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\def\irr{\operatorname{irr}}
\def\Ass{\operatorname{Ass}}
\def\dim{\operatorname{dim}}
\def\hgt{\operatorname{ht}}
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\topmatter
\centerline{ {\bf{ON THE IRREDUCIBLE COMPONENTS OF AN IDEAL}}}
\vskip .35 in
\centerline{ William Heinzer}
\vskip .12 in
\centerline{ Department of Mathematics}
\centerline{ Purdue University}
\centerline{ W. Lafayette, IN 47907}
\centerline{ email: heinzer\@math.purdue.edu}
\vskip .2 in
\centerline{ L. J. Ratliff, Jr.}
\vskip .12 in
\centerline{ Department of Mathematics}
\centerline{ University of California}
\centerline{ Riverside, CA 92521}
\centerline{ email:  ratliff\@ucrmath.ucr.edu}
\vskip .2 in
\centerline{ Kishor Shah}
\vskip .12 in
\centerline{ Department of Mathematics}
\centerline{ Southwest Missouri State University}
\centerline{ Springfield, MO 65804}
\centerline{ email:  kis100f\@wpgate.smsu.edu}
\abstract
Let  $I$  be an  $M$ -primary ideal in a local ring  $(R,M)$
and let $\irr(I)$ denote
the set of irreducible components of  $I$, where  an ideal  $q$  is
an {\it irreducible component } of  $I$  if $q$  occurs as a
factor in some decomposition of  $I$  as an irredundant intersection
of irreducible ideals.  We give several
characterizations of the ideals in  $\irr(I)$ and  show that
if  $J$  is an ideal between  $I$  and an irreducible component of $I$,
then $J$ is the intersection of ideals in  $\irr(I)$. We also
exhibit examples showing that there may exist irreducible ideals
containing   $I$  that contain
no ideal in  $\irr(I)$. Also, we determine necessary and sufficient
conditions that the principal ideal $uR[u, tI]$ of the Rees ring
$R[u, tI]$ have a unique cover, and apply this to the study of the
form ring of $R$ with respect to  $I$.
\endabstract
\endtopmatter
\document

 \baselineskip 18pt \vsize = 8.5 true in {{\bf 1. INTRODUCTION}}.
{\it The following notation is fixed for this paper:} $(R,M)$  is a
local ring with identity  $1$ $\ne$ $0$, and $I$  is an open ($=$
$M$-primary) ideal in  $R$.  Our  terminology is generally the same
as that in [M], [N], and [ZS].

Irreducible ideals have interested us ever since
we learned that each ideal in a Noetherian ring is a finite
intersection of irreducible  ideals.
This is a classical result of Emmy Noether [No, Satz II, p. 33],
and is the first  of four different types of decomposition
considered by Noether in [No].
\footnote {An informative discussion of
Noether's work in [No] is presented by Robert Gilmer in [Gi].}
Noether's work stimulated
an important classical paper on irreducible ideals
by  Wolfgang Gr\"obner [Gr].
\footnote {Gr\"obner in his paper thanks E. Noether for direction and valuable advice.}
However, apparently since 1934 few papers have been
devoted to the study of irreducible ideals and the
decomposition of ideals as a finite intersection
of irreducible ideals. Our purpose here is to begin such a study.
(This study was partly suggested by our work in
[HRS1], [HRS2], and [HRS3], where
we discovered that irreducible ideals
are closely related to the maximal embedded components of an ideal.)
Our results in the present paper show that irreducible ideals have some
interesting and useful (and, perhaps, unexpected) properties.

In Section 2 we give several characterizations of the irreducible
components of  $I$, and then show that  $n_{irr}(I)+1$  is an upper
bound on the number  $n(I)$  of ideals in a decomposition of  $I$
as an irredundant intersection of irreducible ideals, where
$n_{irr}(I)$ $=$ $\min \{\ell(q/I)$; $q$  is an irreducible ideal in
$R$  that contains  $I\}$.

In Section 3 it is shown in (3.2) that each ideal  $J$ $\in$ $\bold
I (I)$ $=$ $\{J$; $I$ $\subseteq$ $J$ $\subseteq$ $q$  for some
ideal  $q$ $\in$ $\irr(I)\}$  is the intersection of the ideals in
$\irr(I)$  that contain  $J$.  Then we characterize the maximal
reducible ideals in  $\bold I (I)$  and also show that the ideals in
$\bold I (I)$  that are minimal with respect to properly containing
$I$  are the covers of  $I$.  Also, we briefly consider the concept
of an ideal  $J$  being irreducibly related to  $I$  (where $J$  is
irreducibly related to  $I$  in case  $J$  is a finite intersection
of ideals in  $\irr(I)$).

In Section 4 the ideal structure of the Artinian Gorenstein local
ring  $R$ $=$ $F[x,y]$ $=$ $F[X,Y]/(X^3,Y^3)$  (where  $F$  is
the field  $\{0,1\}$)  is considered.  The main result
shows that the ideal  $(x+y)R$  is an irreducible
ideal that contains  $x^2y^2R$, but contains no irreducible
component of  $x^2y^2R$.  A large part of this section
is devoted to describing how the computer program Macaulay \cite{BS}
was used in developing this example.

The main result in Section 5 gives a useful characterization of
when the principal ideal  $uR[u,tI]$  has a unique cover, and this
is then used to show that if  $R$  is an Artinian Gorenstein
local ring, then  $uR[u,tI]$  is irreducible if and only
if  $R[u,tI]$  is Gorenstein if and only if the form
ring  $\bold F (R,I)$  of  $R$  with respect to  $I$  is
Gorenstein.

Finally, in Section 6 we give three examples of rather ``bad''
behavior of irreducible ideals.  All three examples are in a regular
local ring  $R$  of altitude two.  The first shows that the set
$\bold S$ $=$ $\{I$; $n(I)$ $\le$ $n_{irr}(I)\}$  is nonempty.  The
next is an example of an infinite descending chain of open ideals
$I_1 \supset I_2 \supset \dots$  in  $R$  and an infinite set
$\bold Q$  of irreducible ideals  $q_i$  in  $R$  such that, for all
positive integers  $n$  and  $k$, $irr(I_{n+k}) \cap \bold Q$ $=$
$\{q_1 ,\dots,q_{n+k}\}$  and  $I_{n+k}$ $=$ $q_n \cap q_{n+k}$.
The final example shows that if  $k$ $<$ $m$  are positive integers,
then there exists an open ideal  $I$  in  $R$  such that  $n(I)$ $=$
$m$ and there exists an ideal  $J$ $\in$ $\bold I (I)$  such that
$J$  is the irredundant intersection of  $m+k$  ideals in
$\irr(I)$.



\bigskip
\bigskip
{\bf{2.  THE IDEALS IN  irr(I).}} In this section we give several
definitions that are needed in what follows, recall several facts
concerning irreducible ideals that have previously appeared in the
literature, then give several characterizations of the ideals in
$\irr(I)$, and then show that $n_{irr}(I)+1$  is an upper bound on
$n(I)$.

We begin with several definitions.

\bigskip
\noindent
{\bf{(2.1)  DEFINITION.}}
Let  $J$  be an ideal in a Noetherian ring  $A$.  Then:

\noindent
{\bf{(2.1.1)}}
$J$  is $\bold {reducible}$ in case there exist ideals  $K$  and  $L$  in
$A$  that properly contain  $J$  such that  $J$ $=$ $K \cap L$.  $J$  is
$\bold {irreducible}$  in case  $J$  is not reducible.
An ideal  $q$   is an $\bold {irreducible~ component}$
of  $J$  in case  $q$  appears as a factor in a decomposition
of  $J$  as an irredundant intersection of irreducible ideals (see (2.2.3)).

\noindent {\bf{(2.1.2)}} $\irr(J)$ $=$ $\{q$; $q$  is an irreducible
component of  $J\}$, and  $\bold I (J)$  denotes the set of all
intersections of the ideals in  $\irr(J)$  (excluding  $A$, the
empty intersection).

\noindent
{\bf{(2.1.3)}}
$n(J)$  denotes the number of ideals in
a decomposition of  $J$  as an irredundant intersection of
irreducible ideals (see (2.2.3) and (2.2.4)).

\noindent {\bf{(2.1.4)}} If  $A$  is local, then $n_{irr}(J)$ $=$
$\min \{\ell(q/J)$; $q$  is an irreducible ideal in  $A$  that
contains  $J\}$, where  $\ell(q/J)$  denotes the length of the  $A$
-module  $q/J$.

\noindent
{\bf{(2.1.5)}}
If  $A$  is local, then
$\bold S$ $=$ $\{J$; $n(J)$ $\le$
$n_{irr} (J)\}$.  (The letter  S  is an abbreviation for ``short'' - the
ideals  $J$  in  $\bold S$  have a shorter irreducible decomposition
than  $n_{irr}(J) + 1$  (see (2.5.3).)

\noindent
{\bf{(2.1.6)}}
An ideal  $K$  is a  $\bold{cover}$  of  $J$  in case  $J$ $\subset$ $K$  and
there exist no ideals between  $J$  and  $K$.  (In this case,
$K/J$ $\cong$ $A/N$  for some maximal ideal  $N$  in  $A$,
and it then follows that  $K$ $=$ $(J,b)A$  for
some  $b$ $\in$ $N$  and  $NK$ $\subseteq$ $J$.)  Also,
$K$  is
an  $\bold {irreducible}$ $\bold {cover}$  of  $J$  in
case  $K$  is an irreducible
ideal and a cover of  $J$.
\bigskip
Concerning (2.1.4), let  $\bold J$ $=$ $\{q$; $q$  is an irreducible
ideal in  $R$  that contains  $J\}$, so  $\irr(J)$ $\subseteq$
$\bold J$, so  $n_{irr}(J)$ $\le$ $\min \{\ell(q/J)$; $q$ $\in$
$\irr(J)\}$.  If  $J$  is an open ideal, then an interesting
question is whether this inequality is always an equality.  We show
in (4.1) that there may exist ideals that are minimal in  $\bold J$
that are not in  $\irr(J)$. This indicates that there is a
possibility that $n_{irr}(J)$ strictly less than $\min \{\ell(q/J)$;
$q$ $\in$ $\irr(J)\}$   may be achievable in an appropriate example.

A number of known results concerning irreducible ideals will be
frequently used below, so we briefly summarize them here.
\bigskip
\noindent
{\bf{(2.2)  REMARK.}}
Let  $I$  be an open ideal in a local ring  $(R,M)$.  Then:

\noindent
{\bf{(2.2.1)}}
[ZS, Theorem 34, p. 248]  $I$  is irreducible if
and only if  $I$  has a unique cover (and then
its unique cover is  $I:M$).

\noindent
{\bf{(2.2.2)}}
If  $I$, $J$, and  $q$  are open ideals in  $R$  such
that  $I$ $\nsubseteq$ $J$  and  $q$  is maximal with respect to
containing  $I$  and not containing  $J$, then  $q$  is irreducible.

\noindent
{\bf{(2.2.3)}}
[No, Satz II and Satz IV]  $I$  is a finite intersection of irreducible ideals,
and if the intersection is irredundant, then the number of such
ideals is the same for each such representation of  $I$.

\noindent {\bf{(2.2.4)}} [HRS2, (3.3.3)]  $n(I)$ $=$
$\dim_{R/M}(S(R/I))$, where  $S(R/I)$  is the $\bold{socle}$
$(0):(M/I)$  of  $R/I$; see [SV, p. 69].)

\noindent {\bf{(2.2.5)}} [HRS2, (3.2)]  If  $q$ $\in$ $\irr(I)$  and
$J$ $\nsubseteq$ $q$  is an ideal between  $I$  and  $I:M$, then
$(q/I) \cap (J/I)$  is a codimensional one subspace of the  $R/M$
vector space  $J/I$.

\noindent {\bf{(2.2.6)}} [HRS3, (3.2)]  There are no containment
relations among the ideals in  $\irr(I)$.

\demo{Proof}
The proofs of these, except for (2.2.2), are given
in the cited references.

