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\newcommand{\heading}[1]{\begin{center}\small\bf #1\end{center}}
\begin{document}

\begin{slide}
\heading{PROJECTIVELY FULL IDEALS IN }

\heading{ NOETHERIAN RINGS }

\vspace{0.2in}


\begin{center}
William Heinzer
\end{center}

\begin{center}
Purdue University
\end{center}

\vspace{0.2in}

\begin{center}
joint with Catalin Ciuperca, Jack Ratliff and Dave Rush
\end{center}
\end{slide}




\begin{slide}

\heading{SETUP AND DEFINITIONS}

\medskip

Let $I$ be a regular proper ideal of a Noetherian ring $R$.

\medskip

{\bf DEFINITION}. An ideal $J$ of $R$ is {\bf projectively
equivalent }

\medskip

to $I$ if there exist positive integers $m$ and $n$ such that
$I^m$ and $J^n$

\medskip

have the same integral closure, i.e., $(I^m)_a = (J^n)_a$.


\medskip

{\bf NOTATION}. Let $\mathbf P(I)$ denote the set of integrally
closed

\medskip

ideals projectively equivalent to $I$.

\medskip

{\bf FACT}. The set $\mathbf P(I)$ is  linearly ordered and
discrete with

\medskip

respect to inclusion.





\end{slide}


\begin{slide}

\heading{MORE SETUP AND DEFINITIONS}



Since $\mathbf P(I)$ is discrete and since $J \in \mathbf P(I)$
implies $(J^n)_a \in \mathbf P(I)$,

\medskip

there is naturally associated to $I$ and $\mathbf P(I)$ a numerical
semigroup,

\medskip


i.e., a subsemigroup $S(I)$ of the additive semigroup  of
nonnegative

\medskip

integers $\N_0$ that contains all sufficiently large integers.

\medskip

{\bf DEFINITION}. The set $\mathbf P(I)$ is said to be {\bf
projectively full}

\medskip
if $S(I) = \N_0$, or equivalently,  if every element of $\mathbf
P(I)$ is the integral

\medskip

closure of a power of the largest element $K$ of $\mathbf P(I)$,
i.e., every

\medskip

element of $\mathbf P(I)$ has the form $(K^n)_a$, for some
positive integer $n$.

\medskip


If this holds, the ideal $K$ is said to be {\bf projectively
full}.


\end{slide}



\begin{slide}
\begin{center}
\heading{MAIN GOALS}
\end{center}

\medskip

{\bf 1}.  Describe ideals $K$ that are projectively full.

\medskip

{\bf 2}.  Describe ideals $I$ such that $\mathbf P(I)$ is
projectively full.

\medskip

{\bf 3}.  Can things be improved by passing to an integral
extension?

\medskip

{\bf THEOREM}. If $R$  contains the field of rational numbers,
then

 there exists a finite free integral extension ring $A$  of
$R$ such that

$\mathbf P(IA)$  is projectively full; and if $R$ is an integral
domain, then

there also exists a finite integral extension domain $B$  of $R$
such

that $\mathbf P(IB)$  is projectively full.

\end{slide}


\begin{slide}
\begin{center}
\heading{EXAMPLES}
\end{center}

{\bf EXAMPLE 1}. Let $(R,M)$ be a Noetherian local ring having

the property
$$a \in M^i \setminus M^{i+1}  \text{ and } b \in M^j
\setminus M^{j+1}  \implies ab \not\in M^{i+j+1},
$$
then $M$ is projectively full. Thus if the associated graded ring

$$
\mathbf G(R, M) = R/M \oplus M/M^2 \oplus \cdots \oplus
M^n/M^{n+1} \oplus \cdots
$$
is an integral domain, then $M$ is projectively full.

\medskip

{\bf EXAMPLE 2}. Let $k$ be a field and let $R = k[[x^2, x^3]]$.
Then

the maximal ideal $M = (x^2, x^3)R$ is not projectively full. The

numerical semigroup $S(M)$ is generated by $2$ and $3$.

