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\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}