Single Spot Ideals of Codimension~3 and long Bourbaki Sequences
Yukihide Takayama
Let $K$ be a field and $S = K[x_1,\ldots, x_n]$ be a polynomial ring. A single spot ideal $I \subset S$ is a graded ideal whose local cohomology $\Coh{i}{S/I}$, $i<\dim S/I$ and $\mm = (x_1,\ldots, x_n)$, only has non-trivial value $N$, a finite length module, at $i = \depth S/I$. We consider characterization of single spot ideals in terms of (long) Bourbaki sequences. The codimension~2 case has been fairly well investigated. In this paper, we focus on the codimension~3 case.
math0303384.tex
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\begin{document}

\title{Single Spot Ideals of Codimension~3 and 
long Bourbaki Sequences}
\author{Yukihide Takayama}
\address{Yukihide Takayama, Department of Mathematical
Sciences, Ritsumeikan University, 
1-1-1 Nojihigashi, Kusatsu, Shiga 525-8577, Japan}
\email{takayama@se.ritsumei.ac.jp}
%\date{}


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\begin{abstract}
Let $K$ be a field and $S = K[x_1,\ldots, x_n]$ be a polynomial ring.
A single spot ideal $I \subset S$ is a graded
ideal whose local cohomology $\Coh{i}{S/I}$, $i<\dim S/I$ and 
$\mm = (x_1,\ldots, x_n)$, only has
non-trivial value $N$, a finite length module, at $i = \depth S/I$. We
consider characterization of single spot ideals in terms of (long)
Bourbaki sequences. The codimension~2 case has been fairly well
investigated.  In this paper, we focus on the codimension~3 case.
\end{abstract}

\maketitle
%%%%%%%%%%%%%%%%%%%%%%%
\section*{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%
Let $S = \poly{K}{x}{n}$ be a polynomial ring over a field $K$ with 
the standard grading and let
$\mm = (x_1,\ldots, x_n)$. All the modules and ideals in this paper
are graded. 
A finitely generated $S$-module $M$ is
called a {\em generalized Cohen-Macaulay $(CM)$ module} if the local
cohomoloy module $\Coh{i}{M}$ has finite length for all $i<\dim(M)$. 
A ring $R$ is a {\em generalized CM ring} if it is a generalized CM
$R$-module. An ideal $I\subset S$ is called a {\em generalized CM
ideal} if $S/I$ is a generalized CM ring. If a generalized CM module
$M$ satisfies $\dim M = \dim S$, it is called {\em maximal}.

In this paper, we are interested in generalized CM ideals. 
In particular, single spot ideals. An ideal $I\subset S$ is called a 
{\em single spot ideal} of type $(t,N)$ where $t = \depth S/I$ and $N$ 
is a finite length
$S$-module if the local cohomology only has a non-trivial value $N$ at
dimension $t$, i.e.,
\[
    \Coh{i}{S/I} = \left\{
			\begin{array}{ll}
				0 & \mbox{for all $i<\dim S/I$ with $i\neq t$} \\
                                N & \mbox{if $i=t$}
			\end{array}
                   \right.
\]
Let $I\subset S$ be a generalized CM ideal of $\codim I = r$ $(r\geq 2)$. 
Then by Corollary~1.3 \cite{HT2}
we have a long Bourbaki sequence
\[
	0\To F_{r-1} \To \cdots \To F_1 \To M \To I \To 0
\]
with $S$-free modules $F_i$ $(i=1,\ldots, r-1)$ and $M$ is a maximal generalized CM 
module whose local cohomology is as follows 
%
\begin{equation}
\label{localcohomologies}
	\Coh{i}{M}  \iso 
\left\{
\begin{array}{ll}
  \Coh{i}{I} \iso \Coh{i-1}{S/I}   & \mbox{if }i < n-r+1 \\
0                              & \mbox{if } i=n-r+1\\
%unspecified & \mbox{if } i=n-1
\end{array}
\right.
\end{equation}
%
In this sence, the ideal $I$ is approximated by $M$.
Notice that the value of $\Coh{i}{M}$ for $i=n-r+2,\ldots, n-1$ are
irrelevant to this approximation. However,
the construction given in the proof of Corollary~1.3
\cite{HT2} (and also Lemma~1.3 \cite{Ama1} in a slightly different
situation) always makes the module
$M$ such that $\Coh{i}{M} =0$ for $i=n-r+2,\ldots, n-1$.
In this paper, we are interested in
long Bourbaki sequences with approximation modules $M$ such that
$\Coh{i}{M}$ $(i=n-r+2,\ldots, n-1)$  are not always trivial,
and study the case of codimension~3, namely the case of 
$\Coh{n-1}{M}=N$ where $N$ is a non-trivial finite length module.
Notice that in the case of codimension~2 we always have $\Coh{n-1}{M}=0$.

First of all, we will give a characterization of a maxmal generalized
CM module $M$ whose local cohomology is
\begin{equation}
\label{lc-condition-codim3}
\Coh{i}{M} = \left\{
		\begin{array}{ll}
		    K  & \mbox{if $i=t+1$}\\
                    N  & \mbox{if $i=n-1$} \\
                    0  & \mbox{if $i<n-2$, $i\ne t+1$}
		\end{array}
             \right.
\end{equation}
in terms of the first syzygy of $M$. See Theorem~\ref{theorem:main1}.
Then we consider the special case of $M = E_{t+1}\dirsum E_{n-1}(d)$ where
$E_j$ denotes the $j$th sygyzy module of the field $K$ over $S$. 
We will use the notation $M(d)$, for a graded module $M$ and $d\in\ZZ$,
such that $M(d)_i= M_{d+i}$ (the $d+i$th component of $M$) for all $i\in\ZZ$.
%
Our question is 
how we can construct a long Bourbaki sequence 
\begin{equation}
\label{the:bourbaki-sequence}
   0 \To F\To G \To E_{t+1}\dirsum E_{n-1}(d) \To I(c) \To 0
\end{equation}
of {\em non-trivial} type. Here a {\em trival type} construction is as follows. First
construct a long Bourbaki sequence
\[
   0 \To F' \namedTo{f} G' \namedTo{f} E_{t+1} \namedTo{\phi} I(c) \To 0
\]
according to the method given in the proof of Corollary~1.3 \cite{HT2}
(or Lemma~1.3 \cite{Ama1}). Then make the direct sum
\[
   0 \To F'\dirsum K_n(d) \namedTo{f\dirsum\partial_n}
 G'\dirsum K_{n-1}(d) \namedTo{g\dirsum\partial_{n-1}}
 E_{t+1}\dirsum E_{n-1}(d)  \namedTo{\phi} I(c) \To 0
\]
where $(K_\bullet, \partial_\bullet)$ is the Koszul complex of the sequence 
$x_1,\ldots, x_n$ over $S$. We will denote a base $x_{i_1}\wedge\cdots\wedge x_{i_k}$
($1\leq i_1 < \cdots < i_k\leq n$) of the Koszul complex of the sequence
$x_1,\ldots, x_n$ by $e_{\{i_1,\ldots,i_k\}}$ or $e_{i_1\cdots i_k}$.   
Notice that, in the trivial type Bourbaki sequence
$E_{n-1}(d)$ does not contribute to $I$ via $\phi$. 


It is well known that a (short) Bourbaki sequence $0 \to F \to M \To I
\to 0$ is constructed by finding '(graded) basic elements' in $M$.
See \cite{Bour,Kunz} for the standard basic element theory
and \cite{HT2,F} for graded version. 
However, there is no comparative notion  for long Bourbaki 
sequences. We give a simple answer to this problem in the case of 
$E_{t+1}\dirsum E_{n-1}(d)$ (and $M= E_{t+1}$).
We will give a characterization of long Bourbaki sequences
(\ref{the:bourbaki-sequence}) in terms of elemets from $K_{t+1}\dirsum
K_{n-1}$ (from $K_{t+1}$) 
and from $K_{n-t-1}\dirsum K_{1}$ (from $K_{n-t-1}$) satisfying certain
conditions, which suggests a consruction of the long Bourbaki sequences.  See
Theorem~\ref{theorem:construction} and~\ref{theorem:construction2}. 
In particular, non-trivial type
construction is characterized by an additional condition on the
elements from $K_{t+1}\dirsum K_{n-1}$
(Theorem~\ref{theorem:non-trivialcond}).

However, the existence of a long Bourbaki sequence
(\ref{the:bourbaki-sequence}) only means that $I$ is a single spot
ideal of codimension less than or equal to $3$. We give a numerial
condition to assure $\codim I = 3$. See
Theorem~\ref{theorem:numerical}.  Finally, we give some examples. 

For a module $M$ ($\ne K$), 
we will denote the $j$th syzygy module over $S$ by $\Omega_j(M)$.
Also we use two kinds of duals,
$(-)^\vee = \Hom_S(-,K)$ and $(-)^*= \Hom_S(-, S(-n))$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Approximation Modules of Single Spot Ideals of type $(t, K(-c))$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section, we consider 
approximation modules $M$ of codimension~3 single spot ideals $I\subset S$
of type $(t, K(-c))$ in long Bourbaki sequences
\[
   0\To F \To G \To  M \To I(c) \To 0.
\]
If we restrict ourself to the case of $\Coh{n-1}{M}=0$, we have 
$M = E_{t+1}\dirsum H$ for some free $S$-module $H$ 
according to Herzog, Takayama \cite{HT2} and Amasaki \cite{Ama1}.
We will now consider the general case. 


\begin{Theorem}
\label{theorem:main1}
Let $M$ be a maximal generalized CM module over $S$ and 
consider its first sygyzy:
\[
   0 \To \Omega_{1}(M) \To F \To M \To 0
\]
and let $g_1,\ldots, g_l$ be a minimal set of generators 
of $\Omega_{1}(M)$ whose
degrees are $a_1,\ldots, a_l$.
Also let $N$ be a finite length module over $S$,
which may be $0$.
Then the following are equivalent.
\begin{enumerate}
 \item [$(i)$]
   For $t \leq n-4$, we have
\[ \Coh{i}{M} =
	\left\{
	\begin{array}{ll}
	   K   & \mbox{if } i=t+1 \\
	0 & \mbox{if } i\leq n-2, i\ne t+1\\
	   N   & \mbox{if } i=n-1 
	\end{array}
      \right.
\]
\item [$(ii)$]
\begin{enumerate}
\item [$(a)$] $\Omega_1(M) \iso \Dirsum_{i=1}^lS(-a_i)/E_{t+3}$, and
\item [$(b)$] $\Omega_1(M)^* \iso N^\vee + F^*/M^*$ 
\end{enumerate}
\end{enumerate}
\end{Theorem}
%
% PROOF %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\begin{proof}
We first prove $(i)$ to $(ii)$. Let $F_\bullet$ be a minimal free resolution of $M$
over $S$:
\[
 F_\bullet : 
	0\To F_{n-t-1} \namedTo{\varphi_{n-t-1}}
	     F_{n-t-2} \namedTo{\varphi_{n-t-2}}
	         \cdots\namedTo{\varphi_3}
             F_2       \namedTo{\varphi_2}
	     F_1       \namedTo{\varphi_1} 
             F_0       \namedTo{\varphi_0}
               M       \To 0.
\]
By taking the dual, we have 
\[
 0 \To F_0^*      \namedTo{\varphi_1^*} 
       F_1^*      \namedTo{\varphi_2^*}
       F_2^*      \namedTo{\varphi_3^*}
       \cdots     \namedTo{\varphi_{n-t-2}^*} 
       F_{n-t-2}^*\namedTo{\varphi_{n-t-1}^*}
       F_{n-t-1}^*\To 0.
\]
Then by local duality the  $j$th cohomology of this complex is
\[
   \Ext_S^j(M, S(-n)) \iso \Coh{n-j}{M}^\vee
   =   \left\{
	\begin{array}{ll}
                N^\vee & \mbox{if $j=1$} \\
                0      & \mbox{if $j\geq 2, j\ne n-t-1$}\\
		K      & \mbox{if $j=n-t-1$} 
	\end{array}
\right.
\] 
for $j\geq 1$. Thus 
\[
0\To \Im\varphi_2^* \To F_2^* \To\cdots\To F_{n-t-1}^*\To K \To 0
\]
is exact and $F_2^*$ to $F_{n-t-1}^*$ part is a begining of a minimal
free resolution of $K$, which is isomorphic to a begining of the Koszul
complex $(K_\bullet,\partial_\bullet)$ of the sequence $x_1,\ldots,
x_n$. Namely,
\[
     F_{n-t-1}^* \iso K_0,\quad\ldots\quad, F_2^* \iso K_{n-t-3} \qquad\mbox{and}\quad 
     \Im\varphi_2^* \iso E_{n-t-2}.
\]
On the other hand, we have
$N^\vee \iso \Ker\varphi_2^*/\Im\varphi_1^*$ 
and 
$E_{n-t-2}\iso \Im\varphi_2^* \iso  F_1^*/\Ker\varphi_2^*$.
Now set $U := \Coker\varphi_1^* = F_1^*/\Im\varphi_1^*$. Then
\[
U/N^\vee \iso (F_1^*/\Im\varphi_1^*)/(\Ker\varphi_2^*/\Im\varphi_1^*)
 \iso  F_1^*/\Ker\varphi_2^* = E_{n-t-2}.
\]
Thus  we have
\[
      0 \To N^\vee \To U \To E_{n-t-2} \To 0
\]
Taking the dual, we have
\[
    0\To E_{n-t-2}^* \To U^* \To (N^\vee)^*.
\]
Since $N$ has finite length, $N^\vee$ has also finite length by Matlis
duality, so that $(N^\vee)^* =0$. Also $E_{n-t-2}^* \iso E_{t+3}$
by selfduality of Koszul complex. Thus we have $U^* \iso E_{t+3}$. 
Then by dualizing  the exact sequence
\begin{equation}
\label{prop:new1-seq1}
  0\To M^* \To F_0^* \namedTo{\varphi_1^*}F_1^* \To U \To 0
\end{equation}
we have
\[
	0\To E_{t+3} \To F_1 \namedTo{\varphi_1} F_0 \To M\To 0.
\]
This proves $(ii) (a)$. 
Now from the short exact sequence
\[
    0 \To \Omega_1(M) \To F \namedTo{\varphi} M \To 0
\]
we have the long exact sequence 
\[
   0 \To M^* \namedTo{\varphi^*}
         F^* \To 
         \Omega_1(M)^* \To
         N^\vee \To 0
\]
since we have $\Ext_S^1(M,S(-n))\iso \Coh{n-1}{M}^n = N^\vee$ by local
duality. This proves $(ii) (b)$.

