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\begin{document}
\title[the leading ideal of a complete intersection]{The leading ideal of a complete
intersection \\ of height two}
% 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@math.skku.ac.kr}
\thanks{Mee-Kyoung Kim is supported by the Korea Science and Engineering Foundation
(R04-2003-000-10113-0).}
\
\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 Noetherian local ring and let $I = (f,
g)$ be an ideal in $S$ generated by a regular sequence $f, g$ of
length two. Assume that the associated graded ring $\gr_{\n}(S)$ of
$S$ with respect to $\n$ is a UFD. We examine generators of the
leading form ideal $I^*$ of $I$ in $\gr_{\n}(S)$ and prove that
$I^*$ is a perfect ideal of $\gr_{\n}(S)$, if $I^*$ is 3-generated.
Thus, in this case, letting $R = S/I$ and $\m = \n/I$, if
$\gr_{\n}(S)$ is Cohen-Macaulay, then $\gr_{\m}(R) =
\gr_{\n}(S)/I^*$ is Cohen-Macaulay. As an application, we prove
that if $(R, \m)$ is a one-dimensional Gorenstein local ring of
embedding dimension 3, then $\gr_{\m}(R)$ is Cohen-Macaulay if the
reduction number of $\m$ is at most 4.
\end{abstract}
\baselineskip 20 pt
\section{Introduction}
\begin{setting} \label{1.1}
{\em Let $(S,\frak{n})$ be a Noetherian 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 $\m = \n/I$.
Let
$$\mathrm{R}'(\n) = \sum_{i \in \Bbb Z}\n^i t^i \subseteq S[t,t^{-1}]\ \ \ \text{and}
\ \ \mathrm{R}'(\m) = \sum_{i \in \Bbb Z}\m^i t^i \subseteq R[t,t^{-1}]$$
denote the extended Rees algebras of $\frak{n}$ and $\frak{m}$ respectively, where $t$ is an indeterminate.
Let $$\gr_{\n}(S) = \mathrm{R}'(\n)/t^{-1}\mathrm{R}'(\n) \ \ \ \text{and} \ \
\gr_{\m}(R) = \mathrm{R}'(\m)/t^{-1}\mathrm{R}'(\m).$$ Then the canonical
map $S \to R$ induces the homomorphism $\varphi : \gr_{\n}(S) \to \gr_{\m}(R)$ of the associated graded rings.
We put $$I^*= \mathrm{Ker}~(\gr_{\m}(S) \overset{\varphi}\to \gr_{\m}(R)).$$
Then the ideal $I^*$ is generated by the initial forms of elements of $I$ and $\gr_{\m}(R)
\cong \gr_{\n}(S)/I^*$. We assume that $G = \gr_{\n}(S)$ is a UFD. Hence $\hgt_GI^* = \grade_GI^* = 2$.}
\end{setting}
We are interested in determining generators for $I^*$ and thereby
obtaining conditions in order that $\gr_{\m}(R)$ be Cohen-Macaulay.
The goal of the paper is to prove Theorem \ref{1.2}, the proof of
which is given in Section \ref{sec2}.
\begin{thm} \label{1.2}
Assume notation as in Setting \ref{1.1}, so, in particular,
$\gr_{\n}(S)$ is a UFD. If $I^*$ is $3$-generated, then $I^*$ is a
perfect ideal of $\gr_{\n}(S)$. Therefore if $\gr_{\n}(S)$ is
Cohen-Macaulay, then $\gr_{\m}(R) = \gr_{\n}(S)/I^*$ is
Cohen-Macaulay.
\end{thm}
As an immediate corollary to Theorem \ref{1.2}, we have
\begin{cor} \label{1.3}
With notation as in Setting \ref{1.1}, if $(S, \n)$ is a regular
local ring and $I^*$ is $3$-generated, then $\gr_{\m}(R)$ is
Cohen-Macaulay.
\end{cor}
In Section \ref{sec3} we discuss some consequences of Theorem
\ref{1.2}.
\begin{notation}
{\em Let $G = \gr_{\n}(S)$. For each $f \in S$ let $o(f) = \sup \{i
\in \Bbb Z \mid f \in \n^i \}$, the order of $f$. We put
\begin{eqnarray}
f^* = \left\{
\begin{array}{ll}
\overline{ft^i} & \hbox{ if } f \ne 0 \hbox{ and } i=o(f), \\
0 & \hbox{ if } f=0
\end{array}
\right.\nonumber
\end{eqnarray}
and call it the {\it initial form } of $f$, where $\overline{ft^i}$
denotes the image in $G$ of $ft^i \in \n^it^i$ in
$\mathrm{R}'(\n)$. Then for all $f, g \in S$ we have
\par
$o(fg)= o(f) + o(g), \ \ (fg)^* = f^* g^*$, \par
$o(f+g) \geq \min \{ o(f), o(g) \} $, and \par
$o(f+g) = \min \{ o(f), o(g) \}$ if $o(f) \ne o(g)$. \par}
\end{notation}
With this notation the following two simple examples illustrate
the situation we are considering. In both examples we let $S =
k[[x, y, z]]$ be the formal power series ring in the three variables
$x, y, z$ over a field $k$.
