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\begin{document}

\title*{Catenary local rings with geometrically \\ normal formal
fibers}

\titlerunning{Catenary local rings}
% allows abbreviation of title, if the full title is too long
% to fit in the running head
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\author{William Heinzer
\inst {1}
\and 
Christel Rotthaus
\inst{2}
\and Sylvia Wiegand\inst{3}}
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\authorrunning{W. Heinzer, C. Rotthaus, and S. Wiegand}
% if there are more than two authors,
% please abbreviate author list for running head


\institute{Department of Mathematics, Purdue
 University,
 West Lafayette, IN 47907-1395 USA
\and
Department of Mathematics, 
Michigan State University, East Lansing,
 MI 48824-1027, USA
\and 
Department of Mathematics and Statistics,
 University of Nebraska,
 Lincoln, NE 68588-0323, USA}

\maketitle              % typesets the title of the contribution

\begin{abstract}
{We discuss relations between the catenary property and
geometrically normal formal fibers. We present for each
integer $n \ge 2$ an example of a catenary
Noetherian local integral  domain of dimension $n$
which has geometrically regular formal fibers and
is not universally catenary.
These examples are obtained by means of a construction developed
in our previous articles
which uses power series rings,
homomorphic images and intersections. }
\end{abstract}

\section{ Introduction}

We are happy to dedicate this paper to Shreeram S. Abhyankar 
in celebration of his seventieth birthday. In his mathematical work
Ram has opened up many avenues. In the present paper we are 
pursuing one of these related to power series and completions. 
\footnote{The authors would like to thank the National Science
Foundation and the National Security Agency for
support for this research. In addition they are grateful for
the hospitality and cooperation of Michigan State, Nebraska
and Purdue, where several work sessions on this research were conducted.} 


A Noetherian ring $R$ is said to be {\it catenary} if, for
every pair of comparable prime ideals $P \subseteq Q$ of $R$, every saturated
chain of prime ideals from $P$ to $Q$ has the same
length \cite{A, page 11}. The ring $R$ is  {\it universally
catenary} if every finitely generated $R$-algebra is catenary. 
A Noetherian local ring
$(R,\m)$ with $\m$-adic completion $\widehat R$ has {\it geometrically
normal} (respectively {\it geometrically regular) formal fibers} if for each prime
$P$ of $R$ and for each finite algebraic extension $k'$ of
the field $k(P) := R_P/PR_P$, the ring   $\widehat R \otimes_R k(P)
\otimes_{k(P)} k'$ is normal (respectively regular).

In this paper we investigate the catenary property in Noetherian local
rings having  geometrically normal
formal fibers. In Example~4.2 we apply a technique from our earlier papers to 
construct, for each integer $n \ge 2$, an example of a 
catenary Noetherian local integral domain of dimension $n$
with  geometrically regular formal fibers which is not 
universally catenary. 

Let $(R,\m)$ be a Noetherian local ring.
We denote the Henselization of $R$ by
$R^h$.  We say $(R,\m)$ is
{\it  formally equidimensional},  or in other terminology
{\it quasi-unmixed},  provided
all the minimal primes of the $\m$-adic completion
$\widehat R$ have the same dimension.  A theorem of Ratliff
(Theorem 2.1) which is crucial for our work states that
 $R$ is universally catenary if
and only if
$R/\p$ is  formally equidimensional for each minimal prime
$\p$ of $R$ \cite{Ra, Theorem 2.6}.

Section 2  of this paper contains several results concerning
conditions for  a Noetherian local ring $(R,\m)$
to be  universally catenary. First, Ratliff's theorem leads
to the
observation (Proposition 2.2) that a Henselian Noetherian local
ring having geometrically normal formal fibers is universally
catenary.

Suppose now that $(R,\m)$ is a Noetherian local integral domain
having geometrically normal formal fibers.
It follows (Corollary 2.3) that $R^h$ also has geometrically
normal formal fibers and thus by (2.2) is universally catenary; 
moreover, if the derived normal ring $\overline{R}$ is
local, then $R$ is universally catenary. In Theorem 2.6, we show 
 $R$ is universally catenary  if and only if  the   set
$\Gamma
$ is empty, where  $$
\Gamma : = \{ W \in \Spec(R^h) \, | \,  \dim(R^h/W)
< \dim(R/(W \cap R)) \}.
$$
We also observe  that $\Gamma$ has a ``going down" property.

In Theorem 2.7 we prove  for $R$ as above that 
$R$  is catenary   but is not universally catenary
if and only if $\Gamma$ is nonempty and each prime $W$ in $\Gamma$
has dimension one. Thus, as we observe in Corollary~2.8,  if $R$ 
is catenary but not universally catenary, this is
signaled by the existence of dimension one minimal primes of the
$\m$-adic
completion $\widehat R$ of $R$. If $R$ is catenary, each 
minimal prime of $\widehat R$ having dimension different from
$\dim(R)$  must have dimension one.

In Section~3 we provide examples to illustrate the results
of Section~2. We apply a construction involving power 
series, homomorphic images and intersections.
The use of power series to construct
interesting examples of Noetherian integral domains has a 
rich history \cite{Az}, \cite{BR1}, \cite{BR2}, \cite{H},
\cite{O1}, \cite{O2}, \cite{R1}, \cite{R2}, \cite{R3}.  
We give a brief review of relevant notation and 
results from our
earlier papers describing this technique \cite{HRW1},
\cite{HRW2}, \cite{HRW3}, \cite{HRW4}. 
The construction begins with a Noetherian domain which
may  be  taken to be a 
``standard"  Noetherian
domain such as a polynomial ring  in several indeterminates over a field.
In Theorem~3.6 we extend this
construction by proving  that in certain
circumstances it is possible to transfer the flatness, Noetherian
and computability properties of integral domains 
associated with ideals $I_1,\dots, I_n$ to the
integral domain associated with 
their intersection $I = I_1 \cap \cdots \cap I_n$.