For (2.2.2), it is clear that every ideal that contains  $q$
must contain  $J$, so  $q+J$  is the unique cover of  $q$, so
(2.2.2) follows from (2.2.1), \qed
\enddemo

\bigskip
In (2.3) we give two useful characterizations of the ideals in
$\irr(I)$.
\bigskip
\noindent
\proclaim{(2.3)  THEOREM}
Let  $I$  be an open ideal in a local ring  $(R,M)$.  Then the following
are equivalent for an ideal  $q$  in  $R$:

\noindent {\bf{(2.3.1)}} $q$ $\in$ $\irr(I)$.

\noindent
{\bf{(2.3.2)}}
$q$  is irreducible, $I$ $\subseteq$ $q$, and  $I:M$ $\nsubseteq$ $q$.

\noindent
{\bf{(2.3.3)}}
$q$  is an ideal that is maximal with respect
to:  (a)  containing some ideal  $J$  that
contains  $I$; and, (b)  not containing  $I:M$.

\endproclaim

\demo{Proof} It is shown in [HRS2, (3.4)] that an irreducible ideal
$q$  in  $R$  is in  $\irr(I)$  if and only if  $I$ $\subseteq$ $q$
and  $I:M$ $\nsubseteq$ $q$, so (2.3.1) $\Leftrightarrow$ (2.3.2).

If (2.3.2) holds, then  $I$ $\subseteq$ $q$, so  $I:M$ $\subseteq$
$q:M$  and  $q:M$  is the unique cover of  $q$, by (2.2.1).  Therefore
every ideal that properly contains  $q$  must contain  $I:M$, so  $q$  is
maximal with respect to (2.3.3)(a) (with  $J$ $=$ $I$) and
(2.3.3)(b), so (2.3.2) $\Rightarrow$ (2.3.3).

Finally, if (2.3.3) holds, then  $q$  is irreducible, by (2.2.2), so
since  $I$ $\subseteq$ $J$ $\subseteq$ $q$  and  $I:M$ $\nsubseteq$
$q$, [HRS2, (3.4)] shows that  $q$ $\in$ $\irr(I)$, hence (2.3.3)
$\Rightarrow$ (2.3.1), \qed
\enddemo
\bigskip
\noindent \proclaim{(2.4)  COROLLARY} $\Irr(I)$ $=$ $\{q$; $q$  is
an irreducible ideal in  $R$, $I$ $\subseteq$ $q$, and  $q\cap
(I:M)$  is covered by  $I:M\}$ $=$ $\{q$; $q$  is an ideal in  $R$
that is maximal with respect to: (a)  containing an ideal  $J$  that
contains  $I$; and, (b)  intersecting $I:M$  in an ideal that is
covered by  $I:M\}$.
\endproclaim

\demo{Proof}
This readily follows from (2.3) and (2.2.5), \qed
\enddemo

\bigskip
In (2.5) we note a relation between  $n(I)$  and  $n_{irr}(I)$  (see
(2.1.3) and (2.1.4)).
\bigskip
\proclaim{ (2.5)  PROPOSITION  (2.5.1)}
If  $I$  is irreducible,
then  $n(I)$ $=$ $1$  and  $n_{irr}(I)$ $=$ $0$, so  $I$ $\notin$ $\bold S$
(see (2.1.5)).

\noindent {\bf{(2.5.2)}} If  $I$  has an irreducible cover  $q$  and
if  $I$  is reducible, then  $q$ $\in$ $\irr(I)$, $n(I)$ $=$ $2$,
and  $n_{irr}(I)$ $=$ $1$, so  $I$ $\notin$ $\bold S$.

\noindent
{\bf{(2.5.3)}}
It is always true that  $n(I)$ $\le$ $n_{irr}(I) + 1$.
\endproclaim
\demo{Proof}
(2.5.1) is clear.

For (2.5.2) assume that  $I$  is reducible and that  $q$  is an
irreducible cover of  $I$, and let  $Q$  be an ideal in  $R$  that
is maximal with respect to containing  $I$  and not containing  $q$.
(Such an ideal  $Q$  exists, since  $I$  is reducible.)  Then $Q$
is irreducible, by (2.2.2), and  $I$ $=$ $q \cap Q$, hence  $q$, $Q$
$\in$ $\irr(I)$. Therefore it follows that  $n(I)$ $=$ $2$  and
$n_{irr}(I)$ $=$ $1$ $=$ $\ell(q/I)$, so  $I$ $\notin$ $\bold S$.

For (2.5.3), let  $n _{irr}(I)$ $=$ $k$,
let  $q$  be an irreducible ideal
in  $R$  such that  $I$ $\subseteq$ $q$  and  $\ell(q/I)$ $=$ $k$,
and let  $q$ $=$ $q_0 \supset \dots \supset q_k$ $=$ $I$  be
a (maximal) chain of ideals is  $R$  of length  $k$  between  $q$  and  $I$.
Then, for  $i$ $=$ $1,\dots,k$, $q_{i-1}$  covers  $q_i$, so
if  $q^{(i)}$  is an ideal in  $R$  that
is maximal with respect to containing  $q_i$  and not containing  $q_{i-1}$,
then (2.2.2) shows that  $q^{(i)}$  is irreducible,
and  $q^{(i)} \cap q_{i-1}$ $=$ $q_i$.  Therefore it follows
that  $I$ $=$ $q_0 \cap q^{(1)} \cap \dots \cap q^{(k)}$,
so  $n(I)$ $\le$ $k + 1$ $=$ $l_{irr}(I) + 1$, \qed
\enddemo
\bigskip
Note that no irreducible  $M$-primary ideal is in the set  $\bold S$
of (2.1.5), by (2.5.1), and a similar statement holds for each
reducible  $M$-primary ideal that has an irreducible cover, by
(2.5.2). So it is natural to wonder if either  $\bold S$  is empty
or if the following ``converse'' of (2.5.2) holds:  if  $I$  is the
irredundant intersection of two irreducible ideals, then $I$  has an
irreducible cover.  In (6.2.1) and (6.2.2) we give examples of when
this converse holds, but (6.2.3) shows that it does not hold in
general, so  $\bold S$  is not empty.
\bigskip
\bigskip
{\bf{3.  THE IDEALS IN  I(I).}} The main result in this section,
(3.1), shows that if  $I$ $\subseteq$ $J$, if  $K$  is a cover of
$J$, and if  $I:M$ $\nsubseteq$ $J$, then there exists  $q$ $\in$
$\irr(I)$  such that  $q \cap K$ $=$ $J$.  An immediate consequence
of this is (3.2), which characterizes the ideals in  $\bold I (I)$
by showing that these ideals are precisely the ideals between  $I$
and an arbitrary irreducible component of  $I$.  Then we briefly
consider some consequences of (3.2), including the relation of
irreducibly related (see (3.6)).
\bigskip
The proof of (3.1) is somewhat similar to the proof of (2.12) in [HRS3].
\bigskip
\proclaim{ (3.1)  THEOREM} Let  $J$  be an ideal in  $R$  such that
$I$ $\subseteq$ $J$  and  $I:M$ $\nsubseteq$ $J$, and let  $K$  be a
cover of  $J$.  Then there exists $q$ $\in$ $\irr(I)$  such that
$q\cap K$ $=$ $J$.
\endproclaim

\demo{Proof}
Note first that if  $I$  is irreducible, then  $J$ $=$ $I$  (since  $I:M$
$\nsubseteq$ $J$  and  $I:M$  is the unique cover of  $I$  (by (2.2.1))),
so it follows that  $K$ $=$ $I:M$  and we may take  $q$ $=$ $I$.

Therefore it may be assumed that  $I$  is reducible.  Then since
$I:M$ $\nsubseteq$ $J$, there exists an ideal  $q''$  in  $R$  that
is maximal with respect to containing  $J$  and not containing
$I:M$. Then $q''$ $\in$ $\irr(I)$, by (2.3.3) $\Rightarrow$ (2.3.1),
and since  $K$  covers  $J$  it follows that either $q''\cap K$ $=$
$J$  (as desired), or  $K$ $\subseteq$ $q''$.

Therefore it may be assumed that  $K$ $\subseteq$ $q''$.  Then
$I:M$ $\nsubseteq$ $K$, since  $I:M$ $\nsubseteq$ $q''$, so it
follows that  $J\cap(I:M)$ $\subseteq$ $K\cap(I:M)$ $\subset$ $I:M$.
If  $J\cap(I:M)$ $\subset$ $K\cap(I:M)$, then let  $t$ $\in$
$(K\cap(I:M)) - J$. Then if  $q'$  is an ideal in  $R$  that is
maximal with respect to containing  $J$  and not containing  $t$,
then it follows that  $I:M$ $\nsubseteq$ $q'$  (so  $q'$ $\in$
$\irr(I)$)  and  $K$ $\nsubseteq$ $q'$, so  $q' \cap K$ $=$ $J$, as
desired.

Therefore it may be assumed that  $J\cap(I:M)$ $=$ $K\cap(I:M)$.
Then since  $K$  covers  $J$, there exists  $x$
$\in$ $K-J$  such that  $xM$ $\subseteq$ $J$.  Also,
since  $K\cap(I:M)$  is properly contained in  $I:M$, there
exists  $y$ $\in$ $(I:M)-K$, so let  $K'$ $=$ $(J,x+y)R$.
Then  $(x+y)M$ $\subseteq$ $J$, so  $K'$  covers  $J$,
and  $K'$ $\ne$ $K$  (since  $x$ $\in$ $K$  and  $y$
$\notin$ $K$).

Suppose  $J\cap(I:M)$ $\subset$ $K'\cap(I:M)$
and let  $z$ $\in$ $(K'\cap(I:M)) - J$.  Then  $z$
$=$ $j+r(x+y)$  for
some  $j$ $\in$ $J$  and for some  $r$ $\in$ $R$.
Then  $r$  is a unit, since  $(x+y)M$ $\subseteq$ $J$  and  $z$
$\notin$ $J$.  Also, $z-ry$ $\in$ $I:M$  (since  $z$, $y$ $\in$
$I:M$)  and  $j+rx$ $\in$ $K$  (since  $j$ $\in$ $J$ $\subseteq$
$K$  and  $x$ $\in$ $K$), so  $z-ry$ $=$ $j+rx$ $\in$
$K\cap(I:M)$ $=$ $J\cap(I:M)$.  Therefore  $j+rx$ $\in$ $J$,
and  $j$ $\in$ $J$ and  $r$  is a unit, hence  $x$ $\in$ $J$,
and this contradicts the choice of  $x$.

Therefore it follows that  $J\cap(I:M)$ $=$ $K'\cap(I:M)$.
Therefore let  $q$  be an ideal in  $R$  that is maximal
with respect to containing  $K'$  and not containing  $y$  (where  $y$
$\in$ $(I:M)-K$).  Then  $I:M$ $\nsubseteq$ $q$  (since  $y$
$\in$ $(I:M)-q$), so  $q$ $\in$ $irr(I)$.  And  $K$
$\nsubseteq$ $q$, since  $(J,x+y)R$ $=$ $K'$ $\subseteq$
$q$  and  $y$ $\notin$ $q$  (so  $x$ $\notin$ $q$  and  $x$ $\in$ $K$).
Therefore, since  $K$  covers  $J$  it follows that  $q\cap K$ $=$ $J$, \qed
\enddemo

\bigskip
\proclaim{ (3.2)  COROLLARY} Assume that  $I$  is reducible, let $q$
$\in$ $irr(I)$, and let  $J$  be an ideal in  $R$  such that  $I$
$\subseteq$ $J$ $\subseteq$ $q$.  Then  $J$ $\in$ $\bold I(I)$. In
fact, $J$  is the (possibly redundant) intersection of $\ell(q/J)+1$
ideals in  $\irr(I)$. Therefore  $\bold I (I)$ $=$ $\{J$; $J$  is an
ideal in  $R$  such that  $I$ $\subseteq$ $J$ $\subseteq$ $q$  for
some  $q$ $\in$ $\irr(I)\}$ $=$ $\{J$; $J$  is a finite intersection
of ideals in  $\irr(I)\}$.
\endproclaim

\demo{Proof} Let  $\ell(q/J)$ $=$ $k$  and let  $q$ $=$ $q_0$
$\supset$ $q_1$ $\supset \dots \supset$ $q_k$ $=$ $J$  be a maximal
chain of ideals between  $q$  and  $J$.  Then  $I:M$ $\nsubseteq$
$q_0$, by (2.3.1) $\Rightarrow$ (2.3.2), so  $I:M$ $\nsubseteq$
$q_i$  for  $i$ $=$ $1,\dots,k$, and  $q_{i-1}$  covers  $q_i$, so
(3.1) shows that there exists  $q^{(i)}$ $\in$ $\irr(I)$  such that
$q^{(i)}\cap q_{i-1}$ $=$ $q_i$. Therefore it follows that  $J$ $=$
$q_0 \cap q^{(1)} \cdots \cap q^{(k)}$, hence  $J$  is the
intersection of  $\ell(q/J)+1$  ideals in  $\irr(I)$, so  $J$ $\in$
$\bold I (I)$.

