% ------------------------------------------------------ % University College London - Department of Mathematics % MSci Mathematics Project: M400 (2001-2002) % Title: Skew Fields of Fractions % Author: Dierk Philipp Fahr % Supervisor: Dr. Mark L Roberts % ------------------------------------------------------ \documentclass[12pt,a4paper,titlepage]{article} \pagestyle{plain} % -------------------------- Theorems & Definitions -------------------------- \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newenvironment{proof}{\noindent \textnormal{\textbf{Proof:}}}{\begin{flushright}$\mathbf{Q.E.D.}$\end{flushright}} % Proofs \newtheorem{defn}{Definition}[section] % numbering afresh %- OR ? %\newtheorem{defn}[thm]{Definition} % numbering depending on the theorem \newenvironment{rem}{\noindent \textnormal{\textbf{Remark:}}}{} % Remarks %\newcounter{remark} % Remarks should be counted from 1 for each section. %\newtheorem{rem}[remark]{Remark} % ------------------ Commands, Environments & Abbreviations ------------------ \newcommand{\foff}{field of fractions} \newcommand{\ID}{integral domain} \newcommand{\IFF}{if and only if} \newcommand{\ex}{\noindent \textbf{Example: }} \newcommand{\exs}{\noindent \textbf{Examples: }} \newcommand{\rems}{\noindent \textnormal{\textbf{Remarks:}}} %\newcommand{%\remafresh}{\setcounter{remark}{0}} % Reset Remark counter to 1 %\newcommand{\qedproof}{\begin{flushright}$\mathbf{Q.E.D.}$\end{flushright}} \newcommand{\R}{\mathbf{R}} % ring R in bold \newcommand{\mbf}[1]{$\mathbf{#1}$} % bold math \newcommand{\ca}[1]{\mathcal{#1}} % written math \newcommand{\MAT}{\mathcal{M(\mathbf{R})}} % matrix set M(R) \newcommand{\MATn}{\mathcal{M}_{n}(\mathbf{R})} % matrix set Mn(R) \newcommand{\MATRIX}[4]{\left(\begin{array}{cc} #1 & #2 \\#3 & #4 \end{array}\right)} % A 2x2 matrix \newenvironment{deflist}[1]{\newcounter{#1}\begin{list}{\textnormal{\textsf{(#1.\arabic{#1})}}}{\usecounter{#1}}}{\end{list}} % Create list of definitions with names #1.1, #1.2, etc. \newenvironment{numlist}[1]{\newcounter{#1}\begin{list}{\textnormal{\textbf{(\roman{#1})}}}{\usecounter{#1}}}{\end{list}} % Create list of definitions with number i, ii, etc. \begin{document} % -------------------------- TITLE & CONTENTS -------------------------- \title{Skew Fields of Fractions} \author{Dierk Philipp Fahr} \date{\today} \thanks{\\University College London\\Department of Mathematics\\ \\MSci Mathematics Project - M400\\Supervisor: Dr. Mark L Roberts} \maketitle \pagenumbering{roman} % Start page numbering at roman i here for the table of contents. \tableofcontents \newpage \pagenumbering{arabic} % Start page numbering at 1 here. % -------------------------- INTRODUCTION -------------------------- \section{Introduction} The aim of this project is to find conditions for the existence of an embedding from a non-commutative ring into a skew field. Skew fields are also known as division rings, or simply non-commutative fields. To determine which ring occurs as a subring of a field, i.e.~whether a \foff\ exists for a ring $\R$, it is important to study methods of embedding rings into fields. Every non-zero commutative ring which is an \ID\ has a unique \foff. In the non-commutative case the absence of zero-divisors is still a necessary, but not a sufficient condition, as found by A.~Malcev in 1937~\cite{Malcev}. In the commutative case each element of the \foff\ can be written in the form a/b, where $a,b \in \R$ and $b\neq0$. In the general case however, such a representation cannot be found, as a sum of fractions cannot generally be brought to a common denominator. This problem is overcome by inverting matrices rather than elements.\\ In more detail, this project will lead to a characterization of rings with a so-called \emph{fully inverting} homomorphism to a field. We will see that these rings are precisely \emph{Sylvester domains}. We will find sufficient conditions for a ring to have a universal \foff. If it exists it is unique, in comparison to a \foff, of which there might be more than one. The conclusion will be that every semifir has a universal \foff.\\ The project is structured as follows: The method obtaining the unique \foff\ of a commutative ring will be briefly outlined in chapter~\ref{comidFOFF}. The technique called \emph{localization} will be described in chapter~\ref{Localization}, and we will use a similar construction on sets of matrices in chapters~\ref{SingularKernel}. \emph{Ore domains} will be defined in~\ref{Orecase}, and it will shown that they have a unique \foff. Chapter~\ref{FullMatrices} will look closely at the concept of \emph{full matrices}. When embedding a ring into a field we will be inverting certain sets of matrices leading to \emph{matrix ideals}. These will be defined in chapter~\ref{MatrixIdeals} and together with chapter~\ref{UniversalFOFF}, in which we define Sylvester domains and semifirs, this forms the core of the project. Finally, in chapter~\ref{Examples}, we will give examples with proofs of classes of rings, which are embeddable into a field. We will also mention Malcev's counter-example in chapter~\ref{Malmore} and show that there are \ID s with zero localization.\\ The main sources of reference used for the completion of this project, were the Algebra Volumes II and III by Prof.~Paul Moritz Cohn~(\cite{Cohn2},~\cite{Cohn3}), and his classic and fundamental book ``Free Rings and their Relations''~\cite{CohnFRTR}. % -------------------------- BASIC DEFINITIONS -------------------------- \subsection{Basic definitions}\label{Basicdefinitions} %\remafresh \begin{defn} % RING A \emph{ring} is a non-empty set $\R$, on which there are two laws of composition, addition $(+)$ and multiplication $(\cdot)$. These binary operations are defined such that for all $r,s \in \R,\ r+s \in \R$ and $r\cdot s \in \R$. The latter is simply written as the product $rs$. Under addition\/ $\R$ forms an abelian group and moreover the following are satisfied: \begin{deflist}{R} \item Associativity: $\ \forall\ r,s,t \in \R,\ (rs)t=r(st)$ \item Identity: $\ \exists\ 1\in \R$, such that $\ \forall\, r \in \R,\ r1=1r=r$ \item Distributivity: $\forall\, r,s,t \in \R$, \begin{eqnarray*} (r+s)t & = & rt+st\\ r(s+t) & = & rs+rt \end{eqnarray*} \end{deflist} If $ab=ba$ for all $a,b \in \R$, then\/ $\R$ is called a \emph{commutative} ring.\\ Moreover if 1=0, then $\R$ is the \emph{zero ring} denoted simply by a\/ $0$, which has only one element. \end{defn} \begin{defn} % INTEGRAL DOMAIN A ring\/ $\R$ is called an \emph{integral domain}, if \begin{numlist}{ID} \item $\R \neq 0$ \item $\forall\, a,b\in\R,\ ab=0 \Rightarrow a=0 \mbox{ or } b=0$. \end{numlist} \end{defn} \ex % INTEGRAL DOMAIN The standard example of rings which fail to be \ID s arise as matrix rings. Take $\R$ to be any (nontrivial) ring, and let $\ca{M}_{2}(\R)$ be the set of all $2\times2$ matrices over $\R$. Consider the following two matrices: \[ r=\MATRIX{0}{0}{1}{0},\ s=\MATRIX{0}{0}{0}{1},\] then clearly $rs=0$, but both $r,s\neq0$. We note, that this ring is not commutative either because $sr\neq0$.\\ % M(R) Throughout we will use the notation $\MATn$ to denote the set of all $n\times n$ matrices with entries in $\R$, and $\MAT$ to be the set of all square matrices (of any size). Nevertheless the word ``square'' will be stressed explicitely most of the time, since we nearly always consider our matrices to be square. % UNIT We call an element $u \in\R$ a \emph{unit} (or \emph{invertible}) in $\R$, if there is an element $w\in\R$, such that $uw$=$1$ and $wu$=$1$. If such an element $w$ exists, it is unique and called the inverse of $u$, usually written $u^{-1}$. \begin{defn} % FIELD A \emph{(skew\footnote{`Skew fields' are also called `division rings'.}) field} is a non-zero (non-)commutative ring\/ \mbf{K}, in which every non-zero element is a unit of\/ \mbf{K}. \end{defn} \ex % H HAMILTON QUATERNIONS The first skew field and most elementary example was discovered by Sir W.~R.~Hamilton in 1843, called the \emph{quaternions}. It can be defined by $$ \mathbf{H}=\{a1+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}\mid a,b,c,d \in \R\},$$ where the elements $1,\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy the following identities: $$\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-1;\quad \mathbf{i}\mathbf{j}=\mathbf{k},\, \mathbf{j}\mathbf{k}=\mathbf{i},\, \mathbf{k}\mathbf{i}=\mathbf{j};\quad \mathbf{j}\mathbf{i}=-\mathbf{k},\, \mathbf{k}\mathbf{j}=-\mathbf{i},\, \mathbf{i}\mathbf{k}=-\mathbf{j}.