\end{slide}

\begin{slide}
\begin{center}
\heading{DISCUSSION}
\end{center}
With $R = k[[x^2, x^3]]$ as in Example 2, $R$ is not integrally
closed.

Things improve by passing to the integral extension $R[x] =
k[[x]]$.

For each regular proper ideal $I$ of $R$, $\mathbf P(IR[x])$ is
projectively full.

\medskip

Want to give  an example of a normal local domain $(R,M)$ such


that $M$ is not projectively full?

\medskip
{\bf EXAMPLE 3}.  Let $F$ be a field and let $x, y, z, w$ be
variables. Let

$$ R_0 = F[x, y]_{(x, y)}\,\text{  and  } \, R = \frac{R_0[z, w]}{(z^2 - x^3 - y^3, \,\, w^2
- x^3 + y^3)}.
$$

If $2$ and $3$ are units of $F$, then $R$ is a 2-dim normal local
domain,

and the maximal ideal $M = (x,y, z, w)R$ is not projectively full.
\end{slide}





\begin{slide}
\begin{center}
\heading{SOME HISTORY}
\end{center}

\medskip

The concept of projective equivalence of ideals and the study of

ideals projectively equivalent to $I$ was introduced by Samuel in

\medskip

{\em Some  asymptotic properties of powers of ideals}, Annals of
Math

56 (1952), 11-21.

\medskip


and further developed by Nagata in

\medskip

{\em Note on a paper of Samuel concerning asymptotic properties

of ideals}, Mem. Coll. Sci. Univ. Kyoto, Ser. A Math. 30 (1957),

165-175.

\medskip

\medskip

\end{slide}


\begin{slide}
\begin{center}
\heading{MORE HISTORY}
\end{center}



Rather than `projectively equivalent', Hartmut  G\"ohner uses the

term `asymptotically equivalent' in

{\em Semifactoriality and Muhly's condition $(N)$  in
two-dimensional

local rings}, J. Algebra 34 (1975), 403-429.

\medskip
G\"ohner mentions that the expression `projective asymptotic

equivalence' is used by David Rees in

{\em Valuations associated with ideals (II)}, J. London Math. Soc.
36 (1956), 221-228.

and by H. T. Muhly in

{\em On the existence of asymptotically irreducible ideals}, J.
London Math. Soc. 40 (1965), 99-107.




\end{slide}








\begin{slide}
\begin{center}
\heading{RESULTS OF REES }
\end{center}

Let $I$ be a regular proper ideal of a Noetherian ring $R$.

For each  $x \in R$,   let $v_I(x)$ $=$ $\max \{k \in \N  \mid x
\in I^k\}$, and define


  $$\overline v _I (x)=\lim _{k \rightarrow \infty}
(\frac{v_I(x^k)}{k}).$$

Rees established that:

(a)  $\overline v _I(x)$  is well defined;

(b)  for each  $k$ $\in$ $\N$ and  $x$ $\in$ $R$, $\overline v _I
(x)$ $\ge$ $k  \iff x$ $\in$ $(I^k)_a$

(c) there exist discrete valuations  $v_1,\dots,v_g$  defined on
$R$,   and

positive integers $e_1,\dots,e_g$ such that, for each $x$ $\in$
$R$,

$\overline v _I(x) = \min \{\frac{v_i(x)}{e_i} \mid i =
1,\dots,g\}$.



\end{slide}


\begin{slide}
\begin{center}
\textbf{Describing the Rees Valuation Rings}
\end{center}

For simplicity, assume $R$ is a Noetherian integral domain with
field

of fractions $F$. Let  $t$  be an indeterminate,  The {\bf Rees
ring } of $R$

with respect to $I$ is the graded subring

$$
\mathbf R = R[t^{-1}, It] = \bigoplus_{n \in \Z}I^nt^n
$$
of the Laurent polynomial ring $R[t^{-1}, t]$. The integral
closure

$B$ of $R[t^{-1}, It]$ is a Krull domain, and $B_P$ is a DVR for
each

minimal prime $P$ of $t^{-1}B$. The set $\Rees I$ of {\bf Rees
valuation

rings} of $I$ is  precisely the set of rings  $V = B_P \cap F$,
where $P$ is

a minimal prime of $t^{-1}B$.