Finally, since  $R/I$  has finite length, it follows that the ideals
in  $\bold I (I)$  are {\it{finite}} intersections of the ideals in
$\irr(I)$, so the final statement follows from what was shown in the
preceding paragraph, \qed
\enddemo
\bigskip
\proclaim{ (3.3)  COROLLARY} Assume that  $I$  is reducible and let
$m$ $=$ $\min(\{\ell(M/q)$; $q$ $\in$ $\irr(I)\})$. Then $card(\{q$;
$q$ $\in$ $\irr(I)$  and  $\ell(M/q)$ $=$ $m\})$ $\ge$ $2$.
\endproclaim
\demo{Proof} Let  $q_1$ $\in$ $\irr(I)$  such that  $\ell(M/q_1)$
$=$ $m$.  Then since $\ell(q_1/I)$  is finite, it is clear that
there exists an ideal $J$  in  $R$  such that  $q_1$  covers  $J$
and  $I$ $\subseteq$ $J$.  Now  $J$  is reducible, by (2.2.6), so
$n(J)$ $\ge$ $2$.  But since  $\ell(q_1/J)$ $=$ $1$, (3.2) shows
that there exists an ideal  $q_2$ $\in$ $\irr(I)$  such that  $J$
$=$ $q_1 \cap q_2$. Then it follows that  $\ell(M/q_2)$ $\le$
$\ell(M/J) - 1$ $=$ $\ell(M/q_1)$ $=$ $m$ $\le$ $\ell(M/q_2)$  (this
last inequality by the definition of  $m$), hence  $\ell(M/q_2)$ $=$
$m$ $=$ $\ell(M/q_1)$, \qed
\enddemo
\bigskip
The next result lists several properties of the ideals in  $\bold I (I)$.
\bigskip
\noindent
\proclaim{(3.4)  PROPOSITION}
The following statements hold for an open reducible ideal  $I$:

\noindent {\bf{(3.4.1)}} The maximal elements in  $\bold I(I)$  are
the elements in  $\irr(I)$.

\noindent {\bf{(3.4.2)}} The maximal reducible ideals in  $\bold
I(I)$  are the ideals  $J$ $\in$ $\bold I (I)$  such that  $n(J)$
$=$ $2$  and  $J$  is covered by all ideals  $q$ $\in$ $\irr(I)$
that contain  $J$.

\noindent
{\bf{(3.4.3)}}
$I$  is the minimum element in  $\bold I (I)$.

\noindent
{\bf{(3.4.4)}}
The ideals  $J$  in  $\bold I (I)$  that are minimal with respect
to properly containing  $I$  are the ideal covers of  $I$, and for
each such ideal  $J$  it holds that  $n(J)$ $\ge$ $n(I) - 1$.

\noindent {\bf{(3.4.5)}} If  $J$ $\in$ $\bold I (I)$  and if  $J$
$=$ $Q_1 \cap \dots \cap Q_h$  is an arbitrary decomposition of  $J$
as an intersection of irreducible ideals, then at least one  $Q_i$
is in  $\irr(I)$.

\noindent {\bf{(3.4.6)}} If  $J$ $\in$ $\bold I (I)$, if  $J$ $=$
$Q_1 \cap \dots \cap Q_h$  is an arbitrary decomposition of  $J$  as
an intersection of irreducible ideals, and if  $\ell((I:M)/(J \cap
(I:M)))$ $=$ $k$, then at least  $k$  of the  $Q_i$  are in
$\irr(I)$.

\noindent
{\bf{(3.4.7)}}
An ideal  $J$  in  $R$  is in  $\bold I (I)$  if and
only if  $I$ $\subseteq$ $J$  and  $I:M$ $\nsubseteq$ $J$.
\endproclaim

\demo{Proof} (3.4.1) is clear by the definitions of  $\bold I (I)$
and  $\irr(I)$.

For (3.4.2) let  $J$ $\in$ $\bold I (I)$  such that  $n(J)$ $=$ $2$
and $J$  is covered by all ideals  $q$ $\in$ $\irr(I)$  such that
$J$ $\subseteq$ $q$.  To see that  $q$  is a maximal reducible ideal
in  $\bold I (I)$  let  $q'$ $\in$ $\bold I (I)$  such that  $q$
$\subset$ $q'$  and let  $q''$ $\in$ $\irr(I)$  such that  $q'$
$\subseteq$ $q''$.  Then  $q$ $\subset$ $q''$, so $\ell(q''/q)$ $=$
$1$, by hypothesis, hence  $q'$ $=$ $q''$, so it follows that  $q$
is a maximal reducible ideal in  $\bold I (I)$.

Conversely, let  $q$  be a maximal reducible ideal in  $\bold I
(I)$. Then  $q$ $=$ $q_1 \cap \dots \cap q_k$  for some ideals $q_1,
\dots,q_k$  in  $\irr(I)$.  Assume this intersection is irredundant.
Then  $k$ $=$ $2$, since otherwise  $q$ $\subset$ $q_1 \cap q_2$ and
$q_1 \cap q_2$  is reducible and is in  $\bold I (I)$.  Also, if
$q$ $\subset$ $q'$ $\in$ $\irr(I)$, and if  $\ell(q'/q)$ $>$ $1$,
then there exists an ideal  $q''$  in  $R$  such that  $q$ $\subset$
$q''$ $\subset$ $q'$, so $q''$ $\in$ $\bold I (I)$, by (3.2), and
$q''$  is reducible, by (2.2.6), and this contradicts the choice of
$q$.  Therefore (3.4.2) holds.

It is clear that  $I$  is the minimum element in  $\bold I (I)$,
so (3.4.3) holds.

For (3.4.4) let  $J$  be a cover of  $I$.  Then  $I$ $\subset$ $J$
$\subseteq$ $I:M$. If  $J$ $=$ $I:M$, then since every cover of  $I$
is contained in  $I:M$  it follows that  $I:M$  is the unique cover
of  $I$, so  $I$  is irreducible by (2.2.1), and this contradicts
the hypothesis that  $I$  is reducible. Therefore  $J$ $\subset$
$I:M$, so there exists an ideal  $q$ $\in$ $\irr(I)$  such that  $J$
$\subseteq$ $q$, by (2.3.3) $\Rightarrow$ (2.3.1), so  $J$ $\in$
$\bold I (I)$, by (3.2), and it then readily follows that  $J$  is
minimal in  $\bold I (I)$ with respect to properly containing  $I$.

For the converse let  $J$ $\in$ $\bold I (I)$  be minimal with
respect to properly containing  $I$. Then since every ideal properly
between  $I$  and  $J$  is in  $\bold I (I)$, by (3.2), it follows
that  $\ell(J/I)$ $=$ $1$, hence  $J$  is a cover of  $I$.
Therefore, if  $n(J)$ $=$ $k$  and  $J$ $=$ $q_1 \cap \dots \cap
q_k$  is a decomposition of  $J$  as an irredundant intersection of
irreducible ideals, then (3.1) shows that there exists  $q$ $\in$
$\irr(I)$  such that  $q_1 \cap \dots \cap q_k \cap q$ $=$ $I$,
hence  $n(I)$ $\le$ $n(J) + 1$.

For (3.4.5), note that  $I:M$  is not contained in any ideal in
$\irr(I)$, by (2.3.1) $\Rightarrow$ (2.3.2), so it follows from
(3.2) that  $I:M$ $\nsubseteq$ $J$.  Therefore  $I:M$ $\nsubseteq$
$Q_i$  for some  $i$ $=$ $1,\dots,h$, and  $I$ $\subseteq$ $J$
$\subseteq$ $Q_i$, so  $Q_i$ $\in$ $\irr(I)$  by (2.3.2)
$\Rightarrow$ (2.3.1).

For (3.4.6) let  $j$  such that  $Q_1 ,\dots ,Q_j$  are in $\irr(I)$
and  $Q_{j+1} ,\dots,Q_h$  are not in  $\irr(I)$, so  $j$ $\ge$ $1$
by (3.4.5).  Let  $J_0$ $=$ $I:M$  and for  $i$ $=$ $1,\dots,j$  let
$J_i$ $=$ $Q_i \cap J_{i-1}$. Then (2.2.5) shows that either  $J_i$
is a codimensional one subspace of  $J_{i-1}$ or  $J_i$ $=$
$J_{i-1}$.  Therefore since  $\ell((I:M)/(J \cap (I:M)))$ $=$ $k$
it follows that  $j$ $\ge$ $k$.

Finally, for (3.4.7), if  $J$  is an ideal in  $R$  such that  $I$
$\subseteq$ $J$  and  $I:M$ $\nsubseteq$ $J$, then (2.3.3)
$\Rightarrow$ (2.3.1) shows that there exists  $q$ $\in$ $\irr(I)$
such that  $J$ $\subseteq$ $q$, hence  $J$ $\in$ $\bold I (I)$  by
(3.2).

And, if  $J$ $\in$ $\bold I (I)$, then  $I$ $\subseteq$ $J$
$\subseteq$ $q$  for some  $q$ $\in$ $\irr(I)$, by (3.2), and  $I:M$
$\nsubseteq$ $q$, by (2.3.1) $\Rightarrow$ (2.3.2), hence  $I:M$
$\nsubseteq$ $J$, \qed
\enddemo

\bigskip
The next remark generalizes (3.3).
\bigskip
\noindent {\bf{(3.5)  REMARK.}} It follows from (3.4.2) that if  $J$
is a maximal reducible ideal in  $\bold I (I)$, then  $\ell (M/q)$
$=$ $\ell(M/J) - 1$  for all ideals  $q$ $\in$ $\irr(I)$  that
contain  $J$.  And there exist at least two such ideals  $q$, by
(3.2).
\bigskip
We next breifly consider a new relation (called ``irreducibly
related'') between two open ideals in  $R$.
\bigskip
\noindent {\bf{(3.6)  DEFINITION.}} If  $I$  and  $J$  are open
ideals in $R$, then it will be said that  $J$  is {\bf{irreducibly
related to}} $I$  in case  $J$  is the (finite) intersection of
ideals in  $\irr(I)$. We will denote this relation by  $J~\underline
{ir}~I$.

\bigskip
\proclaim{ (3.7)  PROPOSITION}
The following hold for the
relation ``irreducibly related'':

\noindent
{\bf{(3.7.1)}}
The ideals that are irreducibly related to  $I$  are the ideals in
$\bold I (I)$.

\noindent
{\bf{(3.7.2)}}
$I~\underline{ir}~I$  for all open ideals  $I$, so the relation is reflexive.

\noindent
{\bf{(3.7.3)}}
If  $I~\underline{ir}~J$  and  $J~\underline{ir}~I$, then  $I$ $=$ $J$, so the relation is anti-symmetric.

\noindent
{\bf{(3.7.4)}}
The relation is not transitive.

\noindent
{\bf{(3.7.5)}}
If  $I$, $J$, and  $K$  are open ideals in $R$  such
that  $J~\underline{ir}~I$, $K~\underline{ir}~J$, and  $I:M$ $=$ $J:M$, then  $K~\underline{ir}~I$.