Next we prove $(ii)$ to $(i)$. By $(ii)(a)$ we have a $S$-free resolution
of $M$:
\[
   0\To K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}}
          K_{t+3}\namedTo{\partial_{t+3}}
        F_1\namedTo{\varphi_1} F_0 \namedTo{\varphi_0} M \To 0
\]
where $F_0$ and $F_1$ are $S$-free modules. By taking the dual, we have
the complex
\[
    0\To 
      M^*    \namedTo{\varphi_0^*}
      F_0^* \namedTo{\varphi_1^*} F_1^* \namedTo{\partial_{t+3}^*}
	K_{t+3}^* \namedTo{\partial_{t+4}^*}\cdots\namedTo{\partial_n^*}
       K_n^* \To 0
\]
Then by local duality and selfduality of Koszul
complex we compute 
\[
    \Coh{i}{M} \iso \Ext_S^{n-i}(M,S(-n))^\vee
                =\left\{
			\begin{array}{ll}
			    K   & \mbox{if $i = t+1$} \\
                            0   & \mbox{if $i\leq n-2$, $i\ne t+1$}
			\end{array}
                \right.
\]
Now by dualizing the exact sequence
\[
    K_{t+3}\namedTo{\partial_{t+3}} F_1 \namedTo{\varphi_1} \Ker\varphi_0 \To 0
\]
we have 
\[
   0\To (\Ker\varphi_0)^* \To F_1^* \namedTo{\partial_{t+3}^*} K_{t+3}^*,
\]
so that we have $\Omega_1(M)^* =(\Ker\varphi_0)^*\iso \Ker(\partial_{t+3}^*)$.
Then by the condition $(ii)(b)$ we compute
\begin{eqnarray*}
\Coh{n-1}{M}^\vee 
      &\iso &\Ext_S^{1}(M, S(-n)) = \Ker \partial_{t+3}^* / \Im \varphi_1^*
\\
      &\iso & \Omega_1(M)^* / (F_0^*/\varphi_0^*(M^*))
\\
      &  = & N^\vee
\end{eqnarray*}
as required.
\end{proof}

\begin{Corollary}
Let $M$ be a maximal generalized CM module satisfying Theorem~\ref{theorem:main1} (i).
Then its minimal free resolution is in the form of 
\[
0\to K_n\namedTo{\partial_n}
     K_{n-1}\namedTo{\partial_{n-1}}\cdots\namedTo{\partial_{t+4}}
     K_{t+3}\namedTo{\partial_{t+3}}
     F_1\namedTo{\varphi}
     F_0 \to M \to 0.
\]
\end{Corollary}
Notice that $F_1$ is a $S$-free module containing a submodule isomorphic to
$E_{t+3}$.

\begin{Example}\label{example:1}
Let $M = E_{t+1}\dirsum E_{n-1}$. 
Then $\Coh{i}{M}$ is as in Theorem~\ref{theorem:main1} $(i)$
with $N=K$.
Since $M$ has a minimal free resolution
\[
  0\To K_n \namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}}
       K_{t+3}\namedTo{\partial_{t+3}}
       K_{t+2}\dirsum K_{n} \namedTo{\partial_{t+2}\dirsum \partial_n}
       K_{t+1}\dirsum K_{n-1}\namedTo{\partial_{t+1}\dirsum \partial_{n-1}} M \To 0,
\]
we have $\Omega_1(M) = E_{t+2}\dirsum E_{n} \iso G/E_{t+3}$
where $G = K_{n+2} \dirsum E_{n}$. Thus we have Theorem~\ref{theorem:main1} $(ii) (a)$.
On the other hand, 
we have $\Omega_1(M)^* = E_{t+2}^*\dirsum E_{n}^* \iso E_{n-t-1}\dirsum S$
by selfduality of Koszul complex and $E_{n} \iso S(-n)$.
Again by selfduality we have
$(K_{t+1}\dirsum K_{n-1})^*/M^*
=
(K_{t+1}\dirsum K_{n-1})^*/(E_{t+1}\dirsum E_{n-1})^*
= (K_{n-t-1}\dirsum K_1)/ (E_{n-t}\dirsum E_2)
= (K_{n-t-1}/E_{n-t})\dirsum (K_1/E_2)
= E_{n-t-1} \dirsum E_1
= E_{n-t-1}\dirsum \mm$.
Thus $\Omega_1(M)^*/((K_{t+1}\dirsum K_{n-1})^*/M^*) \iso S/\mm \iso K \iso K^\vee$ and 
we obtain Theorem~\ref{theorem:main1} $(ii)(b)$.
\end{Example}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Long Bourbaki sequences with Approximation
Module $E_{t+1}\dirsum E_{n-1}(d)$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In the last chapter, we considered approximation modules $M$ satisfying 
the condition of Theorem~\ref{theorem:main1}(i). We now focus on a 
special case of $M = E_{t+1}\dirsum E_{n-1}(d)$, and investigate the 
long Bourbaki sequences.

We will use the following well known result frequently without
refering it. First of all, we will give a proof for the readers'
convenience.

\begin{Lemma} 
\label{lemma:wellknown}
For the $t$-th syzygy module of $K$ over $S$, we 
have 
\[
\rank E_t =\binom{n-1}{t-1}
\]
\end{Lemma}
\begin{proof}
Let $K_i$ be the $i$th Koszul complex. Then we have
an exact sequence
\[
  0\To K_n\namedTo{\partial_n} K_{n-1}\namedTo{\partial_{n-1}}\cdots\namedTo{\partial_t} K_t \To E_t \To 0 
\]
so that 
\[
	\rank(E_t) =\sum_{i=t}^{n}(-1)^{i-t}\rank(K_i)
		   = (-1)^t\sum_{i=t}^{n}(-1)^i\binom{n}{i} 
= (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}.
\]
%\]
%By the binomial coefficients theorem, we have 
%\[
%	0 = (1-1)^n =\sum_{i=1}^{n}(-1)^i\binom{n}{i}
%		    = \sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
%	            + \sum_{i=t}^n(-1)^i\binom{n}{i}
%\]
%so that 
%\[
%	\rank(E_t) = (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}.
%\]
Now set $\alpha(n,t) := \rank(E_t)$. 
By a straightforward calculation, we have
$\alpha(n,t) - \alpha(n-1, t) = \alpha(n-1,t-1)$.
%We compute
%\begin{eqnarray*}
%\lefteqn{\alpha(n,t) - \alpha(n-1,t)} \\
%     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
%	   - (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i} \\
%     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
%          - (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
%          +  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i-1}\\
%     &   & \mbox{by }
%		\binom{n}{i} = \binom{n-1}{i} + \binom{n-1}{i-1} \\
%     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i-1}
%       =  (-1)^{t-1}\sum_{i=1}^{t-1}(-1)^i\binom{n-1}{i-1} \\
%     & = & (-1)^{t-1}\sum_{j=0}^{t-2}(-1)^{j+1}\binom{n-1}{j} 
%      =    (-1)^{t-2}\sum_{j=0}^{t-2}(-1)^{j}\binom{n-1}{j} 
%      =  \alpha(n-1, t-1)\\
%     &   & \mbox{by changing the index $j = i -1$}\\
%\end{eqnarray*}
Thus we have 
\begin{eqnarray*}
\alpha(n,t) &= &\alpha(n-1,t) + \alpha(n-1, t-1) \\
            & =& (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i}
              + (-1)^{t-2}\sum_{i=0}^{t-2}(-1)^i\binom{n-1}{i} 
	     = \binom{n-1}{t-1}.
\end{eqnarray*}
as required.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Characterization of long Bourbaki Sequences}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


For $J, K \subset [n]=\{1,2,\ldots, n\}$ with $J\cap K =\emptyset$ we
define $\sigma(J, K) = (-1)^i$ where $i = \sharp\{ (j,k)\in J\times K
\mid j > k\}$. Then we have $x_J\wedge x_K = \sigma(J,K)x_{J\cup K}$.

Now a long Bourbaki sequence with approximation module $E_{t+1}\dirsum
E_{n-1}(d)$ is characterized by suitable sequences from
$K_{t+1}\dirsum K_{n-1}$ and its dual. Namely,

\begin{Theorem}
\label{theorem:construction}
Following are equivalent.
\begin{enumerate}
\item [$(i)$] We have a long Bourbaki sequence
\[
 0 \To \Dirsum_{i=1}^p S(-a_i) 
   \To \Dirsum_{i=1}^q S(-b_i) 
   \To  E_{t+1}\dirsum E_{n-1}(d)
   \To I(c) \To 0 \quad (exact)
\]
where  $I \subset S$ is a graded ideal.
\item [$(ii)$]
We have 
$\beta_1,\ldots,\beta_q \in K_{t+1}\dirsum K_{n-1}(d)\backslash E_{t+2}\dirsum E_n(d)$
and $\varphi = (a,b)$ with
$a \in {\cal A}$ and $b\in {\cal B}$,  
where 
\begin{eqnarray*}
{\cal A}& =& \left\langle
		\sum_{j=1}^{n-t}
			(-1)^{j+1}\sigma(L\backslash\{i_j\}, 
([n]\backslash L)\cup\{i_j\})
				x_{i_j} e^*_{([n]\backslash L)\cup\{i_j\}}
		\mid L = \{i_1,\ldots,i_{n-t}\}\subset [n]
            \right\rangle \\
{\cal B} &= &\langle
		(-1)^ix_j e^*_{[n]\backslash\{i\}} -
		(-1)^jx_i e^*_{[n]\backslash\{j\}}
		\mid 1\leq i < j \leq n
            \rangle,
\end{eqnarray*}
such that 
\begin{enumerate}
\item [$(a)$] 
      $\varphi :  K_{t+1}\dirsum K_{n-1}(d)\to S(-n)$ is 
      a degree '$n+c$' homomorphism and 
      $\Ker(\varphi)
	= \langle\beta_1,\ldots\beta_q\rangle + E_{t+2}\dirsum E_n(d)$, and 
\item [$(b)$] we we have the following diagram, with $p = q - n + 2 -\binom{n-1}{t}$
\[
\begin{CD}
     @.    0              @.            0         \\
 @.        @VVV                         @VVV      \\
 0  @>>>  \Ker Res(\beta)         @>>>  \Ker\beta  @>>> 0\\
 @.        @VVV                         @VVV             \\
 0  @>>>  \Dirsum_{i=1}^pS(-a_i)  @>>>  \Dirsum_{i=1}^qS(-b_i) \\
 @.       @V{Res(\beta)}VV              @V{\beta}VV \\
0 @>>> \langle\beta_1,\ldots,\beta_q\rangle
      \cap E_{t+2}\dirsum E_n(d) 
                               @>>>  \langle\beta_1,\ldots,\beta_q\rangle \\
 @.        @VVV                         @VVV \\
     @.    0            @.               0   \\
\end{CD}
\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccc}
%     & 0                       &      &  0                       &
%\\      
%     & \da                     &      &  \da                     &
%\\
%0\To & \Ker Res(\beta)         & \To  & \Ker\beta                &\To 0
%\\    
%     & \da                     &      &  \da                     &
%\\
%0\To & \Dirsum_{i=1}^p S(-a_i) & \To  & \Dirsum_{i=1}^q S(-b_i)  & 
%\\
%     & \hspace*{1.4cm}\da Res(\beta)         
%                               &      & \hspace*{0.5cm}\da \beta &
%\\
%0\To & \langle\beta_1,\ldots,
%       \beta_q\rangle\cap 
%       E_{t+2}\dirsum E_n(d)   & \To  & \langle\beta_1,\ldots,
%                                         \beta_q\rangle          & 
%\\
%     & \da                     &      &  \da                     &
%\\
%     &  0                      &      &   0                      &
%\end{array}
%\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $\beta(g_i) = \beta_i$ for all $i$ with $g_1,\ldots, g_q$ the free basis 
of $\Dirsum_{i=1}^q S(-b_i)$, and $Res(-)$ denotes the restriction of maps.
\end{enumerate}
\end{enumerate}
In this case, we have $I = \varphi(K_{t+1}\dirsum K_{n-1}(d))(-c)$
\end{Theorem}
%
%
\begin{proof}
We first prove $(ii)$ to $(i)$. 
First notice that,  by the selfduality of Koszul complex, we have
%
\begin{eqnarray*}
E_i & \iso &  E^*_{n-i+1}\iso\partial^*_{n-i+1}(E^*_{n-i+1}) \\
& =   & \left\{ \sum_{k=1}^{i}(-1)^{k+1} 
                   \sigma(J\backslash \{j_k\}, [n]-(J\backslash\{j_k\}))
                    x_{j_k} e^*_{[n]-(J\backslash\{j_k\})}
: J = \{j_1,\ldots, j_i\}\subset [n]
\right\}.\\
\end{eqnarray*}
%
See \cite{BH} Chapter 1.6. Thus ${\cal A} = \partial^*_{t+1}(E_{t+1}^*)$ and 
${\cal B} = \partial^*_{n-1}(E_{n-1}^*)$. Then, there exists 
$\bar{a} \in E_{t+1}^*$ and $\bar{b}\in E_{n-1}^*$ such that 
$a = \bar{a}\circ \partial_{t+1}$ and $b = \bar{b}\circ\partial_{n-1}$.
Then by $(a)$ we have the diagram
\begin{equation}
\label{construction:proof-1}
\begin{CD}
0 @>>> \langle\beta_1,\ldots,\beta_q\rangle
       + E_{t+2}\dirsum E_n(d) @>>> K_{t+1}\dirsum K_{n-1}(d) @>{\varphi}>> S(c) \\
@.     @V{\bdel}VV                 @V{\bdel}VV                         @|   \\
0 @>>> \langle\bdel\beta_1,\ldots,\bdel\beta_q\rangle
                              @>>> E_{t+1}\dirsum E_{n-1}(d)@>{\phi}>>  S(c) \\
@.      @VVV                        @VVV                                @.  \\
   @.    0                    @.     0                       @.            
\end{CD}
\end{equation}
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{equation}
%\label{construction:proof-1}
%\begin{array}{cccccc}
%0\To & \langle\beta_1,\ldots,\beta_q\rangle + E_{t+2}\dirsum E_n(d) &
% \To  & K_{t+1}\dirsum K_{n-1}(d) &
% \namedTo{\varphi} & S(c) \\
% & \hspace*{0.3cm}\da\bdel && \hspace*{0.3cm}\da\bdel  &     & || \\
%0\To & \langle\bdel\beta_1,\ldots,\bdel\beta_q\rangle &
% \To  & E_{t+1}\dirsum E_{n-1}(d) &
% \namedTo{\phi} & S(c) \\
%     & \da &      & \da  &     &  \\
%     &  0  &      &  0   &     &   \\
%\end{array}
%\end{equation}
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $\phi = (\bar{a},\bar{b})$ and $\bdel := \partial_{t+1}\dirsum \partial_{n-1}(d)$.
On the other hand, we have 
\begin{eqnarray*}
 \Ker\bdel\circ\beta & = & 
   \{ \sum_{i=1}^q h_i g_i \mid \bdel(\sum_{i=1}^q h_i\beta_i) = 0, h_i\in S \} 
\\                   & = &
   \{ \sum_{i=1}^q h_i g_i \mid \sum_{i=1}^q h_i\beta_i\in\Ker\bdel = E_{t+2}\dirsum E_n(d), h_i \in S\}
\\                   & = & 
   \{ \sum_{i=1}^q h_i g_i \mid \beta(\sum_{i=1}^q h_ig_i)
                                 \in\langle\beta_1,\ldots, \beta_q\rangle
				\cap  E_{t+2}\dirsum E_n(d)\}
\\                   & \iso & \Dirsum_{i=1}^p S(-a_i)
\end{eqnarray*}
where the last isomorphism is by $(b)$. 
Notice that let $u\in\Dirsum_{i=1}^q S(-b_i)$ be such that
$\beta(u)\in\langle\beta_1,\ldots,\beta_q\rangle\cap
E_{t+2}\dirsum E_{n}(d)$. 
Then $u$ must be in $\Dirsum_{i=1}^p S(-a_i)$. In fact,
by (b) we can choose 
$v\in \Dirsum_{i=1}^pS(-a_i)$ such that
$Res(\beta)(v) = \beta(u)$. Thus $u- v \in\Ker\beta 
\iso \Ker Res(\beta)$, and we have $u \in v + \Ker Res(\beta)
\subset \Dirsum_{i=1}^pS(-a_i)$ as required.
Then by (\ref{construction:proof-1}) we obtain
\[
  0 \To \Dirsum_{i=1}^p S(-a_i) \To \Dirsum_{i=1}^q S(-b_i)
	\namedTo{\bdel\circ\beta} E_{t+1}\dirsum E_{n-1}(d) 
        \namedTo{\phi} S(c)\qquad (exact)
\]
and since $p = q - n + 2 -\binom{n-1}{t}$ we know that $\rank\;\Im\psi =1$
so that we have 
\[
  0 \To \Dirsum_{i=1}^p S(-a_i) \To \Dirsum_{i=1}^q S(-b_i)
	\namedTo{\bdel\circ\beta} E_{t+1}\dirsum E_{n-1} 
        \namedTo{\phi} I(c)\To 0\qquad (exact)
\]
for the ideal $I := \phi(E_{t+1}\dirsum E_{n-1}(d))(-c)
                =  \varphi(K_{t+1}\dirsum K_{n-1}(d))(-c)$
as required.