We apply these concepts in Examples 4.1 - 4.3 
to produce Noetherian local domains which are not universally
catenary.  In Remark~4.4, we specifiy precisely  
which  of these rings are catenary. These domains
illustrate the results of Section 2, because 
in Section 5  we prove that they have
geometrically regular formal fibers.

The books of Matsumura \cite{M1}, \cite{M2} and the book of
Nagata \cite{N2} are good references for
our terminology. 

We would like to thank M. Brodmann and R. Sharp for
raising a question on catenary/universally catenary rings
which motivated our work in this paper.

\section{ Geometric normality of  formal fibers}

Throughout this
section $(R,\m)$ is a Noetherian local ring, usually a domain.

We use the following interesting result proved by Ratliff in
 \cite{Ra, Theorem 2.6} relating the universally catenary 
property to properties of the completion: 

\result{2.1 Theorem.   (Ratliff)}  A Noetherian local ring $(R, \m)$ 
is universally catenary if and only if $R/\p$ is 
formally equidimensional for every minimal prime ideal 
$\p\in\Spec(R)$.  


\result{2.2 Proposition. } If $(R,\m)$ is  a Henselian 
Noetherian local ring having  geometrically  normal
formal fibers, then $R$ is universally catenary,  and for
each $P \in \Spec R$, the extension  $P\widehat R$ of
$P$ to the $\m$-adic completion of $R$ is also prime. 


\begin{proof} By Theorem 2.1, to show $R$ is
universally catenary, it suffices to show 
every minimal prime $\p$ of  $R$ is formally 
equidimensional.  By passing from $R$ to $R/\p$, we may
assume that $R$ is an integral domain. We  prove
that $\widehat R$ is also an integral domain, so,
in particular, the zero ideal of $R$ is formally
equidimensional.  Since $R$ has normal formal fibers,
the completion $\widehat R$ of $R$ is reduced. 
Hence the derived normal ring $\overline{R}$ of $R$
is a finitely generated  $R$-module \cite{N2, (32.2)}.
Moreover, since $R$ is Henselian, $\overline{R}$  is local \cite{N2, (43.12)}.

The completion $\widehat {\overline{R}}$ of $\overline{R}$ is
$\widehat R \otimes_R \overline{R}$ \cite{N2, (17.8)}.
Since the formal fibers of $R$ are geometrically
normal, the formal fibers of $\overline{R}$ are also
geometrically normal. It follows that 
$\widehat{\overline{R}}$ is
normal \cite{M2, Corollary, page~184},
and hence an integral domain because $\overline{R}$ is local.
Since $\widehat R$ is a flat $R$-module, $\widehat R$
is a subring of $\widehat{\overline{R}}$. Therefore
$\widehat R$ is an integral domain  and so 
$R$ is formally equidimensional.\qed 
\end{proof}


\result{2.3 Corollary.} Suppose $R$ is a Noetherian local domain
having  geometrically
normal  formal fibers. Then
\begin{description}
\item[(1)] $R^h$ is universally catenary. 
\item[(2)] If the derived normal ring  $\overline{R}$ of $R$ is
again local, then $R$ is universally catenary.
\end {description}
In particular,
if $R$ is a normal Noetherian local domain having geometrically
normal formal fibers, then $R$ is universally catenary.

\begin{proof} For item (1), the Henselization $R^h$ of $R$ is again
a Noetherian local domain having  geometrically normal formal 
fibers. For (2), by \cite{N2, (43.20)}, 
$R$ is formally equidimensional and hence
universally catenary. \qed
\end{proof}

We use the following result relating the catenary property to
the height of maximal ideals of the derived normal ring.

\result{2.4 Proposition.}
Let $(R,\m)$ be a Noetherian local domain of dimension $d$ and
let $\overline{R}$ be the derived normal ring of $R$. If 
$\overline{R}$ contains a maximal ideal $\overline{\m}$ 
with $\hgt(\overline{\m}) = r \not\in \{1,d\}$, then $R$ 
is not catenary.

\begin{proof} Since $\overline{R}$ has only finitely many
maximal ideals \cite{N2, (33.10)}, there exists $b \in \overline{\m}$
such that $b$ is in no other maximal ideal of $\overline{R}$.
Let $R' = R[b]$ and $\m' = \overline{\m} \cap R'$. 
By the Going Up Theorem\cite {M2,(9.3)},  
$\hgt(\m') = r \not\in \{1, d \}$. 
Since $R'$ is a finitely 
generated $R$-module there exists a nonzero $a \in \m$
such that $aR' \subseteq R$. It follows that $R[1/a] = R'[1/a]$.
The maximal ideals of $R[1/a]$ have the form $PR[1/a]$, 
where $P \in \Spec(R)$ is maximal with respect to not
containing  $a$. Since there
are no prime ideals strictly between $P$ and $\m$ 
\cite{M2, (13.5)}, if $\hgt(P) = h$, then there exists in
$R$ a saturated chain of prime ideals through $P$ of
length $h+1$. Thus to show $R$ is not catenary, it suffices to establish the
existence of a maximal ideal of $R[1/a]$ having height 
different from $d-1$. 
Since $R[1/a] = R'[1/a]$, the maximal ideals of $R[1/a]$
correspond to $P' \in \Spec(R')$ maximal with respect to
not containing  $a$.
Since $\hgt(\m') > 1$, there exists $c \in \m'$ such that $c$ is
not in any minimal prime of $aR'$ nor in any maximal ideal of $R'$
other than $\m'$.
Hence there exist prime ideals of $R'$ containing $c$ and not
containing $a$. Let $P' \in \Spec(R')$ be maximal with respect
to $c \in P'$ and $a \not\in P'$. 
Then  $P' \subseteq \m'$, so $\hgt(P') \le r-1 < d-1$. Therefore $R$
is not catenary. \qed
\end{proof}

\result{2.5 Remark.}
For $(R,\m)$ a Noetherian local domain, it is well known that the
maximal ideals of the derived normal ring $\overline{R}$ of $R$ 
are in one-to-one
correspondence with the minimal primes of the 
Henselization 
$R^h$  of $R$ \cite{N2, (43.20)}. Moreover, if a
maximal ideal $\overline{\m}$ of $\overline{R}$
corresponds to a minimal prime $Q$ of $R^h$,
then the derived normal ring of $R^h/Q$ is the
Henselization of $\overline{R}_{\overline{\m}}$
\cite{N2, Ex. 2, page~188}, \cite{N1}. Therefore
$\hgt(\overline{\m}) = \dim(R^h/Q)$.

