\documentclass[10pt]{amsart} \usepackage[cmtip,all]{xy} \usepackage{amssymb,latexsym, xspace, enumerate} \usepackage{amssymb,amscd} \usepackage{graphicx} \usepackage{showkeys} \usepackage{amsfonts} \usepackage{amsmath} \usepackage{amsthm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \newtheorem{question}[theorem]{Question} \newtheorem{remark}[theorem]{Remark} \newtheorem{setting}[theorem]{Setting} \newtheorem{claim}[theorem]{Claim} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{discussion}[theorem]{Discussion} \def\alert#1{\smallskip{\hskip\parindent\vrule% \vbox{\advance\hsize-2\parindent\hrule\smallskip\parindent.4\parindent% \narrower\noindent#1\smallskip\hrule}\vrule\hfill}\smallskip} \setlength{\textwidth}{5.5in} \setlength{\textheight}{8.5in} \setlength{\abovedisplayskip}{14pt} \setlength{\belowdisplayskip}{14pt} \setlength{\abovedisplayshortskip}{14pt} \setlength{\belowdisplayshortskip}{14pt} \setlength{\oddsidemargin}{.5in} \setlength{\evensidemargin}{.5in} \setlength{\topmargin}{-.25in} \DeclareMathOperator\injdim{injdim} \DeclareMathOperator\mSpec{mSpec } \DeclareMathOperator{\m}{\mathbf m} \DeclareMathOperator{\n}{\mathbf n} \DeclareMathOperator{\gr}{grade} \DeclareMathOperator{\Ann}{Ann } \DeclareMathOperator{\Ass}{Ass } \DeclareMathOperator{\Soc}{Soc } \DeclareMathOperator{\ord}{ord } \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Proj}{Proj} \DeclareMathOperator{\Reg}{Reg} \DeclareMathOperator{\epd}{epd} \DeclareMathOperator{\hpd}{hpd} \DeclareMathOperator{\xpd}{xpd} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\C}{\mathcal C} \DeclareMathOperator{\hgt}{ht} \DeclareMathOperator{\depth}{depth} \DeclareMathOperator{\trdeg}{trdeg} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\grade}{grade} \DeclareMathOperator\projdim{projdim} \DeclareMathOperator{\Supp}{Supp} \DeclareMathOperator{\Rees}{Rees} \DeclareMathOperator{\Quot}{Quot} \DeclareMathOperator{\Min}{Min} \DeclareMathOperator{\Gr}{gr} \DeclareMathOperator{\Rad}{Rad} \DeclareMathOperator{\bl}{Bl } \DeclareMathOperator{\an}{\mathbb{A}^n} \DeclareMathOperator{\pn}{\mathbb{P}^n} \DeclareMathOperator{\a0an}{a_0, a_1, \ldots, a_n} \DeclareMathOperator{\pt}{\mathcal{O}} \DeclareMathOperator{\r1}{R_1} \begin{document} \title[Notes on blowups]{Notes on blowups} \maketitle \date{\today } \baselineskip 18 pt I would like to expand on these notes from last summer as class notes for what we cover this fall. \section{May 19, 2009} \subsection{Motivation} \begin{proposition} Let $(R, \m)$ be a two-dimensional regular local ring with $R/ \m = k = \overline{k}$ algebraically closed. Let $I$ be a complete $\m$-primary ideal in $R$ and $\lambda(\m/I) = 1$. Then $I$ is projectively full. \end{proposition} \begin{definition} \begin{enumerate} \item Let $I$ and $J$ be any ideals in a commutative ring $R$. Then $I$ and $J$ are \textit{projectively equivalent} if there exist positive integers $m$ and $n$ such that $\overline{I^m} = \overline{J^n}$. \item A regular ideal $I$ in a Noetherian ring $R$ is \textit{projectively full} if the only integrally closed ideals which are projectively equivalent to $I$ are of the form $\overline{I^n}$ for a positive integer $n$. \end{enumerate} \end{definition} Let $(R, \m)$ be a two-dimensional Muhly local domain, i.e. a two-dimensional integrally closed Noetherian local domain with $k = \overline{k}$ and $\Gr_{\m} (R)$ an integrally closed domain. \begin{remark} If $(R, \m)$ is a two-dimensional Muhly local domain, then the blowup $\bl_{\m}(R)$ at $\m$ is a desingularization of $R$. We want to understand the blowup. \end{remark} \subsection{Affine Varieties (Material from Chapter 1 of Hartshorne)} Let $k$ be a fixed algebraically closed field. Define \textit{affine $n$-space over $k$} to be the set of all $n$-tuples of elements of $k$, denoted $\mathbb{A}^n_k = \mathbb{A}^n$. Let $p \in \an$ be a point. Write $p = (a_1, \ldots, a_n)$, $a_i \in k$ the coordinates of $p$. Let $A = k[x_1, \ldots, x_n]$. We can regard any $f \in A$ as a function $f: \an \rightarrow k$ by $f(p) = f(a_1, \ldots, a_n)$. Also for any $f \in A$ define the \textit{zero set of $f$} to be $$Z(f) = \{ p \in \an | f(p) = 0 \}.$$ More generally, if $T \subset A$ define $$Z(T) = \{p \in \an | f(p) = 0 \mbox{ for all } f \in T \}$$ to be the zero set of $T$. \begin{definition} A subset $Y$ of $\an$ is an algebraic set if there exists $T \subset A$ such that $Y = Z(T)$. \end{definition} \begin{proposition} Let $\{Y_i\}_i$ be a collection of algebraic sets in $\an$. Then \begin{enumerate} \item $Y_i \cup Y_j$ is an algebraic set \item $\cap_i Y_i$ is an algebraic set \item $\an$ and $\emptyset$ are algebraic sets \end{enumerate} \end{proposition} We define a topology on $\an$ where the open sets are complements of algebraic sets, called the \textit{Zariski topology}. \begin{definition} A nonempty subset $Y$ of a topological space $X$ is \textit{irreducible} if it cannot be expressed as the union of two proper subsets, each of which is closed in $Y$. \end{definition} \begin{example} Any non-empty open subset $Y$ of an irreducible space $X$ is irreducible and dense in $X$. \end{example} \begin{example} Consider the Zariski topology on $\mathbb{A}^1$. Let $Y \subset \mathbb{A}^1$ be an algebraic set. Then there exists a subset $T$ of $A$ such that $Y = Z(T) = Z(f)$ for some $f \in A$ since $A = k[x]$ is a PID. Since $k$ is algebraically closed we can write $f = c(x - a_1) \cdots (x - a_t)$ such that $a_i \in k$. Hence $Y = Z(f) = \{a_1, \ldots, a_t\}$. Hence we can see that $\mathbb{A}^1$ is irreducible. \end{example} \begin{definition} \begin{enumerate} \item An \textit{affine algebraic variety} (or an affine variety) is an irreducible closed subset of $\an$. \item An open subset of an affine variety is a \textit{quasi-affine variety}. \end{enumerate} \end{definition} \begin{definition} Let $Y \subset \an$ be any subset. Define the ideal of $Y$ in $A$ to be $$\mathcal{I}(Y) = \{f \in A | f(p) = 0 \mbox{ for all } p \in Y\}.$$ \end{definition} \begin{proposition} \begin{itemize} \item[(a)] If $T_1 \subset T_2$ are subsets of $A$, then $Z(T_1) \supset Z(T_2)$. \item[(b)] If $Y_1 \subset Y_2$ are subsets of $\an$, then $\mathcal{I}(Y_1) \supset \mathcal{I}(Y_2)$. \item[(c)] Let $Y_1$ and $Y_2$ be subsets of $\an$, then $\mathcal{I}(Y_1 \cup Y_2) = \mathcal{I}(Y_1) \cap \mathcal{I}(Y_2)$. \item[(d)] Let $I$ be any ideal in $A$, then $\mathcal{I}(Z(I)) = \sqrt{I}$. \item[(e)] Let $Y$ be a subset of $\an$, then $Z(\mathcal{I}(Y)) = \overline{Y}$. \end{itemize} \end{proposition} Note that part (d) above is due to Hilbert's Nullstellensatz. \begin{corollary} There is a one-to-one, inclusion reversing correspondence $$\{ \mbox{algebraic sets in } \an \} \leftrightarrow \{ \mbox{radical ideals in }A \}.$$ Furthermore, an algebraic set is irreducible if and only if its ideal is prime. \end{corollary} \begin{example} $\an$ is irreducible. \end{example} \begin{example} Let $f_n$ be an irreducible polynomial in $A = k[x_1, \cdots, x_n]$. Then $(f_n)$ is a prime ideal in $A$ and hence $Z(f_n)$ is irreducible. We call $Z(f_n)$ an (affine) curve for $n = 2$, surface for $n = 3$, and hypersurface for $n > 3$. \end{example} \begin{example} A maximal ideal $m$ of $A = k[x_1, \cdots, x_n]$ corresponds to a minimal nonempty irreducible closed subset of $\an$, which must be a point. \end{example} \begin{definition} If $Y \subset \an$ is an affine algebraic set, the \textit{affine coordinate ring of Y} is $A(Y) = A/\mathcal{I}(Y)$. \end{definition} Remark: If $Y$ is an affine variety then $A(Y)$ is a domain and a finitely generated $k$-algebra. Conversely, any finitely generated $k$-algebra $B$ which is a domain is the affine coordinate ring of some affine variety. \section{May 21, 2009} \subsection{Projective Varieties (Hartshorne, Section 1.2)} Let $k$ be a fixed algebraically closed field. Define \textit{projective $n$-space over $k$} to be the set of equivalence classes of $(n+1)$-tuples, $(a_0, a_1, \ldots, a_n)$, of elements in $k$ with not all $a_i$ zero under the equivalence relation $$(a_0, a_1, \ldots, a_n) \sim \lambda(a_0, a_1, \ldots, a_n)$$ for $\lambda \in k - \{0\}$, denoted $\mathbb{P}^n_k = \pn$. Let $P \in \pn$ be a point. Write $$P = [a_0, a_1, \ldots, a_n] = \{(a_0, a_1, \ldots, a_n) | (a_0, a_1, \ldots, a_n) \sim \lambda (a_0, a_1, \ldots, a_n), \lambda \in k - \{0\}\}.$$ This set is called the \textit{set of homogeneous coordinates of $P$}. \begin{example} \begin{enumerate} \item Consider $\mathbb{P}^0$. For all $a_0 \neq 0$, $[a_0] = a_0[1]$ so that $\mathbb{P}^0$ is a point. \item Consider $\mathbb{P}^1$. Let $[a_0, a_1] \in \mathbb{P}^1$. Then if $a_0 \neq 0$, we have $[a_0, a_1] = [1, a_1/a_0]$. Otherwise $a_0 = 0$ but then we must have $a_1 \neq 0$ so that $[a_0, a_1] = [0, a_1] = a_1[0,1]$. We can think of projective 1-space as the collection of lines through the origin. \end{enumerate} \end{example} Let $S = k[x_0, x_1, \ldots, x_n]$ be a graded ring with $\deg x_i = 1$ for all $i$. Let $f \in S$ be any homogeneous polynomial of degree $d$. Then we can define the \textit{zero set of $f$} to be $$Z(f) = \{ P \in \pn | f(p) = 0 \}.$$ Notice that this definition is well-defined since $f$ is a homogeneous polynomial. Similarly for $T \subset S^h$ where $S^h$ is the set of all homogeneous polynomials in $S$ we define $$Z(T) = \{P \in \pn |f(P) = 0 \mbox{ for all } f \in T\}.