$$ The quaternions are clearly not commutative, and it is easily verified that each non-zero element is invertible.\\ % EMBEDDING \noindent Finally we call an injective homomorphism an \emph{embedding}. % -------------------------- FIELD OF FRACTIONS -------------------------- \subsection{Field of fractions}\label{comidFOFF} %\remafresh In this section we will prove the well-known fact that if $\R$ is a commutative integral domain, then it is possible to embed $\R$ in a \foff, as defined below: \begin{defn}\label{defn:comfoff} % FIELD OF FRACTIONS A \emph{\foff} for the commutative ring\/ $\R$ is a field\/ $\mathbf{K}$ containing a subring\footnote{A subring \mbf{S} of a ring $\R$ is a subset which is a ring under the operations of $\R$, and the multiplicative identity of \mbf{S} coincides with that of $\R$.}\/ $\mathbf{R'}$ isomorphic to $\R$, such that every element of\/ $\mathbf{K}$ can be expressed in the form $r \over s$ (or $rs^{-1}$), for $r,s \in \mathbf{R'}$, where $s\neq 0$. We usually identify $\R$ with its image under an embedding $\R \rightarrow \mathbf{K}$. \end{defn} % Embedding Z in Q The procedure by which one obtains the rational field \mbf{Q} from the rings of integers \mbf{Z} (i.e.~the embedding of \mbf{Z} into \mbf{Q}) extends easily to any commutative \ID\ and produces the (unique) \foff. To formally construct fractions with numerator and denominator in an \ID\ $\R$ we will consider ordered pairs $(r,s)$, where $r$ represents the numerator and $s$ represents the denominator, assuming $s\neq0$. We know that for example in $\mathbf{Q}=\{{a\over b}:a,b\in\mathbf{Z},b\neq0\}$, two fractions $r \over s$ and $t \over u$ may be equal even though their corresponding numerators and denominators are not equal. We can express the fact that ${r \over s} = {t \over u}$ by writing $ru=st$. This leads to the introduction of equivalence classes, as used in the proof of the following: \begin{thm} % COM ID => FofF Every commutative \ID\ has a \foff. \end{thm} %- \newpage \begin{proof} % COM ID => FofF Let $\R$ be a commutative integral domain, and let \mbf{S} be the set of all ordered pairs $(r,s)$, where $r,s\in\R$ and $s\neq 0$. Define a relation $\sim$ on \mbf{S} by \[ (r,s) \sim (t,u) \Leftrightarrow ru=st.\] An easy verification shows that $\sim$ is an equivalence relation. This is where we need our commutative ring to be an integral domain, because the verification that the relation is transitive involves cancellation, i.e.~the fact that the ring has no zero-divisor other than zero. We denote the equivalence class of (r,s) by [r,s] and are left to show, that the set \mbf{K} of equivalence classes will provide the \foff. In order to do this, we define two binary operations on the equivalence classes: \begin{eqnarray*} [r,s]+[t,u]&=&[ru+ts,su] \\ \,[r,s][t,u]&=&[rt,su] \end{eqnarray*} Then the straightforward verification of the following properties yields that \mbf{K} is the wanted \foff: \begin{enumerate} \item The operations are well defined on \mbf{K}. \item \mbf{K} is indeed a field:\\ We note, that we have the additive zero $[0,1]$ and the multiplicative identity $[1,1].$ The additive inverse is $[-a,b],$ and clearly the multiplicative inverse will be $[b,a].$ \item The map $\R \rightarrow \mathbf{K}$ which sends $r \mapsto [r,1]$ is an injective homomorphism, i.e.~an embedding. \item\label{item:frac} ${[r,s]}={[r,1] \over [s,1]}$ \end{enumerate} \end{proof} % ID => FofF \rems % FofF \begin{enumerate} \item If we identify an element $r\in\R$ with its image $[r,1]$ in \mbf{K}, we have ${[r,s]}={r \over s}$ by~\ref{item:frac}, and this allows us to assume that $\R$ is a subring of \mbf{K} as required by the general definition~\ref{defn:comfoff}. \item It can be shown that for a given commutative \ID\ $\R$, all fields of fractions are isomorphic. We also note without proof, that a finite \ID\ is a field\footnote{Note that by Wedderburn's theorem, it is not possible to give any finite example of a field that is not commutative.}.\\ \end{enumerate} \ex % K[X] If \mbf{K} is a field, then we know that the ring of polynomials $\mathbf{K}[x]$ is an \ID. Applying above theorem shows that we can construct a \foff\ containing $\mathbf{K}[x]$ by considering all fractions of the form $f(x) \over g(x)$, where $f(x),g(x)$ are polynomials with $g(x)\neq0$. This field is called the \emph{field of rational functions} in $x$, and is denoted by $\mathbf{K}(x)$.\\ So far so good: Now we know that for any commutative \ID\ $\R$, we can formally invert the non-zero elements of $\R$ to form a unique \foff. Furthermore we add fractions by taking common denominators, and multiply fractions by multiplying numerators and denominators. Some of this will still work in the following sections, where we first discuss the general procedure of localization, and then use the Ore condition to show that so called Ore domains also have embeddings in fields. % -------------------------- LOCALIZATION -------------------------- \subsection{Localization}\label{Localization} \newcommand{\Rs}{\R_{\mathbf{S}}} %\remafresh The main technical problem is the construction of inverses. We will now discuss the general issues of ``inverting'' a given multiplicative set \mbf{S} of non-zero elements in a (possibly) non-commutative ring $\R$. \emph{Localization} is a very useful tool in commutative rings and thus we look at how much can be extended to the non-commutative case. Also we will use a similar technique in chapter~\ref{SingularKernel} extending it to invert matrices when this is possible. % DEF MULTIPLICATIVE A set $\mathbf{S}\subseteq\R$ is called \emph{multiplicative}, if $1\in\mathbf{S}$ and $a,b \in \mathbf{S} \Rightarrow ab\in \mathbf{S}$. \begin{defn}\label{sinverting} % S-inverting Let \mbf{S} be a subset of a ring\/ $\R$. An\/ \emph{\mbf{S}-inverting} homomorphism is a homomorphism $f:\R \rightarrow \R'$, such that for all $s\in \mathbf{S},\ f(s)$ is a unit in the ring\/ $\R'$. \end{defn} \begin{thm}[Localization]\label{univpict} % UNIVERSAL LOCALIZATION Let $\R$ be a ring, $\mathbf{S}\subseteq\R$ a multiplicative subset. Then there exists a ring, denoted by $\Rs$, with an \mbf{S}-inverting homomorphism $\lambda:\R \rightarrow \Rs$, with the following universal property: For any \mbf{S}-inverting homomorphism $f: \R \rightarrow \R'$, there is a unique ring homomorphism $f': \Rs \rightarrow \R'$, such that $f=f'\circ\lambda.$ \end{thm} % DIAGRAMM R->Rs->R' UNIVERSAL LOCALIZATION $$\begin{array}{ccc} \R & \stackrel{\lambda}{\longrightarrow} & \Rs \\ & f \searrow & \downarrow f' \\ &&\R' \end{array}$$ % PROOF UNIVERSAL LOCALIZATION The universal property\footnote{\emph{cf.} Eilenberg's \emph{Abstract Inverse Category Theory} or see Mac~Lane, \emph{Categories for the Working Mathematician.}} guarantees the uniqueness of $\lambda:\R \rightarrow \Rs$ and thus determines $\Rs$ up to isomorphism. The proof goes by adjoining elements $s'$, such that for all $s\in\mathbf{S},\ ss'=s's=1.$ Then $\lambda(s)$ is invertible with inverse $s'$, and $\lambda$ is thus an \mbf{S}-inverting homomorphism. Refer to theorem~\ref{matrixlocaliz} for details in the case of matrices. The proof of the universal property is best found in Skew Fields~\cite{CohnSkew}, chapter 1.3. \begin{defn} % LOCALIZATION Rs The above constructed ring\/ $\Rs$ is called the \emph{localization} of\/ $\R$ at\/ \mbf{S} or equivalently the \emph{universal \mbf{S}-inverting ring.