\noindent
{\bf{(3.7.6)}}
If  $J$  is an ideal in $R$, then  $J~\underline{ir}~I$  if and
only if  $I$ $\subseteq$ $J$  and  $I:M$ $\nsubseteq$ $J$.
\endproclaim

\demo{Proof}
(3.7.1) follows immediately from (3.2) and the definitions.

(3.7.2) is clear from the definition, and (3.7.3) follows immediately
from (3.7.1).

For (3.7.4) it suffices to give an example of ideals  $H,J,K$  such
that  $J~\underline{ir}~H$, $K~\underline{ir}~J$, and  $K$  is not irreducibly related to  $H$.
For this, let  $(L,N)$  be a local ring, let  $H$, $J$, and  $q$  be
open ideals such that  $H$ $\subset$ $J$ $\subset$ $q$ $\in$
$irr(H)$ $\nsupseteq$ $irr(J)$, and let  $K$ $\in$ $irr(J) - irr(H)$.  Then
(3.2) and (3.7.1) show that  $J~\underline{ir}~H$, $K~\underline{ir}~J$,
and  $K$  is not irreducibly related to  $H$, hence this
relation is not transitive.

For (3.7.5), let  $I,J,K$  be open ideals in  $R$  such
that  $J~\underline{ir}~I$, $K~\underline{ir}~J$, and  $I:M$ $=$ $J:M$.
Then  $K$ $=$ $Q_1 \cap \dots \cap Q_h$  for a finite number
of ideals  $Q_1,\dots,Q_h$  in  $irr(J)$, and for  $j$
$=$ $1,\dots,h$  we have  $I$ $\subseteq$ $J$ $\subseteq$ $Q_j$  and  $I:M$
$=$ $J:M$ $\nsubseteq$ $Q_j$, so  $Q_j$ $\in$
$irr(I)$, by (2.3.2) $\Rightarrow$
(2.3.1).  Therefore  $K~\underline{ir}~I$.

Finally, (3.7.6) follows immediately from (3.4.7) and (3.7.1), \qed
\enddemo
\bigskip
\bigskip
{\bf {4.  MINIMAL IRREDUCIBLES NEED NOT BE IRREDUCIBLE COMPONENTS.}}
In this section we consider the ideal structure of the
ring  $L$ $=$ $F[X,Y]/(X^3,Y^3)$  (where  $F$  is the field of two elements),
and use it to show that it is possible for an irreducible
ideal  $q$  in a local ring  $R$  to contain an open ideal  $I$  and
yet not contain any irreducible component of  $I$.  (In this
regard, it is shown in [HRS4, Corollary 6.5] that if $I$ is a
monomial ideal in a Gorenstein local ring $R$ and if $Q$ is
minimal in the set $\{q$; $I$ $\subseteq$ $q$  and  $q$
is an irreducible monomial ideal in  $R \}$, then
$Q$ is an irreducible component of $I$. (4.1)
shows that the ``monomial'' condition was crucial for this result.)
\bigskip
\proclaim{ (4.1)  THEOREM}
There exist a local ring  $(L,N)$  and open ideals  $J$
$\subset$ $q$  such that  $q$  is irreducible and contains
no irreducible component of  $J$.
\endproclaim
\demo{Proof}
(See (4.2.3) for more details concerning the following proof.)
Let  $F$  be the field of two elements, let  $X$  and  $Y$  be
indeterminates, and let  $L$ $=$ $F[X,Y]/K$, where  $K$
$=$ $(X^3,Y^3)F[X,Y]$, so  $L$  is a finite local ring
(and  $L$  is Gorenstein, since  $K$  is irreducible).
Let  $N$ $=$ $(x,y)L$, where  $x$ $=$ $X+K$  and  $y$ $=$ $Y+K$,
let  $J$ $=$ $x^2y^2L$, and let  $q$ $=$ $(x+y)L$.
Then it is readily
checked that  $(0):(x^2+xy+y^2)L$ $=$ $q$,
so  $q$  is irreducible (by [ZS, Theorem 35, p.250],
since  $(X^3,Y^3)F[X,Y]_{(X,Y)}$  is irreducible).
Also, $J$ $=$ $xy^2L \cap x^2yL$, so  $J$  is
reducible.  Further, $J$ $=$ $N^4$, $J:N$ $=$ $N^3$
$=$ $(x^2y,xy^2)L$, and  $x^2y$ $=$ $x^3+x^2y$ $=$ $x^2(x+y)$  and  $xy^2$
$=$ $xy^2+y^3$ $=$ $y^2(x+y)$, so  $N^3$ $=$ $(x^2y,xy^2)L$
$\subset$ $q$, hence  $q$ $\notin$ $irr(J)$, by (2.3.2) $\Leftrightarrow$ (2.3.1).
Finally, the only ideals properly between  $q$  and  $J$  are
the ideals  $(x^2+xy,y^2+xy)L$, $(x^2+xy)L$, $(xy+y^2)L$,
$(x^2,y^2)L$, $N^3$, $x^2yL$, $xy^2L$, and  $(x^2y+xy^2)L$, and none of
these ideals is irreducible by [ZS, Theorem 35, p. 250]
(since none of them is
the annihilator of a principal ideal; specifically,
$(x^2+xy,y^2+xy)L$ $=$ $(0):(x^2+xy+y^2,xy^2)L$,
$(x^2+xy)L$ $=$ $(0):(x^2,xy+y^2)L$,
$(xy+y^2)L$ $=$ $(0):(x^2+xy,y^2)L$,
$(x^2+y^2)L$ $=$ $(0):(x^2+y^2,xy)L$,
$N^3$ $=$ $(0):N^2$,
$x^2yL$ $=$ $(0):(x,y^2)L$,
$xy^2L$ $=$ $(0):(x^2,y)L$,
and  $(x^2y+xy^2)L$ $=$ $(0):(x^2,x+y,y^2)L$)), \qed
\enddemo
\bigskip
\noindent
{\bf{(4.2)  REMARK.  (4.2.1)}}
It follows, by passing to  $L/J$, that
another way to state (4.1) is that there exists
an Artinian local ring  $(L,N)$  with an irreducible
ideal  $q$  such that  $q$  contains no irreducible
component of zero.

\noindent
{\bf{(4.2.2)}}
The computation to determine the ideal structure of  $L$  in (4.1)
was carried out by the computer program Macaulay.  This
computation also showed that  $(x^2+x+y)L$  is
an irreducible ideal that contains  $J$ $=$ $x^2y^2L$  and that
does not contain any irreducible component of  $J$.  (Craig Huneke
pointed out to us examples of irreducible ideals  $q$
containing  $I$  that fail to contain
an irreducible component of  $I$.  One of his examples is in a
regular local ring  $(R,M)$  of altitude three
with  $M$ $=$ $(x,y,z)R$  and  $I$ $=$ $(x^3,y^3,z^3,xyz)R$.
He argues that with  $K$ $=$ $(x^3,y^3,z^3)R$  and  $f$
$=$ $yx^2+zy^2+xz^2$, it follows that  $K:fR$  is
irreducible, contains  $I$, and fails
to contain an irreducible component of  $I$.  His other
example is the one presented in (4.1).)

\noindent
{\bf{(4.2.3)}}
If  $F$  is the field with two elements,
then there are 256 = $2^8$ = $(1+1)^8$ $=$ $\sum_{i=0}^8 {8 \choose i}$   nonunits in the ring  $L$
$=$ $F[X,Y]/(X^3,Y^3)$  (since there are the
8 monomials  $x$, $y$, $x^2$, $xy$, $y^2$, $x^2y$, $xy^2$, $x^2y^2$)  ranging
from  $0,x, \dots,x^2y^2,x+y, \dots,xy^2 +x^2y^2, x+y+x^2, \dots,x+y+x^2+xy+y^2+x^2y+xy^2+x^2y^2$),
so there are also 256 units (each
being of the form  $f+1$, where  $f$  is a nonunit).
It is straightforward to write a computer program to compute
and store the nonunits in one file and the units in another file.
Then to get a list of
generators of the distinct principal ideals,
Macaulay can quickly compute all unit multiples
of each of the nonunits, and this shows
that each of the 255 nonzero nonunits
is a unit multiple of one of the
following 20 polynomials (14 homogeneous, 4 homogenizable (by
adjusting weights), and 2
nonhomogenizable):  $x$, $y$, $x^2$, $xy$, $y^2$, $x^2y$, $xy^2$,
$x^2y^2$, $x+y$, $x^2+y$, $x+y^2$, $x^2+x+y$, $x^2+xy$, $xy+y^2$,
$x^2+y^2$, $x^2+xy^2$, $x^2y+y^2, x^2+xy+y^2$, $x^2y+x^2+xy+y^2$, $x^2y+xy^2$.  (There is
some symmetry in the generators of
the two nonhomogenizable principal ideals; for example, $(x^2+x+y)L$
$=$ $(y^2+x+y)L$ $=$ $(xy+x+y)L$, and
$(x^2y+x^2+xy+y^2)L$ $=$ $(xy^2+x^2+xy+y^2)L$ $=$ $(x^2y^2+x^2y+x^2+xy+y^2)L$
$=$ $(x^2y^2+xy^2+x^2+xy+y^2)L$.)
[To compute these unit multiples, we created a file (``xy'', say) to be
fed into Macaulay with the ``Macaulay $<$ xy'' command.  It's first few
lines specified: (a)  an output file (with Macaulay's ``monitor'' command);
(b)  the base ring  $A$  (with
Macaulay's ``ring'' command (specifying characteristic
2  and  2  variables)); (c)  the kernel $K$  (with
Macaulay's ``ideal'' command (specifying
the two generators $x^3,y^3$)); and,
(d)  the factor ring  $L$ $=$ $A/K$  (with Macaulay's ``qring'' command).
The next line uses Macaulay's ``poly'' command (to specify
the nonunit polynomial  $f$  whose
unit multiples are desired), and this is followed by 255 pairs of lines
``$poly$ $h$ $\{f\}*(g+1)$''
``$type$ $ h$''
(with  $g$  varying over the 255 nonzero nonunit polynomials),
to compute and display (in the output file)
the 255 unit multiples  $h$ $=$ $f(g+1)$  of  $f$),
and then the file is ended with Macaulay's ``exit'' command
(to exit from Macaulay).
By successively changing the definition of  $f$  in this file, and then
feeding it into Macaulay,
it is readily checked that each of the 255 nonzero nonunit polynomials is
a unit multiple of one of the 20 polynomials listed above;
in this regard, it is useful to
write a short program to keep track of the distinct polynomials
that are unit multiples of these 20 polynomials, since Macaulay's
output from each run on the file  ``xy''  contains 255 polynomials, with
at most 32 distinct ones.
Using this process,
Macaulay shows that there are:  32 unit multiples of each of  $x$, $y$, $x+y$, $x+y^2$, $x^2+y$, $x^2+x+y$; 8 unit multiples of each of  $x^2+xy$, $xy+y^2$, $xy$, $x^2+y^2$;
4 unit multiples of each
of  $x^2$, $y^2$, $x^2+xy^2$, $x^2y+y^2$, $x^2+xy+y^2$, $x^2y+x^2+xy+y^2$;
2 unit multiples of each of  $x^2y$, $xy^2$, $x^2y+xy^2$; and, 1
unit multiple of  $x^2y^2$.]