$$ \begin{definition} A subset $Y$ of $\pn$ is an \textit{algebraic set} if there exists $T \subset S^h$ such that $Y = Z(T)$. \end{definition} \begin{proposition} Let $\{Y_i\}_i$ be a collection of algebraic sets in $\pn$. Then \begin{enumerate} \item $Y_i \cup Y_j$ is an algebraic set \item $\cap_i Y_i$ is an algebraic set \item $\emptyset$ and $\pn$ are algebraic sets \end{enumerate} \end{proposition} Now we can define the \textit{Zariski topology} on $\pn$ by letting the open sets be the complements of the algebraic sets. Notice that by exercise 2.2 on page 11 of Hartshorne for an ideal $a \subset S$ we have $Z(a) = \emptyset$ if and only if $\sqrt{a} = $ either $S$ or the ideal $S_+ = \bigoplus_{d >0} S_d$ if and only if $a \supset S_d$ for some $d > 0$. Hence a proper ideal can have an empty zero set, unlike in affine $n$-space. \begin{definition} \begin{enumerate} \item Let $Y \subset \pn$. Define the \textit{homogeneous ideal} of $Y$ in $S$ to be $$\mathcal{I}(Y) = \{ f \in S | f \mbox{ is a homogeneous and } f(P) = 0 \mbox{ for all } P \in Y\}.$$ \item If $Y$ is an algebraic set in $\pn$, define the \textit{homogeneous coordiate ring} of $Y$ to be $S(Y) = S/ \mathcal{I}(Y)$. \end{enumerate} \end{definition} If $f \in S$ is a linear homogeneous polynomial then $Z(f)$ is a hyperplane. In particular, if $H_i = Z(x_i)$ is the zero set of $x_i$ then $U_i = \pn - H_i$ is an open set for $i = 0, 1, \ldots, n$. Then $\pn$ is covered by the open sets $U_i$: Let $P \in \pn$ and choose $i$ so that $P = [a_0, a_1, \ldots, a_n]$ with $a_i \neq 0$. Then $P \in U_i$. Now define a mapping $\varphi_i: U_i \rightarrow \an$ by $[a_0, a_1, \ldots, a_n] \mapsto (\frac{a_0}{a_i}, \ldots, \hat{\frac{a_i}{a_i}}, \ldots, \frac{a_n}{a_i})$. \begin{proposition} $\varphi_i$ is a homeomorphism for all $i$. \end{proposition} Now notice exercise 2.4 on page 11 of Hartshorne, which says that there is a 1-1 inclusion reversing correspondence between algebraic sets in $\pn$ and homogeneous radical ideals of $S$ not equal to $S_+$. \begin{example} Consider again $\mathbb{P}^1$ and the correspondence $$\{ \mbox{algebraic sets of } \pn \} \leftrightarrow \{I \mbox{ homogeneous ideal in } S = k[x_0, x_1] | \sqrt{I} = I \neq S_+ \}.$$ Under this correspondence we have $$[1,0] \leftrightarrow (x_1)$$ $$[0,1] \leftrightarrow (x_0)$$ $$[1,1] \leftrightarrow (x_0-x_1)$$ $$[1,2] \leftrightarrow (2x_0 - x_1)$$ so for $f \in S^h$ of $\deg f = d$ then $f = g_1 \cdots g_d$ with $g_i$ linear for all $i$. \end{example} \subsection{The Blowup is regular} \begin{proposition} Let $(R, \m)$ be a two-dimensional normal local domain such that the associated graded ring $\Gr_{\m}(R)$ a normal domain. Then $R[\frac{\m}{a}] := A$ is a regular ring for all $0 \neq a \in \m$. Moreover, if $P$ is a maximal ideal of $A$ that contains $a$, then $A_P/aA_P$ is a DVR. Thus $a$ is part of a minimal set of generators of $PA_P$. \end{proposition} \begin{proof} Since $\Gr_{\m}(R)$ is a domain the $\m$-adic order function defines a valuation on $R$. Hence the powers of $\m$ are integrally closed so the Rees ring $R[\m t]$ and the extended Rees ring $R[\m t, t^{-1}]$ are normal domains. Since the graded ring $R[\m t][\frac{1}{at}] = R[\frac{\m}{a}][at, \frac{1}{at}]$ is a normal domain that is a Laurent polynomial ring over its degree zero component $R[\frac{\m}{a}] = A$, it follows that $A$ is a normal domain. To show that $A$ is regular, it remains to show that $A_P$ is a RLR for every $P \in \Spec A$ with $\hgt P = 2$. Notice that $\hgt P = 2$ implies $P \cap R = \m$ and $a \in P$. Let $$N = PR[\frac{\m}{a}][at, \frac{1}{at}] \cap R[\m t, t^{-1}].$$ Then $$R[\m t, t^{-1}] \subset R[\frac{\m}{a}][at, \frac{1}{at}] = R[\m t, t^{-1}][\frac{1}{at}]$$ so $NR[\frac{\m}{a}][at, \frac{1}{at}] = PR[\frac{\m}{a}][at, \frac{1}{at}]$. Hence $$ R[\m t, t^{-1}]_N = A[at, \frac{1}{at}]_{PA[at, \frac{1}{at}]} = A_P(at).$$ Now $A_P$ is a two-dimensional normal local domain, and $A_P$ is a RLR if and only if $A_P(at)$ is a RLR. Since $at$ is a unit of $A_P(at)$, the principal ideals generated by $a$ and $t^{-1}$ are equal in $A_P(at)$. It suffices to show that $R[\m t, t^{-1}]_N$ is a RLR, and that $R[\m t, t^{-1}]_N/t^{-1}R[\m t, t^{-1}]_N$ is a DVR. By hypothesis $$\Gr_{\m}(R) \cong R[\m t, t^{-1}]/(t^{-1}R[\m t, t^{-1}])$$ is a normal domain. Since $a \in P$, we have $t^{-1} \in NR[\m t, t^{-1}]$. Hence $N/(t^{-1})$ is a height one prime in $R[\m t, t^{-1}]/(t^{-1}R[\m t, t^{-1}])$. Therefore $R[\m t, t^{-1}]_N/(t^{-1}R[\m t, t^{-1}]_N)$ is a DVR. We conclude that R$[\m t, t^{-1}]_N$ is a RLR. \end{proof} \section{May 26, 2009} \subsection{Morphisms (Section 1.3 of Hartshorne)} \begin{definition} \begin{enumerate} \item (Affine Case) Let $Y$ be a quasi-affine variety of $\an$. A function $f: Y \rightarrow k$ is \textit{regular at a point $p \in Y$} if there exists an open neighborhood $U$ of $p \in U \subset Y$ and there exist polynomials $g$ and $h$ in $A = k[x_1, \ldots, x_n]$ such that $h$ is nowhere zero on $U$ and $f = g/h$ on $U$. We say $f$ is \textit{regular on $Y$} if $f$ is regular at every point of $Y$. \item (Projective Case) Let $Y$ be a quasi-projective variety of $\pn$. A function $f: Y \rightarrow k$ is \textit{regular at a point $p \in Y$} if there exists an open neighborhood $U$ of $p \in U \subset Y$ and there exist homogeneous polynomials $g$ and $h$ of the same degree in $S = k[x_0, \ldots, x_n]$ such that $h$ is nowhere zero on $U$ and $f = g/h$ on $U$. We say $f$ is \textit{regular on $Y$} if $f$ is regular at every point of $Y$. \end{enumerate} \end{definition} \begin{remark} \begin{enumerate} \item Regular functions are continuous when we identify $k$ with $\mathbb{A}^1$. \item If $f$ and $g$ are regular functions on a variety $X$ and $f = g$ on some nonempty open subset $U \subset X$, then $f = g$ on $X$. Proof: $f= g$ on $X$ if and only if $(f - g)(Q) = 0$ for all $Q \in X$ if and only if $X = Z(f - g)$, but $U \subset Z(f - g) \subset X$ and $U$ is dense in $X$ so $\overline{U} = X \subset Z(f -g) \subset X$. \end{enumerate} \end{remark} \begin{definition} A variety over $k$ is any affine, quasi-affine, projective, or quasi-projective variety. If $X$ and $Y$ are two varieties, a \textit{morphism} is a continuous map $\phi:X \rightarrow Y$ such that for every open set $V \subset Y$ and for every regular function $f: V \rightarrow k$ we have that $f \circ \phi: \phi^{-1}(V) \rightarrow k$ is regular. $$\xymatrix{{\phi^{-1}(V)}\ar@{-->}[rr]\ar@{->}[rd]&{\phi}&{V}\ar@{->}[ld]\\{}&{k}&{}}$$ \end{definition} \begin{remark} If $\phi$ and $\psi$ are morphisms then $\phi \circ \psi$ is a morphism. Also a morphism $\phi:X \rightarrow Y$ is an isomorphism if there exists a morphism $\psi:Y \rightarrow X$ such that $\psi \circ \phi = id_X$ and $\phi \circ \psi = id_Y$. Also if a morphism is an isomorphism then it is bijective and bicontinuous, but the converse is not true in general as seen in the example below. \end{remark} \begin{example} Exercise 3.2(a): Let $\phi: \mathbb{A}^1 \rightarrow \mathbb{A}^2$ is defined by $t \mapsto (t^2, t^3)$. Then $\phi$ is bijective and bicontinuous, but $\phi$ is not an isomorphism since $A(X) \cong k[T]$ and $A(Y) \cong k[x, y]/(y^2 - x^3)$ by Corollary 3.7 in Hartshorne: \begin{corollary} If $X$ and $Y$ are affine varieties, then $X$ and $Y$ are isomorphic if and only if $A(X)$ and $A(Y)$ are isomorphic as $k$-algebras. \end{corollary} \end{example} \begin{definition} Let $Y$ be a variety. Define $\mathcal{O}(Y)$ to be the ring of all regular functions on $Y$. For $p \in Y$ define $\mathcal{O}_{p, Y}$ or simply $\mathcal{O}_p$ to be the ring of germs of regular functions on $Y$ near $p$, i.e. elements of $\mathcal{O}_p$ are a pair $$ where $U$ is an open subset of $Y$ containing $p$ and $f$ is a regular function on $U$ where we identify $$ and $$ if $f = g$ on $U \cap V$. \end{definition} Note that $\mathcal{O}_p$ is a local ring with maximal ideal $\m_p = \{ \in \mathcal{O}_p: f(p) = 0\}$ as follows: If $ \in \mathcal{O}_p$ with $f(p) \neq 0$ then $$ is a unit since $1/f$ is a regular function on $V:=U - (U \cap Z(f))$. \begin{definition} Let $Y$ be a variety. Define the function field of $Y$, denoted $K(Y)$, to be the set of all equivalence classes of pairs $$ where $U$ is a nonempty open set in $Y$, $f$ is regular on $Y$, and we identify $$ and $$ if $f = g$ in $U \cap V$. Elements of $K(Y)$ are callled rational functions on $Y$. \end{definition} Notice that $K(Y)$ is a field and that $\mathcal{O} \subset \mathcal{O}_p \subset K(Y)$ and if replace $Y$ by an isomorphic variety then the corresponding rings are isomorphic so they are invariants up to isomorphism. \begin{theorem} Let $Y \subset \an$ be an affine variety with affine coordinate ring $A(Y)$. Then \begin{enumerate} \item $\mathcal{O} \cong A(Y)$ \item for $p \in Y$, let $\m_p \subset A(Y)$ be the ideal of functions vanishing at $p$. Then $p \mapsto \m_p$ gives a 1-1 correspondence between points of $Y$ and maximal ideals of $A(Y)$. \item $K(Y) \cong Quot(A(Y))$ so $K(Y)$ is a finitely generated field extension of $k$ and $trdeg_k K(Y) = \dim Y$. \end{enumerate} \end{theorem} Question: Let $Y$ be any affine or quasi-affine variety. Is $Y$ a quasi-projective variety? Answer: See Hartshorne page 12 exercise 2.9 to see that every affine variety is a quasi-projective variety. \section{June 2, 2009} \subsection{An example where $(R,\m)$ is a two-dimensional normal local domain and $\Gr_{\m}(R)$ is a domain (but not normal) and the blowup of $R$ at $\m$ is regular.} \begin{example} Let $S = k[X, Y, Z]_{(X, Y, Z)}$ and $f = X^3 - Y^2Z +Z^4 \in S$. Define $R = S/fS$ and let $\n = (X, Y, Z)S$ denote the maximal ideal of $S$ and $\m = (x, y, z)$ denote the maximal ideal of $R$. We will show that the blowup of $R$ at $\m$ is regular by showing that the following are regular rings $$R \left[ \frac{y}{x}, \frac{z}{x} \right], R \left[ \frac{x}{y}, \frac{z}{y} \right], R \left[ \frac{x}{z}, \frac{y}{z} \right].