\end{slide}





\begin{slide}
\begin{center}
\heading{THE REES  INTEGERS OF $I$}
\end{center}
Let $(V_1, N_1), \ldots, (V_g, N_g)$ be the Rees valuation rings
of $I$. The

integers $(e_1, \ldots, e_g)$, where $IV_i = N_i^{e_i}$, are the
{\bf Rees integers }

of $I$.

{\bf PROPOSITION}. A sufficient condition for $I$ to be
projectively

full is that $\gcd(e_1, \ldots, e_g) = 1$. This is not a necessary
condition.

\medskip

{\bf EXAMPLE 4}. Let $(R,M)$ be a 2-dimensional regular local ring

with $M = (x, y)R$. The ideal $I = (x, y^2)R$ is integrally closed
with

unique Rees valuation ring $V = R[x/y^2]_{MR[x/y^2]}$. The
integrally

closed ideals projectively equivalent to $I$ are precisely the
powers

$I^n$ of $I$. Thus $I$ is projectively full. The maximal ideal of
$V$ is

$N = yV$ and $IV = N^2$, so the gcd is two not one.

\end{slide}


\begin{slide}

\heading{Projective Equivalence and Rees Valuation Rings}

Recall that ideals $I$ and $J$ are projectively equivalent if

$(I^m)_a = (J^n)_a$ for some $m,n \in \N$. If $I$ and $J$ are
projectively

equivalent, then $\Rees I = \Rees J$. The converse holds if $I$ or
$J$

has only one Rees valuation ring, but fails in general. Steve

McAdam, Jack Ratliff and Judy Sally show in


{\em Integrally closed projectively equivalent ideals}, in
Commutative

Algebra, MSRI Pub. 15, 1988, 391-405

that if $I$ and $J$ are projectively equivalent, then the Rees
integers

of $I$ and $J$ are proportional. The converse also holds:  if

$\Rees I = \Rees J$ and the Rees integers of $I$ and $J$ are
proportional,

then $I$ and $J$ are projectively equivalent.


\medskip

\end{slide}


\begin{slide}
\begin{center}
\textbf{Projectively full ideals of a 2-dim RLR}
\end{center}

\medskip

{\bf EXAMPLE 4}. (continued) $(R,M)$ is a 2-dim regular local ring

\medskip

with $M = (x, y)R$. \,\, \, Zariski's theory of unique
factorization of

\medskip

complete (= integrally closed) ideals of $R$ as  finite products
of

\medskip

simple complete ideals implies $\mathbf P(I)$ is projectively full
for each

\medskip


nonzero proper ideal $I$ of $R$. The ideal $I$ has a unique Rees

\medskip



valuation ring if and only if $I$ is a power of a simple complete
ideal.

\medskip



If $I$ factors as \qquad $I = I_1^{f_1} \cdots I_g^{f_g}$,

\medskip

where $I_1, \ldots, I_g$ are distinct simple complete ideals, then
$I$ is

\medskip

projectively full if and only if $\gcd(f_1, \ldots, f_g) = 1$.





\end{slide}



\begin{slide}
\begin{center}
\textbf{Projective fullness and Rees integers }
\end{center}

\medskip

{\bf EXAMPLE 4}. (continued) $(R,M)$ is a 2-dim regular local ring

\medskip


with $M = (x, y)R$, and $I = I_1^{f_1} \cdots I_g^{f_g}$ is the
factorization of the

\medskip


$M$-primary ideal $I$ as a product of distinct simple complete
ideals.

\medskip


How do the integers $f_1, \ldots, f_g$ relate to the Rees integers
of $I$?

\medskip


The simple complete $M$-primary ideals of $R$ are in one-to-one

\medskip



correspondence with the prime divisors birationally dominating
$R$.