} \end{defn} \rems % Rs=0 \begin{enumerate} \item At this point we importantly note that the classical case of embedding a commutative \ID\ $\R$ into its \foff\ corresponds to the localization of $\R$ at the multiplicative set $\R\setminus\{0\}$. \item Proposition~\ref{univpict} is, in fact, true for any subset $\mathbf{S}\subseteq\R$. For convenience we took \mbf{S} to be multiplicative, and we also note that if $0\in\mathbf{S}$ then $\Rs=0$, $ker(\lambda)=\R$, and so everything gets mapped to zero. Leaving this trivial case, we may suppose that $0 \not\in\mathbf{S}$. \item Moreover, and contrary to the commutative case, $\Rs$ may be the zero ring even though $\R\neq0 \mbox{ and } 0\not\in\mathbf{S}$. Even when $\R$ is an \ID\ this can happen, and an example of this can be found in chapter~\ref{Examples}. \end{enumerate} The problem of the localization $\Rs$ in the non-commutative case is that we do not have the nice normal form property for its elements as in the commutative case. Hence we will introduce additional conditions on the set \mbf{S} so that we can find a kind of ``normal form''. As noted above, we will assume it to be multiplicative and that $0\not\in\mathbf{S}$. Recall that in the commutative case every element in $\Rs$ may then be written as a fraction $a \over s$ (or $as^{-1}$), for $a\in\R,\, s\in\mathbf{S}$. Also ${a \over s}={a' \over s'}$ \IFF\ $as't=a'st$, for some $t\in\mathbf{S}$. % KERNEL This helps us to determine the kernel: \begin{equation}\label{orekernel} ker(\lambda)=\{ a\in \R : at=0 \mbox{ for some } t\in\mathbf{S} \} \end{equation} Ore's idea now consists in asking under what circumstances the elements of $\Rs$ have this form, when commutativity is not assumed. So we want to express an element $s^{-1}a$, where $a\in\R,\, s\in\mathbf{S}$, as $a_{1}s_{1}^{-1}\ (a_{1}\in\R,\, s_{1}\in\mathbf{S}).$ Thus we want to look at the equation $as_{1}=sa_{1}$. More precisely we have $as_{1}1^{-1}=sa_{1}1^{-1}$ and therefore $as_{1}t=sa_{1}t$, for some $t\in\mathbf{S}$. This leads to conditions set out by Ore, which are discussed next. % -------------------------- ORE CASE -------------------------- \subsection{Ore case}\label{Orecase} Let us study Ore's localization theory in this section, developed by Ore in the early 1930s. Here we will find the necessary and sufficient conditions for constructing the (Ore) localization $\Rs$ with respect to a given multiplicative set $\mathbf{S} \subseteq \R$. Ore took the set \mbf{S} to be the multiplicative set of all non zero-divisors in $\R$ leading to the definition of so called Ore domains, which we will mention later. \begin{thm}[Ore condition]\label{locallore} % ORE CONDITION Let \mbf{S} be a subset of a ring $\R$, such that \begin{deflist}{O} \item \mbf{S} is multiplicative. \item For any $a \in \R,\, s\in\mathbf{S},\quad s\R\cap a\mathbf{S} \neq \emptyset.$ \item For any $a\in \R,\, s\in\mathbf{S},\quad sa=0 \Rightarrow at=0,\ \mbox{ for some } t\in\mathbf{S}.$ \end{deflist} Then the localization $\R_{\mathbf{S}}$ may be constructed as follows:\\ Define an equivalence relation $(a,s)\sim(a',s')$ on $\R\times\mathbf{S}$, whenever $$au=a'u',\, su=s'u' \in \mathbf{S}, \mbox{ for some } u,u' \in \R.$$ The quotient set\/ $\R\times\mathbf{S}/\sim$ is then the localization of\/ $\R$ at\/ \mbf{S}. In particular, the elements of\/ $\Rs$ may be written as fractions ${a \over s}={as^{-1}}$ and $ker(\lambda)$ is given by the above equation~\textnormal{(\ref{orekernel})}. \end{thm} The proof is a series of detailed and quite technical verifications and might best be followed in chapter 9, Algebra Vol.~3~\cite{Cohn3}. For a quicker method my advice is to look at T.Y.~Lam's proof in~\cite{Lam}, chapter 10. Although not needed here, the following terminology is widely used: \begin{defn} % ORE SET A multiplicative set\/ $\mathbf{S}\subseteq\R$ satisfying \textnormal{(O.2)} is called a \emph{right Ore set}. If\/ \mbf{S}\/ satisfies \textnormal{(O.3)} it is said to be \emph{right reversible}. If all Ore conditions \textnormal{(O.1)-(O.3)} hold, \mbf{S} is called a \emph{right denominator set}. \end{defn} % ORE DOMAINS \subsubsection*{Ore domains}\label{Oredomains} When our multiplicative set $\mathbf{S}\subseteq\ca{Z}(\R)$\footnote{$\ca{Z}(\R)$ is called the \emph{centre} of $\R$ defined by $\ca{Z}(\R)=\{z\in\R\mid zx=xz,\, \forall\, x\in\R\}$, i.e.~all those elements which commute.}, and in particular when $\R$ is commutative, \textnormal{(O.1) \& (O.2)} hold automatically. If $\R$ is an \ID, \textnormal{(O.3)} can be omitted, and if in addition $\mathbf{S}=\R\setminus\{0\}$ then \textnormal{(O.2)} gives us\footnote{This equation is sometimes referred to as the \emph{right Ore condition} for \ID s.}: \begin{equation}\label{oredomain} % ORE DOMAIN CONDITION \forall\, a,b \in \R\setminus\{0\},\quad a\R\cap b\mathbf{R} \neq0. \end{equation} \begin{defn} % ORE DOMAIN An \ID\ $\R$ satisfying equation~\textnormal{(\ref{oredomain})} is called a \emph{right Ore domain}. \end{defn} \begin{thm}\label{thmOREdomain} % ORE DOMAIN Let $\R$ be an \ID\ such that for all $\,a,b \in \mbox{$\R\setminus\{0\}$},\\ a\R\cap b\mathbf{R} \neq0.$ Then the localization of $\R$ at $\R\setminus\{0\}$ is a field\/ \mbf{K} and the natural homomorphism $\lambda : \R \rightarrow \mathbf{K}$ is an embedding. \end{thm} % ORE DOMAIN \begin{proof} % ORE DOMAIN If $\mathbf{S}=\R\setminus\{0\}$ and equation~\textnormal{(\ref{oredomain})} holds, then in that case $\Rs$ (applying theorem~\ref{locallore}) is a field, since when $a,b\in\R\setminus\{0\}$, then $ab^{-1}$ has the inverse $ba^{-1}.$ Moreover $\lambda$ is injective, as $ker(\lambda)=0$ by equation~\textnormal{(\ref{orekernel})} for an \ID. It is therefore an embedding. \end{proof} \noindent This means that every right Ore domain has a \foff. Moreover: \begin{prop} % Uniqueness of Ore domains The \foff\ of a right Ore domain is unique up to isomorphism. \end{prop} \begin{proof} % Uniqueness of Ore domains Suppose $\R$ is a right Ore domain and $\R'$ is any \foff\ of $\R$. Then we have an embedding $f:\R\rightarrow\R'$. Let $\mathbf{S}=\R\setminus\{0\}$. Then we have a homomorphism $f': \Rs \rightarrow \R'$, which is an embedding, because $\Rs$ is a field by theorem~\ref{thmOREdomain}. The image of $\Rs$ under $f'$ is a field containing $\R$ and hence must be equal to $\R'$, because $\R'$ was a \foff. (It is helpful to refer to the diagram of proposition~\ref{univpict} in the section on localization.) Therefore $f'$ is an isomorphism and this determines the \foff\ of a right Ore domain uniquely. \end{proof} Now we are another step further. So far we have seen that commutative \ID s and Ore domains are embeddable in a field. Note that the absence of zero-divisors \begin{equation} xy=0\ \Rightarrow \ x=0 \mbox{ or } y=0\quad (1\neq 0) \end{equation} is a necessary condition for the embeddability of a ring in a field, which is also sufficient in the commutative case. This is not true in general. As mentioned in the introduction, A.~Malcev~\cite{Malcev} has shown in his famous counter-example (see chapter~\ref{MalcevCounter}), that if $\R$ is any \ID, an embedding into a field need not always exist. Let us finally remark that all definitions on denominator sets, Ore conditions and Ore domains are left-right symmetric and that a left and right Ore domain is simply called an \emph{Ore domain}. \pagebreak Further interesting consequences of the Ore condition and its link to Noetherian domains can be found in the examples chapter~\ref{NoetherianBezout}. % -------------------------- EPIC R-FIELDS -------------------------- \section{Epic $\R$-Fields}\label{EpicRFields} \newcommand{\ERF}{epic $\R$-field} %\remafresh In this chapter we will study \ERF s, which are more general than fields of fractions, as they are fields generated by the image of $\R$ with a homomorphism $\R \rightarrow \mathbf{K}$. Later we will see that there is an interesting correspondence to prime matrix ideals, which we will define in chapter~\ref{MatrixIdeals}. \subsection{Commutative case}\label{comcaseepic} % DEF COMMUTATIVE IDEAL Let us first look at a ring homomorphism from a commutative ring into a field. Recall that a nonempty subset \mbf{I} of a commutative ring $\R$ is called an \emph{ideal} of $\R$, if it is a subgroup under addition and $ra\in\mathbf{I}$ for all $a\in \mathbf{I}$ and $r\in\R$. A \emph{prime ideal}\footnote{In the non-commutative setting such an ideal is said to be \emph{completely prime}. We will define a \emph{prime ideal} in the non-commutative case in chapter~\ref{MIPropt}.} is a proper ideal \mbf{I} of $\R$, such that if $ab\in\mathbf{I}$ then $a\in \mathbf{I} \mbox{ or } b \in\mathbf{I}$, for all $a,b\in\R$. Recall further that an ideal $\mathbf{I}$ in a commutative ring $\R$ is prime, \IFF\ $\R/\mathbf{I}$ is an \ID.\footnote{We also have that $\mathbf{I}$ is maximal \IFF\ $\R/\mathbf{I}$ is a field.