$$ \end{example} \begin{proposition} Let $(S, \n)$ be a $d$-dimensional regular local ring. The blowup of $S$ at $\n$ is regular. \end{proposition} Proof: Let $\n = (x_1, \ldots, x_d)$. We will show that $S[\frac{\n}{x_1}]$ is a regular ring then by symmetry we will have $S[\frac{\n}{x_i}]$ is regular for $1 \leq i \leq d$ and since these rings cover the blowup we will be done. Let $p \in \Spec(S[\frac{\n}{x_1}])$ and $q = p \cap S$. Suppose $q = \n$. Then $\n S[\frac{\n}{x_1}] = x_1 S[\frac{\n}{x_1}]$. Now $$S \left[ \frac{\n}{x_1} \right] = S \left[ \frac{x_2}{x_1}, \frac{x_3}{x_1}, \ldots, \frac{x_d}{x_1} \right] \cong \frac{S[T_2, \ldots, T_d]}{(x_1T_2 - x_2, \ldots, x_1T_d - x_d)}$$ so that $$\frac{S \left[\frac{\n}{x_1} \right]}{\n S \left[\frac{\n}{x_1} \right]} \cong (S/ \n)[T_2, \ldots, T_d]$$ which is clearly regular. Hence $$\frac{S \left[\frac{\n}{x_1} \right]_p}{x_1 S \left[\frac{\n}{x_1} \right]_p}$$ is regular which implies that $S[\frac{\n}{x_1}]_p$ is a regular ring since $x_1 S \left[\frac{\n}{x_1} \right]_p$ is a height one principal ideal. Now suppose that $q \neq \n$. Then $x_1$ is not in $p$ so $S[\frac{\n}{x_1}]_p = S[\frac{1}{x_1}]_{pS[\frac{\n}{x_1}]_p}$ and since $S[\frac{1}{x_1}]$ is regular $S[\frac{\n}{x_1}]_p$ is regular. Therefore we can conclude that $S[\frac{\n}{x_1}]$ is regular. $\blacksquare$ \medskip Now since we have that $S = k[X, Y, Z]_{(X, Y, Z)}$ is regular we will use the following version of the Jacobian criterion to show that the rings $$R \left[ \frac{y}{x}, \frac{z}{x} \right], R \left[ \frac{x}{y}, \frac{z}{y} \right], R \left[ \frac{x}{z}, \frac{y}{z} \right]$$ are regular. \begin{theorem} [Matsumura, Theorem 30.4(ii)] \label{4.3} Let $R$ be a regular local ring and $P \in \Spec(R)$. Let $I \subset P$ be an ideal of $R$ with $\hgt IR_P = r$. If $D_1, \ldots, D_r \in Der(R)$ and $f_1, \ldots, f_r \in I$ are such that $\det (D_if_j) \notin P$ then $$IR_P = (f_1, \ldots, f_r)R_P \mbox{ and } R_P/IR_P \mbox{ is regular.}$$ \end{theorem} We have shown that $$S \left[ \frac{Y}{X}, \frac{Z}{X} \right], S \left[ \frac{X}{Y}, \frac{Z}{Y} \right], S \left[ \frac{X}{Z}, \frac{Y}{Z} \right]$$ are regular rings. Then $S[\frac{Y}{X}, \frac{Z}{X}] \rightarrow R[\frac{y}{x}, \frac{z}{x}]$ is surjective with kernel $K_1$ of height 1. Let $p \in \Spec(R[\frac{y}{x}, \frac{z}{x}])$. Then there exists $P \in \Spec(S[\frac{Y}{X}, \frac{Z}{X}])$ such that $K_1 \subset P$ and $P/K_1 = p$ and $$\frac{S[\frac{Y}{X}, \frac{Z}{X}]_P}{(K_1)_P} \cong R[\frac{y}{x}, \frac{z}{x}]_p.$$ We show that the left hand side above is a regular ring by using Theorem~\ref{4.3}. . Notice that $\hgt K_1S[\frac{Y}{X}, \frac{Z}{X}]_P = 1$. Then choose $f_1 \in I$ to be $f_1 = 1 - (\frac{Y}{X})^2(\frac{Z}{X}) +X(\frac{Z}{X})^4$. Also let $D:A \rightarrow A$ with $A = S[\frac{Y}{X}, \frac{Z}{X}]$ to be the derivation with $X \mapsto 1$, $Y/X \mapsto 0$, and $Z/X \mapsto 0$. Then $Df_1 = D(X(\frac{Z}{X})^4) = (\frac{Z}{X})^4$. If $(\frac{Z}{X})^4 \in P$ then $1 \in P$ since $f_1 \in P$ as well. Hence $Df_1 \notin P$ so by the theorem we have regularity. We similarly show that the other rings are regular, but I don't have time to type that up right now :-p \section{June 4, 2009} \subsection{An example where $(R,\m)$ is a two-dimensional normal local domain such that $\Gr_{\m}(R)$ is a domain (not normal) and the blowup of $R$ at $\m$ is NOT regular.} \begin{example} Let $S = k[X, Y, Z]_{(X, Y, Z)}$ and $f = X^3 - Y^2Z +Z^5 \in S$. Define $R = S/fS$ and let $\n = (X, Y, Z)S$ denote the maximal ideal of $S$ and $\m = (x, y, z)$ denote the maximal ideal of $R$. This example is very similar to the one given on June 2, but we will show that in this case the blowup of $R$ at $\m$ is not regular. It suffices to find an element $ 0 \neq a \in m$ and $p \in \Spec(R[\frac{\m}{a}])$ such that $R[\frac{\m}{a}]_p$ is not a regular local ring. Recall as in the previous example that the natural map $S[\frac{X}{Z}, \frac{Y}{Z}] \mapsto R[\frac{x}{z}, \frac{y}{z}]$ is surjective with kernel $K$, a prime ideal of height one. Hence it is enough to find a $Q \in \Spec(S[\frac{\n}{Z}])$ such that $K \subset Q$ and $\displaystyle{\frac{S[\frac{\n}{Z}]_Q}{K_Q}}$ is not regular. Note that $K$ is not a maximal ideal since $R[\frac{x}{z}, \frac{y}{z}]$ is not a field. Let $Q = (\frac{X}{Z}, \frac{Y}{Z}, Z)S[\frac{X}{Z}, \frac{Y}{Z}]$. Then $Q$ is a height 3 maximal ideal that contains $K$ and is generated by a regular sequence. Since $S[\frac{X}{Z}, \frac{Y}{Z}]_Q$ is a 3-dimensional regular local ring, it is a UFD. Hence $K_Q$, a height one prime ideal, is principal. In fact, letting $X_1 = \frac{X}{Z}$ and $Y_1 = \frac{Y}{Z}$ we can write $f = (X_1Z)^3 - (Y_1Z)^2Z + Z^5 = Z^3(X_1^3 - Y_1^2+Z^2)$. Then $g = X_1^3 - Y_1^2+Z^2$ is an irreducible element in $K$ and hence $K_Q = gS[\frac{\n}{Z}]_Q$. Now $$\frac{S[\frac{\n}{Z}]_Q}{K_Q} = \frac{S[\frac{\n}{Z}]_Q}{gS[\frac{\n}{Z}]_Q}$$ is a two-dimensional Noetherian local domain with maximal ideal $\frac{QS[\frac{\n}{Z}]_Q}{gS[\frac{\n}{Z}]_Q}$. Since $g \in Q^2$ we have the following isomorphism letting $A = S[\frac{\n}{Z}]$ $$\frac{QA_Q}{gA_Q}/\frac{Q^2A_Q}{gA_Q} \cong \frac{QA_Q}{Q^2A_Q}.$$ Hence by Nakayama we see that $QA_Q$ is minimially generated by 3 elements and therefore $A_Q$ is not a regular ring. \end{example} \subsection{Hartshorne, sections 1.3 and 1.4} \begin{proposition} Let $X$ be any variety and $Y$ an affine variety. Then there exists a natural bijective mapping of sets $$\alpha: \Hom(X,Y) \rightarrow \Hom(A(Y), \mathcal{O}(X)).$$ \end{proposition} \begin{corollary} If $X$ and $Y$ are affine varieties, then $X \cong Y$ if and only if $A(X) \cong A(Y)$. \end{corollary} Note that by exercise 3.9 we see that $S(X)$ is not invariant under isomorphism for a projective variety $X$. \begin{definition} \begin{enumerate} \item Let $X$ and $Y$ be varieties. A rational map $\phi: X \rightarrow Y$ is an equivalence class of pairs $$ where $U$ is a non-empty open subset of $X$ and $\phi_U$ is a morphism of $U$ to $Y$ and where $ \sim $ if $\phi_U = \phi_V$ on $U \cap V$. \item The rational map $\phi$ is dominant if for some (and hence every) pair $$ the image of $\phi_U$ is dense in $Y$. \item A birational map $\phi: X \rightarrow Y$ is a rational map with an inverse rational map, i.e, there is a rational map $\psi:Y \rightarrow X$ such that $\psi \circ \phi = 1_X$ and $\phi \circ \psi = 1_Y$. \end{enumerate} \end{definition} Note that $\an \times \mathbb{A}^m \cong \mathbb{A}^{n+m}$ but in general $\pn \times \mathbb{P}^m \cong \mathbb{P}^{n+m}$. However by the Segre embedding given in exercise 3.14 $\pn \times \mathbb{P}^m$ can be identified with a subvariety of $\mathbb{P}^N$ where $N = nm +n+m$. Also by exercise 3.16, if $X \subset \pn$ and $Y \subset \mathbb{P}^m$ are respectively quasi-projective or projective varieties then so is $X \times Y$ in $\mathbb{P}^N$. \subsection{The Blowup of $\an$ at the origin $\pt = (0, \ldots, 0)$} First consider $\mathbb{A}^n \times \mathbb{P}^{n-1}$ as a quasi-projective variety. Now if $x_1, \ldots, x_n$ are the affine coordinates of $\an$ and $y_1, \ldots, y_n$ are the homogeneous coordinates of $\mathbb{P}^{n-1}$ we define the \textit{blowup of $\an$ at $\pt$} as follows: Let $X$ be the closed subset of $\an \times \mathbb{P}^{n - 1}$ defined by the equations $x_iy_j = x_jy_i$ for $i, j = 1, 2, \ldots, n$, i.e., $$X = \{(x_1, \ldots, x_n, y_1, \ldots, y_n) \in \an \times \mathbb{P}^{n - 1} |x_iy_j = x_jy_i \mbox{ for } i,j = 1, 2, \ldots, n\}.$$ The blowup map $\phi$ is given as the composition of the embedding of $X$ in $\an \times \mathbb{P}^{n - 1}$ followed by the projection onto $\an$ as in the diagram below: $$\xymatrix{{X}\ar@{->}[r]\ar@{-->}[rd]&{\an \times \mathbb{P}^{n - 1}\ar@{->}[d]}\\{}&{\an}}$$ Properties of $X$: \begin{enumerate} \item If $P \in \an$ with $P \neq \pt$ then we have the following: \begin{enumerate} \item $\phi^{-1}(P)$ consists of a single point. \item $\phi: X - \phi^{-1}(\pt) \rightarrow \an - \pt$ is an isomorphism. \item The points of $\phi^{-1}(\pt)$ are in one-to-one correspondence with the set of lines through $\pt$ in $\an$. \item $X$ is irreducible. \end{enumerate} \item If $P \in \an$ with $P = \pt$ then $\phi^{-1}(P) \cong \mathbb{P}^{n-1}$. \end{enumerate} \section{June 16 and 18, 2009} \subsection{The degree function of Rees} Let $(R,\m)$ be a Noetherian local integral domain and let $I$ be an $\m$-primary ideal. In \cite{R}, Rees introduced the interesting concept of the degree function $d_I(x)$ that is defined for each nonzero $x \in \m$ to be the multiplicity of the image of the ideal $I + xR$ in the ring $R/xR$. Rees proves that there exist a finite set $\{v_1, \ldots, v_k\}$ of discrete valuations centered on $\m$ such that for each nonzero $x \in \m$ one has $$ d_I(x) = \sum_{i = 1}^k d_i(I)v_i(x), $$ where the $d_i(I)$ are positive integers called the degree coefficients of $I$. The discrete valuations $v_i$ are now called the Rees valuations of the ideal $I$. We would like to get some understanding of what these integers $d_i(I)$ are in various examples. \begin{example} If $(R,\m)$ is a normal local domain of the form $S/fS$, where $S$ is a 3-dimensional RLR with maximal ideal $(x,y,z)S$ and with residue field of characteristic zero, and if $f = x^n + y^n + z^n$, then the order valuation on $R$ defined by the powers of $\m$ is the unique Rees valuation $v$ of $\m$. Notice that $v(\m) = 