Next we prove $(i)$ to $(ii)$. Given a long Bourbaki sequence
\[
  0 \To \Dirsum_{i=1}^p S(-a_i) 
 	\namedTo{f} \Dirsum_{i=1}^q S(-b_i)
	\namedTo{g}E_{t+1}\dirsum E_{n-1}(d) 
        \namedTo{\phi} I(c)\To 0\qquad (exact)
\]
with a graded ideal $I\subset S$. Then we have $p = q - n + 2 + \binom{n-1}{t}$
since $1 = \rank I(c) = \rank E_{t+1}\dirsum E_{n-1}(d) - q + p$. Also since  $I(c) \subset S(c)$ we have 
\begin{eqnarray*}
\phi & \in &  \Hom_S(E_{t+1}\dirsum E_{n-1}(d), S(c)) \\
& =  &  \Hom_S(E_{t+1}\dirsum E_{n-1}(d), S(-n))(n+c)\\
& =  & (E_{t+1}^*\dirsum E_{n-1}^*(-d))(n+c) 
\iso(\partial_{t+1}^*(E_{t+1}^*)\dirsum \partial_{n-1}^*(E_{n-1}^*(-d)))(n+c).\\
\end{eqnarray*}
Thus there exists a unique $(a,b) \in 
\partial_{t+1}^*(E_{t+1}^*)\dirsum \partial_{n-1}^*(E_{n-1}^*(-d))$ such that 
$\phi\circ\bdel = (a,b)$. Now we set $\varphi = (a,b)$.
Then $E_{t+2}\dirsum E_{n}(d) + N = \bdel^{-1}(\Ker\phi) (\subset K_{t+1}\dirsum
K_{n-1}(d))$ for some module $N(\ne 0)$.
Let $\{\beta_1,\ldots, \beta_q\}$ be a minimal set of generators of $N$. 
Then we have the following diagram:
\[
\begin{CD}
   @.                                           @.      0        @. \\
@.       @.                                             @AAA     @. \\
   @.    \Dirsum_{i=1}^qS(-b_i)                 @>{g}>> E_{t+1}\dirsum E_{n-1}(d)
@>{\phi}>>          I(c) @>>> 0 \\
@.       @V{\beta}VV                                  @A{\bdel}AA
                    @|   \\
0 @>>> \langle\beta_1,\ldots,\beta_q\rangle @>>> K_{t+1}\dirsum K_{n-1}(d) @>{\varphi=(a,b)}>> I(c) \\
@.       @VVV                                         @.               @. \\
    @.     0                                    @.           @.           \\
\end{CD}
\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccccc}
%     &                         &             &    0                    &                %&      &
%\\
%     &                         &             &    \ua                  &                
%&      &
%\\
%     & \Dirsum_{i=1}^q S(-b_i) & \namedTo{g} &  E_{t+1}\dirsum E_{n-1}(d) & 
%\namedTo{\phi} & I(c) & \To 0
%\\
%\\
%     &\hspace*{0.3cm}\da \beta &             & \hspace*{0.3cm}\ua\bdel &   
%\nearrow     &\varphi=(a,b) & 
%\\
%\\
%0\To &  \langle\beta_1,\ldots,
%            \beta_q\rangle     &   \To       &  K_{t+1}\dirsum K_{n-1}(d) &               
%&      &
%\\
%     &    \da                  &             &                         &                
%&      &
%\\
%     &    0                    &             &                         &                
%&      &
%\\
%\end{array}
%\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Then we have obtained $\beta_1,\ldots, \beta_q \in K_{t+1}\dirsum K_{n-1}(d)$ and $(a,b)\in
{\cal A}\times {\cal B}$ satisfying the condition $(a)$ and the numerical condition on $p$, $q$, $t$ and  $n$ in $(b)$. Also we have 
$I = \varphi(K_{t+1}\dirsum K_{n-1}(d))(-c)$.

Now since $\Ker g =\Ker (Res(\bdel)\circ\beta) = \Im f =
\Dirsum_{i=1}^p S(-a_i)$ we readily have the following diagram:
%
\[
\begin{CD}
   @.    0               @.       0          @.    \\
@.      @VVV                      @VVV           @.\\
0  @>>> \Ker\beta\circ f @>>>     \Ker\beta @>>> 0 \\
@.      @VVV                      @VVV           @.\\
0 @>>> \Dirsum_{i=1}^pS(-a_i) 
                         @>{f}>> \Dirsum_{i=1}^qS(-b_i) 
@>{g}>> 
E_{t+1}\dirsum E_{n-1}(d)\\
@.       @.                       @V{\beta}VV     @| \\
0 @>>> \langle\beta_1,\ldots,\beta_q\rangle\cap E_{t+2}\dirsum E_n(d)
                         @>>>     \langle\beta_1,\ldots,\beta_q\rangle
                                                  @>{Res(\bdel)}>>
                                                  E_{t+1}\dirsum E_{n-1}(d)\\
@.       @.                       @VVV             @. \\
    @.                   @.       0               @. \\
\end{CD}
\]
%%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{cccccc}
%     & 0   &            &  0                       &            &
%\\      
%     & \da  &           &  \da                     &            &
%\\
%0\To & \Ker\beta\circ f & \To  & \Ker\beta         & \To             & 0
%\\    
%     & \da  &           &  \da                     &                 & 
%\\
%0\To & \Dirsum_{i=1}^p S(-a_i) &\namedTo{f} & \Dirsum_{i=1}^q S(-b_i)  
%& \namedTo{g}     &  E_{t+1}\dirsum E_{n-1}(d)
%\\
%     &      &            & \hspace*{0.4cm}\da \beta &               &      || 
%\\
%0\To & \langle\beta_1,\ldots,
%       \beta_q\rangle\cap 
%       E_{t+2}\dirsum E_n(d)     & \To       
% & \langle\beta_1,\ldots,\beta_q\rangle & \namedTo{Res(\bdel)} 
%    & E_{t+1}\dirsum E_{n-1}(d)
%\\
%     &   &            &  \da              &                 &
%\\
%     &   &            &   0               &                 &
%\end{array}
%\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Notice that since $\Ker\beta \subset\Ker (Res(\bdel)\circ\beta) 
= \Ker g = \Im f$ we have the exactness of the first row.
Since 
$\Im (\beta\circ f) = \beta (\Ker(Red(\bdel)\circ \beta)) =  \Ker Res(\bdel) 
= \langle\beta_1,\ldots,\beta_q\rangle\cap E_{t+2}\dirsum E_n(d)$, 
we have a well-defined surjection
\[
   \beta\circ f :  \Dirsum_{i=1}^p S(-a_i)
 \To \langle\beta_1,\ldots,\beta_q\rangle
					\cap E_{t+2}\dirsum E_n(d).
\]
Thus we obtained the diagram of $(b)$ as required.
\end{proof}

For any codimension~3 single spot ideal of type $(t, K(-c))$, there exists
a long Bourbaki sequence with approximation module $E_{t+1}\dirsum H$ where 
$H$ is a $S$-free module \cite{HT2,Ama1}. The following is the case 
of $H=0$, which can be proved with the same idea as that of  
Theorem~\ref{theorem:construction}.
%
\begin{Theorem}
\label{theorem:construction2}
Following are equivalent.
\begin{enumerate}
\item [$(i)$] We have a long Bourbaki sequence
\[
 0 \To \Dirsum_{i=1}^p S(-a_i) 
   \To \Dirsum_{i=1}^q S(-b_i) 
   \To  E_{t+1}
   \To I(c) \To 0 \quad (exact)
\]
where  $I\subset S$ is a graded ideal.
\item [$(ii)$]
We have 
$\beta_1,\ldots,\beta_q \in K_{t+1}\backslash E_{t+2}$
and $\varphi\in {\cal A}$
where 
\[
{\cal A} = \left\langle
		\sum_{j=1}^{n-t}
			(-1)^{j+1}\sigma(L\backslash\{i_j\},
 ([n]\backslash L)\cup\{i_j\})
				x_{i_j} e^*_{([n]\backslash L)\cup\{i_j\}}
		\mid L=\{i_1,\ldots,i_{n-t}\} \subset [n] \right\rangle
\]
such that 
\begin{enumerate}
\item [$(a)$] 
       $\varphi : K_{t+1} \to S(-n)$ defines a degree '$n+c$' homomorphism and 
	$\Ker\varphi = \langle\beta_1,\ldots\beta_q\rangle + E_{t+1}$, and 
\item [$(b)$] we we have the following diagram, with $p = q + 1 -\binom{n-1}{t}$
\[
\begin{CD}
    @.    0              @.    0        @.      \\
@.       @VVV                  @VVV           @.\\
0 @>>> \Ker Res(\beta) @>>> \Ker\beta @>>>  0 \\
@.       @VVV                  @VVV           @.\\
0 @>>> \Dirsum_{i=1}^pS(-a_i)
                         @>>>  \Dirsum_{i=1}^qS(-b_i)
                                        @.       \\
@.       @V{Res(\beta)}VV       @V{\beta}VV    @. \\
0 @>>> \langle\beta_1,\ldots,\beta_q\rangle\cap E_{t+2}
                         @>>>  \langle\beta_1,\ldots,\beta_q\rangle
                                               @. \\
@.        @VVV                  @VVV           @. \\
     @.    0             @.     0        @.        \\
\end{CD}
\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccc}
%     & 0                       &      &  0                       &
%\\      
%     & \da                     &      &  \da                     &
%\\
%0\To & \Ker Res(\beta)         & \To  & \Ker\beta                &\To 0
%\\    
%     & \da                     &      &  \da                     &
%\\
%0\To & \Dirsum_{i=1}^p S(-a_i) & \To  & \Dirsum_{i=1}^q S(-b_i)  & 
%\\
%     & \hspace*{1.4cm}\da Res(\beta)         
%                               &      & \hspace*{0.5cm}\da \beta &
%\\
%0\To & \langle\beta_1,\ldots,
%       \beta_q\rangle\cap 
%       E_{t+2}                 & \To  & \langle\beta_1,\ldots,
%                                         \beta_q\rangle          & 
%\\
%     & \da                     &      &  \da                     &
%\\
%     &  0                      &      &   0                      &
%\end{array}
%\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $\beta(g_i) = \beta_i$ for all $i$ with $g_1,\ldots, g_q$ the free basis 
of $\Dirsum_{i=1}^q S(-b_i)$.
\end{enumerate}
In this case, we have $I = \varphi(K_{t+1})(n-c)$.
\end{enumerate}
\end{Theorem}

\begin{Corollary}
\label{cor:depthzero}
There is no codimension~3 single spot ideal $I$ of type $(0, K(-c))$ with 
approximation module $E_{t+1}$ if $t=0$. 
\end{Corollary}
\begin{proof}
Assume that there exists a codimension~3 single spot ideal
$I \subset S$ of type $(0, K(-c))$
fitting into a long Bourbaki sequence
\[
   0 \To F \To G \To E_1 \To I(c) \To 0.
\]
Then by Theorem~\ref{theorem:construction2}, there exist
$\beta_1,\ldots, \beta_q \in K_1\backslash E_2$ and 
$(0\ne) \varphi \in {\cal A}$
such that $\langle\beta_1,\ldots,\beta_q\rangle + E_2 = \Ker(\varphi : K_1\to S(-n))$. Since ${\cal A} =  \langle x_1e^*_1 +\cdots +  x_ne^*_n\rangle$
we must have $\beta_1,\ldots, \beta_q\in E_2$, 
a contradiction.
\end{proof}