Now, since  $(X^3,Y^3)$  is irreducible in  $F[X^3,Y^3]_{(X,Y)}$,
[ZS, Theorem 35, p. 250] shows that an ideal  $J$  in  $L$  is
irreducible if and only if  $(0):J$  is a principal ideal.
Also, $fL$ $=$ $(0):((0):fL)$, so it follows that there are exactly
20 nonzero irreducible ideals in  $L$, namely the
20 nonzero ideals  $(0):fL$.
Using Macaulay's ``quotient'' command to compute  $(0):fL$,
it turns out that the 20 nonzero irreducible ideals are:  $N$ $=$ $(x,y)L$
$=$ $(0):x^2y^2L$;
$(x,y^2)L$ $=$ $(0):x^2yL$; $(x^2,y)L$ $=$ $(0):xy^2L$;
$(x^2,x+y,y^2)L$ $=$ $(0):(x^2y+xy^2)L$; $xL$ $=$ $(0):x^2L$;
$x^2L$ $=$ $(0):xL$; $yL$ $=$ $(0):y^2L$;
$y^2L$ $=$ $(0):yL$; $(x+y)L$ $=$ $(0):(x^2+xy+y^2)L$;
$(x^2+xy+y^2)L$ $=$ $(0):(x+y)L$; $(x+y^2)L$ $=$ $(0):(x^2+xy^2)L$;
$(x^2+xy^2)L$ $=$ $(0):(x+y^2)L$; $(x^2+y)L$ $=$ $(0):(x^2y+y^2)L$;
$(x^2y+y^2)L$ $=$ $(0):(x^2+y)L$; $(x^2+x+y)L$ $=$ $(0):(x^2y+x^2+xy+y^2)L$;
$(x^2y+x^2+xy+y^2)L$ $=$ $(0):(x^2+x+y)L$;
$(x^2,xy+y^2)L$ $=$ $(0):(x^2+xy)L$; $(x^2+xy,y^2)L$ $=$ $(0):(xy+y^2)L$;
$(x^2+y^2,xy)L$ $=$ $(0):(x^2+y^2)L$; $(x^2,y^2)L$ $=$ $(0):xyL$.
(Macaulay's ``quotient'' command readily
computes  $(0):fL$  for the 18 homogenizable polynomials  $f$, and
$(0):fL$  can be
computed for the two remaining
polynomials  $f$ $=$ $x^2+x+y$  and  $f$ $=$ $x^2y+x^2+xy+y^2$  by
having Macaulay compute all the nonunit multiples of each of these
two polynomials  $f$.)

This has introduced eight new (non-principal)
ideals  (namely,
$N$, $(x,y^2)L$, $(x^2,y)L$, $(x^2,x+y,y^2)L$ $=$ $(x^2,x+y)L$  (since  $y^2$
$=$ $x^2 + (x+y)^2$), $(x^2,xy+y^2)L$, $(x^2+xy,y^2)L$, $(x^2+y^2,xy)L$,
and  $(x^2,y^2)L$, and it is
straightforward to find eight additional non-principal
ideals  $(x^2,xy)L$, $(xy,y^2)L$, $(x^2,xy^2)L$, $(x^2y,y^2)L$, $N^2$,
$N^3$ $=$ $(x^2y,xy^2)L$, $(x^2+xy,y^2+xy)L$,
and  $(x^2+xy+y^2,xy^2)L$; note that these 16 non-principal ideals
are all homogeneous.
Now Macaulay can be used to show that there are no additional homogeneous
ideals (by having it compute a standard basis for
each homogeneous ideal obtained by adjoining a homogeneous element
to the 30 homogeneous ideals listed above), and then by hand
checking it can be seen (by adjoining a nonhomogeneous element to
each of these 36
ideals) that there are no additional nonzero proper ideals in  $L$.

Next, to determine which irreducible ideals  $q$  are in  $\irr(J)$
for various ideals  $J$, recall that  $q$ $\in$ $\irr(J)$  if and
only $J$ $\subseteq$ $q$  and  $J:N$ $\nsubseteq$ $q$  (by (2.3.2)
$\Leftrightarrow$ (2.3.1)). With this in mind, Macaulay's
``quotient'' command can be used to compute  $J:N$  for the 34
homogenizable ideals, and then  $J:N$  can be computed for the two
remaining nonhomogenizable (principal) ideals by computing all
nonunit multiples of  $x$  and  $y$  (since $f$ $\in$ $gL:N$  if and
only if  $fx$, $fy$ $\in$ $gL$).

Finally, the containment relations between these 36 nonzero proper ideals
can be determined by having Macaulay
compute nonunit multiples of the generators of the ideals.

This procedure is not readily extendible to  $F[X,Y]/(X^m,Y^n)$  with
$m$  and  $n$  much larger than  3; for example, if  $m$ $=$ $3$  and  $n$ $=$ $4$,
then there are $2^{11}$ nonunit polynomials and 40 nonzero
proper principal ideals, and if  $m$ $=$ $4$ $=$ $n$, then there
are  $2^{15}$  nonunit polynomials.
\bigskip
\bigskip
{\bf {5.  WHEN DOES  uR[u,tI]  HAVE A UNIQUE COVER?}}
In this section we answer this question, and then use it
to characterize when  $uR[u,tI]$  is irreducible,
in the case when  $R$  is Artinian.  (Here,
$R[u,tI]$  is the
{\bf{Rees ring of}}  $R$
{\bf{with respect to}}  $I$,
so  $t$  is an indeterminate and  $u$ $=$ $1/t$.)


\noindent
\proclaim{(5.1)  PROPOSITION} Let  $\bold R$
$=$ $R[u,tI]$  and let  $\bold M$ $=$ $(u,M,tI) \bold R$.
Then the following are equivalent:

\noindent
{\bf{(5.1.1)}}
$\bold M$  is a prime divisor of  $u \bold R$.

\noindent
{\bf{(5.1.2)}}
There exists a nonnegative integer  $k$  and an element  $b$ $\in$ $I^k$  such
that  $u \bold R : bt^k \bold R$ $=$ $\bold M$.

\noindent
{\bf{(5.1.3)}}
There exists a nonnegative integer  $k$  and an element  $b$ $\in$ $I^k$  such
that  $(u,bt^k) \bold R$  covers  $u \bold R$.

\noindent
{\bf{(5.1.4)}}
There exists a nonnegative integer  $k$  and an element  $b$ $\in$ $T_k - I^{k+1}$,
where  $T_k$ $=$ $I^k \cap (I^{k+1}:M) \cap (I^{k+2}:I)$.
\endproclaim

\demo{Proof} Since  $u \bold R$  and  $\bold M$  are homogeneous,
the definition of prime divisor shows that (5.1.1) and (5.1.2) are equivalent.

Also, it readily follows from the definition of an ideal cover that
(5.1.2) and (5.1.3) are equivalent.

Now assume that (5.1.2) holds.  Then it follows that  $b$ $\in$
$I^k - I^{k+1}$, that  $bt^kM$ $\subseteq$ $u \bold R$, and that  $bt^k(tI)$
$\subseteq$ $u \bold R$.  Therefore  $b$ $\in$ $I^k - I^{k+1}$,
$bM$ $\subseteq$ $u^{k+1} \bold R \cap R$ $=$ $I^{k+1}$,
and  $bI$ $\subseteq$ $u^{k+2}\bold R \cap R$ $=$ $I^{k+2}$,
so it follows that (5.1.4) holds, hence (5.1.2) $\Rightarrow$ (5.1.4).

Finally, if (5.1.4) holds, then  $bt^k$ $\in$ $\bold R - u \bold R$,
$bM$ $\subset$ $I^{k+1}$, and  $bI$ $\subseteq$ $I^{k+2}$,
so it follows that  $bt^kM$ $\subseteq$ $t^kI^{k+1}$ $=$
$u(tI)^{k+1}$ $\subseteq$ $u\bold R$, and  $bt^k(tI)$ $\subseteq$
$t^{k+1}I^{k+2}$ $=$ $u(tI)^{k+2}$ $\subseteq$ $u\bold R$.
Therefore it follows that  $bt^k$ $\in$ $u\bold R:\bold M - u \bold R$,
so  $u\bold R : bt^k \bold R$ $=$ $\bold M$, hence (5.1.4) $\Rightarrow$ (5.1.2), \qed
\enddemo
\bigskip
\noindent
{\bf{(5.2)  REMARK.}}
Let  $J$  be an ideal in a local ring  $(R,M)$  and for each
nonnegative integer  $k$  let  $T_k$ $=$
$J^k \cap (J^{k+1}:M) \cap (J^{k+2}:J)$  (as in (5.1.4)).  Then:

\noindent
{\bf{(5.2.1)}}
$J^{k+1}$ $\subseteq$ $T_k$  for all  $k$ $\ge$ $0$.

\noindent {\bf{(5.2.2)}} If  $J^{k+1}$ $\subset$ $T_k$, then  $M$
$\in$ $\Ass(R/J^{k+1})$.

\noindent
{\bf{(5.2.3)}}
If  $J$  is regular, then  $J^{k+2}:J$ $=$ $J^{k+1}$  for all large  $k$,
so  $T_k$ $=$ $J^{k+1}$  for all large  $k$.

\noindent
{\bf{(5.2.4)}}
If  $J$ $=$ $bR$  is a regular principal ideal, then  $T_k$ $=$
$b^{k+1}R$ $=$ $J^{k+1}$  for all  $k$ $\ge$ $0$.

\noindent {\bf{(5.2.5)}} If  $J$  is  $M$-primary and  $J^{k+1}$  is
irreducible, then  $J^{k+1}:M$  is its unique cover, so  $J^{k+1}:M$
$\subseteq$ $J^k \cap (J^{k+2}:J)$, hence  $T_k$ $=$ $J^{k+1}:M$
is principal modulo  $J^{k+1}$.

\demo{Proof}
(5.2.1) is clear by the definition of  $T_k$.

For (5.2.2), if  $J^{k+1}$
$\subset$ $T_k$, then  $J^{k+1}$ $\subset$ $J^{k+1}:M$,
so  $M$ $\in$ $Ass(R/J^{k+1})$.

For (5.2.3),
if  $J$  is regular, then it is shown in the proof of [RR, (2.1)]
that  $J^{k+2}:J$ $=$ $J^{k+1}$  for all large  $k$,
so the definition of  $T_k$  and (5.2.1) show that  $J^{k+1}$ $=$ $T_k$  for all
large  $k$.

For (5.2.4), if  $J$ $=$ $bR$  is regular, then it is
clear that  $J^{k+2}:J$ $=$ $J^{k+1}$  for all  $k$ $\ge$ $0$,
and it readily follows from this and (5.2.1) that  $T_k$ $=$ $J^{k+1}$.

Finally, for (5.2.5), if  $J$  is  $M$-primary and  $J^{k+1}$  is
irreducible, then  $J^{k+1}:M$  is its unique cover, by (2.2.1), so
it follows that  $T_k$ $=$ $J^{k+1}:M$  is principal modulo
$J^{k+1}$, \qed
\enddemo
\bigskip
The next result shows an interesting application of the ideals
$T_k$. (If  $\altitude(R)$ $=$ $0$, then (5.3) characterizes the
ideals in  $R$  such that  $u \bold R$  is irreducible; see (5.4).)
\bigskip

\noindent
\proclaim{(5.3)  THEOREM} Let $J$  be an ideal in  $R$  and
let  $\bold R$ $=$ $R[u,tJ]$.  Then the
following are equivalent:

\noindent
{\bf{(5.3.1)}}
$u \bold R$  has a unique cover.

\noindent
{\bf{(5.3.2)}}
There exists a unique nonnegative integer  $k$  such
that  $T_k$ $\ne$ $J^{k+1}$, and for this  $k$, $T_k$  is a principal
ideal modulo  $J^{k+1}$,
where  $T_k$ $=$ $J^k \cap (J^{k+1}:M) \cap (J^{k+2}:J)$.
\endproclaim

\demo{Proof} Assume first that (5.3.1) holds
and let  $f$ $\in$ $\bold R$  such that  $(u,f)\bold R$  is the
unique cover of  $u\bold R$.  Then  $f$ $\notin$ $u\bold R$  and  $fN$
$\subseteq$ $u \bold R$  for some maximal ideal  $N$  in  $\bold R$,
hence  $u \bold R : f \bold R$ $=$ $N$.  Since  $u \bold R$  is
homogeneous, it follows that  $N$ $=$ $\bold M$ $=$ $(u,M,tJ) \bold R$.
Therefore  $\bold M$  is a prime divisor of  $u\bold R$,
so (5.1.1) $\Rightarrow$ (5.1.3) shows that there exists a nonnegative integer  $k$  and
an element  $b$ $\in$ $J^k$  such that  $(u,bt^k)\bold R$  is a cover
of  $u\bold R$.  Therefore the hypothesis implies that  $(u,f)\bold R$
$=$ $(u,bt^k)\bold R$, so it may be assumed to begin with that  $f$ $=$
$bt^k$  is homogeneous.