1$ and the degree coefficient $d(\m, v) = n$. This gives examples where the degree coefficient $d(\m, v)$ can be any positive integer $n$ that we want. \end{example} \begin{example} Let $(R,\m)$ be a 2-dimensional RLR with maximal ideal $\m = (x,y)R$ and for $n$ a fixed positive integer, let $I = (x, y^n)R$. Then $I$ is a simple complete $\m$-primary ideal with Rees valuation $v$ having the property that $v(x) = n$ and $v(y) = 1$. Notice that $v(I) = n$ and that $I + yR = \m$ and $R/yR$ is a DVR. Therefore $d_I(y) = 1$. Hence we must have $d(I, v) = 1$. Notice also that $e(I)$, the multiplicity of $I$ is $n$. \end{example} \begin{remark} Van Lierde mentions that Rees and Sharp in \cite{RS} prove that if $(R,\m)$ is a 2-dimensional local domain and $I$ is an $\m$-primary ideal, then $$ e(I) = \sum_{i = 1}^k d_i(I)v_i(I), $$ where $v_1, \ldots, v_k$ are the Rees valuations of $I$. \end{remark} \begin{question}\label{6.4} Let $(R,\m)$ be a 2-dimensional regular local ring and let $I$ be a complete $\m$-primary ideal. Let $I = \prod_{i=1}^r I_i^{n_i}$ be the factorization of $I$ as a product of distinct simple complete ideals, and let $v_i$ denote the Rees valuation associated of $I_i$. Is the degree coefficient $d(I, v_i) = n_i$ ? \end{question} We have an afirmative answer of Question~\ref{6.4}, if the residue field $R/\m$ is an algebraically closed field. \begin{proposition}\label{6.41} Let $(R,\m)$ be a 2-dimensional regular local ring with $\m:=(x,y)R$. Assume that $R/\m$ is an algebraically closed field. Let $I \neq \m$ be a simple complete $\m$-primary ideal and let $v$ denote the Rees valuation associated of $I$. Then \begin{enumerate} \item The degree coefficient $d(\m, \ord_{R}) = 1$. \item The degree coefficient $d(I, v) = 1$. \end{enumerate} \end{proposition} \begin{proof} (1) : We have $d_{\m}(\alpha)=d(\m, \ord_{R})\ord_{R}(\alpha)$, for every $0 \neq \alpha \in \m$. Taking $\alpha:=x$, then we have $d_{\m}(x)=e(\frac{\m + xR}{xR})=e(R/xR)=1$, since $R/xR$ is an one dimensional RLR, and $\ord_{R}(x)=1$. Hence $d(\m, \ord_{R})=1$.\\ (2) : By the result of Rees and Sharp \cite{RS}, we have $d_{I}(\m)=d_{\m}(I)$. Hence we have $$ d(I, v)v(\m)=d_{I}(\m)=d_{\m}(I)=d(\m,\ord_{R})\ord_{R}(I)=d(I, v)v(\m), $$ the last equality holds by Lipman's reciprocity theorem \cite{Lipman}. Since $v(\m)$ is a positive integer, hence $d(I, v)=d(\m, \ord_{R})=1$. \end{proof} \begin{corollary}\label{6.42} Let $(R,\m)$ be a 2-dimensional regular local ring with $\m:=(x,y)R$ and let $I$ be a complete $\m$-primary ideal. Assume that $R/\m$ is an algebraically closed field. Let $I = \prod_{i=1}^r I_i^{n_i}$ be the factorization of $I$ as a product of distinct simple complete ideals, and let $v_i$ denote the Rees valuation associated of $I_i$. Then the degree coefficient $d(I, v_i) = n_i$,for all $i=1, \ldots, r$. \end{corollary} \begin{proof} Notice that $$ d(I_j, v_i)=\left \{ \aligned 1,\quad \text {if} \quad j=i\quad \text{ by Proposition~\ref{6.41}}\\ 0, \quad \text {if}\quad j \neq i, \quad \text{ by \cite[Lemma 2.2]{R}}. \endaligned \right \} $$ Hence we have $$ d(I, v_i)=d(\prod_{i=1}^r I_i^{n_i}, v_i)=\sum_{j=1}^r n_i d(I_j, v_i)=n_id(I_i,v_i)=n_i, $$ the second equality holds by \cite[lemma 5.1]{RS}. \end{proof} \subsection{The blowup of an ideal as a projective model} Let $I$ be a nonzero proper ideal of the integral domain $R$, and assume that $a_1, \ldots, a_n$ are nonzero elements of $I$ that generate $I$. The blowup $\bl_{I}(R)$ of $R$ at $I$ may be defined in terms of homogeneous localizations of the graded ring $R[It]$, denoted $\Proj R[It]$. Alternatively, one may consider the affine rings $R_i := R[I/a_i]$ over $R$ and the union of $\Spec R_i, 1 \le i \le n$ over $\Spec R$. As in Zariski-Samuel \cite[page~116]{ZS2}, we regard the blowup of $I$ as a family of local rings. We identify each $P \in \Spec R_i$ with the local ring $R[I/a_i]_P := T$. Notice that the ideal $IT$ is principal. We prove in Proposition~\ref{6.5} other facts that hold in this situation, and use the following definition. \begin{definition} Let $S$ be a subring of a local ring $(T, \n)$. \begin{enumerate} \item The prime ideal $P := S \cap \n$ is called the {\bf center} of $T$ on $S$. \item If $S$ is also local and the center of $T$ on $S$ is the maximal ideal of $S$, then we say that $T$ {\bf dominates} $S$, or that $S$ is {\bf is dominated by} $T$. \end{enumerate} \end{definition} \begin{proposition} \label{6.5} Let $I$ be a nonzero finitely generated ideal of the integral domain $R$, let $a$ and $b$ be nonzero elements of $I$ and let $P \in \Spec R[I/a]$. Then \begin{enumerate} \item The ring $R[I/b] \subseteq R[I/a]_P \iff a/b \in R[I/a]_P \iff a/b$ is a unit of $R[I/a]_P$. \item If $R[I/b] \subseteq R[I/a]_P$, then $R[I/a]_P = R[I/b]_Q $, where $Q := PR[I/a]_P \cap R[I/b]$. \end{enumerate} \end{proposition} \begin{proof} We clearly have $T := R[I/a]_P \supseteq R[I/b]_Q =: S$ and $T$ dominates $S$. The ideal $IS = bS$. Since $T$ dominates $S$, we have $aT = bT = IT$. Thus $a/b \in S$ is a unit of $T$. Since $S$ and $T$ are local rings with $T$ dominating $S$, it follows that $a/b$ is a unit of $S$. Therefore $R[I/a] \subseteq S$. Let $P' := QS \cap R[I/a]$. Then $R[I/a]_{P'}$ is dominated by $S$ and $S$ is dominated by $T$. Since $PT \cap R[I/a] = P$, we have $P' = P$, and thus $R[I/a]_P = R[I/b]_Q $. \end{proof} \begin{corollary} Let $I$ be a nonzero finitely generated ideal of the integral domain $R$ and let $S$ be a local overring of $R$. Then we have: \begin{enumerate} \item $S$ dominates a local ring on the blowup $\bl_{I}(R)$ if and only if $IS$ is a principal ideal of $S$. \item If $IS$ is principal, then $S$ dominates a unique local ring of $\bl_I(R)$. \item Each valuation overring $V$ of $R$ dominates a unique local ring $S \in \bl_{I}(R)$. \end{enumerate} \end{corollary} \begin{proof} For item (1), if $IS$ is principal, then $IS = aS$ for some nonzero $a \in I$. Hence $R[I/a] \subseteq S$. Let $P$ denote the center of $S$ on $R[I/a]$. Then $R[I/a]_P \in \bl_I(R)$ is dominated by $S$. On the other hand, if $S$ dominates $T \in \bl_I(R)$, then $IT$ is principal implies $IS$ is principal. Item (2) follows from Proposition~\ref{6.5}.2, and item (3) is immediate from items (1) and (2). \end{proof} \begin{definition} \label{6.6} Let $R$ be an integral domain. \begin{enumerate} \item An extension domain $S$ of $R$ is said to be {\bf birational over} $R$ if $S$ is contained in the field of fractions of $R$. \item Let $I$ be a nonzero finitely generated ideal of $R$. Let $\Sigma$ denote the family of birational local overrings $S$ of $R$ such that $IS$ is principal. We define a partial order $\le$ on $\Sigma$ as follows: $$ \text{Let} \quad S, T \in \Sigma, \, \quad \text{then} \, \quad S \le T \quad \iff \quad T \, \quad \text{dominates} \quad S. $$ \end{enumerate} \end{definition} \begin{proposition} \label{6.7} Let $I$ be a nonzero finitely generated ideal of an integral domain $R$. Then \begin{enumerate} \item A birational local overring $S$ of $R$ is an element of $\bl_{I}(R)$ if and only if $IS$ is a principal ideal and $S$ is a minimal element with respect to the partial order of Definition~\ref{6.6}.2. \item If $T$ is a birational local overring of $R$ and $IT$ is principal, then there exists a unique local ring $S$ such that $T$ dominates $S$ and $S$ is a minimal element with respect to the partial order of Definition~\ref{6.6}.2. \end{enumerate} \end{proposition} \begin{proof} For the proof of (1), if $IS$ is principal, then $IS = aS$ for some nonzero $a \in I$. Hence $R[I/a] \subseteq S$. Let $P$ denote the center of $S$ on $R[I/a]$. Then $S$ birationally dominates $R[I/a]_P$, and the minimality of $S$ in $\Sigma$ implies that $R[I/a]_P = S$. Hence $S \in \bl_I(R)$. On the other hand, if $S \in \bl_I(R)$, then $S = R[I/a]_P$ for some nonzero $a \in I$ and $P \in \Spec R[I/a]$. Suppose $T \in \Sigma$ with $T \le S$. Then $IT = bT$ for some nonzero $b \in I$. Let $Q$ denote the center of $T$ on $R[I/b]$. Then $R[I/b]_Q \le T \le S$, and Proposition~\ref{6.5} implies that $R[I/b]_Q = S$. Hence $S = T$ is minimal in $\Sigma$. \end{proof} \begin{example} \label{6.9} Let $(R,\m)$ be a 2-dimensional regular local ring and let $\m = (x,y)R$. The maximal ideals of $R[y/x]$ of height two have the form $(x, f(y/x))R[y/x]$, where $f(Z) \in R[Z]$ is a monic polynomial whose image in $(R/\m)[Z]$ is irreducible. In particular $P = (x, \frac{y}{x} - a)R[y/x]$ is a maximal ideal of height two. If $a \not\in \m$, then $R[x/y] \subseteq R[y/x]_P$. Let $Q = PR[y/x]_P \cap R[x/y]$. By Proposition~\ref{6.5}, $R[y/x]_P = R[x/y]_Q$. Notice that $Q = (y, \frac{x}{y} - a^{-1})R[x/y]$. \end{example} \section{June 21 and 23, 2009} \subsection{ The integrally closed $\m$-primary ideals adjacent to $\m$ in a two-dimensional Muhly local domain $(R,\m)$ of embedding dimension three.} According to Debremaeker in \cite{D1}, if $(R,\m)$ is a two-dimensional Muhly local domain, then the integrally closed $\m$-primary ideals adjacent to $\m$ are in one-to-one correspondence with the set of immediate local quadratic transformations of $R$. We give a method for finding these ideals in the case where $R$ is the quotient of a 3-dimensional regular local ring, $(S, \n)$, by an irreducible element, $f$. First consider Example~\ref{7.1}. \begin{example} \label{7.1} Let $S = k[X, Y, Z]_{(X, Y, Z)}$ be the localized polynomial ring over an algebraically closed field $k$ and let $\n = (X, Y, Z)S$. Let $f = XZ - Y^2 \in S$ and consider the two-dimensional Muhly local domain $R = S/fS$ with maximal ideal $\m = (x,y,z)R$. Now consider $R_1 = R[\frac{\m}{x}]$. We want to determine the maximal ideals of $R_1$ that contain $\m R_1 = xR_1$. Let $$T_1 = \{p \in \mSpec(R_1): \m R_1 \subset p \}.