\begin{Remark}
In fact, the above Corollary holds for any codimension $\geq 2$. 
Here is an outline of the proof. We consider a straightforward extension of 
Theorem~\ref{theorem:construction2}(a) to any codimension $r$.
In this case we consider long Bourbaki sequences
\[
   0\To F_{r-1} \To F_{r-2}\To\cdots\To F_{1}\To E_1 \namedTo{\varphi} I(c)\To 0.
\]
and $\varphi$ is determined by a nonzero element from
${\cal A} = \langle x_1e^*_1 +\cdots +  x_ne^*_n\rangle$.
\end{Remark}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Long Bourbaki sequences of non-trivial type}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $n\geq 4$ and $t\leq n-4$ and 
consider a long Bourbaki sequence
\begin{equation}
\label{bourbakiseq0}
    0 \To F\namedTo{f} G \namedTo{g} E_{t+1}\dirsum E_{n-1}(d) \namedTo{\phi} I(c) \To 0
\end{equation}
with $c\in\ZZ$ and  $S$-free modules $F$ and $G$.
From this sequence, we construct the following
diagram where the second row is the minimal free resolution of $M =
E_{t+1}\dirsum E_{n-1}(d)$ and the third row is the mapping cone
$C(\alpha,\beta)$ of a chain map $\alpha\dirsum\beta$, which is a free
resolution of $I(c)$. 
%
\begin{equation}
\label{mappingcone}
\begin{array}{ccccccccl}
    &     &     &           &            &         &             &      0        &        \\
    &     &     &           &            &         &             &    \da        &        \\
    &  0  & \To &  F        &\namedTo{f} & G       & \namedTo{g} & \Ker\phi      & \to 0  \\
%
    &     &     & \alpha\da &            &\beta\da &             &  \da          & \      \\
%
0\to K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}}
    &K_{t+3} &\namedTo{\partial_{t+3}}&K_{t+2}
& \namedTo{\begin{array}{c}
\partial_{t+2}\\ 
\dirsum\partial_n
\end{array}}
 & K_{t+1} & \namedTo{\begin{array}{c} \partial_{t+1}\\
\dirsum\partial_{n-1}
\end{array}} & M & \to 0 \\
%
    &       &    &\dirsum K_n(d)&    &\dirsum K_{n-1}(d)&    &\hspace*{0.3cm}\da\phi  &  \\
%
0\to K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}} 
& K_{t+3} &\namedTo{\zeta} & K_{t+2}&\namedTo{\rho} & K_{t+1}&  \namedTo{\begin{array}{c}
\phi\circ(\partial_{t+1}\\ \dirsum\partial_{n-1})
\end{array}} &I(c) &\to 0 \\
%
  &\dirsum F&    &\dirsum K_n(d)&    &\dirsum K_{n-1}(d)&    &  \da  &  \\
%
  &         &    &\dirsum G  &    &               &    &   0   &  \\
\end{array}
\end{equation}
where
\[
  \begin{array}{cccc}
     \rho:  & K_{t+2}\dirsum K_n(d) \dirsum G & \To  & K_{t+1}\dirsum K_{n-1}(d) \\
      & (a, b, c)  & \longmapsto & (\partial_{t+2}(a),\partial_n(b))- \beta(c)  
  \end{array}
\]
and 
\[
  \begin{array}{cccc}
     \zeta:  & K_{t+3} \dirsum F & \To  & K_{t+2}\dirsum K_{n}(d)\dirsum G \\
      & (a, b)  & \longmapsto & (\partial_{t+3}(a), 0, f(b)) - \alpha(b)  
  \end{array}
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $p_1 : K_{t+1}\dirsum K_{n-1}(d) \to K_{t+1}$ and $p_2 : K_{t+1}\dirsum
K_{n-1}(d)\to K_{n-1}(d)$ be the first and the second projections. 
From the diagram (\ref{mappingcone})  we know 
    $\Ker\phi = \Im g = (\partial_{t+1}\dirsum \partial_{n-1})\circ \beta(G)$
and then by considering the ranks of the modules in the short exact sequence
\begin{equation}
\label{kerVarphi}
   0 \To \Ker\phi
     \To  E_{t+1}\dirsum E_{n-1}(d)
     \namedTo{\phi} I(c)
     \To 0
\end{equation}
we have
\begin{equation}
\label{rankcond1} 
    \rank(\Ker\varphi) = n-2 +\binom{n-1}{t}.
\end{equation}
On the other hand, we have 
\begin{equation}
\label{inclusion}
(\partial_{t+1}\circ p_1\circ\beta)(G)
         \dirsum (\partial_{n-1}\circ p_2 \circ \beta)(G)
  \supset
(\partial_{t+1}\dirsum\partial_{n-1})\beta(G)
 = \Ker\phi.
\end{equation}
Thus we have
\begin{equation}
\label{rankcond2}
 \rank {\cal I}_{t+1} +\rank{\cal I}_{n-1} \geq n-2 + \binom{n-1}{t}.
\end{equation}
where  ${\cal I}_{t+1} := (\partial_{t+1}\circ p_1\circ \beta)(G) (\subseteq E_{t+1})$ and 
${\cal I}_{n-1} := (\partial_{n-1}\circ p_2\circ \beta)(G) (\subseteq E_{n-1}(d))$.
Since $\rank E_{t+1} = \binom{n-1}{t}$ and $\rank E_{n-1}(d) = n-1$ we know from 
(\ref{rankcond2}) that 
\[
(\rank {\cal I}_{t+1}, \rank {\cal I}_{n-1}) 
= (\binom{n-1}{t} - 1, n-1), 
  (\binom{n-1}{t}, n-1), \mbox{ or }
  (\binom{n-1}{t}, n-2).
\]
Under this situation, we have
%
\begin{Lemma}
%\label{theorem:non-trivialcond}
\label{chainmapBETA}
Following are equivalent.
\begin{enumerate}
\item [$(i)$]  long Bourbaki sequence $(\ref{bourbakiseq0})$ is 
               of non-trivial type
\item [$(ii)$] For any free basis $\{m_i\}_{i=1}^q$ of $G$
there exists an index $i$ such that
$\beta(m_i) \notin K_{t+1}$ and $\beta(m_i)\notin K_{n-1}(d)$.
\end{enumerate}
\end{Lemma}
%
%
%
\begin{proof}
We will prove $(i)$ to $(ii)$. 
We assume that for all $i$ we have either
$\beta(m_i)\in K_{t+1}$ or $\beta(m_i)\in K_{n-1}(d)$ and 
will deduce  a contradiction.
First of all, we have equality in (\ref{inclusion}), and then
from (\ref{rankcond2}) we have 
\[
    n-2 + \binom{n-1}{t} = \rank {\cal I}_{t+1} +  \rank {\cal I}_{n-1}.
\]
Thus, we have 
$(\rank {\cal I}_{t+1}, \rank {\cal I}_{n-1}) = 
(\binom{n-1}{t}-1, n-1)$ or $(\binom{n-1}{t}, n-2)$.
Also, since
$\Ker\phi = {\cal I}_{t+1}\dirsum {\cal I}_{n-1}$, we have by (\ref{kerVarphi})
\begin{equation}
\label{prop:I(c)}
I(c) \iso (E_{t+1}/{\cal I}_{t+1})\dirsum (E_{n-1}(d)/{\cal I}_{n-1})
\end{equation}
%
\begin{description}
\item [case $(\rank {\cal I}_{t+1}, \rank {\cal I}_{n-1}) = (\binom{n-1}{t}-1, n-1)$:]
Since we have
$\rank E_{n-1}(d)/{\cal I}_{n-1} =\rank E_{n-1}(d) - \rank {\cal I}_{n-1} = 0$,
$E_{n-1}/{\cal I}_{n-1}$ is 0 or a torsion-module. But since $I(c)$ is torsion
free, we must have $E_{n-1}(d) = {\cal I}_{n-1}$ by (\ref{prop:I(c)}). Thus 
$\Ker\phi = {\cal I}_{t+1}\dirsum E_{n-1}(d)$ and then the Bourbaki sequence 
(\ref{bourbakiseq0}) must be of trivial-type
\[
     0\To F'\dirsum K_n \namedTo{f'\dirsum\partial_n}
          G'\dirsum K_{n-1}\namedTo{g'\dirsum\partial_{n-1}}
          E_{t+1}\dirsum E_{n-1}(d)\namedTo{\phi} I(c)\To 0
\]
where $0\to F' \to G' \to {\cal I}_{t+1} \to 0$ is a $S$-free resolution 
of ${\cal I}_{t+1}$, a contradiction.


\item [case $(\rank {\cal I}_{t+1}, \rank {\cal I}_{n-1}) 
= (\binom{n-1}{t}, n-2)$:]
In this case we have $\rank E_{t+1}/{\cal I}_{t+1} =0$.
Since $E_{t+1}/{\cal I}_{t+1}\subset I(c)$ and $I(c)$ is 
torsion-free, we must have $E_{t+1}/{\cal I}_{t+1} =0$. 
Thus $\Ker\phi = E_{t+1}\dirsum {\cal I}_{n-1}$ and 
the Bourbaki sequence (\ref{bourbakiseq0}) is 
\begin{eqnarray*}
     0\To K_n\dirsum U_n \namedTo{\partial_n\dirsum d_n}
                \cdots   \namedTo{\partial_{t+2}\dirsum d_{t+2}}
           & K_{t+1}\dirsum U_{t+1} &  \\
 \namedTo{\partial_{t+1}\dirsum d_{t+1}}
          &E_{t+1}\dirsum E_{n-1}& \namedTo{\phi} I(c)\To 0.\\
\end{eqnarray*}
But then we must have $t \geq n-2$, which contradicts to the 
assumption that $t\leq n-4$.
\end{description}
Now we show $(ii)$ to $(i)$. 
Assume that  (\ref{bourbakiseq0}) is of trivial type. Then
we must have $\beta(G) = p_1(\beta(G))\dirsum p_2(\beta(G))$. From
this we immediately obtain the required result.
\end{proof}

From Lemma~\ref{chainmapBETA}, we immediately have
%
\begin{Theorem}
\label{theorem:non-trivialcond}
Under the situation of Theorem~\ref{theorem:construction},
the long Bourbaki sequence is of non-trivial type if and only if
\begin{enumerate}
\item [$(i)$] the condition Theorem~\ref{theorem:construction} $(ii)$  holds,
              and
\item [$(ii)$] the submodule $N:= \langle\beta_1,\ldots, \beta_q\rangle$
              of $K_{t+1}\dirsum K_{n-1}(d)$ cannot be decomposed in the form
              of 
	      $N = A\dirsum B$ for some  $(0\ne) A\subset K_{t+1}$ and 
              $(0\ne )B \subset K_{n-1}(d)$
\end{enumerate}
\end{Theorem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Numerical Characterizations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Existence of long Bourbaki sequence as in
Theorem~\ref{theorem:construction} and~\ref{theorem:construction2}
only implies that $I$ is a single spot ideal of codimension at most
$3$.  To assure that the codimension is exactly $3$, we need
additional condition. In this section, we give a numerical condition
to assure $\codim I = 3$ for long Bourbaki sequences with approximation
modules $E_{t+1}\dirsum E_{n-1}(d)$.

We assume $n\geq 4$ and $t\leq n-4$, and let $I\subset S$ be a
graded ideal fitting into a long  Bourbaki sequence
\begin{equation}
\label{bourbakiseq}
    0 \To F\namedTo{f} G \namedTo{g} M \namedTo{\phi} I(c) \To 0
\end{equation}
with $M = E_{t+1}\dirsum E_{n-1}(d)$, $F = \Dirsum_{i=1}^p S(-a_i)$ and 
$G=\Dirsum_{i=1}^q S(-b_i)$ are $S$-free modules and $c\in\ZZ$.
As in Theorem~\ref{theorem:construction} we have 
\begin{equation}
\label{rankcondition}
q  = p  + \binom{n-1}{t} + n - 2.
\end{equation}

%Now from the sequence (\ref{bourbakiseq}) we construct the following
%diagram where the second row is the minimal free resolution of $M =
%E_{t+1}\dirsum E_{n-1}$ and the third row is the mapping cone
%$C(\alpha,\beta)$ of a chain map $\alpha\dirsum\beta$, which is a free
%resolution of $I(c)$. We denote the Koszul complex of the sequence
%$x_1,\ldots,x_n$ by $(K_\bullet,\partial_\bullet)$.
%%
%\begin{equation}
%\label{mappingcone}
%\begin{array}{ccccccccl}
%    &     &     &      &            &  &             &      0        &        \\
%    &     &     &      &            &  &             &    \da        &        \\
%    &  0  & \To &  F   &\namedTo{f} & G & \namedTo{g} & \Ker\varphi   & \to 0  \\
%%
%    &     &     & \alpha\da &    & \beta\da &        &  \da         & \      \\
%%
%0\to K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}}
%    &K_{t+3} &\namedTo{\partial_{t+3}}&K_{t+2}
%& \namedTo{\begin{array}{c}
%\partial_{t+2}\\ 
%\dirsum\partial_n
%\end{array}}
% & K_{t+1} & \namedTo{\begin{array}{c} \partial_{t+1}\\
%\dirsum\partial_{n-1}
%\end{array}} & M & \to 0 \\
%%
%    &       &    &\dirsum K_n&    &\dirsum K_{n-1}&    &  \da  & \varphi \\
%%
%0\to K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+4}} 
%& K_{t+3} &\namedTo{\zeta} & K_{t+2}&\namedTo{\rho} & K_{t+1}&
%      \namedTo{\begin{array}{c}\varphi\circ(\partial_{t+1}\\ \dirsum\partial_{n-1})\end{array}} &I(c) &\to 0 \\
%%
%  &\dirsum F&    &\dirsum K_n&    &\dirsum K_{n-1}&    &  \da  &  \\
%%
%  &         &    &\dirsum G  &    &               &    &   0   &  \\
%\end{array}
%\end{equation}
%where
%\[
%  \begin{array}{cccc}
%     \rho:  & K_{t+2}\dirsum K_n \dirsum G & \To  & K_{t+1}\dirsum K_{n-1} \\
%      & (a, b, c)  & \longmapsto & (\partial_{t+2}(a),\partial_n(b))- \beta(c)  
%  \end{array}
%\]
%and 
%\[
%  \begin{array}{cccc}
%     \zeta:  & K_{t+3} \dirsum F & \To  & K_{t+2}\dirsum K_{n}\dirsum G \\
%      & (a, b)  & \longmapsto & (\partial_{t+3}(a), 0, f(b)) - \alpha(b)  
%  \end{array}
%\]
%

Now from the sequence (\ref{bourbakiseq}),  we construct 
the mapping cone $C(\alpha, \beta)$ as in (\ref{mappingcone}).
The cone gives a $S$-free resolution $F_\bullet$ of the residue ring $S/I$.
\[
F_\bullet: 0\to F_{n-t}\to\cdots\to F_1 \to F_0 \to S/I \to 0
\]
where 
\begin{eqnarray*}
F_0 & = & S \\
F_1 & = & K_{t+1}(-c)\dirsum K_{n-1}(d-c) = S(-t-1-c)^{\beta_1}\dirsum S(-n+1+d-c)^n  \\
%
F_2 & = & K_{t+2}(-c)\dirsum K_n(d-c)\dirsum G(-c) \\
    & = & S(-t-2-c)^{\beta_2}\dirsum S(-n+d-c)\dirsum \Dirsum_{i=1}^{q}S(-b_i-c)\\
%
F_3 & = & K_{t+3}(-c)\dirsum F(-c) = S(-t-3-c)^{\beta_3}\dirsum \Dirsum_{i=1}^{p}S(-a_i-c)\\
%
F_i & = & K_{t+i}(-c) =S(-t- i -c)^{\beta_{i}} 
\qquad  (4\leq i\leq n-t)\\
\mbox{with} && \beta_i = \binom{n}{t+i}\quad i=1,\ldots, n-t.
\end{eqnarray*}
Notice that this resolution is minmal if and only if matrix
representations of $\alpha$ and $\beta$ only have their entries from
$\mm$.