\medskip



Thus the Rees valuation rings of $I$ are  $(V_1, N_1), \ldots,
(V_g, N_g)$, where

\medskip


$(V_j, N_j)$ corresponds to $I_j$. If $I_jV_j = N_j^{c_j}$, then
the Rees integers

\medskip



of $I$ are \,\,  $e_1 = c_1f_1, \, \, \ldots,\,\,  e_g = c_gf_g$.

\end{slide}











\begin{slide}
{\bf EXAMPLE 4}. (continued) $(R,M)$ is a 2-dim regular local ring

with $M = (x, y)R$. Let $I = (x^2, xy^2, y^3)R$. Notice that

$J = (x^2, y^3)R$ is a reduction of $I$ and $JI = I^2$, so the
reduction

number $r_J(I) = 1$. Let
$$V = R[\frac{xy^2}{x^2}, \frac{y^3}{x^2}]_{MR[\frac{xy^2}{x^2}, \frac{y^3}{x^2}]} = R[\frac{y^2}{x},
\frac{y^3}{x^2}]_{MR[\frac{y^2}{x}, \frac{y^3}{x^2}]}.
$$
One sees that $V$ is a valuation ring with maximal ideal

$N = (y^2/x)V$, and $I$ is a simple complete ideal. The ideals of

$R$ that are contracted from $V$ descend as follows:

$M = N^2 \cap R \supsetneq (x, y^2)R = N^3 \cap R \supsetneq M^2 =
N^4 \cap R \supsetneq (x, y^2)M$

$= N^5 \cap R \supsetneq I = N^6 \cap R $. The ideals in $\mathbf
P(I)$ are precisely the

ideals $I^m = N^{6m} \cap R$, for $m \in \N$.

\end{slide}



\begin{slide}
\heading{THE ASSOCIATED GRADED RING  $\mathbf G(R,I)$}

\medskip

{\bf PROPOSITION}. If $\mathbf G(R,I) = R/I \oplus I/I^2 \oplus
\cdots \oplus I^n/I^{n+1} \oplus \cdots$

has a minimal prime $p$ such that $p$ is its own $p$-primary
component

of $(0)$, then $I$ has a Rees integer equal to one. Therefore $I$
is

projectively full.

\medskip

More can be said using the Rees ring $\mathbf R = R[t^{-1}, It]$,
and the

identification $\mathbf G(R,I) = \mathbf R/t^{-1}\mathbf R$. \,\,
Let $\mathbf R'$ denote the integral

closure of $\mathbf R$.

\medskip

{\bf PROPOSITION}. The ideal $I$ has a Rees integer equal to one
if

and only if $t^{-1}\mathbf R'$ has a minimal prime $p$ such that
$t^{-1}\mathbf R'_p = p\mathbf R'_p$.



\end{slide}


\begin{slide}


{\bf EXAMPLE 5}. An example of a 2-dim normal local domain

\medskip

$(R,M)$ such that $M$ is projectively full and the associated
graded

\medskip

ring $\mathbf G(R,M)$ is not reduced. Let $F$ be an algebraically
closed  field

\medskip

with $\Char F = 0$,
 and let $R_0$ be a $2$-dim regular local domain with

 \medskip

 maximal
 ideal $M_0 = (x, y)R_0$ and coefficient field $F$, e.g.,

 \medskip

 $R_0 = F[x, y]_{(x, y)}$, or $R_0 = F[[x, y]]$.
 Then
$V_0 = R_0[y/x]_{xR_0[y/x]}$

\medskip

is the unique Rees valuation ring of $M_0$. Let

\medskip

\qquad $R = R_0[z]$, \quad where  \quad $z^2 =  x^3 + y^j$  \quad
with  \quad $j \ge 3$.

\medskip

 It is readily checked that
 $R$ is $2$-dim normal local with maximal

 \medskip

 ideal $M =
 (x, y, z)R$. Notice
 that $I = (x, y)R$ is a reduction of $M$

 \medskip

 since $z$ is integral over $I$.