$$ We can write $R_1 = R[\frac{y}{x}, \frac{z}{x}]$ as a homomorphic image of $S_1 = S[\frac{\n}{x}]$. The kernel of the map, call it $K$, will be a prime ideal of height one. Notice that letting $y_1 = \frac{y}{x}$ and $z_1 = \frac{z}{x}$ in $S_1$ we have $f = xz - y^2 = x(z_1x) - (y_1x)^2 = x^2(z_1 - y_1^2)$ and $z_1 - y_1^2 \in K$ is irreducible. Hence there is a one-to-one correspondence between $T_1$ and $$T_2 = \{p \in \mSpec(S_1):(\n, z_1 - y_1^2)S_1 \subset p\}.$$ Now since $S$ is a regular local ring with maximal ideal $\n = (X, Y, Z)S$ we have that $$S_1 \cong \frac{S[T_1, T_2]}{(xT_1 - y, xT_2 - z)}$$ by \cite{SH} so there is a one-to-one correspondence between $T_2$ and $$T_3 = \{p \in \mSpec(S[T_1, T_2]): (xT_1 - y, xT_2 - z, T_2 - T_1^2, \n)S[T_1, T_2] \subset p\}.$$ Now if $p \in \mSpec(S[T_1, T_2])$ with $\n S[T_1, T_2] \subset p$ then this will correspond to a maximal ideal of $k[T_1, T_2]$, where $k = S/ \n$ is algebraically closed. Hence we are reduced to finding the maximal ideals of $k[T_1, T_2]$ that contain $T_2 - T_1^2$. Since $k$ is algebraically closed these maximal ideals of $k[T_1, T_2]$ are of the form $(T_1 - a, T_2 - a^2)$. This corresponds to ideals in $R_1$ of the form $n_a = (x, \frac{y}{x} - a, \frac{z}{x} - a^2)R_1$. Also the ideal in $R$ that corresponds to $n_a$ is $(x^2, y - xa, z - xa^2)R$ according to \cite{D1}. To find all the immediate quadratic transforms of $R$ (or equivalently all the integrally closed $\m$-primary ideals adjacent to $\m$) we would also need to consider $R_2 = R[\frac{\m}{y}]$ and $R_3 = R[\frac{\m}{z}]$. However since $(x,z)R$ is a reduction of $\m R$ it suffices to consider $R_3$ along with $R_1$ as above. Since $x$ and $z$ play symmetric roles in $R$ we know that the maximal ideals of $R_3$ that contain $\m R_3$ are of the form $(z, \frac{x}{z} - a^2, \frac{y}{z} - a)R_3$ which correspond to the ideals $(z^2, x - a^2z, y - az)R$ in $R$. \end{example} Example~\ref{7.1} indicates how we can find the immediate quadratic transforms of a two-dimensional Muhly local domain $R$ with maximal ideal $\m = (x,y,z)$ that is the quotient of a 3-dimensional regular local ring $S$ with maximal ideal $\n = (X, Y, Z)$ by an irreducible element $f(X, Y, Z)$. We summarize this method here: Let $$R_1 = R[\frac{\m}{x}], R_2 = R[\frac{\m}{y}], R_3 = R[\frac{\m}{z}].$$ We seek to find the maximal ideals of these rings that contain $\m$. We consider $R_1$. Let $Y_1 = \frac{Y}{X}$ and $Z_1 = \frac{Z}{X}$. Then $f(X, Y, Z) = f(X, Y_1X, Z_1X) = X^df_1(X, Y_1, Z_1)$. Then $$R_1 \cong S[\frac{Y}{X}, \frac{Z}{X}] / (f_1(X, Y_1, Z_1))$$ and as in the agrument above the maximal ideals of $R_1$ that contain $\m R_1$ correspond to maximal ideals of $k[T_1, T_2]$ that contain $f_1(X, T_1, T_2)$ where $k = R/ \m$. Similarly for $R_2$ and $R_3$. \begin{remark} With $(R,\m)$ as in Example~\ref{7.1}, the ideals $(x, y)R := P_1 $ and $(y, z)R := P_2$ are height-one prime ideals of $R$ such that $R/P_i$ is a DVR. It follows that $(x^n, y, z)R$ and $(z^n, x, y)R$ are integrally closed ideals for every positive integer $n$. Debremaeker considers ideals having this form in \cite{D2}. \end{remark} \begin{question} \label{7.3} Let $(R,\m)$ be a two-dimensional local domain. What are necessary and sufficient conditions for there to exist a prime ideal $P$ of $R$ such that $R/P$ is a DVR? \end{question} \begin{remark} If $(R,\m)$ is a two-dimensional local UFD and if there exists $P \in \Spec R$ such that $R/P$ is a DVR, then $R$ is regular. Thus, with $S = k[X, Y, Z]_{(X, Y, Z)}$ as in Example~\ref{7.1}, if $R = S/fS$, where $f = X^2 + Y^3 + Z^5$ or $f = X^2 + Y^3 + Z^7$, then $R$ is a two-dimensional normal local domain that does not have any prime ideals $P$ such that $R/P$ is a DVR. \end{remark} \begin{remark} Let $S = k[X_1, \ldots, X_n]$ be a polynomial ring in $n$ variables over an algebraically closed field $k$. If $I$ is a homogeneous ideal of $S$ such that $\dim S/I \ge 1$, then there exists a prime ideal $P$ of $S$ such that $I \subseteq P$ and $S/P$ is a polynomial ring in one variable over $k$. This follows because a graded integral domain of dimension one over $k$ is a polynomial ring, hence $S/P$ is a polynomial ring of dimension one over $k$ for every homogeneous prime ideal $P$ of $S$ such that $\dim S/P = 1$. This provides lots of examples of two-dimensional local domains for which the answer to Question~\ref{7.3} is affirmative. If $I$ is a homogeneous prime ideal of $S$ such that $\dim S/I = 2$ and $R$ is the localization of $S$ at its graded maximal ideal, then there exist infinitely many prime ideals $Q$ of $R$ such that $R/Q$ is a DVR. \end{remark} \section{June 30 and July 2, 2009} Let $S = k [[X,Y,Z]]$ and $n = (X,Y,Z)S$ be a power series ring over a field $k$ and its maximal ideal. Let $f = X^3+Y^2 Z + Z^4 + Y^5$ be in $S$ and $R = S/fS$. We would like to understand the question ~\ref{7.3}. To this end, we want to find maximal ideals of $R[\frac{m}{x}]$ that contain $m$. We have a surjective homomorpism, \[ R[ \frac{y}{x}, \frac{z}{x}] \twoheadleftarrow S[ \frac{Y}{X}, \frac{Z}{X} ] \] Set $Y_1 = \frac{Y}{X}$ and $Z_1 = \frac{Z}{X}$. Then $f = X^3 + Y^2 X^2 Z_1 X + Z_1^4 X^4 + Y_1^5 X^5 = X^3 (1 + Y_1^2 Z_1 + Z_1^4 X + Y_1^5 X^2)$. Let $f_1 = 1 + Y_1^2 Z_1 + Z_1^4 X + Y_1^5 X^2$. First, we look at those maximal ideals of $S[ \frac{Y}{X}, \frac{Z}{X} ]$ that contain $(n, f_1)$. By the following isomorphisms \begin{align*} \frac{ S[T_1,T_2] }{(X T_1 - Y, X T_2 - Z)} &\cong S [ \frac{Y}{X}, \frac{Z}{X} ], \\ \frac{ S[T_1, T_2] }{(n, X T_1 - Y, X T_2 -Z, f_1)} &=\frac{ S[T_1, T_2] }{(X,Y,Z, X T_1 - Y, X T_2 -Z, 1 + Y_1^2 Z_1 + Z_1^4 X + Y_1^5 X^2)} \\ &=\frac{ S[T_1, T_2] }{(X,Y,Z, 1 + Y_1^2 Z_1 )} \\ &\cong \frac{ k[T_1, T_2] }{ ( 1 + Y_1^2 Z_1 ) } \end{align*} Those maximal ideals in $S[ T_1,T_2]$ that contain $(n,f_1)$ correspond to those maximal ideals in $k[T_1, T_2]$ that contain $f_1' = 1 + T_1^2 T_2$. Since $T_1 = 1$ and $T_2 = -1$ is a root of $f_1'$ the maximal ideal $(T_1 -1, T_2 +1)$ contains $f_1'$ and this ideal correspoens to $(X, \frac{Y}{X} -1, \frac{Z}{X} +1)$ in $S$. Now let $(x^2, y-x, z + x) := I$ and $(y-x, z+x) := J$ in $R$. \begin{remark} $J = (y-x, x+z)$ is not a reduction of $m$. Suppose that it is. Because $J \subset I \subset m$, $I$ is a reduction of $m$. This implies $\bar{I} = m$. But $\bar{I} = I \subsetneq m$. This is a contradiction. \end{remark} \subsection{Minimal reduction $J$ of $I$ with certain conditions} By the result of Debremaeker, $\Rees(I) = \{ \text{ord}_R, \text{ord}_{R_1} \}$ where$R_1 := R[ \frac{m}{x}]_M, M = (x, \frac{y}{x} -1, \frac{z}{x} + 1)$ and $m_1 = (x, \frac{y}{x} -1)$. \[ \begin{array}{c|ccc|ccc} & x & y & z & x & y-x & z + x\\ \hline \ord_R & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \ord_{R_1} & 1 & 1 & 1 & 1 & 2 & \ge 2 \end{array} \] If we change our $f$ to $f = XY-Z^2$ then we ge the following table. \[ \begin{array}{c|ccc} & x & z & y \\ \hline v_1 = \ord_R & 1 & 1 & 1 \\ \hline v_2 = \ord_{R_1} & 1 & 2 & 3 \end{array} \] Now $m_1 = (x,\frac{z}{x})$ since $\frac{x y}{x^2} = \frac{z^2}{x^2}$ implies $\frac{y}{x} = (\frac{z}{x})^2$. $I = (x^2, y, z)$. We want to find a two generated reduction $J = (a,b)$ of $I$ such that $I = (a,b,c_i)$ for $i = 1,2, v_1(c_1) > v_1(a) = v_1(b)$ and $v_2(c_2) > v_2(a) = v_2(b)$. \begin{example} We revisit Example~\ref{7.1}. Let $S = k[X,Y,Z]_{(X,Y,Z)}$ where $k$ is a field. Let $f = XZ - Y^2$ and $R = S/fS$ and $m = (x,y,z)R$ where $x,y,z$ denote the images of $X,Y,Z$ in $R$. Further assume $k$ is algebraically close and char $k = 0$. Then $(R,m)$ is a 2-dimensional Muhly local domain. Then, \\ (1) Since $(x,z)$ is a minimal reduction of $m$, $Bl_m(R) = Spec (R[ \frac{m}{x}]) \cup Spec(R[\frac{m}{z}])$. \\ (2) Consider $R[ \frac{m}{x} ]$, the first quadratic transformation of $R$. Find all maximal ideals $m_a$ of $R[ \frac{m}{x}]$ that contain $m$. \[ m_a = (x, \frac{y}{x} - a, \frac{z}{x} -a^2) R[ \frac{m}{x}] \qquad \forall a \in k \] Let $R_{1_a} = R[ \frac{m}{x} ]_{m_a}$. This is a 2-dimensional regular local ring with maximal ideal $m_{1_a} := (x, \frac{y}{x} -a ) R_{1_a}$. \\ (3) Set $I_{1_a} = x m_{1_a} \cap R$. We showed that $I_{1_a} = (x^2, y - a x, z - a^2 x)R$. \\ (4) Consider the simplest case: Take $a = 0$. Then $I = (x^2, y, z)R$. Then $Rees(I) = \{ ord_R, ord_{R_1} \}$ where $R_1 = R[ \frac{m}{x} ]_{(x,\frac{y}{x}, \frac{z}{x})}$. \begin{question} Find a minimal reduction $J = (a,b)$ of $I$ and $c,d \in I$ such that \\ (i) $I = (a,b,c) = (a,b,d)$ and \\ (ii) $\ord_R (c) > \ord_R(a) = \ord_R(b),\ ord_R(d) >\ ord_{R_1} (a) =\ ord_{R_1} (b)$. \end{question} \begin{remark} Such a reduction cannot always be chosen \end{remark} We start with the following table \[ \begin{array}{c|ccc} & x & y & z \\ \hline \ord_R & 1 & 1 & 1 \\ \hline \ord_{R_1} & 1 & 2 & 3 \end{array} \] Consider $J = (x^2-z, y-z)R$. Then (i) $(J, x^2) = (J, z) = (x^2, y, z) = I$ and (ii) $+$ (iii) \[ \begin{array}{c|ccc} & x^2 - z & y - z & x^2 \\ \hline ord_R & 1 & 1 & 2 \\ \hline ord_{R_1} & 2 & 2 & 3 \end{array} \] Because $\ord_{R_1} ( \frac{x^2 - z}{x} ) = \ord_{R_1} ( x - \frac{z}{x} ) = 1$ implies $\ord_{R_1} (x^2-z) = 2$ and $\ord_{R_1} ( \frac{y-z}{x} ) = \ord_{R_1} ( \frac{y}{x} - \frac {z}{x} ) = 1$. Claim: $J$ is a minimal reduction of $I$. Since $\lambda (R/J) < \infty$ $J$ is generated by a SOP of $R$. Hence $e(J) = \lambda(R/J)$. \begin{remark} Let $(A,n)$ be