\begin{exam} \label{1.3}
{\em Let $R = k[[w^5, w^6, w^9]]$ be the subring of the
formal power series ring $k[[w]]$ and define the
homomorphism $\phi :S \to R$ of $k$-algebras by $\phi(x) = w^5$, $\phi(y) = w^6,$
and $\phi(z) = w^9$. Then the ideal
$I = \mathrm{Ker}~\phi$ is generated by $f = z^2 - y^3$ and
$g = zy - x^3$, whence $R$ is a complete intersection of dimension one. We have
$\gr_{\n}(S) = k[x^*, y^*, z^*]$,
$f^* = z^{*2}$, and $g^* = z^*y^*$.
Let $h = yf - zg = zx^3 - y^4$. Then $h^* = z^*x^{*3} - y^{*4}$. Let
$$
J = (f^*, g^*, h^*) = (z^{*2}, z^*y^*, z^*x^{*3} - y^{*4}) \subseteq I^*.
$$
Then the Hilbert series of the graded ring $\gr_{\n}(S)/J$ is
$$
\frac{1+2t+t^2+t^3}{1-t} = 1+3t+4t^2 + 5t^3 + 5t^4 + \cdots + 5t^n + \cdots
$$
and these values are the same as those in the Hilbert series of
$\gr_{\m}(R) = \gr_{\n}(S)/I^*$, so that $J = I^*$. The reduction
number of $\m = (w^5, w^6, w^9)$ with respect to the principal
reduction $(w^5)$ is $3$ and the relation type of $\gr_{\m}(R)$ is
4. The ideal $I^*$ has grade 2 and is generated by the $2 \times 2$
minors of the following matrix
$$
\begin{bmatrix}
y^* & z^{*} & 0 \\
-x^{*3} & -y^{*3} & z^* \\
\end{bmatrix}.
$$
Hence, by the theorem of Hilbert-Burch \cite[Theorem 1.4.17]{BH}, $I^*$ is a perfect ideal
and $\gr_{\n}(S)/I^* = \gr_{\m}(R)$ is a Cohen-Macaulay ring.
}
\end{exam}
\begin{exam} \label{1.4}
{\em
Let $R = k[[w^6, w^7, w^{15}]]$ be the subring of
the formal power series ring $k[[w]]$ and consider the
homomorphism
$\phi : S \to R$ of $k$-algebras defined by $\phi(x) = w^6$, $\phi(y) = w^7$,
and $\phi(z) = w^{15}$. Then
$I = \mathrm{Ker}~\phi$ is generated by $f = z^2 - x^5$
and $g = zx - y^3$, whence $R$ is a complete intersection of dimension one. We have
$\gr_{\n}(S) = k[x^*, y^*, z^*]$,
$f^* = z^{*2},$ and $g^* = z^*x^*$.
Let $h = xf - zg = zy^3 - x^6$.
Then $h^* = z^*y^{*3}$ and
$ (f^*, g^*, h^*) = (z^{*2}, z^*x^*, z^*y^{*3}) \subsetneq I^*.$
The inclusion is strict, since $\hgt_{\gr_{\n}(S)}I^*= 2$
and $z^*$ is a common factor of $f^*$, $g^*$, and $h^*$. We have
$o(f) = o(g) = 2$ and $o(h) = 4$. Let
$ h_1 = xh - y^3g = y^6 - z^7 \in I$. Then $h_1^* = y^{*6}.$
We put
$$
J = (z^{*2}, z^*x^*, z^*y^{*3}, y^{*6} ) \subseteq I^*.