} \begin{prop}\label{PRIMEKERNEL} % KERNEL <-> PRIME IDEAL Let\/ \mbf{I} be an ideal of the commutative ring\/ $\R$. \mbf{I} is a prime ideal \IFF\/ \mbf{I} is the kernel of a ring homomorphism $\phi:\R\rightarrow\mathbf{K}$, where\/ \mbf{K} is a field generated by the image of\/ $\R$. \end{prop} \begin{proof} % KERNEL <-> PRIME IDEAL \noindent ($\Leftarrow$) If $\mathbf{I}=ker(\phi)$, then by the first isomorphism theorem $\R/\mathbf{I}$ is isomorphic to a subring of the field \mbf{K} containing $1=\phi(1)$. It follows that $\R/\mathbf{I}$ is an \ID, and hence \mbf{I} is a prime ideal. \noindent ($\Rightarrow$) If \mbf{I} is a prime ideal, then it is the kernel of the composite mapping defined by the canonical homomorphism $\R \rightarrow\R/\mathbf{I}$ followed by the embedding of $\R/\mathbf{I}$ into its field of fractions. This can be done (see chapter~\ref{comidFOFF}) as $\R/\mathbf{I}$ is a commutative \ID. \end{proof} % LOOK AHEAD PRIME MATRIX IDEAL The later introduced notion of a prime matrix ideal corresponds to the concept of a prime ideal in a commutative ring. Moreover, prime matrix ideals can be used to describe the homomorphisms of general rings into fields, just as prime ideals are used in the commutative case. % -------------------------- IBN -------------------------- \subsection{Invariant basis number}\label{IBN} Let $\R$ be any ring and \mbf{F} a free left $\R$-module on a basis $\ca{B}$. If $\ca{B}$ is finite, we say that \mbf{F} has \emph{unique rank} if any two bases of \mbf{F} have the same number of elements. The number of these elements is called the \emph{rank} of \mbf{F}. Over a field every module is free, of unique rank, as is known from linear algebra. \begin{defn}\label{IBNDef} % IBN A ring $\R$ is said to have \emph{invariant basis number} (IBN) if every free $\R$-module has unique rank. \end{defn} % IBN In terms of matrices, let $B$ be the $m \times n$ matrix describing the surjective homomorphism $\beta : \R^{m}\rightarrow\R^{n}$, and $A$ be the $n\times m$ matrix so that $AB=I_{n}$. We can then restate the condition for IBN as follows: \begin{prop} % MATRIX IBN For any ring $\R$ the following conditions are equivalent: \begin{numlist}{matrixibn} \item $\R$ has invariant basis number (IBN). \item $\R^{m}\cong \R^{n} \Rightarrow\ m=n.$ \item If $A$ is an $n\times m$ matrix, $B$ an $m\times n$ matrix and $AB=I,\ BA=I$, then $m=n$. \end{numlist} \end{prop} \begin{proof} % MATRIX IBN (ii) is basically a restatement of the definition of IBN. By the above, the isomorphism\/ $\R^{m}\cong \R^{n}$ is described by matrices as in (iii), so the condition for IBN is that $m=n$. \end{proof} \noindent Let us also prove the following: \begin{prop}\label{bothIBN} % R->S, both IBN Given a ring homomorphism $f: \R \rightarrow \mathbf{S}$. If\/ \mbf{S} has IBN, then so does\/ $\R$. \end{prop} \begin{proof} % R->S, both IBN Suppose that $\R$ does not have IBN and let $A, B$ be a pair of non-square matrices over $\R$ such that $AB=I,\ BA=I$. Applying $f$, we obtain a pair of non-quare matrices $f(A),f(B)$ satisfying the same equations, hence IBN also fails for \mbf{S}. The converse proves what we wanted. \end{proof} \begin{rem} % Lacking IBN Most rings have IBN. In particular, any field and (by the above proposition) any ring with a homomorphism to a field has IBN. An example of a ring without IBN is the trivial ring in which $1=0$, so the only module is $0$ and $0^{m}\cong 0^{n}$. We usually exclude this happening. Another example lacking IBN is given in the next section. \end{rem} % -------------------------- LOCAL RINGS -------------------------- \subsection{$\R$-fields and local rings}\label{localrings} \begin{defn} % EPIC R-FIELD Let\/ $\R$ be ring. An\/ \emph{$\R$-ring} is a ring\/ \mbf{K} with a homomorphism $\R \rightarrow \mathbf{K}$. An $\R$-ring which is a field is called an \emph{$\R$-field}.\\ By an \emph{\ERF} we understand an $\R$-field which is generated by the image of $\R$. \end{defn} \begin{rem} % FOFF. Note that we can define a \foff\ as follows: If \mbf{K} is an \ERF{}\footnote{In the term ``\ERF'', the word \emph{epic} comes from the fact that the canonical map is then an epimorphism. It can be shown that a homomorphism $f$ from a ring to a field is an epimorphism \IFF\ the field is generated by $im(f)$ (see chapter 4.1 in~\cite{CohnSkew}).} for which the canonical map $\R \rightarrow \mathbf{K}$ is injective, \mbf{K} is called a \emph{\foff} of $\R$.\\ \end{rem} If $f:\mathbf{K} \rightarrow \mathbf{L}$ is a ring-homomorphism between \ERF s, then $f$ is injective. This is because its kernel is an ideal in a field, i.e.~must equal to $0$ or the whole field. But as $im(f)$ is a subfield of \mbf{L} containing the image of $\R$, we must have $\mathbf{L}=im(f)$, as \mbf{L} is an \ERF. Hence $ker(f)=0$ and thus $f$ is injective. This shows, that it is too restrictive to consider ring-homomorphism between \ERF s alone. We will later see, that for a given ring $\R$ the \ERF s form a category, with so-called specializations as morphisms between the objects. We need the following: \begin{defn} % LOCAL HOM & LOCAL RING A \emph{local ring} is a ring $\mathbf{A_{0}}$, in which the non-units form an ideal\/ \mbf{m}. The quotient ring\/ $\mathbf{A_{0}}/\mathbf{m}$ is then a field, called the \emph{residue-class field} of $\mathbf{A_{0}}$. A \emph{local homomorphism} between rings \mbf{A} and\/ \mbf{B} is a homomorphism $f:\mathbf{A_{0}}\rightarrow\mathbf{B}$, where $\mathbf{A_{0}} \subseteq \mathbf{A}$ is a subring, which maps non-units to non-units. \end{defn} \begin{rem} % LOCAL HOMs When we are working with $\R$-rings, a local homomorphism is understood to have a domain which includes the image of $\R$. We note that if \mbf{B} is a field, the non-units in $\mathbf{A_{0}}$ form an ideal (which is $ker(f)$), so that $\mathbf{A_{0}}$ is then a local ring. \end{rem} \begin{defn}\label{specialization} % SPECIALISATION Two local homomorphisms between rings \mbf{A} and\/ \mbf{B} are said to be \emph{equivalent}, if there is a subring of\/ \mbf{A} on which both are defined, and on which they agree and again define a local homomorphism. This is an equivalence relation and an equivalence class of local homomorphisms between \ERF s is called a \emph{specialization}. \end{defn} %- \newpage \exs % EXAMPLES Here are two examples, showing that there are rings for which there does not exist $\R$-fields. \begin{enumerate} \item If $\R=0$, or $\R$ is any simple ring with zero-divisors, such as a matrix ring over a field, then any non-zero homomorphism $\R \rightarrow \mathbf{K}$ must be injective, which is impossible if \mbf{K} is a field. \item If $\R$ is a ring or even an \ID\ lacking IBN, then any $\R$-ring is again without IBN and therefore cannot be a field. Such examples of \ID s without IBN can be constructed\footnote{See the proof of theorem~\ref{MALCEVVV} in chapter~\ref{Rszero}, and for more detail the end of chapter 2.11 in~\cite{CohnFRTR}, p.~149.}, and thus there exist \ID s without \ERF s. Of course, if $\R$ is commutative then it has IBN and if it is also an \ID, it can be even embedded in a \foff\ as seen before. \end{enumerate} % COMMUTATIVE CASE OF CONSTRUCTION OF FOFF and LOZALIZATION Let us now come back to commutative rings and take a look at $\R$-fields in this case. Let $\R$ be a commutative ring and \mbf{K} an \ERF. Then \mbf{K} is again commutative, being generated by the image of $\R$. Now let \mbf{p} be the kernel of the natural mapping $\R \rightarrow \mathbf{K}$. Then \mbf{p} is a prime ideal as seen in proposition~\ref{PRIMEKERNEL}. \mbf{K} can be constructed from $\R$ in two ways: \begin{numlist}{CASES} % TWO WAYS TO CONSTRUCT FOFF. \item We can form the well known quotient ring $\R/\mathbf{p}$, which is an \ID, as \mbf{p} is a prime ideal. \mbf{K} is then obtained as the \foff\ of $\R/\mathbf{p}$. \item Instead of putting the elements in \mbf{p} equal to 0, we can make the elements outside of \mbf{p} invertible. This is done by forming the localization $\R_{\mathbf{p}}$. This is a local ring, and its residue-class field is isomorphic to \mbf{K}. \end{numlist} This can be illustrated by the following commutative diagram: % DIAGRAMM R->R/p->K TWO WAYS TO CONSTRUCT A FOFF. $$\begin{array}{ccc} \R & \longrightarrow & \R_{\mathbf{p}} \\ \downarrow & \searrow & \downarrow \\ \R/\mathbf{p} & \longrightarrow & \mathbf{K} \end{array}$$ Here the two triangles correspond to the two methods of constructing \mbf{K}. The familiar route via $\R/\mathbf{p}$ does unfortunately not