\end{slide}

\begin{slide}
{\bf EXAMPLE 5}  (continued). Since $I = (x, y)R$ is  a reduction

\medskip

of $M$, every Rees valuation ring of $M$ is an   extension  of
$V_0$.

\medskip

Let $V$ be a Rees valuation ring of $M$ and let $v$ denote the

\medskip

normalized valuation with value group $\Z$ corresponding to $V$.

\medskip


Then $v(x) = v(y)$ and the image of $y/x$ in the residue field of
$V$

\medskip


is transcendental over $F$. Since $ z^2 = x^3 + y^j$ \,\,and
\,\,$j \ge 3$,

 we have
$$
2v(z) = v(z^2) = v(x^3 + y^j) = 3v(x). $$ We conclude that  $v(x)
= 2$ and $v(z) = 3$. Therefore $V$ is ramified

\medskip

over $V_0$.  This implies  that $V$ is the unique extension of
$V_0$ and

\medskip

thus the unique Rees valuation ring of $M$.

\end{slide}


\begin{slide}
{\bf EXAMPLE 5}  (continued). For each positive integer $n$, let

\medskip

$I_n = \{ r \in R \, | \, v(r) \ge n \}$. Thus $I_2 = M$. Since
$V$ is the unique

\medskip

Rees valuation ring of $M$, we have $I_{2n} = (M^n)_a$ for each $n
\in \N$.

\medskip

To show $M$ is projectively full,  we prove that  $V$ is not the

\medskip

unique Rees valuation ring of $I_{2n+1}$ for each $n \in \N$.

\medskip

Consider the inclusions

$$ M^2 \subseteq I_4 \subset (z, x^2, xy, y^2)R := J \subseteq I_3 \subset
M. $$


Since $\lambda(M/M^2) = 3$ and since the images of $x$ and $y$ in
$M/M^2$

\medskip

are $F$-linearly independent, \,\,  $J = I_3$ and $M^2 = I_4 =
(M^2)_a$.

\end{slide}

\begin{slide}
{\bf EXAMPLE 5}  (continued). Since $x^3 = z^2 - y^j$ and $j \ge
3$,

\medskip

$L = (z, y^2)R$ is  a reduction of $I_3 = (z, x^2, xy, y^2)R$.
\,\, Indeed,

\medskip

$(x^2)^3 \in L^3$ and $(xy)^3 \in L^3$ implies $x^2$ and $xy$ are
integral over $L$.

\medskip

It follows that $V$ is not a Rees valuation of $I_3$, \, for $zV
\ne y^2V$.

\medskip

Consider $M^3 \subset I_3M \subseteq I_5 \subset I_4 = M^2$. Since
the images of

\medskip

$x^2, xy, y^2, xz, yz$ in $M^2/M^3$ are an $F$-basis, it follows
that

\medskip

$I_3M = I_5$ and $M^3 = (M^3)_a = I_6$. Proceeding by induction,

\medskip

we assume $M^{n+1} = (M^{n+1})_a = I_{2n+2}$, and consider

\medskip



$ \quad \quad M^{n+2} \subset I_3M^n \subseteq I_{2n+3} \subset
M^{n+1} = I_{2n+2}.$

\medskip

\end{slide}



\begin{slide}
{\bf EXAMPLE 5}  (continued). Since the images in
$M^{n+1}/M^{n+2}$

\medskip

of $\{x^ay^b \, | \, a + b = n+1 \} \cup \{zx^ay^b \, | \, a + b =
n \}$ is an $F$-basis,

\medskip


 $\lambda(M^{n+1}/M^{n+2}) = 2n+3$, \quad and the inequalities

 \medskip

$\lambda(M^{n+1}/I_{2n+3}) \ge n+2$ \quad and \quad
$\lambda(I_3M^n/M^{n+2}) \ge n+1$

\medskip


imply $I_3M^n = I_{2n+3}$ \quad and \quad $M^{2n+2} =
(M^{2n+2})_a$.