Now we compute the Hilbert series $Hilb(S/I, \lambda)$ of $S/I$. We have 
\begin{equation}
\label{hilbertSeries}
    Hilb(S/I, \lambda) = \frac{ Q(\lambda)}{(1-\lambda)^n}
\end{equation}
with
\begin{eqnarray*}
\lefteqn{Q(\lambda) =  \sum_{i,j}(-1)^i\beta_{i,j}\lambda^j} \\
           &=& 1 - n \lambda^{n-1+c-d} + \lambda^{n+c-d} 
                + \sum_{i=1}^q\lambda^{b_i+c}
- \sum_{i=1}^p \lambda^{a_i+c}
+ (-1)^t\lambda^c\sum_{i=t+1}^n(-1)^i\binom{n}{i}\lambda^i
\end{eqnarray*}
where $\beta_{i,j}$ are as in $F_i = \Dirsum_{j}S(-j)^{\beta_{ij}}$, $(i=0,\ldots, n-t)$.
(see Lemma~4.1.13 \cite{BH}).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Numerial condition for $\codim I \geq 3$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Since we have $\Coh{i}{S/I} = \Coh{i+1}{M}$ for $0\leq i \leq n-4$ by
(\ref{localcohomologies}), we know that $\dim S/I \geq n-3$, i.e.,
$\codim I \leq 3$. To assure that $\codim I \geq 3$ we must have $Q(1)
= Q'(1) = Q''(1) = 0$ (see Corollary~4.1.14(a) \cite{BH}).

\begin{Proposition}
\label{Q(1)=0}
$Q(1) = 0$ holds for all $n,t,c$ and $p$.
\end{Proposition}
\begin{proof}
We compute using (\ref{rankcondition})
\[
 Q(1) = \binom{n-1}{t} - (-1)^{t+1}\sum_{i=t+1}^{n}(-1)^i\binom{n}{i} 
      = \binom{n-1}{t} - \rank E_{t+1} 
      = 0.
\]
where the last equation follows from the Koszul resolution of $E_{t+1}$:
\[
   0\To K_n\To K_{n-1}\To\cdots\To K_{t+1}\To E_{t+1}\To 0\quad\mbox{(exact)}.
\]
\end{proof}

\begin{Proposition}
\label{Q'(1)=0}
$Q'(1) = 0$ holds if and only if 
\[
   \sum_{i=1}^q b_i  - \sum_{i=1}^p a_i 
= n^2 -(2+d)n + c + d + \binom{n-2}{t-1}
                         + \binom{n-1}{t}t.
\]
\end{Proposition}
\begin{proof}
We compute using (\ref{rankcondition})
\[
Q'(1) = -n^2 + (2+d)n - c - d + \sum_{i=1}^q b_i - \sum_{i=1}^p a_i 
           -(-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i.
\]
and we know 
\begin{equation}
\label{usefulequation}
(-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i
=    (-1)^t\sum_{i=0}^t(-1)^i\binom{n}{i}i
=  \binom{n-2}{t-1} + \binom{n-1}{t}t
\end{equation}
where the last equation is given in Example~2.3 \cite{HT2}.
\end{proof}

Now before we go further, we need to show a combinatorial equation.
\begin{Lemma}
\label{closedF}
\[
(-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2 
= \binom{n-1}{t}(t+1)^2 - \binom{n-2}{t}(2t+1) - 2\binom{n-3}{t-1}
\]
\end{Lemma}
\begin{proof}
 Let $A = (-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2$.
A straightforward computation using the binomial coefficient theorem and (\ref{usefulequation}) shows that 
\begin{eqnarray*}
0& = & \sum_{i=2}^{n}(-1)^i\binom{n}{i}i(i-1)  \\
& = & (-1)^{t+1}A 
    + \sum_{i=2}^{t}(-1)^i\binom{n}{i}i(i-1)
    -(-1)^{t+1}\left[ \binom{n-2}{t-1} + \binom{n-1}{t}t\right]
\end{eqnarray*}
Thus we have
\begin{equation}
\label{intermedAns}
A =  \binom{n-2}{t-1} + \binom{n-1}{t}t
    + (-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n}{i}i(i-1)
\end{equation}
Now we will compute the last term.
Set
\[
\alpha(n,t) = (-1)^{t}\sum_{i = 2}^{t}(-1)^i\binom{n}{i}i(i-1).
\]
and we compute
\begin{eqnarray*}
\lefteqn{\alpha(n,t) + \alpha(n-1,t-1)}\\
&= & 2t^2\binom{n-1}{t}  -
2(-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n-1}{i}i
     + 2(-1)^{t}(n-1) - \alpha(n-1, t).
\end{eqnarray*}
Also we have
\[
	\alpha(n-1, t-1) + \alpha(n-1, t)
	= t(t-1)\binom{n-1}{t}
\]
Thus by using (\ref{usefulequation})
\begin{eqnarray*}
\alpha(n,t)
& = & t(t+1)\binom{n-1}{t} -
2(-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n-1}{i}i
	+  2(-1)^{t}(n-1) \\
& = & t(t+1)\binom{n-1}{t} - 2
	\left[
		t\binom{n-2}{t} + \binom{n-3}{t-1}
              + (-1)^{t}(n-1)
	\right]\\
&   &
	+  2(-1)^{t}(n-1) \\
& = & t(t+1)\binom{n-1}{t} - 2t\binom{n-2}{t} -
2\binom{n-3}{t-1}\\
%& = & t(t-1)\binom{n-1}{t} + 2t\binom{n-2}{t-1} - 2\binom{n-3}{t-1}
\end{eqnarray*}
Substiting this into  (\ref{intermedAns}), we obtain the desired result.
\end{proof}

\begin{Proposition}
\label{Q''(1)=0}
$Q''(1)=0$ holds if and only if 
\begin{eqnarray*}
\sum_{i=1}^q b_i^2  - \sum_{i=1}^p a_i^2
&=& n^3 - (3+2d)n^2  + (d^2+4d+1)n - c^2 - d^2  \\
& &
 + \binom{n-1}{t}(t+1)^2 
    - \binom{n-2}{t}(2t+1) - 2\binom{n-3}{t-1}
\end{eqnarray*}
\end{Proposition}
\begin{proof}
We compute 
\begin{eqnarray*}
Q''(1) &=& c + d -c^2 + d^2 - 2cd + (4c-3-d^2-5d+2cd)n + (4-2c+2d)n^2 - n^3 \\
       & & -(2c-1)\left[\binom{n-2}{t-1} + \binom{n-1}{t}t\right] 
             + \sum_{i=1}^{q}b_i^2 - \sum_{i=1}^p a_i^2\\
       & &   + (2c -1)\left(\sum_{i=1}^{q}b_i - \sum_{i=1}^{p}a_i\right) 
             + (-1)^t\sum_{i=t+1}^{n}(-1)^i\binom{n}{i}i^2 \\
       &=&  c^2 +d^2 -(d^2+4d+1)n + (3+2d)n^2 - n^3  \\
       & &  + \sum_{i=1}^q b_i^2  - \sum_{i=1}^p a_i^2
            + (-1)^t\sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2\\ 
\end{eqnarray*}
where the last equation is by  Proposition~\ref{Q'(1)=0}. 
Then by Lemma~\ref{closedF} we obtain the desired result.
\end{proof}

To summerize, we obtain 
\begin{Theorem}
\label{theorem:numerical}
Let $n\geq 4$ and $t\leq n-4$. Assume that we have the following long Bourbaki sequence
\[
0\To \Dirsum_{i=1}^p S(-a_i)\To \Dirsum_{i=1}^q S(-b_i)\To E_{t+1}\dirsum E_{n-1}(d)\To I(c)\To 0
\]
with $I \subset S$ a graded ideal and $c\in\ZZ$. Then we have $\codim
I \leq 3$ and the equality holds if and only if
\begin{enumerate}
\item $q = p + \binom{n-1}{t} + n - 2$;
\item $\displaystyle{\sum_{i=1}^q }b_i - \displaystyle{\sum_{i=1}^p} a_i
 = n^2 - (2+d)n + c + d + \binom{n-2}{t-1} + \binom{n-1}{t}t$;  
\item and \begin{eqnarray*}
\displaystyle{\sum_{i=1}^q} b^2_i - \displaystyle{\sum_{i=1}^p} a^2_i
               &=& n^3 - (3+2d)n^2 + (d^2+4d+1)n - c^2 - d^2 \\
               & & + \binom{n-1}{t}(t+1)^2 -\binom{n-2}{t}(2t+1) - 2\binom{n-3}{t-1}
      \end{eqnarray*}
\end{enumerate}
\end{Theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Degrees of minimal generators}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%In (\ref{hilbertSeries}), we have
%\[
%Q(\lambda) = 1 + h_{\alpha_1 + c}\lambda^{\alpha_1 +c} +
%		\cdots + h_{\alpha_l + c}\lambda^{\alpha_l + c}
%\]
%where
%\[
%\alpha_l =\max\{n, a_1,\ldots, a_p, b_1,\ldots, b_q\}
%         = \max\{n, a_p\}
%\]
%where $a_1\leq \cdots a_p$ and $b_q\leq\cdots\leq b_q$ are 
%degrees of the basis of $F$ and $G$ as in the previous subsection.
%Thus we have
%
%\begin{Proposition}
%$Hilb(S/I, \lambda) = 
%\frac{1+ h_1\lambda + \cdots + h_N\lambda^N}{(1-\lambda^{n-3})}$
%where $N =\max\{n,a_p\} + c - 3$, and the least degree of the minimal
%generator of $I$ is $\max\{n, a_p\} + c - 2$.
%\end{Proposition}


%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%
\section{Examples}
%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%

\begin{Example}
\label{example1}
We first give an application of Theorem~\ref{theorem:construction2}. Namely,
a single spot ideal $I$ with approximation module $E_{t+1}$. 
Let $t = 1$ and $n =6$. Then ${\cal A} = \{A_1,A_2,A_3,A_4,A_5,A_6\}
 = \partial_2^*(E_2^*) \subset K_2^*$ where 
\begin{eqnarray*}
A_1 & = &  x_1 e^*_{16} + x_2 e^*_{26} + x_3 e^*_{36} + x_4 e^*_{46} + x_5 e^*_{56}
\\
A_2 & = & -x_1 e^*_{15} - x_2 e^*_{25} - x_3 e^*_{35} - x_4 e^*_{45} + x_6 e^*_{56}
\\
A_3 & = &  x_1 e^*_{14} + x_2 e^*_{24} + x_3 e^*_{34} - x_5 e^*_{45} - x_6 e^*_{46}  
\\
A_4 & = & -x_1 e^*_{13} - x_2 e^*_{23} + x_4 e^*_{34} + x_5 e^*_{35} + x_6 e^*_{36}  
\\
A_5 & = &  x_1 e^*_{12} - x_3 e^*_{23} - x_4 e^*_{24} - x_5 e^*_{25} - x_6 e^*_{26}  
\\
A_6 & = &  x_2 e^*_{12} + x_3 e^*_{13} + x_4 e^*_{14} + x_5 e^*_{15} + x_6 e^*_{16}.
\end{eqnarray*}
Now let $a \in {\cal A}$ and $\beta_i$ ($i=1,\ldots, 6$) be as follows:
\begin{eqnarray*}
a & = & x_6 A_1 -x_5 A_2 + x_4 A_3 \\
  & = & x_1x_4 e^*_{14} + x_1x_5 e^*_{15} +x_1x_6 e^*_{16} + x_2x_4 e^*_{24}
       x_2x_5 e^*_{25} + x_2x_6 e^*_{26} +x_3x_4 e^*_{34} + x_3x_5 e^*_{35}\\
  &   & + x_3x_6 e^*_{36} \\
\beta_1 & = & e_{12},\quad \beta_2 = e_{13},\quad \beta_3 = e_{23},
                     \quad \beta_4 = e_{45},\quad \beta_5 = e_{46},
                     \quad \beta_6 = e_{56}
\end{eqnarray*}
Then we have
\[
 \Ker(a : K_2\dirsum K_5 \to S)
  = \langle\beta_1,\ldots,\beta_6\rangle + E_3
\]
and, for the map $\beta : \Dirsum_{i=1}^6 S(-2) \to K_2\dirsum K_5$ such that
$\beta(m_i) = \beta_i$ $(i=1,\ldots, 6)$ where $\{m_i\}$ is a free basis, 
we obtain the diagram
\[
\begin{CD}
   @.    0      @.   0     \\
@.       @VVV        @VVV  \\
0 @>>> \left\langle
\begin{array}{c}
          x_3m_1-x_2m_2+x_1m_3,\\
          x_6m_4-x_5m_5+x_4m_6 
         \end{array}
         \right\rangle
                @>>> \langle m_1,\ldots,m_6\rangle \\
@.       @V{Res(\beta)}VV
                      @V{\beta}VV \\
0 @>>> \langle\beta_1,\ldots,\beta_6\rangle\cap E_3
                 @>>> \langle\beta_1,\ldots,\beta_6\rangle \\
@.       @VVV          @VVV  \\
    @.   0       @.     0    \\
\end{CD}
\]
%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccc}
%  &    & 0                         &       &    0                         \\
%  &    & \da                       &       &  \da                         \\
%0 &\To &\left\langle
%         \begin{array}{c}
%            x_3 m_1 - x_2 m_2 + x_1 m_3,\\
%            x_6 m_4 - x_5 m_5 + x_4 m_6 
%         \end{array}                         
%         \right\rangle                      
% &   \To & \langle m_1,\ldots, m_6\rangle \\
%&&\hspace*{1.3cm}\da Res(\beta)&& 
%\hspace*{0.5cm} \da\beta                     \\
%0 &\To & \langle\beta_1,\ldots,\beta_6 \rangle\cap E_3
%   &\To 
%&\langle\beta_1,\ldots,\beta_6\rangle   \\
%  &    &   \da  &   &  \da \\
%  &    &   0    &   &   0  \\
%\end{array}
%\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and $a$ defines a degree $0$ homomorphism from $E_2$ to $S$.
Then we obtain the long Bourbaki sequence
\[
   0 \To S^2(-3) \namedTo{f} S^6(-2) \namedTo{g} E_2 \namedTo{\varphi} I \To 0
\]
where 
\[
\begin{array}{cccc}
f : & S^2(-3) = S n_1 \dirsum S n_2 & \To &  S^6(-2) = S m_1 \dirsum\cdots\dirsum S m_6 \\
    &           n_1                 &\longmapsto  &  x_3 m_1 - x_2 m_2 + x_1 m_3 \\
    &           n_2                 &\longmapsto  &  x_6 m_4 - x_5 m_5 + x_4 m_6 \\
    &                               &             &                               \\
g : &  S^6(-2) = S m_1 \dirsum\cdots\dirsum S m_6 & \To  & E_2 \\
    &           m_i                 &\longmapsto  &  \partial_2(\beta_i) \qquad (i=1,\ldots,6) \\
\mbox{and} &                        &             &                               \\
\varphi &   E_2               & \To         & I  \\
        & \partial_2(e_{ij})  & \longmapsto & x_ix_j \\
        &                     &\mbox{for }& (i,j) = (1,2),(1,3),(2,3),(4,5),(4,6),(5,6)\\
        & \partial_2(e_{ij})  & \longmapsto & 0 \\
        &                     &\mbox{for }& (i,j) \ne (1,2),(1,3),(2,3),(4,5),(4,6),(5,6)\\
\end{array}
\]
and we obtain 
$I = (x_1x_4,x_1x_5, x_1x_6,x_2x_4,x_2x_5,x_2x_6,x_3x_4, x_3x_5, x_3x_6)
       = (x_1,x_2,x_3)(x_4,x_5,x_6)$, a codimension~3 
single spot ideal of type $(1,K)$.
\end{Example}