$$
Then the Hilbert series of $\gr_{\n}(S)/J$ is given by
$$
\frac{1+2t+t^2+t^3+t^5}{1-t} = 1+3t+4t^2 + 5t^3 + 5t^4 + 6t^5 + \cdots + 6t^n + \cdots
$$
and these values are the same as those in the Hilbert series of
$\gr_{\m}(R) = \gr_{\n}(S)/I^*$, so that $J = I^*$. The reduction
number of $\m = (w^6, w^7, w^{15})$ with respect to the principal
reduction $(w^6)$ is $5$ and the relation type of $\gr_{\m}(R)$ is
6. The ring $\gr_{\m}(R)$ is not Cohen-Macaulay. This is implied by
the gap in the numerator of the Hilbert series, and can be deduced
also from the fact that the ideal $I^*$ has radical $(y^*, z^*)$ and
the ideal $I^* : z^*$ is primary with $\sqrt{I^* : z^*}=(x^*, y^*,
z^*)$. }
\end{exam}
\section{Proof of Theorem \ref{1.2}} \label{sec2}
The purpose of this section is to prove Theorem \ref{1.2}. We assume
notation as in Setting \ref{1.1}. Let $G = \gr_{\n}(S)$ and $J =
I^*$. We choose $f,g \in S$ so that $I=(f,g)$ with $a=o(f) \leq
b=o(g)$. Without loss of generality we may assume that $f^* \not\in
NJ$ and $g^* \not\in NJ + (f^*)$, where $N = G_+$. Hence the
elements $f^*, g^*$ form part of a minimal system of homogeneous
generators of $J$. Notice that if $\hgt_G (f^*,g^*)$ = 2, then the
sequence $f^*, g^*$ is $G$-regular whence $J=(f^*,g^*)$. In what
follows we assume that
$$\hgt_G (f^*,g^*) = 1.$$
\noindent Let $D= \mathrm{GCD}(f^*,g^*)$ and write $f^* = \xi D$,
$g^* = \eta D$, where $D, \xi, \eta$ are homogeneous elements of
$G$ with degree $d>0$, $a-d$, and $b-d$, respectively. Then $\{ \xi,
\eta \}$ is a $G$-regular sequence.
We begin with Lemma \ref{Lem1} which gives some information about
homogeneous elements of $J$ that are not in the ideal $(f^*, g^*)$.
\begin{lem}{\label{Lem1}}
Let $\alpha, \beta \in S$ and $h=\alpha f + \beta g$. Assume that $h^* \not\in (f^*,g^*)$.
Then
\begin{enumerate}
\item[$(1)$] $o(\alpha f) = o(\beta g) < o(h)$.
\item[$(2)$] $o(\alpha) + a = o(\beta) + b$, $o(\alpha) \geq b-d$,
and $o(\beta) \geq a-d$.
\item[$(3)$] $\alpha^*\xi + \beta^*\eta = 0.$
\end{enumerate}
\end{lem}
\begin{proof}
We have $o(h) \geq \min \{ o(\alpha f), o(\beta g) \}$.
If $o(\alpha f) < o(\beta g)$, then $o(h) = o(\alpha f)$
and $h^* = \alpha^* f^* \in (f^*)$, which is impossible.
We similarly have $o(\alpha f)=o(\beta g)$.
Hence $o(h) > o(\alpha f) = o(\beta g)$, because $h^* \not\in (f^*,g^*)$.
Thus $\alpha^* f^* + \beta^* g^* = (\alpha^* \xi + \beta^* \eta )D =0$
whence $\alpha^* \xi + \beta^* \eta = 0$.
Therefore, since the sequence $\xi, \eta$ is $G$-regular, we get
$\alpha^*=- \varphi \eta$ and $\beta^* = \varphi \xi$ for some
homogeneous element $\varphi$ of $G$. Thus $o(\alpha)=\deg \varphi + (b-d)$ and
$o(\beta) = \deg \varphi + (a-d)$, so that
$o(\alpha) + a =o(\beta) + b$, $o(\alpha) \geq b-d$, and
$o(\beta) \geq a-d$, as was claimed.
\end{proof}
The existence of a third generator of the leading ideal $J$ of a
certain form is guaranteed by Proposition \ref{Prop2}.
\begin{prop}{\label{Prop2}}
Assume that the local ring $S$ is $\frak{n}$-adically complete.
Then there exist elements $\alpha, \beta$ of $S$ such that
$o(\alpha)=b-d$, $o(\beta)=a-d$, and
$(\alpha f + \beta g)^* \not\in (f^*, g^*)$.
\end{prop}
\begin{proof}
Assume the contrary.
Let $f_0, g_0 \in S$ with $o(f_0)=a-d$ and $o(g_0) = b-d$
such that $\xi = f_0^*$ and $\eta = g_0^*$.
We are going to construct two sequences $\{ f_i \}_{i=0,1,2,...}$
and $\{ g_i \}_{i=0,1,2,...}$ of elements in $S$ which satisfy the following
conditions:
Let
$h_i = (-\sum_{k=0}^i g_k)f + (\sum_{k=0}^i f_k)g$ for each $i \geq 0$.
Then \par
(1) $h_i \ne 0$, \par
(2) $o(h_i) < o(h_{i+1})$, \par
(3) $o(h_i) -b \leq o(f_{i+1})$ and $o(h_i)-a \leq o(g_{i+1})$ \par
\noindent
for all $i \geq 0$. \par
\noindent
To construct the sequences, firstly we put $h_0 = (-g_0)f + f_0 g$.