generalize to the non-commutative case. We therefore turn to the upper triangle, but we have to be careful, as a \foff\ need not be unique in the non-commutative case. In general an \ERF\ will not be determined by its kernel alone. In the next chapter we will therefore introduce the so-called \emph{singular kernel}, which is the set of matrices mapping to singular matrices. This set will give us more information than only looking at the usual kernel. % -------------------------- SINGULAR KERNEL -------------------------- \subsection{Matrix Localization \& Singular Kernel}\label{SingularKernel} We have seen in chapter~\ref{Localization} how to construct the localization of $\R$ at a subset $S$. ``Matrix Localization" is done in a similar way, but instead of taking a multiplicative set \mbf{S} we now take, as the name suggests, a set of matrices $\mathbf{\Sigma}$ over a ring $\R$. We define: \begin{defn} % Sigma-inverting A homomorphism $f:\R \rightarrow \R'$ is called\/ \emph{$\mathbf{\Sigma}$-inverting} if for all $A\in \mathbf{\Sigma},\ f(A)$, the image of $A$ under $f$, is an invertible matrix over the ring\/ $\R'$. \end{defn} Once more we limit ourselves to square matrices, as only those will play a role for homomorphisms to fields. Let us recall our localization theorem~\ref{univpict} (section~\ref{Localization}) and compare it with the following: \begin{thm}[Matrix localization]\label{matrixlocaliz} % Matrix Locazation Let\/ $\R$ be a ring, $\mathbf{\Sigma}$ a set of matrices over\/ $\R$. Then there exists a ring\/ $\R_{\mathbf{\Sigma}}$ and a homomorphism $\lambda: \R \rightarrow \R_{\mathbf{\Sigma}}$ which is universal $\mathbf{\Sigma}$-inverting in the sense that for all $A\in \mathbf{\Sigma},\ \lambda(A)$ is invertible over\/ $\R'$. \end{thm} \begin{proof} % Matrix Locazation Let $A$ be a $n\times n$ matrix, such that $A=(a_{ij})\in\mathbf{\Sigma}$. For every such matrix we choose $n^{2}$ symbols $a'_{ij}$, which we adjoin to $\R$ with the following matrix relation: $$ AA'=I=A'A,\ \mbox{ where } A'=(a'_{ij}).$$ The resulting ring is denoted by $\R_{\mathbf{\Sigma}}$ and then the natural map $\lambda: \R \rightarrow \R_{\mathbf{\Sigma}}$ is $\mathbf{\Sigma}$-inverting. As required the images of the matrices are then invertible. \end{proof} As in the ``standard'' localization terminology, the ring $\R_{\mathbf{\Sigma}}$ is called the \emph{universal $\mathbf{\Sigma}$-inverting ring} or the \emph{localization} of $\R$ at $\mathbf{\Sigma}$. Furthermore note that every $\mathbf{\Sigma}$-inverting homomorphism from $\R$ to another ring can be factored uniquely by $\lambda$, which we leave unproved here. Coming back to describe \ERF s, we have seen that we need more than elements which map to zero. We will actually look at matrices which become singular. The process of forming fields of fractions, or more generally \ERF s for a ring $\R$, is described in terms of the \emph{singular kernel} as defined below. We will show how any \ERF\ can be characterized by its singular kernel, which will later have a simple description as prime matrix ideal. \begin{defn} % SINGULAR KERNEL Given an $\R$-field\/ \mbf{K} with natural map $\lambda:\R \rightarrow \mathbf{K}$. By the \emph{singular kernel} of\/ \mbf{K} (or of $\lambda$), written $Ker(\lambda)$, we understand the collection of all matrices $A \in \MAT$, which map to singular matrices over \mbf{K}. \end{defn} % LOOK AHEAD If $\ca{P}$ is the set of all such matrices, then we can localize at the complement $\mathbf{\Sigma}$ in $\MAT$. Analogous to $\R_{\mathbf{p}}$ in the commutative case we write the localization as $\R_{\ca{P}}$. We will see in chapter~\ref{TheConstruction} how an \ERF\ can be constructed in terms of its singular kernel, which is a very interesting set of matrices, as we will get different singular kernels for different fields of fractions. The matrices in the complement will become invertible. If $\mathbf{\Sigma}$ is the set of these matrices, then we will also see that the localization $\R_{\mathbf{\Sigma}}$ is a local ring having residue-class field \mbf{K}, which will be equal to the \ERF\ having singular kernel $\ca{P}$. We are left to investigate when a collection of matrices is a singular kernel, just as we can tell when a collection of elements of $\R$ is a prime ideal. Later singular kernels will be characterized in much the same way in which kernels of $\R$-fields are characterized as prime ideals in the commutative case. To conclude this chapter let us define a \emph{universal \ERF} to be an \ERF\, which has every other \ERF\ as specialization. % -------------------------- FULL MATRICES -------------------------- \section{Full Matrices}\label{FullMatrices} %\remafresh % -------------------------- UGN -------------------------- \subsection{Unbounded generating number}\label{UGN} In section~\ref{IBN} we discussed the invariant basis number. A strengthening of IBN is the unbounded generating number, UGN for short: \begin{defn} % UGN A ring $\R$ is said to have \emph{unbounded generating number} (UGN) if for every $n=1,2,\ldots $ there is a finitely generated left $R$-module which cannot be generated by fewer than $n$ elements. \end{defn} Equivalently we can say that $\R$ has UGN if for all $n$, $R^{n}$ cannot be generated by fewer than $n$ elements.\\ \begin{rem} % UGN -> IBN Recall the definition of IBN in terms of matrices (see section~\ref{IBN}). If IBN fails for a ring $\R$, i.e.~$\R^{m} \cong \R^{n} \mbox{ for } m > n$ say, then $\R^{m}$ has a basis of $nS, both UGN Given a ring homomorphism $f: \R \rightarrow \mathbf{S}$. If\/ \mbf{S} has UGN, then so does\/ $\R$. \end{prop} % -------------------------- Inner Rank -------------------------- \subsection{Inner rank}\label{Innerrank} It was briefly mentioned in the introduction that for a general theory of fields it is convenient to invert matrices rather than elements. This means we have to consider which matrices can become invertible under a homomorphism to a field. For this it is sufficient to consider square matrices. If $A\in\MATn$ and $A$ can be written in the form \begin{equation}\label{matrixfact} A=PQ,\ \mbox{ where $P$ is an } n \times r,\ Q \mbox{ an } r\times n \mbox{ matrix}, \end{equation} then it is clear that under any homomorphism from $\R$ to a field we again have such a factorization as in~(\ref{matrixfact}). Hence the image of $A$ cannot have rank greater than $r$, and will not be invertible whenever $rRs->R' UNIVERSAL LOCALIZATION $$\begin{array}{ccc} \R & \stackrel{\mu_1}{\longrightarrow} & \mathbf{K}_1 \\ & \mu_2 \searrow & \downarrow \alpha \\ &&\mathbf{K}_2 \end{array}$$ \noindent ($\Rightarrow$) Assume that a specialization $\alpha:\mathbf{K}_1\rightarrow \mathbf{K}_2$ exists. Take $A\in\mathbf{\Sigma}_2$. Then $\mu_2(A)$ has an inverse (in $\mathbf{K}_2$) which is the image of a matrix $B$ over $\mathbf{K}_1$: $(\mu_2(A))(\alpha(B))=I$. But then $(\mu_1(A))B=I+C$, where $C$ is a matrix such that $\alpha(C)=0$. Then $I+C$ has an inverse and so does $\mu_1(A)$. Therefore $A\in\mathbf{\Sigma}_1$. Since $A\in\mathbf{\Sigma}_2$ was arbitrary, we have the inclusion $\ca{P}_1 \subseteq \ca{P}_2$. \noindent ($\Leftarrow$) This direction of the proof needs to pass through the corresponding localization $\R_{\ca{P}_1},\,\R_{\ca{P}_2}$ respectively. We will first show that there is a ring-homomorphism $\R_{\ca{P}_2}\rightarrow\R_{\ca{P}_1}$.\footnote{Note the reversal of direction compared to the specialization between the \ERF s.} Now assume $\ca{P}_1 \subseteq \ca{P}_2$ holds. Then $\lambda_1:\R\rightarrow\R_{\ca{P}_1}$ is $\mathbf{\Sigma}_2$-inverting and so may be factored by $\lambda_2$ (see the matrix localization theorem~\ref{matrixlocaliz}) to give a homomorphism $\R_{\ca{P}_2}\rightarrow\R_{\ca{P}_1}$. Having this homomorphism, let $S$ be the image of $\R_{\ca{P}_2}$ in $\R_{\ca{P}_1}$, and write $\ca{M}_1$ for the maximal ideal of $\R_{\ca{P}_1}$. The natural homomorphism $\R_{\ca{P}_1}\rightarrow\mathbf{K}_1$ maps $S$ to $S'=S/(S\cap\ca{M}_1)$\footnote{Here it helps to draw a picture which has been left out for both technical and spacing reasons.