\begin{Example}
\label{example2}
We continue to consider the situation in Example~\ref{example1}. 
As an application of Theorem~\ref{theorem:construction}, we can see
that the same ideal fits into a long Bourbaki sequence with approximation
module $E_{t+1}\dirsum E_{n-1} = E_1 \dirsum E_5$.
%
In this case, we must also  consider 
${\cal B} = \{B_{ij} \mid 1\leq i <j \leq 6 \}
 = \partial_5^*(E_5^*) \subset K_5^*$ 
where $B_{ij} = (-1)^{i}x_je_{[6]\backslash i}^*
              - (-1)^{j}x_ie_{[6]\backslash j}^*$.
Then we set $a \in {\cal A}$ as in Example~\ref{example1} and 
\[
b = -x_1^2x_2x_4 B_{14} =x_1^2x_2x_4^2 e^*_{23456} + x_1^3x_2x_4 e^*_{12356}
\in {\cal B}.
\]
Also we set $\beta_1,\ldots, \beta_6$ to be the same as 
those in Example~\ref{example1} and 
$\beta_7 =x_1x_2x_4 e_{14} - e_{23456}$,
$\beta_8 =x_1^2x_2 e_{14} - e_{12356}$,
$\beta_9    = e_{13456}$,
$\beta_{10} = e_{12456}$,
$\beta_{11} = e_{12346}$,
$\beta_{12} = e_{12345}$.
Notice that $\{\beta_i\}_{i=1}^{12}$ satisfies the condition of
Theorem~\ref{theorem:non-trivialcond}.
Then $\varphi = (a,b)$ defines a degree $0$ map on $E_2\dirsum E_5$, and we have
\[
  \Ker\varphi = \langle\beta_1,\ldots,\beta_{12}\rangle + E_3\dirsum E_6
\]
and the diagram
\[
\begin{CD}
   @.    0     @.     0  \\
@.       @VVV        @VVV \\
0 @>>> \left\langle
          \begin{array}{l}
            x_3m_1-x_2m_2+x_1m_3,\;
            x_6m_4-x_5m_5+x_4m_6,\\
          x_1m_7-x_4m_8+x_2m_9-x_3m_{10}-x_5m_{11}+x_6m_{12}
          \end{array}
         \right\rangle
              @>>> \langle m_1,\ldots,m_{12}\rangle \\
@.       @V{Res(\beta)}VV    
                   @V{\beta}VV \\
0 @>>>  \langle\beta_1,\ldots,\beta_6\rangle
          \cap E_3\dirsum E_6
                   @>>> \langle\beta_1,\ldots\beta_{12}\rangle\\
@.        @VVV           @VVV   \\
    @.    0        @.     0
\end{CD}
\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccc}
%  &    & 0           &       &    0                       \\
%  &    & \da         &       &  \da                       \\
%0 &\To &\left\langle
%         \begin{array}{c}
%            x_3 m_1 - x_2 m_2 + x_1 m_3,\\
%            x_6 m_4 - x_5 m_5 + x_4 m_6,\\
%         x_1 m_7 - x_4 m_8 + x_2 m_9 - x_3 m_{10} -x_5 m_{11} + x_6 m_{12}
%         \end{array}                         
%         \right\rangle                      
% &   \To & \langle m_1,\ldots, m_{12}\rangle \\
%   &   & \hspace*{1.3cm}\da Res(\beta)&
%& \hspace*{0.5cm} \da\beta                     \\
%0 &\To & \langle\beta_1,\ldots,\beta_6
%         \rangle\cap E_3\dirsum E_6
%          &   \To 
%& \langle\beta_1,\ldots,\beta_{12}\rangle  \\
%  &   & \da &   & \da \\
%  &   &  0  &   &  0  \\
%\end{array}
%\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Then we have a long Bourbaki sequence of non-trivial type
\[
   0 \To S^2(-3)\dirsum S(-6) 
    \namedTo{f} 
         S^6(-2)\dirsum S^6(-5)
    \namedTo{g} 
         E_2\dirsum E_5
    \namedTo{\varphi} I \To 0
\]
where 
\[
\begin{array}{cccc}
f : & S^2(-3)\dirsum S(-6) = S n_1 \dirsum S n_2 \dirsum S n_3 & \To &  S^6(-2) = S m_1 \dirsum\cdots\dirsum S m_6 \\
    &           n_1                 &\longmapsto  &  x_3 m_1 - x_2 m_2 + x_1 m_3 \\
    &           n_2                 &\longmapsto  &  x_6 m_4 - x_5 m_5 + x_4 m_6 \\
    &           n_3                 &\longmapsto  &  x_1 m_7 - x_4 m_8 + x_2 m_9 \\
    &                               &             & - x_3 m_{10} -x_5 m_{11} + x_6 m_{12} \\
    &                               &             &                               \\
g : &  S^6(-2) = S m_1 \dirsum\cdots\dirsum S m_6 & \To  & E_2\dirsum E_5 \\
    &           m_i                 &\longmapsto  &  \bar\partial(\beta_i) \qquad (i=1,\ldots,12) \\
    &                               &             & \mbox{where }\bar\partial = \partial_2\dirsum\partial_5\\
\mbox{and} &                        &             &                               \\
\varphi &   E_2               & \To         & I  \\
        & \partial_2(e_{ij})  & \longmapsto & x_ix_j \\
        &                     &\mbox{for }& (i,j) = (1,2),(1,3),(2,3),(4,5),(4,6),(5,6)\\
        & \partial_2(e_{ij})  & \longmapsto & 0 \\
        &                     &\mbox{for }& (i,j) \ne (1,2),(1,3),(2,3),(4,5),(4,6),(5,6)\\
        & \partial_5(e_{23456}) & \longmapsto & x_1^2x_2x_4^2 \\
        & \partial_5(e_{12356}) & \longmapsto & x_1^3x_2x_4 \\
        & \partial_5(e_{ijklm}) & \longmapsto & 0\quad \mbox{otherwise} \\
\end{array}
\]
and the ideal $I$ is the same as that in Example~\ref{example1}.
We can also check that this sequence satisfies the numerical condition in Theorem~\ref{theorem:numerical}
\end{Example}


\begin{Example}
By Corollary~\ref{cor:depthzero}, we do not have a long Bourbaki
sequence with an approximation module $E_1$ and a codimension~$3$
generalized CM ideal $I$. However, there are long Bourbaki sequences
with approximation modules $E_1\dirsum E_5(d)$ for $d\in\ZZ$, which 
is an application of Theorem~\ref{theorem:construction}.
%
Let $k=1$ and $n=6$. Then 
\begin{eqnarray*}
{\cal A} & =&  \langle
                   x_1e_1^* + \cdots + x_6 e_6^* 
               \rangle\\
{\cal B} &= & \langle B_{ij} = (-1)^ix_je_{[6]\backslash i}^*
- (-1)^jx_ie_{[6]\backslash j}^*  :
		  1\leq i < j \leq 6 
\rangle.
\end{eqnarray*}
Let $\varphi = (a,b)$ be
\begin{eqnarray*}
a & = & x_1^3e_1^* + x_1^2x_2 e_2^* + x_1^2x_3 e_3^* + 
        x_1^2x_4 e_4^* + x_1^2x_5 e_5^* + x_1^2x_6 e_6^* \\
b & = & -x_2^5 B_{56} + x_6^5 B_{23}\\
  & = & x_2^5x_6e^*_{12346} + x_2^5x_5 e^*_{12345}
         +x_3x_6^5 e^*_{13456} + x_2x_6^5 e^*_{12456} \\
\end{eqnarray*}
and set $\beta_i\in K_1\dirsum K_5(1)$ to be as follows:
\begin{eqnarray*}
\beta_1 & = & - x_6e_{12345}  + x_5 e_{12346} \\
\beta_2 & = &  x_6^5 e_3 - x_1^2 e_{13456} \\
\beta_3 & = &  x_6^5 e_2 - x_1^2 e_{12456} \\
\beta_4 & = &  x_2^4x_5 e_2 - x_1 e_{12345} \\
\beta_5 & = &  x_2^4x_6 e_2 - x_1^2 e_{12346} \\
\beta_6 & = & -x_6^4 e_{12346} + x_2^4 e_{12456} \\
\beta_7 & = & e_{23456} \\
\beta_8 & = & e_{12356}.
\end{eqnarray*}
Notice that $\beta_i \notin E_2\dirsum E_6(1)$ for all $i$, i.e., the
condition in Theorem~\ref{theorem:non-trivialcond} is satisfied.
Then we can check 
\begin{enumerate}
\item $\Ker (\varphi: K_1\dirsum K_5(1)\to S(-6) 
      = \langle \beta_1,\ldots, \beta_8\rangle + E_2\dirsum E_6(1)$ and 
     $\varphi$ is a degree $8$ homomorphism, and 
\item the diagram
\[
\begin{CD}
   @.    0                @.    0                  \\
@.       @VVV             @VVV                     \\
0  @>>>  \Ker Res(\beta)  @>>>  \Ker\beta   @>>> 0 \\
@.       @VVV                   @VVV           @.  \\
0  @>>>  F'               @>>>  G           @.     \\
@.       @V{Res(\beta)}VV       @V{\beta}VV    @.  \\
0 @>>>  \langle\beta_1,\ldots,\beta_8\rangle\cap E_2\dirsum E_6(1)
                          @>>>  \langle\beta_1,\ldots,\beta_8\rangle \\
@.        @VVV                  @VVV           @.   \\
    @.    0               @.    0           @.      \\
\end{CD}
\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\[
%\begin{array}{ccccc}
%     & 0                       &      &  0                       &
%\\      
%     & \da                     &      &  \da                     &
%\\
%0\To & \Ker Res(\beta)         & \To  & \Ker\beta                &\To 0
%\\    
%     & \da                     &      &  \da                     &
%\\
%0\To & F' & \To  &  G  & 
%\\
%     & \hspace*{1.4cm}\da Res(\beta)         
%                               &      & \hspace*{0.5cm}\da \beta &
%\\
%0\To & \langle\beta_1,\ldots,
%       \beta_8\rangle\cap 
%       E_{2}\dirsum E_6(1)   & \To  & \langle\beta_1,\ldots,
%                                         \beta_8\rangle          & 
%\\
%     & \da                     &      &  \da                     &
%\\
%     &  0                      &      &   0                      &
%\end{array}
%\]
%%%%%%% For LaTeX %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where 
\begin{eqnarray*}
G & = & \langle m_1,\ldots, m_8\rangle
=S(-5)\dirsum S^4(-6)\dirsum S(-8)\dirsum S^2(-4) \\
F' & = & \left\langle 
\begin{array}{l}
-x_1^2 m_1 + x_6 m_4 - x_5 m_5,
                 x_2^4 m_3 - x_6^4 m_5 + x_1^2 m_6,\\
                -x_1^2m_1 -x_2 m_2 + x_3 m_3 - x_1^3 m_7 + x_1^2x_4 m_8
             \end{array}
         \right\rangle \\
\Ker\beta & = & \Ker Res(\beta) = \langle -x_1^2 m_1 + x_6 m_4 - x_5 m_5,
                 x_2^4 m_3 - x_6^4 m_5 + x_1^2 m_6
         \rangle \\
\end{eqnarray*}
\end{enumerate}
Thus we have a long Bourbaki sequence
\[
  0 \to F \namedTo{f} G \namedTo{g} E_1\dirsum E_5(1) \namedTo{\phi} I(2) \to 0
\]
where $g(m_i) = \beta_i$, $i=1,\ldots, 8$, and $F = S(-10)\dirsum S^2(-7)
= \langle u, v, w\rangle$ with 
$f(u) = x_2^4 m_3 - x_6^4 m_4 + x_1^2 m_6$,
$f(v) = -x_1^2 m_1 + x_6 m_4 - x_5m_5$
and 
$f(w) = -x_1^2 m_1 - x_2 m_2
+ x_3 m_3 - x_1^3 m_7 + x_1^2x_4 m_8$. The map $\phi$ is as follows:
$\phi(x_i) = x_ix_1^2$ $(i=1,\ldots, 6)$, $\phi(\partial_5(e_{12345})) = x_2^5x_5$,
$\phi(\partial_5(e_{12346})) = x_2^5x_6$, $\phi(\partial_5(e_{12356})) = 0$,
$\phi(\partial_5(e_{12456})) = x_2x_6^5$, $\phi(\partial_5(e_{13456})) 
   = x_3x_6^5$, and $\phi(\partial_5(e_{23456})) = 0$. 
The ideal is $I = \Im\varphi = x_1^2\mm + (x_2^5x_6, x_2^5x_5, x_3x_6^5, x_2x_6^5)$.
Finally we can check that this Bourbaki sequence satisfies the numerical
condition of Theorem~\ref{theorem:numerical}.
\end{Example}



 

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\bibitem{F} H.\ Flenner, Die S\"atze von Bertini f\"ur lokale Ringe, Math.
Ann. {\bf 299}, 97--111, 1977.

%\bibitem{Sch} P.\ Schenzel, Dualisierende Komplexe in der lokalen Algebra
%und Buchsbaum Ringe, LNM~907, Springer-Varlag, 1982.

%\bibitem{SV}J.\ St\"uckrad and W.\ Vogel, Buchsbaum rings and applications,
%Springer-Verlag, 1986.