\medskip

Therefore the ideal $I_{2n+3}$ has a Rees valuation ring different

\medskip

from $V$, and thus is not projectively equivalent to $M$. We

\medskip


conclude that $M$ is projectively full. We have also shown

\medskip

that $M$ is a normal ideal.


\end{slide}

\begin{slide}
\heading{Questions}

\medskip

Let $(R,M)$ be a complete normal Noetherian local domain of

\medskip

altitude two.

\medskip

\begin{enumerate}
\item  What are necessary and sufficient conditions in order that
$M$

\medskip

is projectively full?

\medskip

\item What are necessary and sufficient conditions in order that
$\mathbf P(I)$

\medskip

is projectively full for each nonzero proper ideal $I$ of $R$?

\end{enumerate}





\end{slide}

\begin{slide}


{\bf EXAMPLE 6}. An example of a (complete) 2-dim normal local

\medskip

domain $(R,M)$ such that $M$ is not projectively full.  Let $F$ be
an

\medskip

algebraically closed  field with $\Char F = 0$,
 and let $R_0$ be a $2$-dim

\medskip

 regular local domain with  maximal
 ideal $M_0 = (x, y)R_0$ and

\medskip


 coefficient field $F$, e.g.,
 $R_0 = F[x, y]_{(x, y)}$, or $R_0 = F[[x, y]]$.

 \medskip
 Let $k < i$ be relatively prime positive integers $\ge 2$,
 and let

\medskip

\quad $R = R_0[z, w]$, \quad where \,\, $z^k = x^i + y^i$ \,\,and
\,\,  $w^k = x^i - y^i$.


 \medskip

 It is readily checked that
 $R$ is $2$-dim normal local with maximal

 \medskip

 ideal $M =
 (x, y, z,w)R$. Also $R$ is a free $R_0$-module of rank $k^2$.

 \medskip





\end{slide}


\begin{slide}
{\bf EXAMPLE 6}  (continued). With $R = R_0[z, w]$ as above, we
want

\medskip

 to show  that $M = (x, y, z, w)R$ has a unique Rees
valuation ring

\medskip

and that $M$ is not projectively full. $L = (x, y)R$ is a
reduction of

\medskip

$M$, \,\, for  $z^k \in (x^i, y^i)R \subseteq L^k$ \,\,  and \,\,
$w^k \in (x^i, y^i)R \subseteq L^k$ \, imply

\medskip

$z$ and $w$ are integral over $L$. Thus each Rees valuation ring
$V$ of $M$

\medskip


is an extension of the order valuation ring $V_0 =
R_0[y/x]_{xR_0[y/x]}$ of

\medskip

 $R_0$. To show there exists a unique Rees valuation ring of $M$,
 we

\medskip

 observe that $V$ as an extension of $V_0$ ramifies of  degree \,\,
 $k$ and

 \medskip

 undergoes a residue field extension of degree \, $\ge
 k$.

\end{slide}

\begin{slide}
{\bf EXAMPLE 6}  (continued). To show $V$ ramifies of  degree \,\,
$k$

\medskip
over $V_0$, observe that $kv(z) = v(z^k) = v(x^i + y^i) = iv(x) =
iv(y)$

\medskip

implies $v(z) = i$ and $v(x) = v(y) = k$. Similarly, $v(w) = i$.
Let

\medskip
$N$ denote the maximal ideal of $V$. We have $(z, w)V = N^i$ and

\medskip
$MV = (x, y)V = N^k$. The residue field of $V_0$ is
$F(\overline\tau)$, where $\overline \tau$

\medskip
is the image of $\tau = y/x$ and is transcendental over $F$. Now
$w/z$ is a

\medskip

unit of $V$ and $(\frac{w}{z})^k  = \frac{x^i - y^i}{x^i + y^i} =
\frac{1 - \tau^i}{1 + \tau^i}$. It follows that the residue

\medskip


class of $w/z$ in $V/N$ is algebraic of degree $k$ over
$F(\overline \tau)$. \, This

\medskip
proves that $V$ is the unique Rees valuation ring of $M$.