Now (5.1.3) $\Rightarrow$ (5.1.4) shows that  $b$ $\in$
$T_k-J^{k+1}$, so  $T_k$ $\ne$ $J^{k+1}$.  To see that
$T_k$  modulo  $J^{k+1}$  is principal, let  $c$ $\in$
$T_k - J^{k+1}$.  Then
(5.1.4) $\Rightarrow$ (5.1.3) shows that  $(u,ct^k)\bold R$  covers  $u\bold R$,
so the hypothesis implies that  $(u,ct^k)\bold R$ $=$ $(u,bt^k)\bold R$,
and it then readily follows that  $c$ $=$ $x+vb$  for some  $x$
$\in$ $J^{k+1}$  and unit  $v$  in  $R$.  Therefore  $T_k$  is a
principal ideal modulo  $J^{k+1}$.

Now let  $h$ $\ne$ $k$  be a nonnegative integer and suppose
there exists  $d$ $\in$ $T_h - J^{h+1}$.
Then (5.1.4) $\Rightarrow$ (5.1.3) shows
that  $(u,dt^h)\bold R$  is a cover of  $u\bold R$,
so  $(u,dt^h)\bold R$ $=$ $(u,bt^k)\bold R$.  If  $h$
$<$ $k$, then  $dt^h$ $=$ $u(xt^{h+1}) + (yu^{h-k})(bt^k)$  for
some  $x$ $\in$ $J^{h+1}$  and  $y$ $\in$ $R$, so by cancelling
$t^h$  it follows that  $d$ $\in$ $J^{h+1}$, and this
contradicts the choice of  $d$.  And a similar computation produces the
contradiction that  $b$ $\in$ $J^{k+1}$  if  $h$ $>$ $k$.  Therefore
it follows that  $h$ $=$ $k$, and this contradicts the choice of  $h$,
so the supposition that  $T_h$  properly contains  $J^{h+1}$  leads
to a contradiction.  Therefore  $T_h$ $=$ $J^{h+1}$  for all nonnegative
integers  $h$ $\ne$ $k$, hence (5.3.1) $\Rightarrow$ (5.3.2).

Now assume that (5.3.2) holds and let  $b$ $\in$
$T_k - J^{k+1}$.  Then (5.1.4) $\Rightarrow$ (5.1.3) shows
that  $(u,bt^k)\bold R$  is a cover of  $u\bold R$.
Therefore let  $(u,f)\bold R$  be another cover of  $u\bold R$,
so  $f\bold M$ $\subseteq$ $u\bold R$  and  $\bold M$  and  $u\bold R$  are
homogeneous, so it follows that it may be assumed that  $f$  is
homogeneous, say  $f$ $=$ $ct^h$.  Then (5.1.3) $\Rightarrow$ (5.1.4)
shows that  $c$ $\in$ $T_h - J^{h+1}$, so (5.3.2) shows
that  $h$ $=$ $k$  and
that  $c$ $=$ $x+vb$  for some  $x$ $\in$ $J^{k+1}$  and unit  $v$  in  $R$,
so it follows that  $(u,ct^h)\bold R$ $=$ $(u,bt^k)\bold R$,
hence (5.3.2) $\Rightarrow$ (5.3.1), \qed
\enddemo
\bigskip
\proclaim{(5.4)  COROLLARY} The following statements are equivalent
for an ideal  $I$  in an Artinian local ring  $(R,M)$:

\noindent
{\bf{(5.4.1)}}
$u\bold R$  is irreducible.

\noindent
{\bf{(5.4.2)}}
(5.3.2) holds.

\noindent
{\bf{(5.4.3)}}
$\bold R$  is Gorenstein.

\noindent
{\bf{(5.4.4)}}
The form ring  $\bold F(R,I)$  of  $R$  with respect to  $I$  is Gorenstein.
\endproclaim

\demo{Proof} Since  $\altitude(R)$ $=$ $0$, it follows that
$\altitude(\bold R)$ $=$ $1$, so since  $u\bold R$  is homogeneous
it follows that  $u\bold R$  is primary for  $\bold M$ $=$
$(u,M,tI)\bold R$.  Therefore  $u\bold R$  is irreducible if and
only if  $u\bold R$  has a unique cover (by (2.2.1)), so (5.4.1)
$\Leftrightarrow$ (5.4.2) by (5.3.1) $\Leftrightarrow$ (5.3.2).

It is known that  $\bold R$  is Gorenstein if and only if  $\bold
R_{\bold M}$  is Gorenstein \cite{BH, Prop.~3.1.19, page~94 and
Ex.~3.6.20, page~142}. Since  $\altitude(\bold R_{\bold M})$ $=$
$1$, it follows that  $\bold R_{\bold M}$  is Gorenstein if and only
if  $u\bold R_{\bold M}$  is irreducible, and  $u\bold R_{\bold M}$
is irreducible if and only if  $u \bold R$  is irreducible (since
$\bold M$  is the only prime divisor of  $u\bold R$), so (5.4.1)
$\Leftrightarrow$ (5.4.3).

Finally, $\bold F(R,I)$ $=$ $\bold R(R,I)/u\bold R(R,I)$,
so since  $u\bold R$  is regular it follows that (5.4.3) $\Leftrightarrow$ (5.4.4), \qed
\enddemo

\bigskip
\noindent
{\bf{(5.5)  REMARK.  (5.5.1)}}
If  $I$  is an ideal in an Artinian local ring  $(R,M)$,
then  $\bold R$ $=$ $R[u,tI]$  is Cohen-Macaulay.  Also, if
the equivalent statements of (5.4) hold, then  $R$  is Gorenstein.

\noindent
{\bf{(5.5.2)}}
If  $L$ $=$ $F[X,Y]/(X^3,Y^3)$  with  $F$  the field with two
elements, then the computer
program Macaulay can be used to show (by comparing the ideals  $T_k$  and
$I^{k+1}$  for  $k$ $=$ $0,1,2,3,4$)  that, of the 34 homogenizable
ideals in  $L$, only the following three choices for  $I$  yield
that  $uL[u,tI]$  is irreducible:  $I$ $=$ $xL$; $I$ $=$ $yL$;
and  $I$ $=$ $(x,y)L$.  For the ideal  $I$ $=xL$,
$T_k$ $=$ $I^{k+1}$ $=$ $x^{k+1}L$
for $k$ $\ne$ $2$  (and  $x^{k+1}L$ $=$ $(0)$  for  $k$ $\ge$ $2$),
and  $T_2$ $=$ $(I^3,x^2y^2)L$ $=$ $x^2y^2L$.  (Similar
results hold for  $I$ $=$ $yL$.)  And
for  $I$ $=$ $(x,y)L$, $T_k$ $=$ $I^{k+1}$  for  $k$ $\ne$ $4$,
and  $T_4$ $=$ $x^2y^2L$  and  $I^5$ $=$ $(0)$.
(It should be noted that all three
of these ideals are irreducible,
and for the first two of these ideals  $I$, $I^k$  is
irreducible for all  $k$ $\ge$ $1$.)

\noindent {\bf{(5.5.3)}} Assume that  $\altitude(R)$ $>$ $0$, let
$J$  be an ideal in  $R$, and let  $\bold R$ $=$ $R[u,tJ]$.  Then
neither of the following statements implies the other:  (1)  $u\bold
R$  has a unique cover. (2)  $u\bold R$  is irreducible.

\noindent {\bf{(5.5.4)}} If  $\altitude(R)$ $>$ $0$  and  $u\bold R$
has a unique cover, then  $u\bold R$  is not irreducible (since it
is not even primary).

\noindent
{\bf{(5.5.5)}}
If  $J$  is a regular principal ideal in  $R$, then  $uR[u,tJ]$  does
not have a unique cover.

\demo{Proof}
For (5.5.1), $u$  is a regular element in  $\bold R$
$=$ $R[u,tI]$  and  $\bold M$
$=$ $(u,M,tI)\bold R$  has
height one, so  $\bold R_{\bold M}$  is
Cohen-Macaulay, so [HR, (4.11)] shows that  $\bold R$  is Cohen-Macaulay.
Also, if (5.4.3) holds, that is, if  $\bold R$  is
Gorenstein, then its quotient ring  $R[u,t]$ $=$ $\bold R[1/u]$  is
Gorenstein, so since  $t$  is an indeterminate and  $u$ $=$ $1/t$  it
follows that  $R$  is Gorenstein.

For (5.5.3), if  $(R,M)$  is a regular local ring that
is not a field and  $J$ $=$ $M$,
then  $u\bold R$  is prime, and it is clear that  $u\bold R$  has
no cover, so (2) does not imply (1). To see that (1) does
not imply (2), let $F$ be a field and let
$R = F[[X,Y]]/(X^2,XY) = F[[x,y]]$ and let $J = (x,y)R$. Then
the form ring of $R$ with respect to $J$ is  $R[u, tJ]/(u)$
and is isomorphic to the
graded ring $F[X,Y]/(X^2,XY) = F[x,y]$, where $x^2 = xy = 0$.
Since the ideal (0) of this ring is reducible and  has
the unique cover  $xF[x,y]$, $u\bold R$ is reducible and has the unique
cover  $(u,tx)\bold R$.

For (5.5.4), if  $\altitude(R)$ $>$ $0$  and  $u\bold R$  has a
unique cover, then  $u\bold R$  cannot be irreducible, since
$(u,M,tJ)\bold R$  is a prime divisor of  $u\bold R$  (by (5.1.3)
$\Rightarrow$ (5.1.1)) and  $\hgt((u,M,tJ)\bold R)$ $>$ $1$  (so
$u\bold R$  is not even primary).

Finally, for (5.5.5), it follows from (5.2.4) and (5.3) that
$uR[u,tJ]$  does not have a unique cover, \qed
\enddemo

In passing, it should be noted that (5.5.2) and (5.4.1)
$\Rightarrow$ (5.4.4) show that
the form ring  $\bold F$ $=$ $\bold F (L,(x,y)L)$  is Gorenstein.
Another way to see this is
to note that  $\bold F$ $=$
$F[X,Y,u,tX,tY]/(u,t^3X^3,t^3Y^3)$, and  $u,t^3X^3,t^3Y^3$  is
a regular sequence in the locally regular ring  $F[X,Y,u,tX,tY]$,
\bigskip
\proclaim{(5.6)  COROLLARY}
If  $I$  is an ideal in an Artinian Gorenstein local ring  $(R,M)$  such
that  $uR[u,tI]$  is irreducible, then the integer  $k$  such
that  $T_k$  properly contains  $I^{k+1}$  is the largest
integer  $h$  such that  $I^h$ $\ne$ $0$, and in this case  $T_k$ $=$
$bR$, where  $bR$ $=$ $(0):M$  is the unique cover of zero in  $R$.
\endproclaim

\demo{Proof}
By (5.4.1) $\Rightarrow$ (5.4.2)  let  $k$  be the integer
such that  $T_k$  properly contains  $I^{k+1}$.
Choose  $h$  such that  $I^h$ $\ne$ $(0)$  and  $I^{h+1}$ $=$
$(0)$, so  $T_{h+i}$ $=$ $I^{h+i+1}$ $=$ $(0)$  for
all  $i$ $\ge$ $1$, so  $k$ $\le$ $h$.