%\bibitem{Rot} J.\  Rotman, {\em An introduction to homological algebra},
%Academic Press, 1979.
\end{thebibliography}

\end{document}


\newpage
\appendix
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Selfduality of Koszul Complex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The selfduality map is given by
\[
\begin{array}{cccc}
 \omega_i : & K_i &  \To        &  K_{n-i}^*
\\
            & e_I & \longmapsto & \sigma(I,\bar{I}) e^*_{\bar{I}}
\end{array}
\]
where $\sharp(I) =i$ and $\bar{I} = [n]\backslash I$.
Then we have the following diagram:
\[
\begin{array}{ccccc}
   K_{i-1}^*      &           & \namedTo{\partial_i^*}    &                        & K_i^*
\\
                  & \hspace*{1cm}\searrow  
                              &                           & \nearrow\partial_i^*   &
\\
                  &           &  E_i^*                    &                        &
\\
\hspace*{-1cm}
\omega_{n-i+1}\ua &           &   ||                      &                        & \hspace*{0.7cm}\ua\omega_{n-i}
\\
                  &           &  E_{n-i+1}                &                        &
\\
                  & \partial_{n-i+1}
                    \nearrow 
                              &                           & \searrow               &
\\
   K_{n-i+1}      &           & \namedTo{\partial_{n-i+1}}&                     & K_{n-i}
\end{array}
\]
Then by diagram chasing we have 
\[
\begin{array}{cccl}
                   &                           &            & \sum_{k=1}^{n-i+1}
                                                                   (-1)^{k+1}x_{j_k}
								   \sigma(I\backslash j_k,
									  \bar{I}\cup\{j_k\})
								    e^*_{\bar{I}\cup\{j_k\}}
          
\\
                   &                           & \nearrow\partial_i^*
                                                             &
\\
                   &\{\mbox{element of }E_i^*\}&            &
\\
\\
                   &   \ua                     &            & \hspace*{0.7cm}\ua\omega_{n-i}
\\
\\
                   & \sum_{k=1}^{n-i+1}
                        (-1)^{k+1}x_{j_k}
                          e_{I\backslash j_k}   &            &
\\
          \partial_{n-i+1}
           \nearrow &                           & \searrow   &
\\
           e_I     & \namedTo{\partial_{n-i+1}}&            & \sum_{k=1}^{n-i+1}
                                                                          (-1)^{k+1}x_{j_k}
                                                                            e_{I\backslash j_k} 
\end{array}
\]
Since $E_i^* \namedTo{\partial_i^*} K_i^*$ is injective, an element of $E_i^*$ 
can be described by 
$\sum_{k=1}^{n-i+1}(-1)^{k+1}x_{j_k}\sigma(I\backslash j_k,\bar{I}\cup\{j_k\})e^*_{\bar{I}\cup\{j_k\}}$.
Thus we obtain
\[
E_i^* 
= \langle
\sum_{k=1}^{n-i+1}(-1)^{k+1}x_{j_k}\sigma(I\backslash j_k,\bar{I}\cup\{j_k\})e^*_{\bar{I}\cup\{j_k\}}
\mid I \subset [n], \sharp(I)= n-i+1
  \rangle
\]
          

\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A complete calculation of $\rank E_t$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Lemma} 
%\label{lemma:wellknown}
For the $t$-th syzygy module of $K$ over $S$, we 
have 
\[
\rank E_t =\binom{n-1}{t-1}
\]
\end{Lemma}
\begin{proof}
Let $K_i$ be the $i$th Koszul complex. Then we have
an exact sequence
\[
  0\To K_n\namedTo{\partial_n} K_{n-1}\namedTo{\partial_{n-1}}\cdots\namedTo{\partial_t} K_t \To E_t \To 0 
\]
so that 
\[
	\rank(E_t) =\sum_{i=t}^{n}(-1)^{i-t}\rank(K_i)
		   = (-1)^t\sum_{i=t}^{n}(-1)^i\binom{n}{i} 
= (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}.
\]
For the last equation, we used the binomial coefficient theorem. Namely, we have
\[
	0 = (1-1)^n =\sum_{i=1}^{n}(-1)^i\binom{n}{i}
		    = \sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
	            + \sum_{i=t}^n(-1)^i\binom{n}{i}.
\]
Now set $\alpha(n,t) := \rank(E_t)$. 
We compute
\begin{eqnarray*}
\lefteqn{\alpha(n,t) - \alpha(n-1,t)} \\
     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
	   - (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i} \\
     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
          - (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n}{i}
          +  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i-1}\\
     &   & \mbox{by }
		\binom{n}{i} = \binom{n-1}{i} + \binom{n-1}{i-1} \\
     & = &  (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i-1}
       =  (-1)^{t-1}\sum_{i=1}^{t-1}(-1)^i\binom{n-1}{i-1} \\
     & = & (-1)^{t-1}\sum_{j=0}^{t-2}(-1)^{j+1}\binom{n-1}{j} 
      =    (-1)^{t-2}\sum_{j=0}^{t-2}(-1)^{j}\binom{n-1}{j} 
      =  \alpha(n-1, t-1)\\
     &   & \mbox{by changing the index $j = i -1$}\\
\end{eqnarray*}
Thus we have 
\begin{eqnarray*}
\alpha(n,t) &= &\alpha(n-1,t) + \alpha(n-1, t-1) \\
            & =& (-1)^{t-1}\sum_{i=0}^{t-1}(-1)^i\binom{n-1}{i}
              + (-1)^{t-2}\sum_{i=0}^{t-2}(-1)^i\binom{n-1}{i} 
	     = \binom{n-1}{t-1}.
\end{eqnarray*}
as required.
\end{proof}

\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A Complete Proof of Lemma~\ref{closedF}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Lemma}
\[
(-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2 
= \binom{n-1}{t}(t+1)^2 - \binom{n-2}{t}(2t+1) - 2\binom{n-3}{t-1}
\]
\end{Lemma}
\begin{proof}
 Let $A = (-1)^{t+1}\sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2$.
By the binomial coefficient theorem, we have 
\begin{eqnarray*}
0& = & \sum_{i=2}^{n}(-1)^i\binom{n}{i}i(i-1)  \\
 & = &  \sum_{i=2}^{n}(-1)^i\binom{n}{i}i^2
      - \sum_{i=2}^{n}(-1)^i\binom{n}{i}i \\
 & = &   \sum_{i=t+1}^n(-1)^i\binom{n}{i}i^2
       + \sum_{i=2}^{t}(-1)^i\binom{n}{i}i^2
       - \sum_{i=2}^{n}(-1)^i\binom{n}{i}i \\
 & = & (-1)^{t+1}A 
    + \sum_{i=2}^{t}(-1)^i\binom{n}{i}i(i-1)
    - \sum_{i=t+1}^{n}(-1)^i\binom{n}{i}i \\
 & = & (-1)^{t+1}A 
    + \sum_{i=2}^{t}(-1)^i\binom{n}{i}i(i-1)
    -(-1)^{t+1}\left[ \binom{n-2}{t-1} + \binom{n-1}{t}t\right]
\end{eqnarray*}
The last equation is from (\ref{usefulequation}). Thus we have
\begin{equation}
\label{intermedAns1}
A =  \binom{n-2}{t-1} + \binom{n-1}{t}t
    + (-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n}{i}i(i-1)
\end{equation}
Now we will compute the last term.
Set
\[
\alpha(n,t) = (-1)^{t}\sum_{i = 2}^{t}(-1)^i\binom{n}{i}i(i-1).
\]
and we compute
\begin{eqnarray*}
\lefteqn{\alpha(n,t) + \alpha(n-1,t-1)}\\
&= & t(t-1)\binom{n}{t} -
(-1)^{t}\sum_{i=1}^{t-2}(-1)^i\binom{n-1}{i}i(i+1)\\
&= & t(t-1)\binom{n}{t} \\
&  & -
    (-1)^{t}
	\left[
	\begin{array}{l}
	\displaystyle{\sum_{i=2}^{t}}(-1)^i\binom{n-1}{i}i(i+1)
        - 2(n-1) - (-1)^{t-1}t(t-1)\binom{n-1}{t-1}\\
                 - (-1)^{t}t(t+1)\binom{n-1}{t}
	\end{array}
 	\right]\\
&= & t(t-1)\binom{n}{t} \\
&  &-
    (-1)^{t}
	\left[
	\begin{array}{l}
	\displaystyle{\sum_{i=2}^{t}}(-1)^i\binom{n-1}{i}i(i-1)
        + 2\displaystyle{\sum_{i=2}^{t}}(-1)^{i}\binom{n-1}{i}i    \\
        - 2(n-1) - (-1)^{t-1}t(t-1)\binom{n-1}{t-1}
                 - (-1)^{t}t(t+1)\binom{n-1}{t}
	\end{array}
 	\right]\\
&= & t(t-1)\binom{n}{t} -
	\alpha(n-1, t)
      -2(-1)^{t}\sum_{i=2}^{t}(-1)^{i}\binom{n-1}{i}i
      + 2(-1)^{t}(n-1) \\
&  &
      - t(t-1)\binom{n-1}{t-1}
      + t(t+1)\binom{n-1}{t}
\\
&= & 2t^2\binom{n-1}{t}  -
2(-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n-1}{i}i
     + 2(-1)^{t}(n-1) - \alpha(n-1, t).
\end{eqnarray*}
Also we have
\[
	\alpha(n-1, t-1) + \alpha(n-1, t)
	= t(t-1)\binom{n-1}{t}
\]
Thus by using (\ref{usefulequation})
\begin{eqnarray*}
\alpha(n,t)
& = & t(t+1)\binom{n-1}{t} -
2(-1)^{t}\sum_{i=2}^{t}(-1)^i\binom{n-1}{i}i
	+  2(-1)^{t}(n-1) \\
& = & t(t+1)\binom{n-1}{t} - 2
	\left[
		t\binom{n-2}{t} + \binom{n-3}{t-1}
              + (-1)^{t}(n-1)
	\right]\\
&   &
	+  2(-1)^{t}(n-1) \\
& = & t(t+1)\binom{n-1}{t} - 2t\binom{n-2}{t} -
2\binom{n-3}{t-1}\\
%& = & t(t-1)\binom{n-1}{t} + 2t\binom{n-2}{t-1} - 2\binom{n-3}{t-1}
\end{eqnarray*}
Substiting this into  (\ref{intermedAns1}), we obtain the desired result.
\end{proof}

 
%\newpage
%\section{Note on non torsion-free module}
%
%\begin{Lemma}
%If a $S$-module $M$ is not torsion-free, $\Hom_S(M,S)$ has no injective homomorphism. 
%n\end{Lemma}
%\begin{proof}
%If $M$ has torsion, there exist non-zero $r\in S$ and $m\in M$ such
%that $rm=0$. Then, for arbitrary $f\in \Hom_S(M,S)$, we have $0 =
%f(rm) = r f(m)$ in $S$. Since $S$ is an integral domain, we must have
%$f(m)=0$. Thus $f$ is not injective.
%\end{proof}


\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Dumped ``Main Theorem''}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Theorem}
\label{theorem:main}
Let $p\geq 1$, $n\geq 4$ and $0\leq t \leq n-4$ be integers
and set $q = p + n-2 +\binom{n-1}{t}$. 
Assume that 
we have homogeneous $\beta_1,\ldots, \beta_q \in K_{t+1}\dirsum K_{n-1}$
and a homomorphism
$\iota : \Dirsum_{i=1}^q S(-b_i)\to 
\langle \beta_1,\ldots, \beta_q\rangle$ such that $\iota(m_i) = \beta_i$
$(i=1,\ldots, q)$ where $\{m_i\}_i$ is a free basis.
If there exists $(a,b) \in  E_{t+1}^*\dirsum E_{n-1}^*$ 
satisfying the following conditions:
\begin{itemize}
\item [$(i)$] $\Ker \{ (a,b) : E_{t+1}\dirsum E_{n-1}\to S\}
               \iso \langle \bar\partial\beta_1,\ldots, 
                            \bar\partial\beta_q
                    \rangle$
     where $\bar\partial = \partial_{t+1}\dirsum\partial_{n-1}$

\item [$(ii)$] for the homomorphism
$\rho : K_{t+2}\dirsum K_n\dirsum \Dirsum_{i=1}^{q}S(-b_i) \to K_{t+1}\dirsum K_{n-1}$
such that  $\rho(x,y,m) =  \partial_{t+2}(x) + \partial_n(y) - \iota(m)$,
we have
$\Ker\rho = E_{t+3} \dirsum N$
where $N$ is a free submodule of
$K_{t+2}\dirsum K_n\dirsum \Dirsum_{i=1}^{q}S(-b_i)$
with $\rank F = p$.

\item [$(iii)$]
Let $N\iso \Dirsum_{j=1}^pS(-a_j)$
and  $b_i = \deg \beta_i$ $(i=1,\ldots, q)$. 
Then $a_j, b_j, c, n, t$ satisfy
the numerical conditions in $Theorem~\ref{theorem:numerical}$.

\item [$(iv)$] there exists $0\leq i\leq q$ such that 
	$\beta_i\notin K_{t+1}$ and $\beta_i\notin K_{n-1}$\footnote{I think that 
this condition is not correct. The right one is that for any set $\{\beta_j\}$
of generators there eists an index $i$ such as in $(iv)$. This is crusical when
we consider the case of $t=0$ after the proof of this theorem.}

%\item [$(4)$]
%for the homomorphism $\iota : \Dirsum_{i=1}^q S(-b_i)\to \langle
%\beta_1,\ldots, \beta_q\rangle$ such that $\iota(m_i) = \beta_i$
%$(i=1,\ldots, q)$ where $\{m_i\}_i$ is a free basis, $F:=\Ker
%(\partial_{t+1}\dirsum\partial_{n-1})\circ\iota$ is a $S$-free module
%of rank $p$.
\end{itemize}
then there exists a Bourbaki sequence
\[
 0 \To F \To G \To E_1\dirsum E_{n-1} \To I_{(a,b)}(c)\To 0
\]
for some $S$-free modules $F$ and $G$ and $c\in\ZZ$,
such that $\codim I_{(a,b)} = 3$ and $t = \depth S/I_{(a,b)}$. 
Also $I_{(a,b)}$ does not
have a Bourbaki sequence in the form of 
\[
   0 \To F\To G\To E_{n-1} \To I_{(a,b)}(c) \To 0.
\]
\end{Theorem}

\begin{proof}
We will construct a $S$-free resolution of $I_{(a,b)}(c)$ and 
a Bourbaki sequence will be obtained from the resolution.
First of all, 
let $\varphi = (a,b) : E_{t+1}\dirsum E_{n-1}\to S$ and 
denote $I_{(a,b)}(c)$ simply by $I(c)$. Then, we obtain 
the following diagram by the condition (i):
\begin{equation}
\label{equation:Bourbaki}
\begin{array}{cccccccc}
  &     &  0
  &     &  0
  &     & 
  & \\
%
%
  &     & \uparrow
  &     & \uparrow
  &     & 
  & \\
%
0 & \to & \langle\bdel\beta_1,\ldots,\bdel\beta_q\rangle 
  & \to & E_{t+1}\dirsum E_{n-1} 
  & \namedTo{\varphi} & I(c) 
  & \to 0 \\
%
  &     & \hspace*{0.3cm}\uparrow\bdel
  &     & \hspace*{0.3cm}\uparrow\bdel
  &     & ||
  & \\
%
0 & \to & E_{t+2}\dirsum E_{n} + \langle\beta_1,\ldots, \beta_q\rangle
  & \to & K_{t+1}\dirsum K_{n-1} 
  & \namedTo{\varphi\circ\bdel} & I(c) 
  & \to 0 \\
%
\end{array}
\end{equation}
and the lower row is the part of a resolution of $I(c)$.

Now let $\langle m_1,\ldots, m_q\rangle$ be a 
free basis of $\Dirsum_{i=1}^qS(-b_i)$ and 
consider the surjective homomorphism
\[
\rho :   K_{t+2}\dirsum K_n \dirsum \Dirsum_{i=1}^qS(-b_i)
         \To
         E_{t+2}\dirsum E_n + \langle \beta_1,\ldots, \beta_q\rangle
\]
such that $\rho(x,y,m_i) = (\partial_{t+2}(x), \partial_n(y)) - \beta_i$
for $x\in K_{t+2}$, $y\in K_n$ and $i=1,\ldots, q$. 
Then $\Ker\rho =E_{t+3} \dirsum N$, with $N\iso \Dirsum_{i=1}^pS(-a_i)$ for some 
$a_i\in\ZZ$, by the condition (ii). Then we have a (minimal) free resolution
\[
0\to K_n\namedTo{\partial_n}\cdots\namedTo{\partial_{t+5}}
      K_{t+4}\namedTo{\partial_{t+4}} K_{t+3}\dirsum\Dirsum_{i=1}^pS(-a_i)
  \namedTo{\partial_{t+3}\dirsum\alpha}
                               E_{t+3}\dirsum N \to 0
\]
where $\alpha : \Dirsum_{i=1}^pS(-a_i) \to N$ gives an isomorphism,
and we obtain a free resolution of $I(c)$.