Then $o(f_0) = a-d$ and $o(g_0) = b-d$.
We notice $h_0 \ne 0$, because $b-d = o(g_0) < o(g)=b$
(recall that $f,g$ is a regular sequence).
Hence $h_0^* \in (f^*,g^*)$ by our assumption.
We write $h_0^* = f^* \varphi + g^* \psi$ with $\varphi \in G_{o(h_0)-a}$ and
$\psi \in G_{o(h_0)-b}$.
Let $\varphi = \overline{g_1t^{o(h_0)-a}}$ and $\psi = \overline{(-f_1)t^{o(h_0)-b}}$
with $g_1 \in \n^{o(h_0)-a}$ and $f_1 \in \n^{o(h_0)-b}$.
Then $h_0=g_1f+(-f_1)g + h_1$ for some $h_1 \in \n^{o(h_0)+1}$;
hence
$$h_1 = [-(g_0+g_1)]f + (f_0+f_1)g,$$
where $o(f_1) \geq o(h_0)-b$, $o(g_1) \geq o(h_0)-a$, and $o(h_1)>o(h_0)$.
Because
$$\begin{array}{lll}
\overline{h_0t^{a+b-d}} & =
& \overline{(-g_0)t^{b-d}} \cdot \overline{ft^a} + \overline{f_0t^{a-d}} \cdot \overline{gt^b} \\
& = & (-\eta f^*) + \xi g^* \\
& = & (-\eta \cdot \xi D) + \xi (\eta D) \\
& = & 0, \\
\end{array}$$
we get $o(h_0) > a + b-d$, so that $o(f_1) \geq o(h_0)-b > a-d$ and $o(g_1) \geq o(h_0) > b-d$.
Thus $o(g_0+g_1)= o(g_0)=b-d** a-d$ for all $k \geq 1$).
Thus $(\alpha f + \beta g)^* \not\in (f^*, g^*)$ for some elements $\alpha, \beta$ of $S$ with
$o(\alpha)=b-d$ and $o(\beta)=a-d$.
\end{proof}
\begin{remark}{\label{Lem3}}
{\em Let $\alpha, \beta \in S$ with $o(\alpha) = b-d$ and assume
that $(\alpha f + \beta g)^* \not\in (f^*,g^*)$. Then $\alpha^* =
-\bar{u} \eta$ and $\beta^* = \bar{u} \xi$ for some unit $u$ in $S$.
Hence $\alpha^*, \beta^*$ form a $G$-regular sequence.}
\end{remark}
\begin{proof}
With the same notation as in the proof of Lemma \ref{Lem1} we have
$0 \ne \varphi \in G_0=S/\frak{n}$.
Letting $\varphi = \bar{u}$ with a unit $u$ in $S$, we readily get
$\alpha^* = -\bar{u} \eta$ and $\beta^* = \bar{u} \xi$.
\end{proof}
Let $n = \mu_{G}(J)$ and $k =S/\frak{n}$. In Proposition \ref{Prop4}
(3) we prove the uniqueness of the order of $o(\alpha f + \beta g)$
for the elements $\alpha$ and $\beta$ in $S$ given by Proposition
\ref{Prop2} and the uniqueness of the ideal $(f^*, g^*, h^*)$ as
well, where $h = \alpha f + \beta g$.
\begin{prop}{\label{Prop4}}
Let $\alpha, \beta, \sigma, \tau \in S$ with $o(\alpha) = b-d$. Let $h=\alpha f + \beta g$ and $q=\sigma f + \tau g$.
Assume that $h^* \not\in (f^*,g^*)$. Then the following assertions hold true.
\begin{enumerate}
\item[$(1)$] Assume that $q^* \not\in (f^*,g^*)$.
Then $o(q) \geq o(h)+o(\sigma)-(b-d)$.
\item[$(2)$] Assume that $q^* \not\in (f^*,g^*, h^*)$.
Then $o(q) > o(h)+o(\sigma)-(b-d)$.
\item[$(3)$] Assume that $q^* \not\in (f^*,g^*)$ and $o(\sigma) = b-d$.
Then $o(q)=o(h)$ and $(f^*,g^*, q^*)=(f^*,g^*, h^*)$.
\item[$(4)$] The elements $f^*,g^*, h^*$ form a part of a minimal system of
homogeneous generators of $J$.
\item[$(5)$] Assume that $n \geq 4$ and $I \subseteq \n^2$.
Then writing $J = \bigoplus J_n$, we have \newline
$J \supsetneq (J_i \mid 1 \leq i \leq 5)G$.