a Noetherian local ring. $J,I$ n-primary ideals such that $J \subset I$ is a reduction. Then $e(J) = e(I)$. The converse is true if $(A,n)$ is formally equidimentional i.e. $\dim(\hat{A}) = \dim( \hat{\frac{A}{p}})$. In particular when $(A,n)$ is a Cohen-Macaulay local ring then the converse holds. \end{remark} Show: $e(I) = e(J)$. We have $e(J) = \lambda(R/J) \ge \lambda (R/I)$. $2 = \lambda(R/I) \le e(I)$ and $2 = e(m) < e(I)$ i.e $3 \le e(I)$ because $I \subset m$ is not integral. It is enough to show $e(J) = 3$. Show: $\lambda(R/J) = 3$. \begin{align*} \lambda(R/J) &= \lambda(S/(J,f)S) \\ &= \lambda(S/(x^2-z, y-z, xz - y^2)) \\ &= \lambda(\frac{S/(y-z)}{(x^2-z,xz-z^2)}, \qquad \text{mod } y-z \text{ and set } T= S/(y-z) \\ &= \lambda(\frac{T/(x^2-z)}{x^3-x^4}), \qquad \text{mod } x^2-z \text{ and set } A = T/(x^2-z) \\ &= \lambda(A/(x^3-x^4)) \\ &= 3 \end{align*} \end{example} \section{July 9, 2009} \subsection{Generalize the previous section} We would like to generalize the computation we did in the previous section to all such $a$. Let $S = k[X,Y,Z]_{(X,Y,Z)}$ be a polynomial ring over an algebraically closed field localized at a maximal ideal $n = (X,Y,Z)$ of $S$. Assume that char $k = 0$. Let $R = S/fS$ where $f = X Y - Z^2$ in S. Let $x,y,z$ denote the images of $X,Y,Z$ in $R$. Then \\ (1) $M_a = (x, \frac{y}{x} - a^2, \frac{z}{x} - a), \quad \forall a \in k$ \\ (2) $R_{1_a} = R [ \frac{m}{x} ]_{M_a}$ with $m_a = (x, \frac{z}{x} - a)$ \\ (3) $I_{a} := (x^2, y - a^2x, z - ax)R$ \\ (4) $Rees(I_a) = \{ ord_R, ord_{R_{1_a}} \}$ \\ We would like to find a minimal reduction $J_a = (u_a, v_a)$ of $I_a$ that satisfies the following conditions. \\ (i) $I_a = (u_a, v_a, c_a) = (u_a, v_a, d_a)$ and \\ (ii) $ord_R(u_a) = ord_R( v_a ) < ord_R (c_a)$ and $ord_{R_{1_a}}(u_a) = ord_{R_{1_a}}( v_a ) < ord_{R_{1_a}} (c_a)$. Notice that $ord_R(x) = ord_R(y) = ord_R(z) = 1$ and $ord_{R_{1_a}} (x) = 1, ord_{R_{1_a}}(y-a^2x) = 2, ord_{R_{1_a}}(z - ax) = 2$. Because $ord_{R_{1_a}}( \frac{y - a^2x}{x}) = ord_{R_{1_a}}( \frac{y}{x} - a^2) = ord_{R_{1_a}} ( (\frac{z}{x})^2 -a^2 ) = 1$. Let $J_a = (x^2 -(z - ax), (y - a^2 x) - (z - ax) )$. Then \[ \begin{array}{c|c|c} & x^2 -(z - ax) & (y - a^2 x) - (z - ax) \\ \hline ord_R & 1 & 1 \\ \hline ord_{R_1} & 2 & 2 \end{array} \] Then, setting $I_a = (J_a, x^2)$ satisfies the first part of the condition (ii). We would like to find the element that satisfies the second part. Compute $\lambda(R/J_a)$. \begin{align*} \lambda(R/J_a) &= \lambda(S/(I_a, f)S) \\ &= \lambda(\frac{S}{( X^2 - (Z - aX), (Y - a^2 X) - (Z - aX), XY - Z^2) )} ) \\ &= \lambda(\frac{T}{( X^2 - (Y - a^2 X + a X), XY - (Y - a^2 X + aX)^2 )} ) \quad \text{set } T = S/(Y - a^2 X + aX - Z) \\ &= \lambda(\frac{A}{( X(X^2 + a^2X) - (X^2 + a^2 X - a^2 X + a X)^2)}) \quad \text{set } A = T/(X^2 + a^2 X - Y) \\ &= \lambda(\frac{A}{(1- 2a) X^3 - X^4 )} ) \end{align*} Hence $\lambda(R/J_a) = 3$ if $a \ne 1/2$ and $\lambda(R/J_a) = 4$ if $ a = 1/2$. \section{July 16, 2009} \begin{theorem}[Brodmann] Let $(R,m)$ be a d-dimensional Cohen-Macaulay ring. Let $I$ be complete intersection and generic complete intersection of height h. Then (1) $depth (R[It]) = min \{d, depth(R/I) + 2 \}$ if $h = 0$ and $min \{ d+1, depth(R/I) + h + 2 \}$ if $h > 0$. \\ (2) $depth (gr_I(R)) = min \{d, depth(R/I) + h + 1 \}$. \end{theorem} \begin{theorem} Let $(R,m)$ be a 2-dimensional normal local domain. Let $I$ be an $m$-primary ideal that is almost complete intersection. Suppose there exists a minimal reduction $J$ of $I$ with $\mu(I) = 2$ and $JI = I^2$. Then \\ (1) $gr_I(R)$ is Cohen-Macaulay [VV].\\ (2) $R[It]$ is Cohen-Macaulay [GS]. \\ (3) $F(I)$ is a Cohen-Macaulay ring having minimal multiplicity at its homogenous maximal ideal. \\ (4) If $Min(mR[It]) = \{ Q_1, \dots, Q_n \}$. Then $n \le 2$. \ (5) Suppose $Min( mR[It]) = \{ Q_1, Q_2 \}$. Then $R[It]/Q_i$ is regular for $i = 1,2$. \begin{proof} (4) Let $N = (m,It)R[It]$ and $\bar{n}$ to be its image in $F(I)$. Then $\bar{N}$ is the unique homogeneous maximal ideal of $F(I)$. By (3), \[ edim(F(I)_{\bar{N}} ) - dim( F(I)_{\bar{N}} ) + 1 (= 3 - 2 + 1 ) = e(F(I)_{\bar{N}}) = 2 \] It is not hard to check that $dim (\frac{ R[It]_N } { Q_{i_N} }) =2$. By [Matsumura] 14.7, \begin{align*} e(F(I)_{\bar{N}}) &= \sum_{i=1}^n e( (\frac{N}{Q_i})_{\bar{N}} , (\frac{R[It]}{Q_i})_{\bar{N}}) \lambda(F(I)_{\bar{Q_i}} ) \\ &= \sum_{i=1}^n e( ( \frac{R[It]}{Q_i})_{\bar{N}}) \lambda( (\frac{R[It]}{mR[It]})_{\bar{Q_i}} ) \\ &\Rightarrow e(F(I)_{\bar{N}}) \ge n \\ &\Rightarrow n \le 2 \end{align*} \end{proof} \end{theorem} \begin{example} This is an example of a 2-dimensional normal local domain such that the maximal ideal has two Rees valuations and $Bl_m(R)$ is regular. Let $S = k[X,Y,Z]_{(X,Y,Z)}$ where $k$ is a field. Let $n = (X,Y,Z)S$ and $f = XY - Z^3$ in S. Let $R = S/fS$ and $m = (x,y,z)R$ where $x,y,z$ denote the images of $X,Y,Z$ in $R$. Since $gr_m(R) \cong \frac{k[x,y,z]}{(xy)}$ has two minimal primes, $m$ has two Rees valuations. It remains to show that $Bl_m(R)$ is regular. When we blow up at $m$ that is along the point $(0,0,0)$. Since the original curve is in $C^3$ the blow up is in $C^3 \times P^2$. Let the coordinates of $P^2$ be $[X' : Y' : Z']$. There exists three affine charts that cover the projective space $P^2$. Hence it is enough to check the regularity on them. Let $U_0 = \{X'\ne 0 \}, U_1 = \{Y'\ne 0 \}$ and $U_2 = \{Z'\ne 0 \}$. Let's check on $U_2$ first. Since $Z' \ne 0$ we may assume that $Z' = 1$. Then $f = XY - Z^3$ becomes $f = XZ' YZ' - Z'^3 = Z'^2 (XY - Z')$. The rank of the Jacobian matrix for XY - Z' at $(0,0,0)$ is one. This implies that it is regular on this chart (Hartshorne I.5). On $U_0$, we may set $X' = 1$. Then $f = X' YX' - (ZX')^3 = X'^2 (Y - Z^3 X')$. Similarly the rank of the Jacobian matrix for $(Y - Z^3 X')$ is one. Hence it is regular on $U_0$. By symmetry it is regular on $U_1$. Hence the blow up is regular. \begin{remark} If we change $f = XY - Z^3$ to $f = XY - Z^m$ for $m \ge 4$ then $Bl_m(R)$ is not regular. This can be observed by looking at the chart $U_2$ above. The rank of Jacobian matrix at $(0,0,0)$ is zero. This tells us that $Bl_m(R)$ is not regular. Details can de found in I.5 of Hartshorne's Algebraic Geometry \cite{H}. \end{remark} \end{example} \section{July 21, 2009} \begin{example}\label{11.1} Let $(R_0,\m_0)$ be a 2-dimensional RLR with $\m_0 = (x,y)R_0$. Consider the simple integrally closed ideal $I = (x,y^4)R_0$. The QDT-sequence determined by $I$ is: $$ R_0 \subset R_1 = R_0[x/y, y]_{(x/y, y)} \subset R_1[x/y^2]_{(x/y^2,y)} = R_2 \subset R_2[x/y^3]_{(x/y^3, y)} = R_3 $$ The proper transforms of $I$ in these QDTs are $$ I^{R_1} = (x/y, y^3)R_1, \qquad I^{R_2} = (x/y^2, y^2)R_2, \qquad I^{R_3} = (x/y^3, y)R_3. $$ \end{example} \begin{definition} If $\alpha \subsetneq \beta$ is a birational extension of RLRs, then $\beta$ is said to be {\bf proximate} to $\alpha$ if $\beta \subset V_{\alpha}$, where $V_{\alpha}$ is the valuation ring of $\ord_{\alpha}$. \end{definition} \begin{remark} \label{11.3} In Example~\ref{11.1}, we have $R_1$ is proximate to $R_0$ and $R_2$ is proximate to $R_1$, but $R_2$ is not proximate to $R_0$. It would be nice to cover a proof that if $\alpha_1 \subset \alpha_2 \subset \cdots \subset \alpha_r$ is a sequence of QDTs of 2-dimensional RLRs, then $\alpha_r$ is proximate to $\alpha_{r-1}$ and at most one other of the $\alpha_i$. \end{remark} \begin{question} With the notation of Remark~\ref{11.3}, does there exist an example where $r = 4$ and $\alpha_4$ is proximate to $\alpha_1$ and $\alpha_3$? \end{question} \begin{example} For the RLRs $R_i$ of Example~\ref{11.1}, let $V_i$ denote the valuation ring of the $\ord$ valuation defined by the powers of the maximal ideal of $R_i$. Notice that $x/y^{i+1}$ is residually transcendental for $V_i$. With $R_0$ as in Example~\ref{11.1}, consider the affine components $R_0[y^2/x]$, $R_0[y^3/x]$ and $R_0[y^4/x]$ of the blowups of the integrally closed ideals $(x, y^2)R_0, ~ (x, y^3)R_0$ and $(x, y^4)R_0$. These are all integrally closed domains of dimension two. Let $$ S_2 = R_0[y^2/x]_{(x,y, y^2/x)}, \quad S_3 = R_0[y^3/x]_{(x,y,y^3/x)}, \quad S_4 = R_0[y^4/x]_{(x,y,y^4/x)}. $$ Let $\n_i$ denote the maximal ideal of $S_i$. The associated graded ring $\Gr_{\n_2}(S_2)$ is a normal domain, so the blowup $\bl_{\n_2}(S_2)$ of the maximal ideal of $S_2$ is regular. Notice that the associated graded rings $\Gr_{\n_3}(S_3)$ and $\Gr_{\n_4}(S_4)$ are isomorphic rings having two minimal primes. The blowup $\bl_{\n_3}(S_3)$ is regular while the blowup $\bl_{\n_4}(S_4)$ is not regular. Notice that $\Rees \n_2 = \{V_0\}$ and $\Rees \n_3 = \{V_0, V_1\}$. To compute $\Rees \n_4$, let $z = y^4/x$ and notice that $(x+z, y)S_4$ is a reduction of $\n_4$. The elements of $\Rees \n_4$ are the DVRs $V$ that birationally dominate $S_4$ and for which the element $$ \frac{x+z}{y} = \frac{x + y^4/x}{y} = \frac{x}{y} + \frac{y^3}{x} $$ is a unit of $V$ and is residually transcendental for $V$. These DVRs are $V_0$ and $V_2$. The element $x/y$ is residually transcendental for $V_0$ and $y^3/x$ is in the maximal ideal of $V_0$, while $y^3/x$ is residually transcendental for $V_2$ and $x/y$ is in the maximal ideal of $V_2$. Thus $\Rees \n_4 = \{V_0, V_2 \}$. \end{example} \section{July 23, 2009} \subsection{An equation of integrality} Let $R= k[x,y,z]_{(x,y,z)}$ where $k$ is a field with $chak \; k \ne 3$. Let $f = x^3+y^3+z^3$. Let $S = R/(f)$. Let $I = (x^2, y+x, z)$ be an S-ideal. We would like to see that $I^2$ is not integrally closed. In particular the