Suppose that  $k$ $<$ $h$, so there exists  $c$ $\in$
$I^h$  such that  $c$ $\ne$ $0$.  By (5.1.4)
$\Rightarrow$ (5.1.3) let  $b$ $\in$ $T_k - I^{k+1}$  such
that  $(u,bt^k)\bold R$  is
the unique cover of  $u\bold R$.  Then  $(u,bt^k)\bold R$
$\subseteq$ $(u,ct^h)\bold R$, so  $b$ $\in$ $I^{k+1} + cR$
$\subseteq$ $I^{k+1} + I^h$ $=$ $I^{k+1}$, and this contradicts
the choice of  $b$.  Therefore it
follows that  $h$ $=$ $k$,
so  $I^k$ $\ne$ $(0)$  and  $I^{k+1}$ $=$ $(0)$ $\subset$ $T_k$
$=$ $I^k \cap ((0):M) \cap ((0):I)$ $=$ $(0):M$, \qed
\enddemo
\bigskip
In (5.7) we consider the rings  $R_n$ $=$ $F[X]/(X^n)$  and use (5.4)
to show that  $uR_n[u,tI]$  is irreducible and  $\bold F (R_n,I)$  is
Gorenstein if and only if  $I$ $=$ $X^i$  with  $i$  a divisor of  $n$.
\bigskip
\noindent
{\bf{(5.7)  REMARK.}}
Let $n$ be a positive integer, let $F$ be a field, let  $X$ be an indeterminate,
let $R_n = F[X]/(X^n) = F[x]$, where $x^n = 0$, and let $M_n = xR_n$.
Then  $R_n$  is an Artinian Gorenstein local ring and:

\noindent
{\bf{(5.7.1)}}
For each positive integer $n$ it is true that $uR_n[u, tM_n]$ is
irreducible and the form ring  $\bold F(R_n,M_n)$ of $R_n$
with respect to its maximal ideal $M_n$ is Gorenstein.

\noindent
{\bf{(5.7.2)}} For each even positive integer $n$ it is true
that $uR_n[u, tM^2_n]$ is
irreducible and the form ring  $\bold F(R_n,M^2_n)$ of $R_n$
with respect to the ideal $M^2_n$ is Gorenstein.

\noindent
{\bf{(5.7.3)}} For each odd positive integer $n \ge 3$,
$uR_n[u, tM^2_n]$ is
reducible and the form ring  $\bold F(R_n,M^2_n)$ of $R_n$
with respect to the ideal $M^2_n$ is not Gorenstein.

\noindent
{\bf{(5.7.4)}} More generally, if $i$ is an integer with
$1 \le i \le n$ and $I = M^i_n = x^iR_n$, then
$uR_n[u, tM^i_n]$ is
irreducible and the form ring  $\bold F(R_n,M^i_n)$ of $R_n$
with respect to the ideal $M^i_n$ is Gorenstein if and only
if $n$ is a multiple of $i$.

\demo{Proof}
It is clear that  $R$  is an Artinian Gorenstein local ring,
so it suffices to prove (5.7.4), and for this we consider the two
cases:  (a)  $n$  is a multiple of  $i$; and, (b)  $n$  is not
a multiple of  $i$.

For (a), let  $n$ $=$ $qi$, where  $q$  is a positive integer.
Then for  $j$ $=$ $0,1,\dots,q-2$  it is readily
checked that  $T_j$ $=$ $(x^i)^{j+1}R_n$,
that  $T_{q-1}$ $=$ $x^{iq-1}R_n$ $\supset$ $(0)$ $=$ $(x^i)^qR_n$,
and that  $T_j$ $=$ $(0)$ $=$ $(x^i)^{j+1}R_n$  for  $j$ $\ge$ $q$.
Therefore (5.4.2) holds, so it follows from (5.4) that  $uR_n[u, tM^i_n]$ is
irreducible and that $\bold F(R_n,M^i_n)$ is Gorenstein.

For (b), let  $n$ $=$ $qi+r$, where  $q$  is a nonnegative integer and
$1$ $\le$ $r$ $<$ $i$.  Then it is readily checked
that  $T_{q-1}$ $=$ $x^{iq-i+r}R_n$ $\supset$ $x^{iq}R_n$  and
that  $T_q$ $=$ $x^{n-1}R_n$ $\supset$ $(x^i)^{q+1}R_n$ $=$ $(0)$.
Therefore (5.4.2) does not hold, so it follows from (5.4) that
$uR_n[u, tM^i_n]$ is
not irreducible and that $\bold F(R_n,M^i_n)$ is not Gorenstein, \qed
\enddemo
\bigskip
We close this section with an example of an Artinian Gorenstein
local ring  $(R,M)$  such that  $uR[u,tM]$  is reducible and  $\bold F(R,M)$  is
not Gorenstein.
\bigskip

\noindent
{\bf{(5.8)  EXAMPLE.}}
Let $F$ be a field, let  $X$  be an indeterminate, let
$R = F[X^2,X^3]/(X^5) = F[x^2,x^3]$, where $x^5 = 0$
and $x^n = 0$ for $n \ge 7$, and let  $M = (x^2,x^3)R$.
Then  $R$  is an Artinian Gorenstein local
ring such that $uR[u,tM]$ is reducible
and the form ring  $\bold F(R,M)$ of $R$
with respect to its maximal ideal $M$ is not Gorenstein.

\demo{Proof}
It is clear that  $R$  is an Artinian Gorenstein local ring, so
by (5.4) it suffices to show that (5.4.2) does not hold.
And for this, it is readily checked that  $T_1$ $=$ $x^3R$ $\supset$ $x^4$
$=$ $M^2$  and that  $T_3$ $=$ $x^6$ $\supset$ $(0)$ $=$ $M^4$,
so (5.4.2) does not hold, \qed
\enddemo

\bigskip
\bigskip
{\bf{6.  SOME EXAMPLES.}}
In this section we give several examples of the ``bad'' behavior of the
irreducible components of an ideal, even in
regular local rings of altitude two.  Our first example, (6.2), shows
that  $\bold S$  (see (2.1.5)) is not empty, and to prove (6.2.3) we need
the following result.
\bigskip
\proclaim{ (6.1)  PROPOSITION} Let  $q$  be an irreducible
$M$-primary ideal in a local ring  $(R,M)$, let $q_1$ $=$ $q:M$  be
the unique cover of  $q$, and let  $( \overline R, \overline M)$ $=$
$(R/q,M/q)$.  Then  $q_1$  has an irreducible cover if and only if
$\overline M$  covers a principal ideal.
\endproclaim
\demo{Proof}
Let  $Q$  be a cover of  $q_1$.
Then since  $q_1$  is the unique cover of  $q$, and since the
operation  $\overline I \rightarrow \overline I^\prime$ $=$
$(0): \overline I$  on the set of ideals
$\overline I$  of  $\overline R$  is one-to-one
and reverses inclusion (see [ZS, pp. 247-251]),
it follows that  $\overline {q_1}^\prime$ $=$ $\overline M$  is a
cover of  $\overline Q ^\prime$.  Also, $Q$  is irreducible if and only
if  $\overline Q ^\prime$  is a principal ideal, by [ZS, Theorem 35, p. 250],
and the conclusion readily follows from this, \qed
\enddemo
\bigskip
The following example was discussed following (2.5).
\bigskip
\noindent
{\bf{(6.2)  EXAMPLE.}}
Let  $(R,M = (x,y)R)$  be a regular local ring of altitude two
and let  $n > 1$  and  $m > 1$  be integers.

\noindent {\bf{(6.2.1)}} Let  $I$ $=$ $(x^n,xy,y^m)R$. Then  $q_1$
$=$ $(x^n,y)R$, $q_2$ $=$ $(x,y^m)R$, and  $q_3$ $=$ $(xy,x^{n-1} +
y^{m-1})R$  are in  $\irr(I)$, $I$ $=$ $q_1 \cap q_2$  (so  $n(I)$
$=$ $2$), $\ell(q_1/I)$ $=$ $m-1$, $\ell(q_2/I)$ $=$ $n-1$, and
$n_{irr}(I)$ $=$ $1$ $=$ $\ell(q_3/I)$  (so  $q_3$  is an
irreducible cover of  $I$), so  $I$ $\notin$ $\bold S$  by (2.5.2).

\noindent
{\bf{(6.2.2)}}
If  $m$ $=$ $2$  and  $I$ $=$ $(x^n,x^{n-1}y,y^2)R$,
then  $q_1$ $=$ $(x^{n-1},y^2)R$  is an irreducible cover of  $I$,
so  $n(I)$ $=$ $2$  and
$\ell(q_1/I)$ $=$ $1$ $=$ $n_{irr}(I)$,
so  $I$ $\notin$ $\bold S$  by (2.5.2).
(Similarly, if  $n$ $=$ $2$  and  $I$ $=$ $(x^2,xy^{m-1},y^m)R$,
then  $q_2$ $=$ $(x^2,y^{m-1})R$  is an irreducible cover of  $I$,
so  $n(I)$ $=$ $2$  and
$\ell(q_2/I)$ $=$ $1$ $=$ $n_{irr}(I)$,
so  $I$ $\notin$ $\bold S$  by (2.5.2).)

\noindent {\bf{(6.2.3)}} Let  $n$ $>$ $2$, $m$ $>$ $2$, and  $I$ $=$
$(x^n,x^{n-1}y^{m-1},y^m)R$. Then  $q_1$ $=$ $(x^n,y^{m-1})R$  and
$q_2$ $=$ $(x^{n-1},y^m)R$ are in  $\irr(I)$, $I$ $=$ $q_1 \cap q_2$
(so  $n(I)$ $=$ $2$), $\ell(q_1/I)$ $=$ $m-1$, $\ell(q_2/I)$ $=$
$n-1$, and  $I$  has no irreducible cover, so  $I$ $\in$ $\bold S$.
\demo{Proof} For (6.2.1), it is shown in [HRS4, (2.1.2), (3.1) and
(4.1)] that  $q_1$  and  $q_2$  are in  $\irr(I)$  and that  $I$ $=$
$q_1 \cap q_2$.  Also, it is readily checked that  $I$ $\subset$
$(x^{n-1},xy,y^m)R \subset \cdots \subset (x^2,xy,y^m)R \subset
(x,y^m)R = q_2$  is a saturated chain of ideals between  $I$  and
$q_2$  (so $l(q_2/I)$ $=$ $n-1$)  and that $I$ $\subset$
$(x^n,xy,y^{m-1})R \subset \cdots \subset (x^n,xy,y^2)R \subset
(x^n,y)R = q_1$  is a saturated chain of ideals between  $I$  and
$q_1$  (so $l(q_1/I)$ $=$ $m-1$).  Further, it is readily checked
that  $q_3$  is a cover of  $I$  (so  $l(q_3/I)$ $=$ $1$), and $q_3$
is irreducible (since it is generated by a system of parameters), so
$q_3$ $\in$ $\irr(I)$  by (2.5.2). It therefore follows that  $I$
$\notin$ $\bold S$.

For (6.2.2), it is readily checked that  $q_1$  is an irreducible
cover of  $I$, and the conclusions readily follow from this.

Finally, for (6.2.3), it is shown in [HRS4, (2.1.2), (3.1) and (4.1)]
that  $q_1$  and  $q_2$  are
in  $irr(I)$  and that  $I$ $=$ $q_1 \cap q_2$.  Also,
it is readily checked
that  $I$ $\subset$ $(x^n,x^{n-2}y^{m-1},y^m)R
\subset \cdots \subset (x^n,xy^{m-1},y^m)R \subset (x^n,y^{m-1})R= q_1$  is
a saturated chain of ideals between  $I$  and  $q_1$  (so
$l(q_1/I)$ $=$ $n-1$), and a similar chain shows  $l(q_2/I)$ $=$
$m-1$.
Finally, it follows from (6.1) (with  $q$ $=$ $(x^n,y^m)R$  and  $q_1$
$=$ $I$)  that if  $n$ $>$ $2$  and  $m$ $>$ $2$, then  $I$  has
no irreducible cover (since if  $\overline M$ $=$ $(\overline x,\overline y) \overline R$
covers  $\overline b \overline R$, then  $\overline M$ $=$
$(\overline b ,\overline c )\overline R$  (for some  $c$ $\in$ $M$)  and
$\overline c \overline M$ $\subseteq$ $\overline b \overline R$,
so  $\overline M ^2$ $\subseteq$ $\overline b \overline R$,
hence  $(x^2,xy,y^2 )R$ $\subseteq$ $(x^n,y^m,b)R$,
and this clearly cannot happen if  $n$ $>$ $2$  and  $m$ $>$ $2$), \qed
\enddemo
\bigskip
The examples in (6.2) show some of the things that do not
necessarily hold in a given irreducible decomposition of an
$M$-primary ideal  $I$, as noted in the following remark.
\bigskip
\noindent
{\bf{(6.3)  REMARK.}}
Let  $I$ $=$ $q_1 \cap \cdots \cap q_g$  be an irredundant irreducible
decomposition of  $I$.  Then:

\noindent
{\bf{(6.3.1)}}
(6.2.1) shows that it is possible that  $\ell(q_i/I)$ $>$ $n_{irr}(I)$  for  $i$ $=$ $1,\dots,g$.