\result{2.6 Theorem.} 
Suppose $(R,\m)$ is a Noetherian local integral domain
having geometrically normal formal fibers.  Consider the set $$
\Gamma : = \{ W \in \Spec(R^h) \, | \,  \dim(R^h/W) < \dim(R/(W \cap R)) \}.
$$
\begin{description}
\item[(1)] 
For $\p \in \Spec(R)$, 
$R/\p$ is not universally catenary if and only if 
there exists $W \in \Gamma$ such that $\p = W \cap R$.
The set $\Gamma$ is 
empty if and only if $R$ is universally catenary. 
\item[(2)]
If $\p \subseteq \q$ in $\Spec(R)$ and if there 
exists $W \in \Gamma$ with $W \cap R = \q$, then 
there also exists $W' \in \Gamma$ with $W' \cap R = \p$. 
\item[(3)]
If $W \in \Gamma$ and $Q$ is a minimal prime of 
$R^h$ such that $Q \subseteq W$, then $Q$ is also in $\Gamma$, that is,  
$\dim(R^h/Q) < \dim(R^h) = \dim(R)$.
\end{description}



\begin{proof}
For item (1), we use that the map of  
$R/\p$ to its $\m$-adic completion $\widehat R/\p\widehat R$
factors through $R^h/\p R^h$.  Therefore, by Theorem 2.1,  $R/\p$ is universally
catenary if and only if $R^h/\p R^h$ is equidimensional if and 
only if there does not exist $W \in \Gamma$ with $W \cap R = \p$. 
To prove item (2), observe that if $R/\p$ is universally 
catenary, then $R/\q$ is also
universally catenary \cite{M2, Theorem 31.6}.

It remains to prove item (3). 
Suppose there exists $P \subseteq W$ in $\Spec(R^h)$ 
with $\dim(R^h/P) = \dim(R^h)$.
Let $\w = W \cap R$. Since $R^h$ is flat over $R$
with zero-dimensional fibers, $\hgt(W) = \hgt(\w)$ \cite{M2, Theorem~15.1}.
By Proposition 2.3.1, $R^h$ is universally catenary. 
Therefore $\hgt(W/P) + \dim(R^h/W) = 
\dim(R^h/P)=\dim(R^h)\ge\hgt(W) + \dim(R^h/W)$, and so $\hgt(W/P)
=\hgt(W)$. 
Since $\hgt(\w) + \dim(R/\w) \le \dim(R)=\dim(R^h)$,
it follows that $\dim(R^h/W) \ge \dim(R/\w)$, so
$W \not\in \Gamma$. \qed
\end{proof}

\result{2.7 Theorem.} 
Let $(R,\m)$ and $\Gamma$ be as in Theorem 2.6. 
Then $R$ is catenary but not universally catenary if 
and only if $\Gamma$ is nonempty and each prime 
$W \in \Gamma$ has dimension one. In this case,
each $W \in \Gamma$ is a minimal prime of $R^h$.


\begin{proof} Assume that $R$ is catenary but not universally
catenary. By Theorem~2.6, the set $\Gamma$ is nonempty
and there exist minimal primes $Q$ of $R^h$ such 
that $\dim(R^h/Q) < \dim(R^h)$.
By Remark~2.5, if a maximal 
ideal $\overline{\m}$ of $\overline{R}$ corresponds to 
a minimal prime $Q$ of $R^h$, then $\hgt(\overline{\m}) =
\dim(R^h/Q)$. Since $R$ is catenary,
Proposition~2.4 implies the height of each maximal ideal
of the derived normal ring $\overline{R}$ of $R$ is 
either one or $\dim(R)$. Therefore $\dim(R^h/Q) = 1$
for each minimal prime $Q$ of $R^h$ for which
$\dim(R^h/Q) \ne \dim(R^h)$. Part (3) of Theorem~2.6 
implies each $W \in \Gamma$ is a minimal prime of $R^h$ and
of dimension one.

For the converse, assume that $\Gamma$ is nonempty and 
each prime $W \in \Gamma$ has dimension one. Then 
$R$ is not universally catenary by part (1) of Theorem~2.6
and by part (3) of Theorem~2.6, each prime of $\Gamma$
is a minimal prime of $R^h$ and therefore
lies over $(0)$ in $R$. To show $R$ is catenary, it 
suffices to show for each
nonzero nonmaximal prime ideal $\p$ of $R$ that 
$\hgt(\p) + \dim(R/\p) = \dim(R)$ \cite{M2, Theorem 31.4}.
Let $P \in \Spec(R^h)$ be a minimal prime of $\p R^h$. 
Since $R^h$ is flat over $R$ with zero-dimensional fibers,
$\hgt(\p) = \hgt(P)$. Let $Q$ be a minimal prime of $R^h$ 
with $Q \subseteq P$.  Then $Q \not\in \Gamma$. For by
assumption every prime of $\Gamma$ has dimension one, 
so if $Q$ were in $\Gamma$, then $Q = P$. But $P \cap R = \p$,
which is nonzero, and $Q \cap R = (0)$.  Therefore 
$Q \not\in \Gamma$ and hence $\dim(R^h/Q) = \dim(R^h)$.
Since $R^h$ is catenary, it follows that   
$\hgt(P) + \dim(R^h/P) = \dim(R^h)$.
Since $P \not\in \Gamma$, we have $\dim(R/\p) = \dim(R^h/P)$.
Therefore $\hgt(\p) + \dim(R/\p) = \dim(R)$ and $R$ is
catenary. \qed
\end{proof}