\end{slide}



\begin{slide}
{\bf EXAMPLE 6}  (continued). To show $M$ is not projectively

\medskip

full, notice that $(z^k, w^k)R = (x^i + y^i, x^i - y^i)R$, and
since

\medskip


$\Char F \ne  2$, \,\, $(x^i + y^i, \,  x^i - y^i)R = (x^i,
y^i)R$. \,\, Since $(x^i, y^i)R$


\medskip
 is a reduction of $M^i$, we have $((z^k, w^k)R)_a =
(M^i)_a$. \, Also

\medskip

$((z^k, w^k)R)_a = ((z, w)^kR)_a$. Therefore $(z, w)R$ and $M$ are
\medskip

 projectively equivalent, so $V$ is the unique Rees valuation ring

\medskip

of $(z, w)R$, and $((z, w)R)_a = N^i \cap R$. We have $M = N^k
\cap R$,

\medskip

$M^nV = N^{nk}$ and $(M^n)_a = N^{nk} \cap R$, for each positive
integer $n$.

\medskip

Since $i$ is not a multiple of $k$, $M$ is  not projectively full.


\end{slide}

\begin{slide}
\heading{RATIONAL SINGULARITIES}

 Joe Lipman in his paper

{\em Rational singularities, with applications to algebraic
surfaces and

unique factorization}, Publ. Math. Inst. Hautes \'Etudes Sci.

N$^o$ 36 (1969), 195-279.


\medskip
extended Zariski's theory of complete ideals in 2-dim regular local


 \medskip
rings to 2-dim normal local rings $R$ having a rational
singularity.

\medskip
Lipman proved that $R$ has unique factorization of complete ideals

\medskip
if and only if the completion of $R$ is a UFD. For $R$ having
this

\medskip

property, it follows that $\mathbf P(I)$ is projectively full for
each nonzero

\medskip

proper ideal $I$, e.g., $R = F[[x, y, z]]$, \,\, where $z^2 + y^3
+ x^5 = 0$.




\end{slide}







\begin{slide}
\heading{MORE ON RATIONAL SINGULARITIES}

Let $(R,M)$ be a normal local domain of altitude two. G\"ohner

\medskip


proves that if $R$ has a rational singularity, then the set of
complete

\medskip

 asymptotically irreducible ideals associated to a prime
 $R$-divisor $v$

 \medskip

 consists of the powers of an ideal $A_v$ which is uniquely
 determined

\medskip

 by $v$. In our terminology, this says that if $I$ is a nonzero
 proper

\medskip

 ideal of $R$ having only one Rees valuation ring, then
 $\mathbf P(I)$ is

\medskip

 projectively full. G\"ohner's proof involves choosing a

\medskip

 desingularization $f : X \to \Spec R$ such that $v$ is centered
 on a

 \medskip

 component $E_1$ of the closed fiber on $X$.

\end{slide}

\begin{slide}
\heading{THE CLOSED FIBER 1}
 Let $E_2, \ldots, E_n$ be the other
components of the closed fiber on $X$.

\medskip

Let $E_X$ denote the group of divisors having the form $\sum_{i =
1}^nn_iE_i$,

\medskip

with $n_i \in \Z$, and consider

\medskip

\quad  $E_X^+ = \{D \in E_X \, | \, D \ne 0 \text{ and } (D \cdot
E_i) \le 0  \text{ for all } 1 \le i \le n \}$, \,\, and

\medskip

$E_X^{\#} = \{ D \in E_X \, |\, D \ne 0 \text{ and } O(-D) \text{
is gen.  by its sections over } X \}$.

\medskip
Lipman shows $E_X^{\#} \subseteq E_X^+$ and  equality holds  if
$R$ has a rational

\medskip


singularity. Also, if $D = \sum_in_iE_i \in E_X^+$, then
negative-definiteness

\medskip
of the intersection matrix $(E_i \cdot E_j)$ \,\, implies $n_i \ge
0$ for all $i$.

