + 3mi - 1.
\end{eqnarray*}
Consequently, $t^{\delta} \in \sum_{\alpha = 0}^{m - i - \ell -1}(x, y)^{m + 2i + \ell
- \alpha}{\cdot}z^{\alpha}$.
We write $t^{\delta} = \sum_{\alpha = 0}^{m - i - \ell - 1}\varphi_{\alpha} z^{\alpha}$
with $\varphi_{\alpha} \in (x, y)^{m + 2i + \ell - \alpha}$ such that
$\varphi_{\alpha} \in U_{\delta - \alpha n_3}$. Let us furthermore write
$\varphi_{\alpha} = \sum_{\beta = 0}^{m + 2i + \ell -
\alpha}w_{\alpha, \beta}{\cdot}x^{\beta}y^{m + 2i + \ell - \alpha - \beta}$
with $w_{\alpha, \beta} \in U_{\delta - \alpha n_3 -(\beta n_1 + (m + 2i + \ell - \alpha - \beta)n_2}$.
Then, if $0 \leq \alpha \leq m - i - \ell - 1$, choosing $0 \leq \beta, \gamma \in {\mathbb Z}$
with $\beta + \gamma = m + 2i + \ell - \alpha$, we have
\begin{eqnarray*}
\beta n_1 + \gamma n_2 + \alpha n_3  &=&3m\beta + (3m+1)\gamma + \alpha n_3\\
&\leq& (3m+1)(\beta + \gamma) + \alpha (6m+3)\\
&=&3m^2 + m + 6mi +2i + 3\ell m + \ell + 3m\alpha + 2\alpha\\
&\leq&6m^2 + 3mi - \ell -2 \ \ (\text{since}~\alpha \leq m - i - \ell - 1)\\
&<&\delta = 6m^2 + 3mi -1,
\end{eqnarray*}
whence $\beta n_1 + \gamma n_2 < \delta - \alpha n_3$.
Consequently, $w_{\alpha, \beta} \in M$ for each $\alpha$ and $\beta$, so that
$\varphi_{\alpha} \in M^{m + 2i + \ell - \alpha + 1}$ for all $0 \leq \alpha \leq m - i - \ell - 1$,
whence $t^{\delta} \in M^{m + 2i + \ell + 1}$ as claimed.
\end{proof}
Therefore $t^{\delta} \in M^{2m + i}$, which is however impossible, because
\begin{eqnarray*}
\beta n_1 + \gamma n_2 + \tau n_3 \geq (\beta + \gamma + \tau)n_1 &=& (2m + i){\cdot}3m \\
&=& 6m^2 + 3mi > \delta
\end{eqnarray*} for all $0 \leq \beta, \gamma, \tau \in {\mathbb Z}$ with
$\beta + \gamma + \tau = 2m + i$. Thus the epimorphism $\theta : G/J \to \mathrm{gr}_{\fkm}(R)$ is
injective on the socle of $G/J$, so that $\theta$ is an isomorphism. Hence $I^* = J$.

Because $\mathrm{H}_N^0(G/I^*) =(\overline{X_3}) \cong (G/K)(-1)$,
thanks to the exact sequence
$$0 \to (G/K)(-1) \to G/I^* \to G/(X_2^{3m}, X_3) \to 0$$
of graded $G$-modules, we have$$\mathrm{H}(G/I^*, \lambda) = \lambda{\cdot}\mathrm{H}(G/K, \lambda) +
\displaystyle\frac{1 - \lambda^{3m}}{(1-\lambda)^2}.$$
Therefore
\begin{eqnarray*}
\mathrm{H}(\mathrm{gr}_{\fkm}(R), \lambda)
&=&\displaystyle\frac{\sum_{i=2}^m \lambda^{m  - i+1}(1-\lambda^{3(i-1)})}{1 - \lambda}
+\displaystyle\frac{1 - \lambda^{3m}}{(1 - \lambda)^2}\\
&=&\displaystyle\frac{\sum_{i=2}^{m}\lambda^{m - i +1} - \sum_{i=2}^{m}\lambda^{m + 2i - 2}
+ \sum_{i=0}^{3m-1}\lambda^i}{1 - \lambda}
\end{eqnarray*}
by Proposition \ref{prop2-6}, since $G/K = k[X_2,
X_3]/(X_2^{3i}X_3^{m-i-1} \mid 0 \leq i \leq m - 1).$

\end{proof}








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\end{document}




In Lemma \ref{2.1.1} we record some properties involving associated
graded rings. We relax the hypothesis that $S$ is a regular local
ring in Lemma \ref{2.1.1}.

\begin{lem} \label{2.1.1} Let $(S, \n)$ be a Noetherian local ring,  and
let $w \in \n \setminus \n^{2}$ be such that its leading form $w^*
\in G := \gr_{\n}(S)$ is a regular element of $G$. Let $I \subseteq
\n$ be an ideal of $S$ such that $w^* \not\in I^*$. Consider the
following commutative diagram:
$$
\CD
 S  @>>>   S/wS := \overline{S}  \\
 @V\varphi_1VV      @V\varphi_2VV                   \\
 G := \gr_{\n}(S)  @>>>  G/w^*G := \widetilde{G} \cong
\gr_{\overline \n}(\overline S).
\endCD
$$
Here $\varphi_1$ and $\varphi_2$ denote the canonical maps
associating an element with its leading form in the associated
graded ring, and the identification $\widetilde{G} \cong
\gr_{\overline \n}(\overline S)$ is because $w^*$ is regular on $G$.
We have
\begin{enumerate}
\item
$I^* \widetilde{G} \subseteq (I\overline S)^*$.
\item
$I^*\widetilde {G} = (I\overline S)^* \iff w^*$ is a regular element
on $G/I^*$.
\end{enumerate}
\end{lem}

\demo The commutativity of the displayed diagram implies
$I^*\widetilde{G} \subseteq (I\overline S)^*$. We have
$I^*\widetilde{G} = (I\overline S)^*$ if and only if the leading
ideal in $G/I^*$ of the principal ideal $w + I$ of $S/I$ is
principal and generated by the image of $w^*$ in $G/I^*$. This holds
if and only if $w^*$ is regular on $G/I^*$. \QED