\result{2.8 Corollary.} 
If $R$ has geometrically normal formal fibers and is 
catenary but not universally catenary, then there
exist in the  $\m$-adic completion $\widehat R$ of $R$ 
minimal prime ideals $\widehat q$ such that 
$\dim(\widehat R/\widehat q) = 1$. 



\begin{proof} By Theorem 2.7, each prime ideal $Q \in \Gamma$
has dimension one and is a minimal prime of $R^h$. Moreover, 
$Q\widehat R := \widehat q$ is a minimal prime of $\widehat R$.
Since $\dim(R^h/Q) = 1$, we have $\dim(\widehat R/\widehat q) = 1$. \qed
\end{proof}


\section {A method for constructing examples}

In this section we give
a brief review of relevant notation and
results from our
earlier papers describing a method for constructing examples. 
The construction begins with a Noetherian domain which
may  be  taken to be a 
``standard"  Noetherian
domain such as a polynomial ring  in several indeterminates over a field.
In Theorem~3.6 we  extend our previous results; we use
Theorem~3.6 in Section~4 to  obtain  examples with 
larger dimensions and more 
minimal primes. 

We use the following result from \cite{HRW2}:

\result{3.1 Theorem.} Let $R$ be a Noetherian integral domain
with fraction field $K$. Let $x$ be a nonzero nonunit of $R$ and
let $R^*$ denote the $(x)$-adic completion of $R$. Suppose $I$ is
an ideal of $R^*$ with the property that $\p \cap R = (0)$ for
each $\p \in \Ass(R^*/I)$ , and set  $A:=K\cap (R^*/I)$.
 Then $R\to  (R^*/I)_x$ is flat  if and
only if $A$ is Noetherian and is realizable as a
localization of a subring  of $R_x = R[1/x]$.

For our constructions  we apply Theorem 3.1 and some other results of 
\cite{HRW2} and  \cite{HRW1} to a more specific setting, outlined in (3.2).

\result{ 3.2  Setting and notation for examples.}
Let $k$ be a field, let $n \ge s \in \N$,
and let $x, y_1,\dots,y_n$ be indeterminates over $k$. Let
$R:=k[x, y_1, \dots, y_n]$ and let $R^*$ be the $(x)$-adic
completion of $R$. Suppose $\tau_1, \dots , \tau_s \in
xk[[x]]\subseteq R^*$ are algebraically independent over $k(x,
y_1, \dots, y_n)$. Set $I := (y_1 - \tau_1, \dots , y_s -
\tau_s)R^*$ and $A := k(x, y_1, \dots , y_n) \cap (R^*/I)$.

The domain $A$ can also be expressed as an
intermediate domain between a Noetherian domain and its
$x$-adic completion:
$$
\mbox{(3.3)} \qquad
A := k(x, y_{s+1}, \dots, y_n, \tau_1, \dots, \tau_s) \cap
k[y_{s+1}, \dots, y_n]\,[[x]]. \qquad \qquad
$$
It is convenient to also consider a
{\it local} version of (3.2):

\result{ 3.2$'$  Local setting and notation for examples.}
Here $R$ is the {\it localized} polynomial ring $k[x, y_1, \dots,
y_n]_{(x,y_1, \dots,y_n)}$; otherwise this  
is the same setup as (3.2). 
Again let 
$$
A := k(x, y_1, \dots , y_n) \cap
(R^*/I), \quad \mbox{ where } \quad  I := (y_1 - \tau_1, \dots
, y_s - \tau_s)R^*.
$$
Then $A$ can be expressed as
an intermediate domain between the Noetherian local domain $R$ and
its $x$-adic  completion:
$$
\mbox{(3.3$'$)} \qquad
A := k(x, y_{s+1}, \dots, y_n, \tau_1, \dots, \tau_s) \cap
k[y_{s+1}, \dots, y_n]_{(y_{s+1}, \dots, y_n)}[[x]].\qquad \qquad
$$

The expressions in (3.3) and (3.3$'$) represent a special case of the
construction, a  simpler ``intermediate form''--- so that we need not
pass to a proper homomorphic image of the completion. This was our 
approach to
the construction in \cite{HRW1}.

The following proposition describes the situation for the two settings.

\result{3.4 Proposition.}
(\cite{HRW2, (4.1)}) Assume that $R$, $R^*$, $I$, and $A$ are as in the setting
of (3.2) or (3.2$'$). Then
\begin{description}
\item[(1)]
 The canonical map $\alpha : R \to (R^*/I)_x$ is flat.
\item[(2)]
 With the notation of (3.2), $A$  is Noetherian
of dimension $n - s +1$ and is a localization of a polynomial
ring in $n - s$ variables over a DVR.
\item[(3)]
 $A$ is a nested  union of localizations of
polynomial rings in $n +1$ variables over $k$.
\item[(4)]
 If $k$ has characteristic zero, then $A$ is excellent.
\end{description}


\result{3.5 Examples.} Assume the notation of (3.2) or (3.2$'$).

\noindent {\bf (1)} Let $R:= k[y_1,\dots,y_s,x]$ (that is, $n =
s$ in (3.2)) and let $R^*$ denote the $(x)$-adic completion of $R$. Then
$(R^*/I) \cap K = A$ is the DVR obtained by localizing $U$ at the
prime ideal $xU$. In this example $R_x = U_x$ has dimension $s+1$
and so $\dim(U) = s+1$, while $\dim(R^*/I) = \dim(A) = 1$.

\noindent {\bf (2)} A modification of Example 1  is to take $R$
to be the $(s+1)$-dimensional regular local domain $k[y_1,\dots ,
y_s, x]_{(y_1, \dots, y_s, x)}$. In this case $R_x = U_x$ has
dimension $s$, while we still have $R^*/I \cong k[[x]]$.