\end{enumerate}
\end{prop}
\begin{proof}
Assume that $q^* \not\in (f^*,g^*)$ and let $c=o(\sigma) - (b-d)$.
Then $\sigma^* \xi + \tau^* \eta = 0$ by Lemma \ref{Lem1}.
Choose a unit $u$ in $S$ so that $\alpha^* = -\bar{u} \eta$ and
$\beta^* = \bar{u} \xi$.
Then, since $\sigma^* \xi \bar{u} + \tau^* \eta \bar{u}= 0$, we get
$\sigma^* \beta^* = \tau^* \alpha^*$. Hence $\sigma^* = \alpha^* \delta^*$ and $\tau^* = \beta^* \delta^*$
for some $\delta \in S$ with $o(\delta) =c$, because $\alpha^*, \beta^*$ is a $G$-regular sequence.
Thus $\sigma = \alpha \delta + \sigma_1$ and $\tau = \beta \delta + \tau_1$
for some $\sigma_1, \tau_1 \in S$ with $o(\sigma_1) > o(\sigma)$ and
$o(\tau_1) > o(\tau)$;
\begin{equation} \label{1}
q= h \delta + (\sigma_1 f + \tau_1 g).
\end{equation}
Now let
\begin{eqnarray}
\Lambda = \left\{ o(\sigma' f + \tau' g)
\biggl|
\begin{tabular}{l}
$\sigma', \tau' \in S$ such that \\
$(\sigma' f + \tau' g)^* \not\in (f^*,g^*)$ and $o(\sigma') \geq b-d+c$
\end{tabular} \right\}. \nonumber
\end{eqnarray}
Then $o(q) \in \Lambda$.
Let $n = \min \Lambda$ and put
\begin{eqnarray}
\Gamma = \left\{ o(\sigma')
\biggl|
\begin{tabular}{l}
$\sigma' \in S$ for which there exists $\tau' \in S$ such that \\
$(\sigma' f + \tau' g)^* \not\in (f^*,g^*)$, $o(\sigma') \geq b-d+c$, and
$o(\sigma' f + \tau' g)=n$
\end{tabular} \right\}. \nonumber
\end{eqnarray}
Then $\Gamma \ne \emptyset$ and $\gamma < n-a$ for all $\gamma \in \Gamma$
(cf. Lemma \ref{Lem1} (1)).
Let $\gamma = \max \Gamma$ and choose $\sigma', \tau' \in S$ so that
$(\sigma' f + \tau' g)^* \not\in (f^*,g^*)$, $\gamma = o(\sigma') \geq b-d+c,$
and $o(\sigma' f + \tau' g)=n$.
Let $q' = \sigma' f + \tau' g$.
Then, because $q'^* \not\in (f^*, g^*)$, similarly as in equation (\ref{1}) we have
$$q' = h \delta' + (\sigma_2 f + \tau_2 g)$$
for some $\delta',\sigma_2, \tau_2 \in S$ with
$o(\delta') = o(\sigma') - (b-d)$, $o(\sigma_2) > o(\sigma')$, and
$o(\tau_2) > o(\tau')$.
Let $q''= \sigma_2 f + \tau_2 g$ and assume that $o(q') < o(h \delta')$.
We then have
$$n= o(q') = o(q'') \hbox{ \qquad and \qquad } q'^* = q''^*,$$
whence $q''^* \not\in (f^*,g^*)$.
On the other hand, because $o(\sigma_2) > o(\sigma') \geq b-d+c$,
we get $o(\sigma_2) \in \Gamma$, which is impossible
(recall that $o(\sigma') = \max \Gamma$).
Thus $o(q') \geq o(h \delta')$ and so \begin{eqnarray*}
o(q) & \geq & n = o(q') \geq o(h)+o(\delta') \\
& = & o(h) + o(\sigma') -(b-d) \\
& \geq & o(h) + [(b-d) + c] -(b-d) \\
& = & o(h) + c,
\end{eqnarray*}
as was claimed. This proves assertion (1).
Now assume that $q^* \not\in (f^*,g^*, h^*)$.
Then $o(q) \geq o(h) +c$ by assertion (1), where $c=o(\sigma) -(b-d)$.
Assume $o(q) = o(h)+c$ and write $q=h \delta + (\sigma_1 f + \tau_1 g)$
for some $\delta, \sigma_1, \tau_1 \in S$ with $o(\delta) =c$,
$o(\sigma_1) > o(\sigma)$, and $o(\tau_1) > o(\tau)$ (cf. equation (1)).
We put $q_1 = \sigma_1 f + \tau_1 g$.
Then, because $o(q) = o(h \delta) \geq \min \{ o(h \delta), o(q_1) \}$,
$o(q_1) \geq o(h \delta)$.