element $x(x+y) \in \bar{I^2} \setminus I^2$. Claim: the following equation gives the integrality of $x(x+y)$ over $I^2$. \begin{align} T^2 - (x+y)^2 T + xy(x+y)^2 = 0 \end{align} where $(x+y)^2 \in I^2$ and $xy(x+y)^2 \in I^4$. It is not hard to see that $x(x+y)$ satisfies this equation. Since $y+x \in I$, trivially $(y+x)^2 \in I^2$. It suffices to show that $xy(x+y)^2 \in I^4$. In particular, it is enough to show that $xy(x+y) \in I^3$. \begin{align} x^3+y^3+z^3 &= (x+y)(x^2-xy+y^2) + z^3 \\ &= (x+y)( (x+y)^2 - 3xy) + z^3 \\ &= (x+y)^3 - 3xy(x+y) + z^3 \end{align} Since $x^3+y^3+z^3 = 0$ in $S$. This implies \begin{align} 3xy(x+y) &= (x+y)^3 + z^3 \\ &= (x+y+z)( (x+y)^2 -(x+y)z +z^2) \end{align} Since $x+y, z \in I$, $(x+y+z) \in I$. The same elements show that each term of $(x+y)^2 -(x+y)z +z^2$ is in $I^2$. Hence $xy(x+y) \in I^3$. \\ We can check that $x(x+y) \notin I^2$ by looking at the following equation \begin{align} c_1(y+x)^2 + c_2(y+x)z + c_3 z^3 + c_4 (x^3+y^3+z^3) = x(x+y) \end{align} and showing that there's no $c_i \in k$ that satisfy this equation over $R$. \begin{question} For the $S$-ideal $I = (x^2, x+y, z)S$, what is the reduction number of $I$ with respect to a minimal reduction? I believe $J = (x+y+z, x^2 - z)S$ is a minimal reduction of $I$. Is $JI$ properly contained in $I^2$? \end{question} \subsection{Proximate} With regard to Remark \ref{11.3}, we give the following argument for why $\alpha_r$ is always proximate to $\alpha_{r-1}$: Since $\alpha_r$ is a quadratic transformation of $\alpha_{r-1}$ there exists an $x \in \m_{r-1} - \m_{r-1}^2$ and a maximal ideal $M$ of $\alpha_{r-1}[\frac{\m_{r-1}}{x}]$ such that $\alpha_r = \alpha_{r-1}[\frac{\m_{r-1}}{x}]_M$. Also, $V_{r-1}$, the valuation ring for the order valuation of $\alpha_{r-1}$, is of the form $V_{r-1} = \alpha_{r-1}[\frac{\m_{r-1}}{x}]_{(x)}$, which is a further localization of $\alpha_r$, so clearly $\alpha_r \subset V_{r-1}$. Note that the center of $V_{r-1}$ on $\alpha_r$ is the regular prime $(x)\alpha_r$. Now we want to consider any $\alpha_i$ such that $\alpha_r$ is proximate to $\alpha_i$. Let $P_{r,i}$ be the center of $V_i$ on $\alpha_r$. Then $P_{r,i}$ is a height one prime and we claim that it is a regular prime, i.e. $\alpha_r /P_{r,i}$ is a DVR. As in the argument above, $\alpha_{i+1}$ is proximate to $\alpha_i$ and $\alpha_{i+1}/P_{i+1, i}$ is a DVR. Also since we have the chain of QDTs $$\alpha_1 \subset \ldots \subset \alpha_i \subset \alpha_{i+1} \subset \ldots \subset \alpha_r$$ we have the birational extension $\alpha_{i+1}/P_{i+1, i} \subset \alpha_{r}/P_{r, i}$ where the smaller ring is a DVR so these two rings are equal and our claim is proven. We would still like to understand why $\alpha_r$ is proximate to at most one other $\alpha_i$ and have an example where $r= 4$ and $\alpha_4$ is proximate to $\alpha_3$ and $\alpha_1$ but not $\alpha_2$. Example~\ref{12.1} is, I believe, an example where $\alpha_4$ is proximate to $\alpha_1$ and $ \alpha_3$, but is not proximate to $\alpha_2$. \begin{example} \label{12.1} Let $(R_0, \m_0)$ be a 2-dimensional RLR with $\m_0 = (x, y)R$. Define $R_1 = R_0[y/x]_{(x, y/x)}$. Let $y_1 = y/x$, and define $R_2 = R_1[x/y_1]_{(y_1,x/y_1)}$. Let $x_1 = x/y_1$ and define $R_3 = R_2[x_1/y_1]_({y_1, x_1/y_1)}$. Then $R_3$ is proximate to $R_2$ and $R_0$. This can maybe better be described by regular parameters for the maximal ideals of $R_0, R_1, R_2$ and $R_3$ as $$ (x,y) \to (x, \frac{y}{x}) \to (\frac{x^2}{y}, \frac{y}{x}) \to (\frac{x^3}{y^2}, \frac{y}{x}). $$ Notice that the $\ord$-valuation ring $V_0$ for $R_0$ contains $R_3$. Hence $R_3$ is proximate for $R_0$. Also notice that the $\ord$ valuation for $V_1$ does not contain $R_3$, for $v_1(x) = 1$ and $v_1(y) = 2$. Hence $v_1(x^3/y^2) = - 1$ and $R_3$ is not proximate to $R_1$. \end{example} \section{August 24, 2009} \begin{example} Let $(R, \m)$ be a 2-dim RLR with $\m = (x,y)R$. Then $\bl_{\m}(R) = \Proj (R[\m t])$ is the set of homogeneous localizations of $R[\m t]$ at homogeneous prime ideals other than the unique homogeneous maximal ideal. We have $R[\m t] = R[xt, yt] \subset R[t]$, a polynomial ring graded by $\mathbb{N}$ We can also think of the blowup of $R$ as $\bl_{\m}(R) = \Spec R[y/x] \cup \Spec R[x/y] $, where we regard $\Spec R[y/x]$ as the affine scheme that consists of the local rings $$ \{R[y/x]_P ~| ~ P ~~ \text{ is a prime ideal of} ~~ R[y/x] ~ \}. $$ It is often the case that $R[y/x]_P = R[x/y]_Q$ for some primes $P$ and $Q$ in the respective rings. Indeed, this is true for all prime ideals $P$ of $R[y/x]$ other than the maximal ideal $(x, y/x)R[y/x]$ and all prime ideals $Q$ of $R[x/y]$ other than the maximal ideal $(y, x/y)R[x/y]$. \end{example} Here is a more general example. \begin{example} \label{13.2} Let $R$ be a Noetherian domain with field of fractions $K$ and let $I = (a_1, \ldots, a_n)R$, where each $a_i$ is a nonzero element of $R$. Then $$ \bl_I(R) = \Proj R[It] = \bigcup_{i=1}^n \Spec R[I/a_i]. $$ A local ring $(S, \n)$ is in $\bl_I(R)$ if and only if $S = R[I/a_i]_Q$ for some prime ideal $Q$ of $R[I/a_i]$. \end{example} \begin{question} Let $I$ and $J$ be ideals of the Noetherian domain $R$. When is $\bl_I(R) = \bl_J(R)$? \end{question} \begin{remark} It is always true that $\bl_I(R) = \bl_{I^n}(R)$ for $n$ a positive integer. \end{remark} If $V$ is a valuation domain with $R \subset V$ then $V$ dominates a unique local ring of $\bl_I(R)$. To see that $V$ dominates some local ring of $\bl_I(R)$ observe that $IV = aV$ for some $a \in I$. Hence $R[I/a] \subset V$. Let $\n$ be the maximal ideal of $V$ and let $\n \cap R[I/a] = Q$. Then $R[I/a]_Q$ is dominated by $V$. This is said more clearly in Section 6.2. \begin{definition} Let $S$ be a subring of a local ring $(T, \n).$ \begin{enumerate} \item The prime ideal $P:= \n \cap S$ is called the {\bf center } of $T$ on $S$. \item If $S$ is also local and $\n \cap S$ is the maximal ideal of $S$, then we say $T$ {\bf dominates } $S$. \end{enumerate} \end{definition} With the notation of Example~\ref{13.2}, we have $$ \bl_I(R) = \cup_{i=1}^n \Spec R[I/a_i] = \cup_{0 \neq a \in I} \Spec R[I/a]. $$ Notice that $IR[I/a] = aR[I/a]$ is principal, so $I$ extends to a principal ideal in each local ring of $\bl_I(R)$. By Proposition~\ref{6.5} as repeated below, if $a$ and $b$ are nonzero elements of $I$ and $R[I/b] \subset R[I/a]_Q$, then $R[I/a]_Q = R[I/b]_{Q'}$, where $Q' = QR[I/a]_Q \cap R[I/b]$. \begin{proposition} Let $I$ be a nonzero finitely generated ideal of the integral domain $R$, let $a$ and $b$ be nonzero elements of $I$ and let $P \in \Spec R[I/a]$. Then \begin{enumerate} \item The ring $R[I/b] \subseteq R[I/a]_P \iff a/b \in R[I/a]_P \iff a/b$ is a unit of $R[I/a]_P$. \item If $R[I/b] \subseteq R[I/a]_P$, then $R[I/a]_P = R[I/b]_Q $, where $Q := PR[I/a]_P \cap R[I/b]$. \end{enumerate} \end{proposition} \begin{proof} We clearly have $T := R[I/a]_P \supseteq R[I/b]_Q =: S$ and $T$ dominates $S$. The ideal $IS = bS$. Since $T$ dominates $S$, we have $aT = bT = IT$. Thus $a/b \in S$ is a unit of $T$. Since $S$ and $T$ are local rings with $T$ dominating $S$, it follows that $a/b$ is a unit of $S$. Therefore $R[I/a] \subseteq S$. Let $P' := QS \cap R[I/a]$. Then $R[I/a]_{P'}$ is dominated by $S$ and $S$ is dominated by $T$. Since $PT \cap R[I/a] = P$, we have $P' = P$, and thus $R[I/a]_P = R[I/b]_Q $. \end{proof} Recall Definition~\ref{6.6}. \begin{definition} Let $R$ be an integral domain. \begin{enumerate} \item An extension domain $S$ of $R$ is said to be {\bf birational over} $R$ if $S$ is contained in the field of fractions of $R$. \item Let $I$ be a nonzero finitely generated ideal of $R$. Let $\Sigma$ denote the family of birational local overrings $S$ of $R$ such that $IS$ is principal. We define a partial order $\le$ on $\Sigma$ as follows: $$ \text{Let} \quad S, T \in \Sigma, \, \quad \text{then} \, \quad S \le T \quad \iff \quad T \, \quad \text{dominates} \quad S. $$ \end{enumerate} \end{definition} \begin{question} Let $I$ and $J$ be integrally closed $\m$-primary ideals of a 2-dim RLR $(R, \m)$. Then \begin{enumerate} \item We have $\bl_I(R) = \bl_J(R)$ if and only if $\ldots$ ??? \item We have each $T \in \bl_J(R)$ dominates some $S \in \bl_I(R)$ if and only if $\ldots$ ??? \end{enumerate} \end{question} \begin{definition} Let $I$ and $J$ be nonzero ideals of a Noetherian domain $R$. We say that the blowup $\bl_J(R)$ { \bf dominates } the blowup $\bl_I(R)$ and write $\bl_I(R) \leq \bl_J(R)$ if each $T \in \bl_J(R)$ dominates some $S \in \bl_I(R)$. \end{definition} \section{August 26, 2009} Notes from the paper \cite{HL} of Heinzer and Lantz. Let $(R, \m)$ be a 2-dimensional Noetherian normal local domain. \begin{definition} A {\bf spot } over $R$ is a local domain $(S, \n)$ that is a localization of a finitely generated $R$-algebra (i.e., $S$ is a local ring that is essentially finitely generated over $R$). \end{definition} Consider the normal local domains $(S, \n)$ that are birational over $R$ and are spots over $R$. Other than the field of fractions $K$ of $R$, there are three disjoint classes: \begin{enumerate} \item $S = R_P$, where $P$ is a height one prime of $R$, these are all DVRs, and are called the {\bf essential prime divisors } of $R$. Denote this class by $\epd(R)$. \item $(S, \n)$ is a DVR with $R \subset S \subset K$ and $S$ dominating $R$, i.e., $\n \cap R = \m$ and $S$ is a spot over $R$. These are called the {\bf hidden prime divisors } of $R$, or the prime divisors of the second kind of $R$. Denote this class by $\hpd(R)$. \item The 2-dimensional normal local