With  $R$ as in either (1) or (2), 
each domain $A$ constructed is a directed union of  $(s+1)$-dimensional
regular local domains dominated by $k[[x]]$ and having $k$ as a
coefficient field. In either case, since $(R^*/I)_x$ is a field,
$R\hookrightarrow(R^*/I)_x$ is flat, so we have a
nested union of $(s+1)$-dimensional regular local domains whose
union is Noetherian, in fact a DVR.

\noindent {\bf (3)}  With $R = k[x, y_1, \dots, y_n]_{(x,y_1, \dots,y_n)}$,
a localized polynomial ring in $n+1$ variables, and $d := n-s$,
let $J$ be the ideal $(y_1 - \tau_1, \dots, y_s -
\tau_s)R^*$. Then $R^* \cong
k[y_1,\dots,y_n]_{(y_1,\dots,y_n)}[[x]]$ is
an $n+1$ dimensional regular local domain and $R^*/J \cong
k[y_{s+1}, \dots, y_n]_{(y_{s+1}, \dots,y_n))}[[x]]$ is
a $(d+1)$-dimensional regular local domain. By (3.4.1), $(R^*/J)_x$ is flat
over $R$.
If $V = k[[x]]\cap k(x, \tau_1, \dots ,\tau_s)$, then $V$ is a
DVR and $(R^*/J) \cap K \cong V[y_{s+1}, \dots, y_n]_{(x,y_{s+1}, \dots,
y_n)}$ is a
(d+1)-dimensional regular local domain which is a nested union of
$(n+1)$-dimensional regular local domains.

The following theorem shows that in certain circumstances the
flatness, Noetherian and computability properties of 
the integral domains associated with
ideals $I_1,\dots, I_n$  of $R^*$ as described in Theorem 3.1
transfer to the integral domain associated 
to their intersection $I=I_1\cap \dots \cap I_n$. We
show in Section~5  that the property of  regularity of formal
fibers also transfers in certain cases to the domain associated
with an intersection ideal. 

\result{3.6 Theorem.} Suppose that
$R$ is a Noetherian domain, $x\in R$ is a nonzero nonunit,
$R^*$ is the $(x)$-adic completion of $R$, and $I_1,\dots, I_n$
are ideals of $R^*$ such that, for each $i$ with $1\le i\le n$,
each associated prime of $R^*/I_i$ intersects $R$ in $(0)$.
Suppose that each $(R^*/I_i)_x$ is a flat $R$-module and that the
localizations at $x$ of the $I_i$ are pairwise comaximal; that
is, for all $i \ne j$, $(I_i + I_j)R^*_x = R^*_x$. Let
$I:=I_1\cap \cdots \cap I_n,$ $A := K \cap (R^*/I)$ and, for $i =
1,2,\dots n$, let $A_i := K \cap (R^*/I_i)$. Then 
\begin{description}
\item[(1)]
 Each associated prime of $R^*/I$
intersects $R$ in $(0)$, $(R^*/I)_x$ is flat over $R$, $A$ is
Noetherian, and
  $A^* = R^*/I$ is the $(x)$-adic completion
of $A$. Similarly, $A_i^* = R^*/I_i$ is the $(x)$-adic completion
of $A_i$, for $i = 1,2,\dots n$.
\item[(2)]
$A^*_x  \cong (A_1^*)_x \oplus \dots \oplus (A_n^*)_x$.
If $W \in \Spec(A^*)$ and $x \not\in W$, then $(A^*)_W$ is a
localization of one of the  $A_i^*$.
\item[(3)]
$A \subseteq A_1 \cap \cdots \cap A_n$ and, if $\w \in \Spec A$ with
$x \notin \w$, then $(A_1)_x \cap\dots \cap (A_n)_x \subseteq
A_\w$. In particular, $A_x = (A_1)_x \cap \dots \cap (A_n)_x$.
\end{description}



\begin{proof} For (1), since $\Ass(R^*/(I_1 \cap \dots \cap I_n))
\subseteq \Ass(R^*/I_1) \cup  \dots \cup \Ass(R^*/I_n)$, the
condition on associated primes of  Theorem 3.1
 holds for the ideal $I=I_1
\cap \dots \cap I_n$. The natural $R$-algebra homomorphism
$\pi:R^*\to (R^*/I_1) \oplus \dots \oplus (R^*/I_n)$ has kernel
$I$. Further, the localization of $\pi$ at $x$ is onto because
for each $i \ne j$, $(I_i+I_j)_x=R^*_x $. Thus $(R^*/I)_x \cong
(R^*/I_1)_x \oplus \dots \oplus (R^*/I_n)_x$ is flat over $R$.
Therefore $A$ is Noetherian by Theorem 3.1. By \cite{HRW2,
(2.4.4)}, $A^* = R^*/I$ is the $(x)$-adic completion of $A$.

For (2), the first part is simply that $(R^*/I)_x \cong
(R^*/I_1)_x \oplus \dots \oplus (R^*/I_n)_x$.
%sum is direct since $A^* = R^*/I$ and each $A_i^* = R^*/I_i$.
If $W \in \Spec(A^*)$ and $x \notin W$, then $\pi(W_x)$ is a prime
ideal of $(R^*/I_1)_x\oplus \dots\oplus (R^*/I_n)_x$, so has the form
$(W_i)_x$ in some $i^{\mbox{th}}$ coordinate and $(R^*/I_j)_x$ in
all the others, where $W_i \in \Spec(R^*/I_i)$. It follows that
$A^*_W$ is a localization of some $A_i^*$.