}. Now $S$ is a local ring (as homomorphic image of $\R_{\ca{P}_2}$) and $S\cap\ca{M}_1$ is a proper ideal, so the natural homomorphism $\R_{\ca{P}_2}\rightarrow\mathbf{K}_2$ can be taken via $S'$. %- \newpage \noindent This gives a homomorphism from $S'$ to $\mathbf{K}_2$, which is the required specialization $\alpha$. \end{proof} As mentioned above, this theorem shows that the category of \ERF s, as a partially ordered set, is indeed order-isomorphic to the set of prime matrix ideals over $\R$, partially ordered by inclusion. Of course, if there are specializations $\mathbf{K}_1\rightarrow\mathbf{K}_2$ and $\mathbf{K}_2\rightarrow\mathbf{K}_1$, then $\mathbf{K}_1$ and $\mathbf{K}_2$ are isomorphic as \ERF s. % RADICAL This also means that a specialization between \ERF s can be characterized in terms of the corresponding singular kernels. Moreover, using this correspondence and the mentioned results, we see that there is a universal \ERF\ precisely if there is a least prime matrix ideal. Equivalently we can say that this is the case \IFF\ $\surd \ca{N}$ is prime, where again $\ca{N}$ is the least matrix ideal as constructed above, and the \emph{radical} $\surd \ca{A}$ of a matrix ideal $\ca{A}$ is defined as the set $$ \surd \ca{A} = \bigcap \{\, \ca{P}\ |\ \ca{P} \mbox{ prime } \supseteq \ca{A}\}.$$ This means $\surd \ca{A}$ is the intersection of all prime matrix ideals containing $\ca{A}$, and it can be verified that it is a matrix ideal containing $\ca{A}$. \noindent Let us sum this up: $\R$ has got a \begin{description} \item[\ERF:] if there exists a prime matrix ideal.\footnote{We have already seen in theorem~\ref{PRIMEKERNEL}, chapter~\ref{EpicRFields}, that the \ERF s of a commutative ring $\R$ correspond precisely to the prime ideals of $\R$.} \item[universal \ERF:] if there is a least prime matrix ideal. This is equivalent to saying that $\surd \ca{N}$ is prime\footnote{Note that a commutative ring $\R$ has a universal \ERF\ \IFF\ its nil radical is prime.}, where $\ca{N}$ is the least matrix ideal. \item[\foff:] if there is a prime matrix ideal, which does not contain any non-zero elements. \item[universal \foff:] if there is a least prime matrix ideal, and it does not contain any non-zero elements, i.e.~$\surd \ca{N}$ is prime and $\ca{N}$ does not contain a non-zero element.\footnote{Again comparing to the commutative case: there exists a unique \foff\ for a commutative ring $\R$, when the nil radical is zero, as then $\R$ is an integral domain.} \end{description} Note that if a ring $\R$ has a universal \foff, its singular kernel is the unique least prime matrix ideal of $\R$ and so is contained in the singular kernel of any other \ERF. % COUNTER-EXAMPLE The following (commutative) example shows that not every ring having a universal \ERF\ has a \foff:\\ \ex Let $\R=\mathbf{k}[x]/(x^2)$, where \mbf{k} is a commutative field. Then $\R$ has a universal \ERF, which is simply $\mathbf{k}[x]/(x)$. It cannot have a field of fractions, because $\R$ is not an \ID: $x^2=x\cdot x=0$. Recall theorem~\ref{iffi} and compare it with the following: \begin{eqnarray*} \MATRIX{x}{0}{0}{x}&=&\MATRIX{x}{0}{1}{x}\nabla\MATRIX{0}{0}{-1}{x},\ \mbox{and this is} \\ &=&\left(\begin{array}{c} x \\ 1 \end{array}\right)(1\ x)\nabla\left(\begin{array}{c} 0 \\ 1 \end{array}\right)(-1\ x), \end{eqnarray*} where the matrices on the RHS are non-full. So here we can write a diagonal matrix with non-zero entries on the main diagonal as the determinantal sum of non-full matrices. Thus a \foff\ cannot exist by applying theorem~\ref{iffi}.\\ % LOOKING AHEAD What we now want to show is that every prime matrix ideal occurs as singular kernel of some \ERF. Next chapter will look at this closer and finally prove that there is a correspondence between \ERF s and prime matrix ideals. % -------------------------- THE CONSTRUCTION -------------------------- \section{The Construction}\label{TheConstruction} %\remafresh We will now deal with the rather technical side of the story. Apart of the short section~\ref{result} this chapter can be skipped at first reading, in order to pass on to the very interesting results of chapter~\ref{UniversalFOFF}. We will closely follow P.M.~Cohn's construction as in~\cite{CohnSkew}. Our object will be to show that every prime matrix ideal occurs as singular kernel of some \ERF. % -------------------------- Representing fractions -------------------------- \subsection{Representing fractions}\label{Representing} When embedding a commutative \ID\ into a field, we have seen (in chapter~\ref{comidFOFF}) that each element of the \foff\ can be written in the form $a/b$, where $a,b\in\R$ and $b\neq0$. This cannot always be generalised for a non-commutative ring and therefore we will look closer at the formation of fractions in general rings. We will find a way how to represent fractions with the help of matrices. In the commutative case we assumed a set \mbf{S} to be multiplicative in order to get a nice description of the elements in the localization $\Rs$. In the same way we need to restrict the set of matrices $\mathbf{\Sigma}$ (in our case the complement of a singular kernel) to be able to describe the elements of $\R_{\mathbf{\Sigma}}$. This restriction condition is the following: \begin{defn} % Upper multiplicative A set\/ $\mathbf{\Sigma} \subseteq \MAT$ is called \emph{upper multiplicative} if\/ $1\in\mathbf{\Sigma}$ and, if\/ $A,B \in \mathbf{\Sigma}$ the matrix\/ $\MATRIX{A}{C}{0}{B} \in \mathbf{\Sigma}$, for any matrix\/ $C$ of appropriate size. \end{defn} Also, given a $\mathbf{\Sigma}$-inverting homomorphism $f:\R\rightarrow\R'$, we define the \emph{$\mathbf{\Sigma}$-rational closure} of $f$ as the set of all entries of all matrices $f(A)^{-1}$ for $A\in\mathbf{\Sigma}$. Let us now have a closer look at how to represent elements in $\R'$: \begin{thm} Let\/ $\R$ be any ring,\/ $\mathbf{\Sigma}$ an upper multiplicative set of matrices over\/ $\R$ and\/ $f:\R\rightarrow \R'$ a\/ $\mathbf{\Sigma}$-inverting homomorphism. Then for any $x\in\R'$ the following are equivalent: \begin{numlist}{rationalclosss} \item $x$ lies in the $\mathbf{\Sigma}$-rational closure of\/ $\R$ in\/ $\R'$, \item $x$ is a component of the solution $u$ of an equation $$Au-a=0,\ A\in f(\mathbf{\Sigma}),$$ where $a$ is a column over $im(f)$, \item $x=bA^{-1}c$, where $A\in f(\mathbf{\Sigma})$, $b$ is a row and $c$ a column over $im(f)$. \end{numlist} Moreover, the $\mathbf{\Sigma}$-rational closure of\/ $\R$ in\/ $\R'$ is a subring of\/ $\R'$ containing\/ $im(f)$. \end{thm} \begin{proof} Assume (i) holds. By definition, the $\mathbf{\Sigma}$-rational closure of\/ $\R$ consists of the entries of the inverses of matrices in $f(\mathbf{\Sigma})$. Say $x$ is the $(i,j)$-entry of $A^{-1}$, then $x$ is the $i$th component of the solution of $Au-e_j=0$, where $e_j$ is a column over $im(f)$, having a $1$ in $j$th place. So (ii) holds. Now assume (ii) holds. Then $u_i=e_i^{T}A^{-1}a$, where $e_i^{T}$ denotes the transpose of the column $e_i$. Thus $e_i^{T}=b$, for some row $b$, and since $a$ was a column over $im(f)$, we get (iii). Finally, when (iii) holds, let $x=bA^{-1}c$ then $$\left(\begin{array}{ccc} 1 & b & 0 \\ 0 & A & c \\ 0 & 0 & 1 \end{array}\right)^{-1}=\left(\begin{array}{ccc} 1 & -bA^{-1} & b^{-1}c \\ 0 & A^{-1} & -A^{-1}c \\ 0 & 0 & 1 \end{array}\right),$$ so we have $x$ as an entry on the RHS. But the matrix on the LHS whose inverse is taken, is again in $\mathbf{\Sigma}$, since $\mathbf{\Sigma}$ is upper multiplicative. Thus (iii) $\Rightarrow$ (i), and this shows that the above conditions are equivalent. Let us now first show that $\R_\mathbf{\Sigma}(R')$, which we denote as the $\mathbf{\Sigma}$-rational closure of $\R$ in $\R'$, contains $im(f)$. Take any $c\in im(f)$. Then $c$ satisfies the equation $1u-c=0$, which is of the form as in (ii), taking $A=(1)$. So we are left to prove that $\R_\mathbf{\Sigma}(R')$ is a subring of $\R'$: Suppose $u_i$ is the $i$th component of the solution of the equation as in (ii), and $v_j$ is the $j$th component of the solution of $Bv-b=0$. Then $u_i-v_j$ is the $i$th component of the solution of $$\MATRIX{A}{C}{0}{B}w-\left(\begin{array}{c} a \\ b \end{array}\right)=0,$$ where $C$ has for its $j$th column the $i$th column of $A$ and the rest $0$. So we are left to prove the product $u_iv_j$, which is the $i$th component of the solution of $$\MATRIX{A}{C}{0}{B}w-\left(\begin{array}{c} 0 \\ b \end{array}\right)=0,$$ where $C$ has as its $j$th column $-a$ and the rest $0$. This shows that $\R_\mathbf{\Sigma}(R')$ is closed under subtraction and multiplication, containing $1$. Thus it is a subring of $\R'$, containing $im(f)$. \end{proof} Let $\R$ be any ring and \mbf{K} an \ERF. As we saw in above theorem, any element $p\in \mathbf{K}$ can be expressed in the form $$p=c-uA^{-1}v,\ \mbox{ where } c\in\R,\,u\in\R^n,\,v\in {}^n\R,\,A\in\MATn,$$ and $A$ is not in the singular kernel of \mbf{K}. Thus $p$ is completely determined by the following: \begin{defn} % block A block with elements as described above $$\alpha=\MATRIX{u}{c}{A}{v}$$ is called an \emph{admissible} block for\/ \mbf{K} or\/ \mbf{K}\emph{-admissible}. \end{defn} So we can describe the elements of \mbf{K} either in block form as above, or in equational form (ii) as in the theorem. We will use the block description to prove in the next section, that the localization forms a local ring. However for a proof in the next chapter we need a different notation, which is convenient for the system (ii). Now we take the augmented matrix $(-a\ A)$ as our basic matrix, and shall write our system in the form \begin{equation}\label{syseq} Au=0,\ \mbox{ where $A$ is a $m\times (m+1)$ matrix over $\R$.