domains $(S, \n)$ with $R \subset S \subset K$ and $S$ is a spot over $R$. Denote this class by $\mathcal{S}_2(R)$. \end{enumerate} \begin{remark} Every $S \in \mathcal{S}_2(R)$ dominates $R$. More generally, if $R$ is a Noetherian domain of dimension $n$, then any birational overring of $R$ has dimension $\leq n$. \end{remark} \begin{remark} \begin{enumerate} \item $\epd(R) \cup \hpd(R) \cup \mathcal{S}_2(R) \cup \{K\}$ is the set of all normal local spots birational over $R$. \item If $(S, \n) \in \hpd(R)$, then $\trdeg_{R/ \m} S/ \n = 1$. This follows from the dimension formula: We have $S = R[\theta_1, \ldots, \theta_n]]_{\n \cap R[\theta_1, \ldots, \theta_n]]}$ and $\trdeg S/R = 0$ so $\dim R = \dim S + \trdeg_{R/m}S/n$. Let $(V, N)$ be a DVR with $R \subset V \subset K = Q(R)$ and $N \cap R = \m$. If $V$ is a spot over $R$, then $\trdeg_{R/\m} V/N = 1$. The dimension formula implies that $\trdeg_{R/\m}V/N > 0$ Choose $\theta \in V$ such that the image of $\theta$ in $V/N$ is transcendental over $R/M$. Then $MR[\theta]$ is a nonmaximal prime ideal. It follows that $\hgt MR[\theta] = 1$. Therefore $S$ is a birational overring of the one-dimensional Noetherian local domain $R[\theta]_{MR[\theta]}$. The Krull-Akizuki Theorem implies that $S$ is a localization of an integral extension of $R[\theta]_{MR[\theta]}$. \item If $(S, N) \in \mathcal{S}_2(R)$ then the field $S/N$ is finite algebraic over $R/M$. \end{enumerate} \end{remark} If $(R, M)$ is assumed to be excellent (or more generally, if the $M$-adic completion of $R$ is a domain, i.e. $R$ is analytically irreducible), then every 2-dimensional normal Noetherian local domain $S$ such that $R \subset S \subset K$ is in $\mathcal{S}_2(R)$. \begin{discussion} Fix $(R, M)$ a 2-dimemsional normal Noetherian local domain and $S \in \mathcal{S}_2(R)$. Then we can consider $\epd(R)$, $\hpd(R)$, $\mathcal{S}_2(R)$ and $\epd(S)$, $\hpd(S)$, $\mathcal{S}_2(S)$. \begin{enumerate} \item Clearly $\mathcal{S}_2(S) \subset \mathcal{S}_2(R)$ and $hpd(S) \subset hpd(R)$. \item Also $V \in \epd(R)$ and $S \subset V$ implies $V \in \epd(S)$. For if $S \subset V_p$, let $q = pR_p \cap S$, then $V = R_p \subset S_q = V$. \item If $V \in \epd(S)$ and $MV = V$, then $V \in \epd(R)$. If $S_Q = V$, then either $Q \cap R = M$ or $Q \cap R = P$, where $\hgt P = 1$. If $Q \cap R = P$ then $R_P = S_Q = V$ and $V \in \epd(R)$. Since $MS$ is a nonzero ideal of $S$, $MS$ is contained in at most finitely many height one primes $Q_1, \ldots, Q_n$ of $S$. Then $V_1 = S_{Q_1}, \ldots, V_n = S_{Q_n}$ are precisely the elements of $\epd(S)$ that are not in $\epd(R)$. We say that $V_1, \ldots, V_n$ are the { \bf exceptional prime divisors} of $S/R$ and write $\xpd(S/R) = \epd(S) \cap \hpd(R)$. \item If $S \in \mathcal{S}_2(R)$ and $R \neq S$ then $\epd(S) \cap \hpd(R)$ is a finite nonempty set. \end{enumerate} \end{discussion} \section{August 28, 2009} $(R, M)$ a 2-dimensional normal local domains with $$\epd(R) = \{R_P | P \mbox{ is a height one prime}\}$$ $$hpd(R) = \{V | R \subset V \subset Q(R) \mbox{ and } V \mbox{ is a DVR that dominates } R \mbox{ and } V \mbox{ is a spot over } R\}$$ $$\mathcal{S}_2 = \{S | S \mbox{ is a 2-dimensional normal local domain with } R \subset S \subset Q(R) \mbox{ and } S \mbox{ is a spot over } R \}$$ For $S \in \mathcal{S}_2$, we have $\xpd(S/R) = \epd(S) \cap \hpd(R)$ where $$\xpd(S/R) = \{S_Q | Q \mbox{ is a height one prime of } S \mbox{ and } MS \subset Q\}.$$ \begin{proposition} Assuming $R$ is excellent, let $S_1$ and $S_2$ and $T$ be in $\mathcal{S}_2(R)$. \begin{enumerate} \item If $S_1 \subset T$, then $\xpd(T/S_1) = \emptyset ~~ \iff ~~S_1 = T$. \item Assume $T$ is a localization of the integral closure of $S_1[S_2]$. Then we have \begin{enumerate} \item $\xpd(T/R) \subset \xpd (S_1/R) \cup \xpd(S_2/R)$, \item $\xpd(T/S_2) \subset \xpd(S_1/R)$, and $\xpd(T/S_1) \subset \xpd(S_2/R)$. \end{enumerate} \item If $\xpd (S_1/R) = \xpd(S_2/R)$, then either $S_1 = S_2$ or no element of $\mathcal{S}_2(R)$ dominates both $S_1$ and $S_2$, i.e. $S_1[S_2]$ has dimension less than or equal to 1. \end{enumerate} \end{proposition} Proof: (1)$\Leftarrow$ is clear $\Rightarrow$: Let $N_1$ be the maximal ideal of $S_1$ and $N$ the maximal ideal of $T$. We need to see that if $N_1T$ is $N$-primary then $S_1 = T$. This follows from a form of Zariski's Main Theorem - see Nagata (37.4). (2) Suppose $\xpd(T/R)$ is not a subset of $\xpd(S_1/R) \cup \xpd(S_2/R)$. Then there exists a height one prime $P$ of $T$ such that $T_P \in \xpd(T/R)~\setminus ~ (\xpd(S_1/R) \cup \xpd(S_2/R))$. Hence $P \cap S_1 = N_1$ and $P \cap S_2 = N_2$. Let $Q = P \cap S_1[S_2]$. Notice that $S_1[S_2]/Q$ is generated over $R/M$ by the images of $S_1/N_1$ and $S_2/N_2$ in $S_1[S_2]/Q$. Since $S_1, S_2$ are in $\mathcal{S}_2(R)$, the fields $S_1/N_1$ and $S_2/N_2$ are finite algebraic over $R/M$. Hence the integral domain $S_1[S_2]/Q$ is a field, finite algebraic over $R/M$. Let $P' = P \cap \overline{S_1[S_2]}$ then $T_P = \overline{S_1[S_2]}_{P'}$. Hence $P'$ is a height one non-maximal prime of $\overline{S_1[S_2]}$. But $P \cap S_1[S_2] = Q = P' \cap S_1[S_2]$ and $Q$ is maximal. By the going up theorem $P'$ is maximal. This is a contradicition. Hence $\xpd(T/R) \subset \xpd (S_1/R) \cup\xpd(S_2/R)$. (3) If $V \in \xpd(T/S_2)$ then $V \in \xpd(T/R) \subset \xpd (S_1/R) \cup \xpd(S_2/R)$. Then $V \in \xpd (S_1/R)$ and (3) now follows, for if $S_1 \neq S_2$ and $\dim S_1[S_2] = 2$ then there exists a maximal ideal $N$ of $S_[S_2]$ of height two. We then have $N \cap S_i = N_i$, $i= 1, 2$. This implies the existence of $T \in \mathcal{S}_2(R) $ such that $T$ dominates both $S_1$ and $S_2$. But item (2) then implies that $\xpd(T/S_1) = \emptyset$. $\blacksquare$ \begin{remark} Given $(S_1, N_1)$ and $(S_2, N_2)$ dominating $R$, let $S = S_1 \cap S_2$. \begin{enumerate} \item The ring $S$ has at most 2 maximal ideals $N_1 \cap S$ and $N_2 \cap S$, for each nonunit of $S$ is contained in $(N_1 \cap S) \cup (N_2 \cap S)$. \item If $S$ has two maximal ideals then $\dim S_1 \cap S_2 \leq 1$. For if there exists a maximal ideal $N$ of $S_1[S_2]$ with $\hgt N = 2$, then $N_1 \cap S \subset N \cap S$ and $N_2 \cap S \subset N \cap S$. Hence $N \cap S$ is the unique maximal ideal of $S$. \end{enumerate} \end{remark} \begin{example} Let $(R, M)$ be a 2-dimensional RLR with $M = (x,y)R$. Let $S_1 = R[y^2/x]_{(x, y, y^2/x)}$ and $S_2 = R[x/y]_{(y, x/y)}$. We claim that $R = S_1 \cap S_2$. It suffices to show that every valuation domain dominating $R$ contains either $S_1$ or $S_2$ for $$ R = \bigcap \{V~~ |~~ V \mbox{ is a valuation domain birationally dominating } R ~\}. $$ Now $(x,y)V$ is principal generated either by $xV$ or $yV$. If $(x,y)V = xV$, then $v(y) \geq v(x) > 0$. Hence $v(y^2) > v(x)$, so $v(y^2/x) > 1$ so $V$ dominates $S_1$. If $(x,y)V \neq xV$ then $v(x) > v(y)$ and $v(x/y) > 0$ and $V$ dominates $S_2$. \end{example} \begin{question} Let $(R, M)$ be a 2-dimensional normal local domain and let $S_1$ and $S_2 \in \mathcal{S}_2(R)$. If $R = S_1 \cap S_2$ does it follow that every valuation domain $V$ that birationally dominates $R$ contains either $S_1$ or $S_2$? \end{question} \section{August 31, 2009} Let $(R, M)$ be a 2-dimensional RLR with $M = (x,y)R$. Assume $R/M$ is algebraically closed. Then $R_1 = R[y/x]_{(x, y/x)}$ is a "first" local QDT of $R$. We have $R \subsetneq R_1$ and there are no RLRs $S$ with $R \subsetneq S \subsetneq R_1$. We have $\xpd(R_1/R) = \{ \ord_R\}$. Let $V = \ord R = (\r1)_{M \r1} = (\r1) _{x \r1}$. \begin{question} Which of the local QDTs of $R_1$ are contained in $V$? \end{question} Let $M_1 = (x, y/x)R_1$ and $y_1 = y/x$ then $y = xy_1$. Some local QDTs of $R_1$ are $R_1[y_1/x]_{(x, y_1/x)}$ and $R_1[y_1/x]_{(x, y_1/x - a)}$ where $0 \neq \overline{a} \in R/M_1$. None of these are contained in $V$ since $v(y_1/x) = -1 = v(y_1/x - a)$. But $R_1[x/y_1]_{(y_1, x/y_1)} \subset V$. Hence there is exactly one local first QDT of $R_1$ that is contained in $V$, namely $R_1[x/y_1]_{(y_1, x/y_1)}$. To check this statement: how do we see that $R_1[x/y_1]_{(y_1, x/y_1 - a)}$, with $0 \neq \overline{a} \in R/M_1$, is not contained in $V$? Notice that $x/y_1$ is a unit of $R_1[x/y_1]_{(y_1, x/y_1 - a)}$, but $v(x/y_1) = 1$ so $v(y_1/x) = -1$. Let $R_2 = R_1[x/y_1]_{(y_1, x/y_1)}$ and let $x_1 = x/y_1$ so $x = y_1x_1$. Then $V = \ord R$ is centered on a height one prime $P_2$ of $R_2$, $P_2 = \frac{x}{y_1}R_2$. Notice that $R_2/P_2$ is a DVR. Let $V_1 = \ord R_1$, then $v_1(x) = v_1(y_1) = 1$ and $x/y_1$ is residually transcendental in the residue field of $V_1$. $R_2 \subset V_1$ and $V_1$ is centered on a height one prime $Q_2$ of $R_2$, $Q_2 = y_1R_2$. Note that $R_2/Q_2$ is a DVR and $M_2 = P_2 + Q_2$. \begin{question} Which of the first local QDTs of $R_2$ are contained in $V$ and which are contained in $V_1$? \end{question} Let $M_2 = (y_1, x_1)R_2$ and $v_1(y_1) = 1$, $v_1(x_1) = 0$ and $v(x_1) = 1$, $v(y_1) = 0$. Then \begin{enumerate} \item $R_2[y_1/x_1]_{(x_1, y_1/x_1)} \subset V_1$, but no other first local QDT of $R_2$ is contained in $V_1$. \item $R_2[x_1/y_1]_{(y_1, x_1/y_1)} \subset V$, and no other first local QDE of $R_2$ is contained in $V$. \end{enumerate} \begin{remark} Let $R = R_0 \subset R_1 \subset R_2 \subset \cdots \subset R_n$ be a sequence of local QDTs. Let $V_i$ be the ord-valuation of $R_i$. Then $R_{i+1} \subset V_i$ for $i = 0, 1, \ldots n-1$ and $V_i \in \epd(R_{i+1})$. Let $P_{i+1}$ denote the center of $V_i$ on $R_{i+1}$, then $R_{i+1}/P_{i+1}$ is a DVR. There exists at most one $j$ with $j