Since $R^*/I_i$ is a homomorphic image of $R^*/I$, it follows
that $A \subseteq A_i$ for all $i = 1,2,\dots,n$. Let  $\w \in
\Spec A$ with $x \notin \w$. Since $A^* = R^*/I$ is faithfully
flat over $A$, there exists $\w^* \in \Spec(A^*)$ with $\w^* \cap
A = \w$. Then $x \notin \w^*$ implies $A^*_{\w^*}$ is some
$(A_i^*)_{\w_i^*}$, where $\w_i^* \in \Spec(A_i^*)$. By symmetry,
we may assume $A^*_{\w^*} = (A_1^*)_{\w_1^*}$. Let $\w_1 = \w_1^*
\cap A_1$. Since $A_\w \hookrightarrow A^*_{\w^*}$  and
$(A_1)_{\w_1}  \hookrightarrow (A_1^*)_{\w_1^*}$ are faithfully
flat, we have
$$
A_\w = A^*_{\w^*} \cap K = (A_1^*)_{\w_1^*} \cap K = (A_1)_{\w_1}
\supseteq (A_1)_x.
$$
It follows that $(A_1)_x \cap \dots\cap (A_n)_x \subseteq   A_\w$.
Thus  we have $(A_1)_x \cap \dots\cap (A_n)_x \subseteq  \cap
\{A_\w : \w \in \Spec A \mbox{ and } x \notin \w \} = A_x$. Since
$A_x \subseteq (A_i)_x$, for each $i$, it follows that $A_x =
(A_1)_x \cap\dots\cap (A_n)_x$.  \qed
\end{proof}

\section {Examples which are not universally catenary }

In \cite{HRW2, (4.5)} we give an example of a Noetherian domain
$A$ for which the completion is two dimensional and has exactly
two minimal primes; the first minimal prime has dimension one and
the other has dimension two. Thus $A$ is not universally
catenary. This is done in such a way that $A$ has geometrically
regular formal fibers. We generalize this example in the
following.

\result{4.1 Example.}  We construct a
two-dimensional Noetherian local domain so that the completion has any
desired number of minimal primes of dimensions one and two. For
this, let $R$ be the localized polynomial ring
in three variables  $R:=k[x,y,z]_{(x,y,z)}$, 
where $k$ is a field of characteristic
zero and the field of fractions of $R$ is
$K:=k(x,y,z)$. Then the $(x)$-adic completion of $R$ is
$R^*:=k[y,z]_{(y,z)}[[x]]$. Let $\tau_1,\dots,
\tau_s$,$\beta_1,\beta_2,\dots, \beta_m,\gamma\in xk[[x]]$  be
algebraically independent power series over $k(x)$. Now define
$Q_i:=(z-\tau_i,y-\gamma)R^*$, for $i$ with $1\le i\le r$, and
$P_j:=(z-\beta_j)R^*$, for $j$ with $1\le j\le m$. We apply
Theorem 3.6 with $I_i=Q_i$ for $1\le i\le r$, and $I_{r+j}=P_j$
for $1\le j\le m$. Then the $I_i$ satisfy the comaximality
condition  at the localization at $x$. Let $I:=I_1\cap \dots\cap
I_{r+m}$ and let
 $A:= K\cap ({R^*}/I).$
For $J$ an ideal of R$^*$ containing $I$, let $\bar J$ denote the
image of $J$ in $R^*$/I. Then, for each $i$ with $1\le i\le r$,
$\dim(\,(R^*/I)/\bar Q_i)=\dim(R^*/ Q_i)=1$ and, for each $j$ with
$1\le j\le m$, $\dim(\,(R^*/I)/\bar P_j)=2$. Thus $\widehat A$
contains $r$ minimal primes of dimension one and $m$ of dimension
two.

The integral domain $A$ birationally dominates $R$ and is birationally
dominated by each of the $A_i$. It follows from Corollary 5.3 that
$A$ has geometrically regular formal fibers. Since $\dim(A)=2$, $A$
is catenary.

We show in Example 4.2 that for every integer 
$n\ge 2$ there is a Noetherian local domain $(A,\m)$
of dimension $n$ with geometrically regular formal fibers which is catenary
but not universally catenary.

\result{4.2 Example.}
Let $R = k[x,y_1,\dots, y_n]_{(x,y_1,\dots, y_n)}$ be a 
localized polynomial ring of dimension $n+1$ where $k$ is a field of
characteristic zero. Let 
$ \sigma, \tau_1,\dots,\tau_n\in xk[[x]]$ be 
$n+1$ algebraically independent elements over $k(x)$ 
and consider in $R^* =  k[y_1,\dots, y_n]_{(y_1, \dots,y_n)}[[x]]$ the ideals
$$I_1 = (y_1-\sigma)R^*  \quad \mbox{ and } \quad I_2 = (y_1-\tau_1,\dots,
y_n-\tau_n)R^*.$$
Then the ring
$$A = k(x,y_1,\dots, y_n)\cap (R^*/(I_1\cap I_2))$$
is the desired example. The completion $\widehat A$ of $A$
has two minimal primes $I_1 \widehat A$ having dimension $n$ 
and $I_2 \widehat A$ having dimension one. By 
Corollary 5.3, $A$ has geometrically regular formal fibers.
Therefore the  Henselization $A^h$ has precisely 
two minimal prime ideals $P, Q$ which may be labeled so
that $P\widehat A = I_1\widehat A$ and $Q\widehat A = 
I_2\widehat A$. Thus $\dim(A^h/P) = n$ and $\dim(A^h/Q) = 1$. 
By Theorem~2.7, $A$ is catenary but not universally catenary.

In Example 4.3  we construct,  for each $t\in \N$ and for specified   nonnegative
integers  $n_1,\dots,n_t$  with $n_1 \ge 1$,  a $t$-dimensional  
Noetherian local domain
$A$ that birationally dominates  a $t+1$-dimensional regular local
domain such that the completion of $A$ has,  for each $r$ with $1\le r\le
t$, exactly $n_r$  minimal primes of dimension $t+1-r$.  In
particular, if $n_i > 0$ for some $i \ne 1$,
then $A$ is not universally catenary and is
not a homomorphic image of a regular local domain.
It follows from Remark 2.5 that
the derived normal ring $\overline A$ of $A$ has 
exactly $n_r$ maximal ideals of height $t+1-r$
for each $r$ with $1 \le r \le t$.