\noindent
{\bf{(6.3.2)}}
(6.2.1) shows that it is possible (by
varying  $I$)  that, for  $i$ $=$ $1,\dots,g$, there
is no bound on  $\ell(q_i/I)$, even when  $n_{irr}(I)$  $=$ $2$.
\bigskip
The next two examples were rather a surprise to us. The first of
these shows that, even if  $n(I)$ $=$ $2$, there may exist
arbitrarily long finite chains of ideals in  $\bold I (I)$  each of
which is the intersection of two ideals in  $\irr(I)$, and the
second shows that there may exist ideals in  $\bold I (I)$  that are
the intersection of more than  $n(I)$  elements in  $\irr(I)$.
\bigskip
\proclaim{ (6.4)  PROPOSITION} Let  $(R,M)$  be a regular local ring
of altitude two. Then there exists an infinite chain of $M$-primary
ideals  $I_1 \supset I_2 \supset \cdots$  and an infinite set $\bold
Q$  of irreducible  $M$ -primary ideals    $q_n$  such that for all
positive integers  $n$  and  $k$  it holds that  $I_{n+k}$ is the
irredundant intersection  $q_n \cap q_{n+k}$.  In particular, for
each positive integer  $n$  the ideals in  $\bold Q$  that are in
$\irr(I_n)$  are the ideals  $q_1 ,\dots, q_n$.
\endproclaim
\demo{Proof}
Fix  $x_1$ $\in$ $M - M^2$,
let  $q_1$ $=$ $(x_1,M^2)R$, and let  $I_1$ $=$ $(x_1M,M^2)R$,
so  $I_1$ $=$ $M^2$.
Then:
\medskip
$(a_1)$  $M^2$ $\subseteq$ $I_1$ $\subseteq$ $q_1$  and  $M$
$\nsubseteq$ $q_1$; and,

$(b_1)$  $q_1$ $=$ $(x_1,I_1)R$  is a cover of  $I_1$
\medskip
\noindent
(since  $x_1M$ $\subset$ $M^2$ $=$
$I_1$  and  $q_1$ $=$ $I_1 + x_1R$).
(Therefore if  $y$ $\in$ $M - x_1R$  and if we let  $q$ $=$ $(y,I_1 )R$,
then it is readily checked that  $q_1 \cap q$ $=$ $I_1$,
so  $q_1$  and  $q$  are in  $irr(I_1)$.)

Now let  $z_1$ $\in$ $M - q_1$, let  $x_2$ $=$ $z_1 + x_1$,
let  $q_2$ $=$ $(x_2,M^3)R$, and let  $I_2$ $=$ $(x_2 M,M^3)R$.
Then:
\medskip
$(a_2)$  $M^3$ $\subseteq$ $I_2$ $\subseteq$ $q_2$  and  $M^2$
$\nsubseteq$ $q_2$; and,

$(b_2)$  $q_2$ $=$ $(x_2,I_2)R$  is a cover of  $I_2$.
\medskip
Therefore assume that  $n$ $\ge$ $2$  and that  $z_{n-1}$ $\in$
$M^{n-1} - q_{n-1}$, $x_n$ $=$ $z_{n-1} + x_{n-1}$, $q_n$ $=$
$(x_n,M^{n+1})R$, and  $I_n$ $=$ $(x_n M,M^{n+1})R$  have been defined
so that:
\medskip
$(a_n)$  $M^{n+1}$ $\subseteq$ $I_n$ $\subseteq$ $q_n$  and  $M^n$
$\nsubseteq$ $q_n$; and,

$(b_n)$  $q_n$ $=$ $(x_n,I_n)R$  is a cover of  $I_n$.
\medskip
\noindent
Then let  $z_n$ $\in$ $M^n - q_n$, $x_{n+1}$ $=$
$z_n + x_n$, $q_{n+1}$ $=$ $(x_{n+1},M^{n+2} )R$, and  $I_{n+1}$ $=$
$(x_{n+1}M,M^{n+2} )R$.

Then it is readily checked that:
\medskip
$(a_{n+1})$  $M^{n+2}$ $\subseteq$ $I_{n+1}$ $\subseteq$
$q_{n+1}$  and  $M^{n+1}$ $\nsubseteq$ $q_{n+1}$; and,

$(b_{n+1})$  $q_{n+1}$ $=$ $(x_{n+1},I_{n+1})R$  is a cover of  $I_{n+1}$.
\medskip
Also, for each  $n$  it follows that  $x_n$ $\in$ $M - M^2$.  (For,
$x_1$ $\in$ $M - M^2$; $x_2$ $=$ $x_1 + z_1$  and  $z_1$ $\in$ $M - x_1R$,
so  $x_2$ $\in$ $M - M^2$; and, if  $i > 2$  and  $x_i$ $\in$
$M - M^2$, then  $z_i$ $\in$ $M^i - q_i$  implies
that  $x_{i+1}$ $=$ $x_i + z_i$ $\in$ $M - M^2$.)
Therefore  $q_n$ $=$ $(x_n,M^{n+1})R$  is generated by a system of parameters
(since  $R/x_nR$  is a PID), hence each  $q_n$  is irreducible.

Further, for each  $n$  it follows that  $I_{n+1}$
$\subset$ $I_n$.  (For  $I_{n+1}$
$=$ $(x_{n+1}M,M^{n+2})R$  and  $x_{n+1}M$
$=$ $(x_n + z_n )M$ $\subseteq$ $x_n M + z_n M$ $\subseteq$
$x_n M + M^n M$ $=$ $I_n$, and  $(a_n)$  and  $(a_{n+1})$  show that the containment
is proper.)

Moreover, $(a_n)$  and  $(a_{n+k})$  show
that  $q_n$ $\nsubseteq$ $q_{n+k}$  (since  $M^{n+1}$ $\subseteq$
$q_n$  and  $M^{n+k}$ $\nsubseteq$ $q_{n+k}$).
Therefore, if it is shown that,
for each  $n$  and  $k$, $q_{n+k}$ $\nsubseteq$ $q_n$, then it follows
from  $(b_{n+k})$  that  $q_n \cap q_{n+k}$ $=$ $I_{n+k}$  is
an irredundant intersection.
And it
then follows that  $q_1 ,\dots,q_n$  are in  $irr(I_n)$,
and  $q_{n+i}$ $\notin$ $irr(I_n)$  for all  $i$ $\ge$ $1$, since  $q_{n+i}$  is
a cover of  $I_{n+i}$  and  $I_{n+i}$  is
properly contained in  $I_n$  (so  $I_n$ $\nsubseteq$ $q_{n+i}$).
Therefore it remains to show that  $q_{n+k}$  is
not contained in  $q_n$.

For this, suppose that  $q_{n+k}$ $\subseteq$ $q_n$.
Then it follows from the definition of the ideals  $q_i$  that
$x_{n+k}$ $\in$ $q_n$ $=$ $(x_n,M^{n+1})R$, so there exist  $r$
$\in$ $R$  and  $m$ $\in$ $M^{n+1}$  such that  $x_{n+k}$
$=$ $rx_n + m$.  Also, the definition of the  $x_i$  shows
that  $x_{n+k}$ $=$ $x_n + z_n + z_{n+1} + \cdots + z_{n+k}$,
and the definition of the  $z_i$  shows that  $z_n$ $\notin$ $M^{n+1}$
and that  $z_{n+i}$ $\in$ $M^{n+1}$  for  $i$ $=$ $1,\dots,k$.
Therefore  $x_{n+k}$ $=$ $x_n + z_n + m^\prime$,
where  $m^\prime$ $=$ $z_{n+1} + \cdots + z_{n+k}$ $\in$
$M^{n+1}$.  Therefore
it follows that  $rx_n + m$ $=$ $x_{n+k}$ $=$
$x_n + z_n + m^\prime$,
so  $z_n$ $=$ $(r-1)x_n + m - m^\prime$
$\in$ $(x_n,M^{n+1})R$ $=$ $q_n$, and
this contradicts the choice of  $z_n$ $\in$ $M^n - q_n$.
Therefore  $q_{n+k}$ $\nsubseteq$ $q_n$  for all  $n$  and  $k$, \qed
\enddemo
\bigskip
Concerning the set  $\bold{Q}$  in (6.4), note that the intersection
of each set of more that two elements in  $\bold{Q}$  is redundant.  From
this observation, a natural question is, if  $n(I)$ $=$ $k$,
then is the intersection of each set of  $h$ $>$ $k$  elements in
$irr(I)$  redundant?  The answer is no, as the next result shows.
\bigskip
\proclaim{ (6.5)  PROPOSITION} Let  $(R,M)$  be a regular local ring
of altitude two and let  $k$ $<$ $m$  be positive integers.  Then
there exists an open ideal  $I$  of $R$ such that  $n(I)$ $=$ $m$
and there exist  $m+k$  elements in  $\irr(I)$  whose intersection
is irredundant.
\endproclaim
\demo{Proof}
It is shown in the proof of [HS, (2.1)]  by computing
$Tor(R/I,R/M)$ in two ways via projective resolutions of
$R/I$ and $R/M$ that if  $I$  is an open
ideal in  $R$, then  $n(I)$
$=$ $v(I) - 1$, where  $v(I)$  denotes the number of
elements in a minimal basis of  $I$.
\footnote {See Section 3 of [HRS1] for other applications of this result.}
Also, given positive integers $k < m$,
[HRS4, (3.12)] shows that in every regular local ring
of altitude two
there exists an open ideal  $I$  such
that  $v(I)$ $=$ $m+1$  and  $v(I:M)$ $=$ $m+k+2$.
Therefore there exists an ideal  $J$  in  $R$  such
that  $I$ $\subset$ $J$ $\subset$ $I:M$  and  $v(J)$
$=$ $m+k+1$.  Then it follows that  $n(I)$ $=$ $m$  and
$n(J)$ $=$ $m+k$,
and  $J$  is the (irredundant) intersection of  $m+k$  elements
in  $irr(I)$, by (3.2), \qed
\enddemo
\bigskip
A specific example of ideals  $I$  and  $J$  such that  $I$
$\subset$ $J$ $\subset$ $I:M$  with  $n(I)$ $=$ $m$  and  $n(J)$
$=$ $m+k$  as in (6.5) is the following:  let  $n$
$=$ $m+1$, $s$ $=$ $k+2$, for  $i$ $=$ $1,\dots,s$  let  $f_i$
$=$ $x^{2(i-1)}y^{n+s-2i}$  and  $z_i$
$=$ $x^{2i-1}y^{n+s-2i-1}$, for  $i$ $=$ $s+1 ,\dots,n$  let  $f_i$
$=$ $x^{s+(i-1)}y^{n-i}$  and  $z_i$ $=$
$x^{s+(i-1)}y^{n-1-i}$  (so  $f_i$
$\in$ $z_{i-1}R$  for  $i$ $=$ $s+1,\dots,n$; $z_n$  is not used),
and, finally, let  $I$ $=$
$(f_1,\dots,f_n)R$  and  $J$ $=$
$(f_1,\dots,f_s,z_2,\dots,z_{n-1})R$.
Then the proof of [HRS4, (3.11)] shows that  $I$
$\subset$ $J$ $\subset$ $I:M$, $v(I)$ $=$ $n$  ($=$ $m+1$),
and  $v(J)$ $=$ $s+n-2$  ($=$ $m+k+1$),
so  $n(I)$ $=$ $m$  and  $n(J)$ $=$ $m+k$  (as noted
at the start of the proof of (6.5)).
\bigskip
\centerline{{\bf{ACKNOWLEDGMENT}}}
We thank Craig Huneke for sharing with
us his insight on Theorem~4.1.


\bigskip
\bigskip

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\enddocument
\vfill
\eject