\end{slide}


\begin{slide}
\heading{THE CLOSED FIBER 2}

For if $D \in E_X^+$ and $D = A - B$, where $A$ and $B$ are
effective, then

\medskip

$(A-B \cdot B) \le 0$ and $(A \cdot B) \ge 0$ imply $(B \cdot B)
\ge 0$, \,\, so $B = 0$.

\medskip
Let $v = v_1, v_2, \ldots, v_n$ denote the discrete valuations
corresponding

\medskip

to $E_1, \ldots, E_n$.  Associated with  $D = \sum_in_iE_i \in
E_X^{\#}$ one defines  the

\medskip

complete $M$-primary ideal $I_D = \{ r \in R \, |\, v_i(r) \ge n_i
\text{ for } 1 \le i \le n \}$.

\medskip
This sets up  a one-to-one correspondence between elements of
$E_X^{\#}$

\medskip

and complete $M$-primary ideals that generate invertible
$O_X$-ideals.


\end{slide}

\begin{slide}
\heading{THE CLOSED FIBER 3} Lipman suggested to me the following
proof that $\mathbf P(I)$ is proj.  full

\medskip

for each complete $M$-primary ideal $I$ if $R$ has a rational
singularity.

\medskip

Fix a desingularization $f : X \to \Spec R$ such that $I$ generates
an

\medskip

invertible $O_X$-ideal and let $D = \sum_in_iE_i \in E_X^{\#}$ be
the divisor

\medskip

associated to $I$. Let $g = \gcd \{ n_i \} $. Since $E_X^+ =
E_X^{\#}$, $(1/g) D \in E_X^{\#}$.

\medskip

The ideals $J \in \mathbf P(I)$ correspond to divisors in $E_X^{\#}$
that are integral

\medskip

multiples of $(1/g)D$. Thus if $K$ is the complete $M$-primary ideal

\medskip


associated to $(1/g)D$, then each $J \in \mathbf P(I)$ is the
integral closure of

\medskip

a power of $K$, so $\mathbf P(I)$ is projectively full.


\end{slide}

\begin{slide}
\heading{INTEGRAL EXTENSIONS}

{\bf QUESTION}. Let $I$ be a nonzero ideal of a Noetherian domain
$R$.

\medskip

 Does there always exist a finite integral extension domain $B$ of $R$


\medskip

such that $\mathbf P(IB)$ is projectively full?

\medskip



Let $I = (b_1, \ldots, b_g)R$ be a nonzero regular ideal of the
Noetherian

\medskip

ring $R$, let $R_g = R[X_1, \ldots, X_g]$ and $K = (X^c_1 - b_1,
\ldots, X^c_g - b_g)R_g$,


\medskip

where $c$ is a positive integer. Then $A = R_g/K$ is a finite free

\medskip


integral extension of rank $c^g$ of $R$. Let $x_i = X_i \mod K$.
Then

\medskip

$J = (x_1, \ldots, x_g)A$ is such that $(IA)_a = (J^c)_a$, so $IA$
and $J$ are

\medskip

projectively equivalent.

\end{slide}

\begin{slide}

\heading{MAIN THEOREM}

{\bf THEOREM}. Let $I = (b_1, \ldots, b_g)R$ be as above and let
$(V,N)$

\medskip

be a  Rees valuation ring of $I$. Assume  that  $b_iV = IV = N^c$
for

\medskip

each $i$  with $1 \le i \le g$,  and that $c$ is a unit of $R$. Then
the finite

\medskip

free  integral  extension $A = R_g/K$ is such that $J = (x_1,
\ldots, x_g)A$

\medskip

is projectively  full. Hence $\mathbf P(IA)$ is projectively full.
If $R$ is an

\medskip

integral domain and $p$ is a minimal prime of $A$,  then $B = A/p$

\medskip


is an integral extension domain of $R$ such that $\mathbf P(IB)$ is

\medskip

projectively full.


\end{slide}


\end{document}