\result{4.3 Example.} Let $t\in\N$ and for each
$r$ with $1\le r\le t$, let $n_r$ be a nonnegative integer. 
Assume that $n_1 \ge 1$.  
We construct a $t$-dimensional domain $A$
for which $\widehat A$ has exactly $n_r$ minimal primes of
dimension $t+1-r$ for each $r.$ Let $x, y_1\dots,y_t$  be
indeterminates over a field $k$ of characteristic zero
\footnote{The characteristic zero assumption implies that each
$A_{rj}$  as constructed below is excellent cf. (3.4.4).}. Let $R
= k[x,y_1,\dots,y_t]_{(x,y_1,\dots,y_t)}$, let $R^*=
k[y_1,\dots,y_t]\,[[x]]_{(x,y_1,\dots,y_t)}$ denote the $(x)$-adic
completion of $R$ and let $K$ denote the fraction field of $R$.
For every $r,j,i\in\N$ such that $1\le r\le t$, $1\le j\le n_r $
and $1\le i\le r$, choose   elements $\{\tau_{rji}\}$ of
$xk[[x]]$ which are algebraically independent over
$k(x,y_1,\dots,y_t)$.

For each $r,j$ with $1\le r\le t$ and $1\le j\le n_r$, define
the prime ideal 
$P_{rj}:=(y_1-\tau_{rj1},\dots,y_r-\tau_{rjr})$ of
height $r$ in $R^*$. Then each ideal
$P_{rj}$ in $R^*$ is an example of the type considered in (3.2).

Thus
the $(R^*/P_{rj})_x$ are flat over $R$. Here, for each $r,j$, define
\begin{eqnarray*}
A_{rj} := K\cap (R^*/(P_{rj}) &=
k(x,y_1,\dots,y_t)\cap\frac{k[y_1,\dots,y_t]\,[[x]]_{(-)}}
{(y_1-\tau_{rj1},\dots,y_r-\tau_{rjr})}\\
&\cong k(x,Y_r,\Gamma_{rj})\cap k[Y_r]\,[[x]]_{(-)}
\cong V_{rj}[Y_r]_{(x,Y_r)},
\end{eqnarray*}
where $Y_r:=\{ y_{r+1},\dots,y_t\}$,
$\Gamma_{rj}:=\{\tau_{rj1},\dots,\tau_{rjr}\}$, and
$V_{rj}=k(\Gamma_{rj})\cap k[[x]]$ is a DVR. Then
$A_{rj}$ is a $(t+1-r)$-dimensional regular local domain that is a
nested union of $(t+1)$-dimensional RLRs.

We  take the ideal  $I$ to be the intersection of all the prime
ideals $P_{rj}$. Since the $\tau_{rji}\in xk[[x]]$ are distinct,
the sum of any two of these ideals
$P_{rj}$ and
$P_{mi}$, where we assume
$r\le m$, has radical $(x,y_1,\dots,y_m)R^*$, and thus
$(P_{rj}+P_{mi})R^*[1/x]=R^*[1/x]$. It follows that the
intersection $I$ of the $P_{rj}$ is irredundant and
$\Ass(R^*/I)=\{  P_{rj}\, |\,1\le r\le t, 1\le j\le n_r\}$. Since
$P_{rj} \cap R =  (0)$, $R$ injects into $R^*/I$. Let $A := K
\cap (R^*/I)$.

By Theorem 3.6, $R \hookrightarrow (R^*/I)_x$ is flat,  $A$ is
Noetherian and $A$ is a localization of a subring of
$R[1/x]$. In particular, $A$ birationally the $(t+1)$-dimensional
regular local domain $R$ and the stated properties hold.

\result{4.4 Remark. }  By Theorem~2.7, the ring $A$ constructed
in Example~4.3 is catenary if and only if each minimal prime of 
$\widehat A$ has dimension either one or $t$. By taking $n_r = 0$
for $r \not\in \{1, t\}$ in Example~4.3, we obtain additional examples of 
catenary Noetherian local domains $A$ of dimension $t$ 
having geometrically regular formal fibers for which the 
completion $\widehat A$ has precisely $n_t$ minimal primes of 
dimension one and $n_1$ minimal primes of dimension $t$. 

\result{4.5 Remark. } We would like to thank L. Avramov for
suggesting we consider the depth of the rings  constructed
in Example 4.3. The catenary rings which arise from this
construction all have depth one, but  we can use Example 4.3 to construct,
for each integer $t \ge 3$ and integer $d$ with $2 \le d \le t-1$, 
an example of a noncatenary 
Noetherian local domain $A$  of dimension $t$ and depth $d$ having
geometrically regular formal fibers. The $(x)$-adic completion 
$A^*$ of $A$ has precisely two minimal primes, one of dimension $t$
and one of dimension $d$. To see this with notation
as in Example 4.3, we set $m = t - d + 1$ and take 
$n_r = 0$ for $r \not\in \{1, m \}$ and $n_1 = n_m = 1$. Let
$$
P_1 := P_{11} = (y_1 - \tau_{111})R^* \mbox{ and }
P_m := P_{m1} = (y_1 - \tau_{m11}, \ldots, y_m - \tau_{m1m})R^*.
$$
Consider $A^* = R^*/(P_1 \cap P_m)$ and the short exact 
sequence  
$$
0 \longrightarrow \frac{P_{1}}{P_{1} \cap P_{m}} 
\longrightarrow  \frac{R^*}{P_{1} \cap P_{m}} 
\longrightarrow  \frac{R^*}{P_{1}} \longrightarrow  0.
$$
Since $P_1$ is principal and not contained in $P_m$,
we have $P_1 \cap P_m = P_1P_m$ and 
$P_1/(P_1 \cap P_m)  \cong R^*/P_m$. It follows that
$\depth A = \depth A^* = \depth (R^*/P_m) \,\,\, $
$ = d\,$; 
see for example \cite{K, page 103, ex 14}. 
Moreover, the derived normal ring $\overline A$ of $A$ has
precisely two maximal ideals one of height $t$ and one of
height $d$.