} \end{equation} If we further write $A$ in columns as $A=(A_0\ A_1\ \ldots\ A_m)=(A_0\ A_*\ A_\infty)$, we define $A_\infty=A_m$ as the last column, and $A_*=(A_1\ \ldots\ A_{m-1})$. Moreover: \begin{defn} If~\textnormal{(\ref{syseq})} has a unique solution $u=(u_0\ \ldots\ u_{m+1})\in\R'$, normalized by the condition $u_0=1,\,u_m=p$, then we call~\textnormal{(\ref{syseq})} an \emph{admissible system} and the matrix $A$ and \emph{admissible matrix} of order $m$ for the element $p$ of\/ $\R'$. \end{defn} This means that an $m\times(m-1)$ matrix over $\R$ is admissible if the image of the matrix formed by the last $m$ columns is invertible over $\R'$. \begin{defn}\label{denominatorr} % denominator The last $m$ columns of the matrix $A$, $(A_*\ A_\infty)$, are called the \emph{denominator}, which is of size $m\times m$. We define the first $m$ columns $(A_0\ A_*)$ to be the \emph{numerator}, and $A_*$ to be the \emph{core} of $p$ in the above representation~\textnormal{(\ref{syseq})}. We also write $u=(1\ u_*\ p)^T$. \end{defn} Thus we can say that a system~\textnormal{(\ref{syseq})} is admissible precisely when its denominator is invertible over $\R'$. % -------------------------- Rsigma is a local ring -------------------------- \subsection{$\R_{\mathbf{\Sigma}}$ is a local ring}\label{Rsigmalocalring} There are two aims we are heading for: on the one hand we want define a ring structure on $\R_{\mathbf{\Sigma}}$, and on the other hand we want to find conditions for $\R_{\mathbf{\Sigma}}$ to be a local ring. When they hold, we want to find conditions for the map from $\R$ to the residue-class field of $\R_{\mathbf{\Sigma}}$ to be injective, i.e.~then we have an embedding. Let us start by constructing $\R_{\mathbf{\Sigma}}$. As in the last section we will construct the elements of $\R_{\mathbf{\Sigma}}$ in the form $p-bA^{-1}c$, where $p$ is an element, $b$ a row, $c$ a column and $A$ a matrix in $\mathbf{\Sigma}$. Since $A$ is invertible over $\R_{\mathbf{\Sigma}}$, we can regard $p-bA^{-1}c$ as an element of $\R_{\mathbf{\Sigma}}$. Consider its diagonal sum with $I$, which can be transformed as follows: $$\MATRIX{p-bA^{-1}c}{0}{0}{I}\rightarrow\MATRIX{p-bA^{-1}c}{b}{0}{A}\rightarrow\MATRIX{p}{b}{c}{A}\rightarrow\MATRIX{b}{p}{A}{c}.$$ Here each term is associated to the next over $\R_{\mathbf{\Sigma}}$, and we are interested in the last, which has the advantage of avoiding inversion and to be a matrix defined over $\R$. This means we will consider for any set $\mathbf{\Sigma}$ of square matrices over $\R$ the set $\mathbf{M(\Sigma)}$ of all matrix blocks of the form: \begin{equation}\label{matrixblock} % MATRIXBLOCK a=\MATRIX{a'}{s}{\sigma}{'a}, \end{equation} where $s\in\R,\ a'\in\R^n,\ 'a\in {}^n\R,\ \sigma$ is an $n\times n$ matrix in $\mathbf{\Sigma}$. Note that if $n=0$, then $a=(s)$ as a $1\times1$ matrix. Moreover we will call $\sigma$ the \emph{denominator}. We will be constructing $\R_{\mathbf{\Sigma}}$ as a set of equivalence classes of elements of $\mathbf{M(\Sigma)}$, for any ring $\R$ and upper multiplicative set $\mathbf{\Sigma}$ of square matrices over $\R$. If $\lambda:\R\rightarrow\R_{\mathbf{\Sigma}}$ is the natural homomorphism, then the matrix block given by~(\ref{matrixblock}) is to represent the element $\lambda(s)-\lambda(a')\lambda(\sigma)^{-1}\lambda('a)=p-bA^{-1}c.$ On each matrix~(\ref{matrixblock}) we can perform certain ``elementary'' operations which do not change the element represented. We can \begin{numlist}{superflous} % ELEMENTARY OPERATIONS ! \item add to a given row (column) a left multiple of a later row (a right multiple of an earlier column), \item insert or remove a ``special'' square block $T$, and the row and column block containing it. This block $T$ is such that it lies on the main diagonal of $\sigma$ in~(\ref{matrixblock}) and either the row block or the column block of $T$ is zero apart from $T$ itself. Also any entry below and to the left of $T$ is zero. \end{numlist} Let $a\rightarrow b$ indicate that $b$ is obtained from $a$ by such an elementary operation. Also we will write $a\sim b$ when we can pass from $a$ to $b$ by a series of the above transformations. This is clearly an equivalence relation and we say $a$ and $b$ are \emph{equivalent} if $a\sim b$. The equivalence class containing $a\in\mathbf{M(\Sigma)}$ is written $[a]$ and we denote the set of all such classes by $\R_{\mathbf{\Sigma}}$. Our object will be to define a ring structure on $\R_{\mathbf{\Sigma}}$. For this we will first of all define two operations $\star$ and $\odot$ on $\mathbf{M(\Sigma)}$ as follows (dots $\cdot$ indicate blocks of zeros): \begin{equation} a\star b = \left(\begin{array}{ccc} a' & b' & s+r \\ \sigma & \cdot & 'a \\ \cdot & \rho & 'b \end{array}\right) = \left(\begin{array}{ccc} a' & b' & s \\ \sigma & \cdot & 'a \\ \cdot & \rho & \cdot \end{array}\right) \nabla \left(\begin{array}{ccc} a' & b' & r \\ \sigma & \cdot & \cdot \\ \cdot & \rho & 'b \end{array}\right), \end{equation} \begin{equation} a\odot b = \left(\begin{array}{ccc} a' & sb' & sr \\ \sigma & 'ab' & 'ar \\ \cdot & \rho & 'b \end{array}\right) = \left(\begin{array}{ccc} a' & s & \cdot \\ \sigma & 'a & \cdot \\ \cdot & \cdot & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & \cdot & \cdot \\ \cdot & b' & r \\ \cdot & \rho & 'b \end{array}\right). \end{equation} Let us show that $\R_{\mathbf{\Sigma}}$ is a ring with respect to these operations. $\star$ is clearly commutative, and both $\star$ and $\odot$ are well-defined on the classes, because when $a_1\sim a_2,\ b_1\sim b_2 \Rightarrow (a_1\star b_1) \sim (a_2 \star b_2),\ (a_1\odot b_1) \sim (a_2 \odot b_2)$, as any elementary operation carried out on $a,b$ can also be carried out on $(a\star b),(a\odot b)$. Let us not write out the details of associativity, but note the formulae for the sum and the product of an element $t\in\R$ by a block matrix $a$: $$a\star (t)=(t)\star a=\MATRIX{a'}{s+t}{\sigma}{'a},\ (t)\odot a=\MATRIX{ta'}{ts}{\sigma}{'a},\,a\odot (t)=\MATRIX{a'}{st}{\sigma}{'at}$$ Now taking $t=0$, this gives $( {(0)\star a} )\sim ( {a\star (0)} )\sim a$, $$( {(0)\odot a} )\sim \MATRIX{\cdot}{\cdot}{\sigma}{'a} \sim 0,\ ( {a\odot (0)} )\sim \MATRIX{a'}{\cdot}{\sigma}{\cdot} \sim 0,$$ by using the second elementary operation, and $((1)\odot a)\sim (a\odot(1))\sim a$. This gives us the identities $(0)$ and $(1)$ for the operations $\star$ and $\odot$ respectively. Also with $-a=(-1)\odot a\,$ we can transform $(a\star(-a))\rightarrow (0)$. Hence $(a\star (-a))\sim(0)$ and also $((-a)\star a)\sim(0)$, satisfying the existence of inverses for $\star$. Finally the distributive law $((a\odot c)\star(b\odot c))\sim((a\star b)\odot c)$ holds, and similarly for the other distributive law, as can be checked by carrying out the matrix transformations. So we have a ring $\R_{\mathbf{\Sigma}}$ with a ring homomorphism $\lambda:\R\rightarrow\R_{\mathbf{\Sigma}}$, such that $\lambda:r\mapsto [(r)]$, for $r\in\R$. Now let us show that $\lambda$ is universal $\mathbf{\Sigma}$-inverting. For this we need the following observation: \begin{equation}\label{observation} \left(\MATRIX{a'}{\cdot}{\sigma}{b}\star\MATRIX{a'}{\cdot}{\sigma}{c}\right)\sim\MATRIX{a'}{\cdot}{\sigma}{b+c}, \end{equation} which again can be obtained by applying transformations on the matrix block for the LHS. If $e_j$ denotes the $j$th column of the unit matrix and $e_i^T$ the $i$th row, then the $(ij)$-entry of $\lambda(A)^{-1}$ is represented by the matrix block $$u_{ij}=\MATRIX{-e_i^T}{0}{A}{e_j}.