If $o(q_1) > o(h \delta)$, then we have
$q^* = (h \delta)^* = h^* \delta^* \in (f^*,g^*, h^*)$, which is impossible.
Hence $o(q_1)=o(h \delta) = o(q)$ so that
$q^* = h^* \delta^* + q_1^* \not\in (f^*,g^*,h^*)$.
Consequently $q_1^* \not\in (f^*,g^*)$ and so we get by assertion (1) that
\begin{eqnarray*}
o(h) + c & = & o(h \delta) = o(q_1) \\
& \geq & o(h)+o(\sigma_1) - (b-d) \\
& \geq & o(h) + [o(\sigma)+1] -(b-d) \\
& = & o(h) + c+1,
\end{eqnarray*}
which is absurd. Hence $o(q) > o(h)+c$. This proves assertion (2).
To show assertion (3), thanks to assertion (2), it is enough to check the equality
$o(q)=o(h)$. The inequality $o(q) \geq o(h)$ follows from assertion (1), whence $o(h) = o(q)$ by symmetry.
We now prove assertions (4) and (5).
Let $V=J/NJ$ and choose homogeneous elements
$\delta_1$, $\delta_2$,..., $\delta_n$ of $J$ so that their images
$\bar{\delta_1}$, $\bar{\delta_2}$,..., $\bar{\delta_n}$ in $V$ form
a $k$-basis of $V$.
We may assume $\delta_1 = f^*$, $\delta_2 = g^*$.
Hence $J = (f^*,g^*, \delta_3,..., \delta_n)$.
For each $3 \leq i \leq n$ let $\delta_i = q_i^*$ with $q_i \in I$
and write $q_i = \sigma_i f + \tau_i g$ for some $\sigma_i, \tau_i \in S$.
Then $o(\sigma_i) \geq b-d$ by Lemma \ref{Lem1}. We have $o(q_i) = o(h)$
and $(f^*,g^*, q_i^*) = (f^*,g^*, h^*)$ (resp. $o(q_i) > o(h)$),
if $o(\sigma_i) = b-d$ (resp. if $o(\sigma_i) > b-d$) by assertion (3) (resp. assertion (1)).
Hence $o(q_i) \geq o(h)$.
We may assume $o(q_3) \leq o(q_4) \leq ... \leq o(q_n)$.
Then, because $h^* \in (f^*, g^*, \delta_3,\delta_4,...,\delta_n)$
but $h^* \not\in (f^*, g^*)$, we get $\deg h^* = o(h) \geq \deg \delta_3 = o(q_3)$ so that $o(q_3)=o(h)$,
whence $(f^*, g^*,\delta_3)=(f^*, g^*,h^*)$
by assertion (3). Thus assertion (4) follows.
Suppose that $n \geq 4$.
Then $\delta_4=q_4^* \not\in (f^*,g^*,\delta_3)=(f^*, g^*,h^*)$.
Therefore $o(\sigma_4) > b-d$. Hence by assertion (2) we have
$$
\begin{array}{lll}
\deg \delta_4 & = o(q_4) & \\
& \geq o(h) + [o(\sigma_4)-(b-d)+1] \geq o(h)+2 & \\
& \geq (a+b-d)+3 & \hbox{(by Lemma \ref{Lem1})} \\
& \geq b+4 \geq a+4. &
\end{array}
$$
Consequently, $\deg \delta_4 = o(q_4) \geq 6$, if $I \subseteq \n^2$.
Hence $J \supsetneq (J_i \mid 1 \leq i \leq 5)G$, which completes the proof of Proposition 2.4.
\end{proof}
We are now ready to prove Theorem \ref{1.2} .
\begin{proof}[Proof of Theorem \ref{1.2}]
We may assume that $S$ is complete and $\hgt_G (f^*,g^*) =1$.
Hence $\mu_G(J) = 3$.
Choose $\alpha, \beta \in S$ so that $o(\alpha) =b-d$ and
$(\alpha f + \beta g)^* \not\in (f^*,g^*)$.
Let $h=\alpha f + \beta g$.
Then $J = (f^*,g^*,h^*)$ by Proposition \ref{Prop4} (4).
We furthermore have $h^* \in (\alpha^*, \beta^*)$, because
$\alpha^*, \beta^*$ is a $G$-regular sequence (cf. Remark \ref{Lem3})
and $h \in (\alpha, \beta)$.
Let $h^* = \alpha^* \varphi + \beta^* \psi$ with $\varphi, \psi \in G$.
Then, since $\alpha^* = -\bar{u} \eta$ and $\beta^* = \bar{u} \xi$ for some
unit $u$ in $S$, we see
$$J = I_2 \left( \begin{matrix} \bar{u} \varphi & \bar{u} \psi & D \\
\xi & \eta & 0 \end{matrix} \right) $$
where $D \in G$ is the element such that $f^* = \xi D$ and $g^* = \eta D$.