\section{Regularity of morphisms and geometrical regularity
of  formal fibers}

We show in (5.3) that the ring $A$ of Examples 4.1, 4.2 and 4.3
have geometrically regular formal fibers. 

\result{5.1 Proposition.}  Let $R,\,A$ and $I$ be as in Theorem 3.1.
Suppose that, for each $P\in \mbox{Spec}(R^*/I)$ with $x\notin
P$, the morphism $\psi_P :R_{P\cap R}\longrightarrow (R^*/I)_P$
is regular.  Then 
\begin{description}
\item[(1)]
 $A$ is Noetherian  and the morphism $A \longrightarrow
A^* = R^*/I$ is regular.
\item[(2)]
If $R$ is semilocal with geometrically regular formal fibers
and $x$ is in the Jacobson radical of $R$, then $A$ has
geometrically regular formal fibers.
\end{description}


\begin{proof} Since flatness is a local property (and regularity
of a morphism  includes flatness), the morphism $\psi_x :
R\longrightarrow (R^*/I)_x$ is flat. By Theorem 3.1 and
\cite{HRW2, (2.4.4)}, $A$ is Noetherian with $(x)$-adic
completion $A^* = R^*/I$. Hence $A \longrightarrow A^*$ is flat.

Let $Q\in \mbox{Spec}(A)$, let $k(Q)$ denote the fraction field
of $A/Q$ and  let $Q_0 = Q\cap R$.

\result{Case 1:} $x\in Q$.
Then $R/Q_0 = A/Q = A^*/QA^*$ and the ring
$A^*_{QA^*}/QA^*_{QA^*} = A^*\otimes_A k(Q) = A_Q/QA_Q$ is
trivially geometrically regular over $k(Q)$.

\result{Case 2:} $x\notin Q$. 
Let $k(Q)\subseteq L$ be a finite algebraic field extension. We
show the ring $A^*\otimes_A L$ is regular. Let $W \in \Spec(
A^*\otimes_A L)$ and let $W' = W\cap (A^*\otimes_A k(Q))$. The
prime $W'$ corresponds to a prime ideal $P\in \Spec(A^*)$ with
$P\cap A =Q$. By assumption the morphism
$$ R_{Q_0} \longrightarrow (R^*/I)_P = A^*_P$$
is regular. Since $x\notin Q$ it follows that $R_{Q_0} = U_{Q
\cap U} = A_Q$ and that $k(Q_0)=k(Q)$. Thus the ring
$A^*_P\otimes_{A_Q} L$ is regular. Therefore   $(A^*\otimes_A
L)_W$ which is a localization of this ring is regular.

For part (2), since $R$  has geometrically regular formal fibers,
so has $R^*$  by \cite{R3}. Hence the  morphism $\theta : A^*=
R^*/I \longrightarrow \widehat A = \widehat {(R^*/I)}$ is regular.
By \cite{M1, Thm. 32.1 (i)} and part (1) above, it follows that
$A$ has geometrically regular formal fibers, that is, the
morphism $A \longrightarrow \widehat A$ is regular. \qed
\end{proof}

\result{5.2 Proposition.}  Assume that $R,\,K,\,x,\,R^*$ 
are as in Theorem 3.1 and $n\in\N$.  Let
   $I_1,\dots,I_n$ of $R^*$ be ideals of $R^*$ such that
each associated prime of $R^*/I_i$ intersects $R$ in $(0)$, for
$i = 1,\dots,n$. Let $I := I_1 \cap\dots\cap I_n$. Also assume
\begin{description}
\item[(1)] $R$ is semilocal with geometrically regular formal fibers
and $x$ is  in the Jacobson radical of $R$.
\item[(2)] Each $(R^*/I_i)_x$
is a flat $R$-module and, for each $i\ne j$, the ideals
$I_i(R^*)_x$ and $I_jR^*_x$ are comaximal in $(R^*)_x$.
\item[(3)] For $i = 1,\dots,n$,
$A_i := K \cap (R^*/I_i)$ has geometrically regular formal fibers.
\end{description}
Then  $A := K \cap (R^*/I) = B$ has geometrically regular formal
fibers.


\begin{proof}  Since $R$ has geometrically regular formal fibers,
by (5.1.2), it suffices to show for $W \in \Spec(R^*/I)$ with $x
\not\in W$ that $R_{W_0} \longrightarrow (R^*/I)_W$ is regular,
where $W_0 := W \cap R$. As in (3.1), we have  $(R^*/I)_x =
(R^*/I_1)_x \oplus\dots\oplus (R^*/I_n  )_x$. It follows that
$(R^*/I)_W$ is a localization of  some $R^*/I_i$. Suppose
$(R^*/I)_W = (R^*/I_i)_{W_i}$, for some $i$ with $1\le i\le n$,
and $W_i \in \Spec(R^*/I_i)$. Then $R_{W_0} = (A_i)_{W_i \cap
A_i}$ and $ (A_i)_{W_i \cap A_i} \longrightarrow (R^*/I_i)_{W_i}$
is regular. Thus $R_{W_0} \longrightarrow (R^*/I)_{W}$ is
regular. \qed
\end{proof}

\result{5.3 Corollary.} The rings  $A$ of Examples 4.1, 4.2 and 4.3 
have geometrically regular formal fibers, that is, the
morphism $\phi: A \to \widehat A$ is regular.


\begin{proof} By the definition of $R$ and the observations given
in (5.1), the hypotheses of (5.2) are satisfied.  \qed
\end{proof}


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\end{document}






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