Now we construct a Bourbaki sequence. By (\ref{equation:Bourbaki}), we have
to show that there exists a free resolution of 
$\langle \bdel\beta_1,\ldots, \bdel\beta_q\rangle$ in the following form:
\[
  0\To \Dirsum_{i=1}^pS(-a_i)\To \Dirsum_{i=1}^qS(-b_i)
    \To \langle \bdel\beta_1,\ldots, \bdel\beta_q\rangle
    \To 0
\]
Consider the following diagram:
\begin{equation}
\label{equation:Bourbaki2}
\begin{array}{cccccccc}
  &     &  
  &     &  0
  &     &  0
  & \\
%
%
  &     & 
  &     & \uparrow
  &     & \uparrow
  & \\
%
0 & \to & \Ker\bar\rho
  & \to & \Dirsum_{i=1}^qS(-b_i)
  & \namedTo{\bar\rho} & \langle\bdel\beta_1,\ldots,\bdel\beta_q\rangle 
  & \to 0 \\
%
  &     & 
  &     & \hspace*{1cm}\uparrow\proj
  &     & \hspace*{0.3cm}\uparrow\bdel
  & \\
%
0 & \to & E_{t+3}\dirsum N
  & \to & K_{t+2}\dirsum K_{n}\dirsum\Dirsum_{i=1}^qS(-b_i) 
  & \namedTo{\rho} & E_{t+2}\dirsum E_{n} + \langle\beta_1,\ldots, \beta_q\rangle
  & \to 0 \\
%
\end{array}
\end{equation}
where the lower row is a part of the free resolution of $I(c)$, $\proj$ is the 
projection to the third component and for $m\in \Dirsum_{i=1}^qS(-b_i)$ we define
$\bar\rho(m) = \bdel(\rho(m))$.

Let $\{u_i\}_{i=1,\ldots, p}$ be a free basis of $N$ and set
$u_i = (u_i^{(1)}, u_i^{(2)}, u'_i)$ with $u_i^{(1)}\in K_{t+2}$,
$u_i^{(2)}\in K_n$ and $u'_i \in \Dirsum_{i=1}^qS(-b_i)$ for all $i$.
Then $\proj(E_{t+3}\dirsum N) = \langle u'_1,\ldots, u'_p\rangle 
\subset \Ker\bar\rho$ since we have $\bar\rho(\proj(u_i)) = \bdel(\rho(u_i)) = 0$
for $i=1,\ldots, p$. On the other hand, we have
\begin{eqnarray*}
\Ker\bar\rho 
&=& \{ m=\sum_{i=1}^qh_im_i \in \Dirsum_{i=1}^qS(-b_i) \mid \bdel(\rho(m)) = 0\}
\\
&=& \{ \sum_{i=1}^qh_im_i \in \Dirsum_{i=1}^qS(-b_i) 
\mid (\partial_{t+1}\dirsum\partial_{n-1})(\sum_{i=1}^qh_i\beta_i)=0\}
\\
&=& \{ \sum_{i=1}^qh_im_i \in \Dirsum_{i=1}^qS(-b_i) 
        \mid \sum_{i=1}^qh_i\beta_i\in E_{t+2}\dirsum E_n\} 
\\
&=& \left\{ \sum_{i=1}^qh_im_i \in \Dirsum_{i=1}^qS(-b_i) 
        \mid \begin{array}{l}
\partial_{t+1}(x) + \partial_{n}(y) - \sum_{i=1}^qh_i\beta_i
= 0\\
 \mbox{ for some $x\in K_{t+2}$ and $y\in K_n$}
\end{array}
\right\}
\\
& \subset & \langle u'_1,\ldots, u'_p\rangle 
\end{eqnarray*}
Thus obtain the isomorphism $\Ker\bar\rho\iso \langle u'_1,\ldots,
u'_p\rangle$.  Also we obtain the exact sequence $E_{t+3}\dirsum N
\namedTo{\proj} \Ker\bar\rho \to 0$ and we know $\Ker\proj = E_{t+3}$
from the diagram (\ref{equation:Bourbaki2}).  Thus $\Ker\bar\rho\iso N
\iso \Dirsum_{i=1}^pS(-a_i)$ as required.

Finally, by Theoerm~\ref{theorem:main1} and
Theorem~\ref{theorem:non-trivialcond}, the condition (iii) and (iv)
assure that $\codim I = 3$.
\end{proof}


\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Notes on Theorem~\ref{theorem:main}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{itemize}
\item  Given a Bourbaki sequence
\[
  0 \To F\namedTo{f} G \namedTo{g} E_{t+1}\dirsum E_{n-1}\To I(c)\To 0
\]
and consider the diagram
\[
\cdots\to K_{t+4}\to K_{t+3}\dirsum F \namedTo{\xi}
     K_{t+2}\dirsum K_n\dirsum G \namedTo{\rho} K_{t+1}\dirsum K_{n-1}.
\]
where $\xi(z,u) = (\partial_{t+3}(z), 0, f(u)) - \alpha(u)$.
It is straightforward to check that $\rho\circ \xi = 0$. 
Since $f$ is injective and $\alpha(u)\in K_{t+2}\dirsum K_n$, $\notin G$,
we know that $\Ker\xi =E_{t+4}$. Thus in order for the above 
diagram to be exact we have only to consider a condition for 
$\Ker\rho = \Im\xi$. This implies that we have to consider the chan map
$\alpha$.

\item Consider the exact sequence
\[
0 \To \Im\rho \To K_{t+1}\dirsum K_{n-1}
\namedTo{\varphi\circ (\partial\dirsum\partial)}
I(c) \To 0
\]
where $\Im\rho = E_{t+2}\dirsum E_{n}+\beta(G)$. Then 
we have 
\[
I(c) \iso K_{t+1}\dirsum K_{n-1}/E_{t+2}\dirsum E_n + \beta(G)
\]
Since $E_{t+1}\iso K_{t+1}/E_{t+2}$ and 
$E_{n-1}\iso K_{n-1}/E_n$, we have
\[
I(c) \iso E_{t+1}\dirsum E_{n-1}/\tilde\beta(G)
\]
where $\tilde\beta(G)$ is naturally induced from $\beta(G)$.
On the other hand, $I(c) \iso \varphi(E_{t+1}\dirsum E_{n-1})$ 
by the fist exact sequence. Comparing these two isomorphisms
on $I(c)$ we know that $\varphi$ is essentially determined 
by the chain map $\beta$.

\item The condition $(i)$ is by Theorem~\ref{theorem:non-trivialcond}.
This implies that for arbitrary
non-trivial $(a,b) \in I(c)^*$ we must have $a\ne 0$. This means that
for some $f\in S$ we have $f\mm \subset I$.

\item The condition $(ii)$ is by Section~\ref{section:dualofideal}

\item It is easy to know that a minimal free resolution of $f\mm$ is
isomorphic to $K_\bullet(-d)$ where $K_\bullet$ is the Koszul complex
of sequence $x_1,\ldots, x_n$ and $d = \deg(f)$. Also it is clear that
$\codim f\mm = 1$. Thus $S/f\mm$ is a generalized CM ring of
codimension $1$ and depth $0$.

\item One sufficient condition for $(iii)$ is to postulate that 
there is no relation between $\{ e_{ij}\mapsto \partial(e_{ij})\}$
and $\{m_k \mapsto \beta(m_k)\}$. Precisely, we assume
$\{(a,0,m) \in K_2\dirsum K_n\dirsum G \mid \partial_2(a) =\beta(m)\}$
to  be trivial $\{(0,0,0)\}$. For example, if, for arbitrary $x \in \Im\beta$,
we have $x= (A,B) \in K_1\dirsum K_{n-1}$ with $B\ne 0$, this is satisfied.

\item Perhaps, we may have to restrict ourselves to the case of $c=0$.
One reason for this thought is that if we compute $I(c)^*$, for example
$I(4)^* = \langle x_1^2 e_1^* + \cdots, x_1x_2^3e_{134}^* + \cdots\rangle$ 
as in Example~\ref{badexample}, we do not know explicitly what 
$I^* = \langle x_1^2 e_1^* + \cdots, x_1x_2^3e_{134}^* + \cdots\rangle(4)$ 
is. Thus we cannot explicitly compute generators of $I$.

\item The observation after Lemma~\ref{lemma:dualOfI(c)} shows that 
the minimal free resolution of $I$ is contained in the mapping cone of 
(\ref{mcone}) if and only if $f = (a,b) \in I(c)^*$ is injective.
An example of non-injective $(a,b)$ is given in Example~\ref{badexample}.

\item Perhaps we need an additional condition that $grad(J) =2$.

\item Can this ideal be obtained as a hypersurface section of 
a codimension~2 generalized CM ideal?

\item Maybe the minimal free resolution of $I(c)$ does not contain 
the $0\to K_n\to E_n\to 0$ part. Thus it is a part of 
\[
   \cdots \to K_2\dirsum S \to K_1\dirsum E_{n-1} \to I(c) \to 0
\]
since $E_{n-1} \iso K_{n-1}/E_n$.

\item The hypothesis of the theorem only gives a map
$\beta : K_2\dirsum K_{n}\dirsum G \to K_1\dirsum K_{n-1}$. It may be 
sufficient to construct a free resolution of $I(c)$. But is it 
sufficient for recover the Bourbaki sequence?

\item Do we need to consider the vanishing condition of local cohomology?
We obtain a $S$-free resolution of $I$ as a mapping cone. Thus we can
compute $\Coh{i}{S/I}^\vee \iso \Ext^{n-i}_S(S/I, S(-n))$. By Grothendieck's
vanishing theorem, we must have $\Hom_S(S/I, S(-n)) = \Ext_S^1(S/I, S(-n))
= \Ext_S^2(S/I, S(-n)) = 0$. This is equivalent to 
$\Ker \eta^* = 0$, $\Ker\rho^* \subset \Im\eta^*$ and 
$\Ker\zeta^* \subset \Im \rho^*$ where 
$\eta = \varphi\circ(\partial_1\dirsum \partial_{n-1})$ since 
$\codim I =3$.  This condition does not seem to be easy to check. 
The first two depends on how we define the mysterious map $\varphi$.
The last condition can be described explicitly as 
{\sl ``If $(l,p,q)\in K_2(-c)^*\dirsum K_n(-c)^*\dirsum G(-c)^*$
satisfies $(l,p,q)\circ \alpha = l\circ \partial_3 + q\circ f$,
then there exists $(u,v)\in K_1(-c)^*\dirsum K_{n-1}(-c)^*$ such 
that $(l,p,q) =u\circ \partial_2 + v\circ\partial_n - (u,v)\circ \beta$''}.
But it would be hard to check and we need even consider the chain map $\alpha$ 
and the map $f : F \to G$. On the other hand,
we already consider a numerial condition based on
Hilbert polynomial.  This seems to be enouhg. Is this right?
\end{itemize}

\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Dumped subsection on $\Hom_S(I(c),S)$ \label{section:dualofideal}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $N$ be a finitely generated graded $S$-module of rank one.  If $N$
is torsion-free, then there exists a graded ideal $I \subset S$ and
$c\in \ZZ$ such that $N\iso I(c)$. Thus there exists an injection
$i\in \Hom_S(I(c), S)$.
Conversely, if there exists an injection $f \in \Hom_S(N(-c), S)$
for some  $c\in\ZZ$, then $N(-c)$ is isomorphic to a graded ideal 
$I \subset S$ where $I = \Im f$. Consequently, we have 

\begin{Lemma}
Let $N$ be a finitely generated graded $S$-module. Assume that 
$\rank N =1$. Then, $N$ is torsion-free if and only if 
$\Hom_S(N,S)$ contains an injection.
\end{Lemma}


By taking the dual of the mapping cone (\ref{mappingcone}), we have an exact sequence
\[
\begin{array}{ccc}
       & E_{t+1}^*\dirsum E_{n-1}^* & \\
       & \uparrow & \\
 0 \To &I(c)^* & \To K^*_{t+1}\dirsum K^*_{n-1} \namedTo{\rho^*} K^*_{t+2}\dirsum K^*_n\dirsum G^*
\\
       & \uparrow &  \\
       &    0     &
\end{array}
\]
%
so that, since $\rho^*(a,b) = (\partial^*_{t+2}(a), \partial^*_n(b), -
\beta^*(a,b))$ for $(a,b)\in K^*_{t+1}\dirsum K^*_{n-1}$, we have by
using the self-duality of Koszul complex,
%
\begin{eqnarray*}
I(c)^* & \iso & \Ker\rho^*\\
       & = & \{ (a,b)\in K^*_{t+1}\dirsum K^*_{n-1} 
:  a\in \Ker\partial^*_{t+2}, b\in\Ker\partial^*_n, \beta^*(a,b) =0 \}
\\
        & = &\{(a,b)\in \partial_1^*(E^*_{t+1})\dirsum \partial_{n-1}^*(E^*_{n-1}) : (a,b)\circ \beta = 0\}
\end{eqnarray*}
%
By the selfduality of Koszul complex, we have $E_{t+1}^* \iso E_{n-t}$ and
$E_{n-1}^*\iso E_2$. Thus we can compute further to obtain
%
\begin{equation}
\label{dualofI}
I(c)^*  
     =  \left\{
       (a,b) :  \begin{array}{l}
a \in \langle \sum_{j=1}^{n-t}(-1)^{j+1}\sigma(I, ([n]\backslash I)\cup \{i_j\})
                             x_{i_j}e^*_{[n]-I\backslash i_j} 
		\mid I\subset [n], \sharp I =n-t \rangle,\\
b \in \langle (-1)^ix_ie^*_{[n]\backslash j}
               - (-1)^jx_je^*_{[n]\backslash i}\mid 1\leq i < j\leq n\rangle, \\
(a,b)\circ \beta = 0
\end{array}
         \right\}
\end{equation}
where we denote a $S$-module generated by $m_1,\ldots, m_p$ by
$\langle m_1,\ldots, m_p\rangle$, a basis for the $i$th Koszul module
$K_i$ by $e_I$ with $I\subset [n] = \{1,2,\ldots, n\}$ and $\sharp I =
i$, and the dual basis for $K_i^*$ by $e^*_I$.


By (\ref{mappingcone}), we have 
\[
   0\To \Im\rho \To
         K_{t+1}\dirsum K_{n-1} 
    \namedTo{\varphi\circ(\partial_{t+1}\dirsum\partial_{n-1})} I(c) \to 0
\]
also we have $I(c)^*\subset E_{t+1}^*\dirsum E_{n-1}^*\subset K_{t+1}^*\dirsum K_{n-1}^*$.
Thus we have 


\begin{Lemma}
\label{lemma:dualOfI(c)}
For arbitrary $f\in I(c)^*$, we have
\begin{enumerate}
\item [$(i)$] $\Im f = f(K_{t+1}\dirsum K_{n-1})$ for $f\in I(c)^*$, and 
\item [$(ii)$] $f$ is injective if and only if 
             $\{(x,y) \in K_{t+1}\dirsum K_{n-1} \mid f(x,y) = 0\} \subset \Im\rho$.
\end{enumerate}
\end{Lemma}





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