Thus $J$ is a perfect ideal of $G$, because $\grade_GJ = 2$.
\end{proof}
\begin{dis} \label{5.1} {\em Assume notation as in Setting \ref{1.1}
and also assume that $I \subset \n^2$. Let $\mu(I^*)$ denote the minimal
number of generators of $I^*$.
If $\mu(I^*) = 3$, then
$I^* = (f^*, g^*, h_0^*)G$, where $h_0 = \alpha f
+ \beta g$ and $o( \alpha) = b-d$. We have
$$
2 \le \deg f^* \le \deg g^* < \deg g^* + 2 \le \deg h_0^*,
$$
so $\deg h_0^* \ge 4$. If $\mu(I^*)
\ge 4$, then there exist homogeneous generators for $I^*$ so that
$$
I^* = (f^*, g^*, h_0^*, h_1^*, \ldots, h_r^*)G,
$$
where we have $r = \mu(I^*) - 3$, and
$$
2 \le \deg f^* \le \deg g^* < \deg g^* + 2 \le \deg h_0^* < \deg
h_0^* + 2 \le \deg h_1^* \le \cdots \le \deg h_r^*.
$$
The inequality $\deg h_1^* \ge \deg h_0^* + 2$ is by Proposition
\ref{Prop4} (2). In particular, if $\mu(I^*) \ge 4$, then the relation
type of $\gr_{\m}(R)$ is greater than or equal to 6.
It would be interesting to know whether $\deg
h_i^* + 2 \le \deg h_{i+1}^*$ holds for all $i$ with $0 \le i < r$,
or, if this fails to hold in general, whether $\deg h_i^* + 1 \le
\deg h_{i+1}^*$. An interesting result of Kothari \cite{K} shows
that if $S$ is a 2-dimensional regular local ring containing a
coefficient field, then $\deg h_i^* + 1 \le \deg h_{i+1}^*$ for all
$i$ with $1 \le i < r$. }
\end{dis}
\section{Applications of the theorem} \label{sec3}
Let us give some consequences of Theorem 1.2. We begin with the following.
\begin{cor} \label{5.2}
Let $(R, \m)$ be a $d$-dimensional Gorenstein local ring. Assume that $\m$ is minimally
generated by $d+2$ elements. Then $\gr_{\m}(R)$ is a Cohen-Macaulay ring, if the relation type of $\gr_{\m}(R)$ is
less than or equal to 5.
\end{cor}
\begin{proof} We may assume that $(R,\m)$ is complete. Hence,
thanks to the structure theorem of Cohen (\cite[Theorem A.21]{BH}),
we get $R = S/I$, where $I$ is an ideal of a $(d+2)$-dimensional
regular local ring $(S, \n)$. Because $R$ is a Gorenstein ring and
$\dim R = d$, the ideal $I$ is generated by a regular sequence $f,
g$ of length 2. Let $J = \mathrm{Ker}~(\gr_{\n}(S)
\overset{\varphi}\to \gr_{\m}(R))$, where $\varphi : \gr_{n}(S) \to
\gr_{m}(R)$ denotes the canonical map. We may assume that
$\mu_{\gr_{n}(S)}(J) \geq 3$. Then by Proposition 2.4 (5) the ideal
$J$ is 3-generated, because the relation type of $\gr_{\m}(R)$ is at
most $5$, whence by Theorem \ref{1.2}, $\gr_{\m}(R)$ is a
Cohen-Macaulay ring since the polynomial ring $\gr_{\n}(S)$ is a
UFD.
\end{proof}
\begin{cor} \label{5.3}
Let $(R, \m)$ be a one-dimensional Gorenstein local ring and assume that $\m$ is
minimally generated by $3$ elements. If the reduction number
of $\m$ is less than or equal to $4$, then $\gr_{\m}(R)$ is a
Cohen-Macaulay ring.
\end{cor}
\begin{proof}
The result of Huckaba \cite[Theorem 2.3]{H} shows that in our
setting the relation type of $\gr_{\m}(R)$ is at most one more
than the reduction number of $\m$. Hence by Corollary \ref{5.2} the ring $\gr_{\m}(R)$ is Cohen-Macaulay.
\end{proof}
The example studied in Example \ref{1.4} shows that Corollary
\ref{5.3} may fail if the reduction number of $\m$ is $5$. The
following example is explored by Sally \cite[Example 2.2]{S} and
shows that Corollary \ref{5.2} may fail if we assume that $R$ is a
Cohen-Macaulay (rather than Gorenstein) ring.
\begin{exam} \label{5.4}
{\em
Let $S = k[[x, y, z]]$ be the formal power
series ring with three variables $x, y, z$ over a field $k$ . Let $R = k[[w^4, w^5, w^{11}]]$ be the subring of the
formal power series ring $k[[w]]$ and consider the
homomorphism $\phi :S \to R$ of $k$-algebras defined by $\phi(x) = w^4$, $\phi(y) = w^5,$
and $\phi(z) = w^{11}$. Then $I = \mathrm{Ker}~\phi$ is generated by $xz - y^3$, $yz - x^4$, and
$z^2 - x^3y^2$. We have
$\gr_{\n}(S) = k[x^*, y^*, z^*]$,
$$
I^* = (z^{*2}, z^*y^*, z^*x^*, y^{*4}),
$$
and the ring $\gr_{\m}(R) = \gr_{\n}(S)/I^*$ is not Cohen-Macaulay.
The relation type of $\gr_{\m}(R)$ is $4$ and the reduction
number of $\m$ is $3$.
}
\end{exam}
\begin{cor} \label{5.5}
Let $(R, \m)$ be a one-dimensional Gorenstein local ring and assume that $\m$ is
minimally generated by $3$ elements. If the reduction number $r$
of $\m$ is less than or equal to $4$, then $\gr_{\m}(R)$ is
a Gorenstein ring if and only if $J^r:\m^r = \m^r$, where $J$ is a
reduction of $\m$.
\end{cor}
\begin{proof} By Corollary \ref{5.3}, $\gr_{\m}(R)$ is Cohen-Macaulay.
Therefore all the powers of $\m$ are closed in the sense of Ratliff-Rush. Hence
$\gr_{\m}(R)$ is a Gorenstein ring if and only if $J^r:\m^r = \m^r$ (cf. \cite[Corollary 4.8]{HKU}).
\end{proof}
\medskip
\noindent
{\bf Acknowledgement.} The main question considered in this paper
arose during the collaboration of W. Heinzer and M-K Kim with Bernd Ulrich in \cite{HKU}.
We would like to thank him for his help in developing this material.
\begin{thebibliography}{GGP0}
\bibitem[BH]{BH}{W. Bruns and J. Herzog, {\em Cohen--Macaulay Rings},
Cambridge University Press, Cambridge, 1993.}
\bibitem[HKU]{HKU}{W. Heinzer, Mee-Kyoung Kim and B. Ulrich, The
Gorenstein and complete intersection properties of associated graded
rings, J. Pure Appl. Algebra, to appear.}
\bibitem[H]{H}{S. Huckaba, Reduction numbers for ideals of higher
analytic spread, Math. Proc. Camb. Phil. Soc. {\bf 102}(1987), 49-57.}
\bibitem[K]{K}{S. Kothari,
The local Hilbert function of a pair of plane curves,
Proc. Amer. Math. Soc. {\bf 72} (1978), 439-442.}
\bibitem[S]{S}{J. D. Sally, Tangent cones at Gorenstein singularities,
Compositio Math. {\bf 40} (1980), 167-175.}
\end{thebibliography}
\end{document}
\begin{remark}
{\em Assume notation as in Setting \ref{1.1} and that $h = \alpha f
+ \beta g$ and $q = \sigma f + \tau g$ are elements of $I = (f, g)$,
where $o( \alpha) = b-d$, and $o( \sigma) = b-d+c$ for some integer
$c \ge 1$. An interesting result of Kothari \cite{K} implies that
if $S$ is a $2$-dimensional regular local ring and if $h^* \not\in
(f^*, g^*)$ and $q^* \not\in (f^*, g^*)$, then $\deg h^* \le \deg
q^* -2$. This fails in higher dimension as the following argument
shows. Assume that $G$ is Cohen-Macaulay and let $x \in \n \setminus
\n^2$. Consider the element $q = (\alpha x)f + (\beta x)g$, where
$\alpha, \beta \in S$ are as in Proposition \ref{Prop2} and $h =
\alpha f + \beta g$. Then $o(q) = o(h) + 1 < o(h) + 2$. If $\dim G
\ge 3$, then for some $x \in \n \setminus \n^2$, the element $q^*
\not\in (f^*, g^*)$. For if each $q^* \in (f^*, g^*)$, then with
$N = G_+$ we have $Nh^* \subseteq (f^*, g^*) = D{\cdot}(\xi, \eta)$,
so $N \in \Ass_GG/D{\cdot}(\xi, \eta)$. Since $N \not\in
\Ass_GG/DG$, we have $N \in \Ass_GG/(\xi, \eta)$, from which it
follows that $\dim S = \dim G = 2$, because the sequence $\xi, \eta$
is $G$-regular and $G$ is a Cohen-Macaulay ring. }
\end{remark}
**