$$ Using the observation~(\ref{observation}) with $n$ summands, we then have: $$\sum_j\MATRIX{-e_i^T}{\cdot}{A}{e_j}\odot(a_{jk})=\sum\MATRIX{-e_i^T}{\cdot}{A}{e_ja_{jk}}\rightarrow\MATRIX{-e_i^T}{\cdot}{A}{\sum{e_ja_jk}},$$ and then applying the first and then the second elementary operation we get $$=\MATRIX{-e_i^T}{\cdot}{A}{A_k}\rightarrow\MATRIX{-e_i^T}{\delta_{ik}}{A}{\cdot}\rightarrow(\delta_{ik}),$$ where $A_k$ denotes the $k$th column of $A$. In a similar manner we can also show that $\sum(a_{ij})\odot u_{jk}\sim(\delta_{ik})$. Therefore $[u_{ij}]$ is the $(ij)$-entry of $\lambda(A)^{-1}$, and so $\lambda(A)$ is invertible, which is what we want. We are left to prove the universal property: Let $f:\R\rightarrow\R'$ be a $\mathbf{\Sigma}$-inverting homomorphism and for any element $[a]$ of $\R_\mathbf{\Sigma}$ define $f_1([a])=f(s)-f(a')f(\sigma)^{-1}f('a)$. The RHS of this is unaffected by elementary transformations (just as in the case of $\lambda$), hence $f_1$ is well defined on $\R_\mathbf{\Sigma}$ and is uniquely determined by $f$. Moreover, for any $r\in\R$, we have $f(r)=f_1([(r)])=f_1\lambda(r)$. But this means $f$ has been factored by $f_1$, which shows $\lambda$ to be universal $\mathbf{\Sigma}$-inverting.\\ Summing up, we have shown by the above that the set $\mathbf{M(\Sigma)}$ of all matrix blocks as in~(\ref{matrixblock}), taken modulo the equivalence~$\sim$ defined by the elementary operations (i) and (ii), together with the operations $\star,\,\odot$ as defined above, forms a ring $\R_\mathbf{\Sigma}$. This is the localization of $\R$ at $\mathbf{\Sigma}$, with the natural homomorphism $\lambda:\R\rightarrow \R_\mathbf{\Sigma}$, given by $r\mapsto[(r)]$.\\ We next derive conditions for the localization to be a local ring. We need to remark that a ring $\R$ is local \IFF\ $1\neq0$ and for every $x\in\R$ either $x$ has a left inverse or $1-x$ has a right inverse. For a local ring these conditions are clearly satisfied, but we shall not go into the details for proving the converse. \begin{thm}[$\R_{\mathbf{\Sigma}}$ is a local ring]\label{Rsigmalocalringthm} %Rsigma a local ring Let $\R$ be a ring and $\mathbf{\Sigma}$ a set of full matrices, such that $\mathbf{\Sigma}$ is upper multiplicative, closed under permutations of rows or of columns, while the complement of $\mathbf{\Sigma}$ admits determinantal sums. Then $\R_{\mathbf{\Sigma}}$ is a local ring. \end{thm} \begin{proof} $\R_{\mathbf{\Sigma}}$ is non-trivial, because the element $(1)\in\mathbf{\Sigma}$, so $[(1)]\neq0$.\footnote{This needs some tedious verification, and note that it may be quite difficult to decide when $\R_{\mathbf{\Sigma}}$ is trivial.} Given $x\in\R_{\mathbf{\Sigma}}$, let $x=u_1$ be the first component of the solution of $Au-e_1=0$, and suppose that $x$ has no left inverse. Then the numerator of $u_1$ has no left inverse in $\R_{\mathbf{\Sigma}}$.\footnote{This follows by a generalisation of Cramer's rule, which can be found in~\cite{CohnFRTR}.} Let us write $A=(A_1\ A')$, where $A_1$ is the first column of $A$. Then this numerator is $(e_1\ A')$. We claim that $(A_1-e_1\ A')\in\mathbf{\Sigma}$, since if this were not the case, then $(e_1\ A')$ and $(A_1-e_1\ A')$ both lie in the complement of $\mathbf{\Sigma}$. Hence so does $A=(A_1-e_1\ A')\nabla(e_1\ A)$. But this is a contradiction. Hence $(A_1-e_1\ A')\in\mathbf{\Sigma}$, and we can form the system $(A_1-e_1\ A')v-e_1=0$, which may be written as: \begin{equation}\label{weissnicht} Av-e_1(1+v_1)=0. \end{equation} Since we had $Au-e_1=0$ and by above equation~(\ref{weissnicht}) we get $Av=e_1(1+v_1)=Au(1+v_1)$, so $v=u(1+v_1)$. Equating the first components, we find that $v_1=u_1(1+v_1)$, and so $$(1-u_1)(1+v_1)=1.$$ This shows that $1-x$ has a right inverse, and as mentioned in a remark before the theorem, this shows that $\R_{\mathbf{\Sigma}}$ is a local ring. \end{proof} % -------------------------- The Result -------------------------- \subsection{The Result}\label{result} Having done all this technical construction, we will present the aim we were really heading for in this section. Let us first of all clear off unfinished business. We need to prove: \begin{thm} Let\/ $\R$ be any ring. If\/ $\mathbf{\Sigma}$ is a set of matrices such that the universal localization\/ $\R_\mathbf{\Sigma}$ is a local ring, then the residue-class field of\/ $\R_\mathbf{\Sigma}$ is an \ERF. \end{thm} \begin{proof} Let $\mathbf{\Sigma}$ be a set of matrices such that $\R_\mathbf{\Sigma}$ is a local ring, and let us denote its residue-class field by \mbf{K}. By composing the natural mappings we get a homomorphism $\R\rightarrow \R_\mathbf{\Sigma} \rightarrow \mathbf{K}$. \mbf{K} is generated by the inverses of images of matrices in $\mathbf{\Sigma}$, hence an \ERF. \end{proof} Also let us show that any \ERF\ is indeed determined by its singular kernel. We left this to prove at the end of chapter~\ref{SingularKernel}. Let \mbf{K} be an \ERF, $\mu:\R\rightarrow \mathbf{K}$ the canonical map and $\ca{P}=Ker(\mu)$ its singular kernel. Then by the matrix localization theorem~\ref{matrixlocaliz} (see section~\ref{SingularKernel}) we know that $\mu$ can be factored uniquely by $\lambda:\R\rightarrow\R_{\ca{P}}$ to give another map $f:\R_{\ca{P}}\rightarrow\mathbf{K}$, such that $\mu=\lambda\circ f$: % DIAGRAMM R->Rp->K UNIVERSAL LOCALIZATION $$\begin{array}{ccc} \R & \stackrel{\lambda}{\longrightarrow} & \R_{\ca{P}} \\ & \mu \searrow & \downarrow f \\ &&\mathbf{K} \end{array}$$ Since $\ca{P}$ admits determinantal sums and contains all non-full matrices, then by section~\ref{Rsigmalocalring}, $\R_{\ca{P}}$ is a local ring and \mbf{K} is the residue class field of $\R_{\ca{P}}$. This is what we want and need: \begin{thm}[The Theorem]\label{thetheorem} Let\/ $\R$ be any ring and\/ $\ca{P}$ a prime matrix ideal. Then there is an \ERF\ \mbf{K} with singular kernel\/ $\ca{P}$. Moreover the localization $\R_{\ca{P}}$ is a local ring with residue-class field\/ \mbf{K}. \end{thm} This result finally shows that there is a natural bijection between \ERF s and prime matrix ideals. % --------------------- UNIVERSAL FIELDS OF FRACTIONS --------------------- \section{Universal Fields of Fractions}\label{UniversalFOFF} %\remafresh % -------------------------- SEMIFIRS -------------------------- \subsection{$n$-firs and semifirs}\label{semifirs} % Semifirs An interesting class of rings are $n$-firs and semifirs, which we define here for $n\geq 0$ as follows: \begin{defn}\label{semifir} % n-fir, semifir An \emph{$n$-fir} is a non-zero ring\/ $\R$ in which any submodule on at most $n$ generators of a free right\/ $\R$-module is again free of unique rank. If\/ $\R$ is an $n$-fir for all $n$, it is called a \emph{semifir}. \end{defn} \rems %semifir \begin{enumerate} \item 0-firs are just non-zero rings by convention, whereas a 1-fir is an \ID. 2-firs then are \ID s in which any two elements with a common non-zero right multiple generate a principal right ideal. In particular, a commutative \ID\ is a 2-fir precisely if every 2-generator ideal is principal. Later in section~\ref{NoetherianBezout} we will define Bezout domains. It turns out that a commutative 2-fir is nothing else than a Bezout domain. \item Every commutative 2-fir is a semifir. \item It can be shown that definition~\ref{semifir} is equivalent to saying that $\R$ is a semifir then every finitely generated right ideal is free of unique rank. \end{enumerate} The original name of $n$-firs and semifirs come from a special class of rings called ``free ideal rings'' (firs). The precise definition is as follows: \begin{defn} % FIRS By a \emph{free right ideal ring} or \emph{right fir} for short, we understand a ring with invariant basis number (IBN)\footnote{Having IBN (as defined in section~\ref{IBN}) implies that the ring is non-zero.}, whose right ideals are free, as right\/ $\R$-modules. \emph{Left firs} are defined similarly and a left and right fir is called a \emph{fir}. \end{defn} Again, we will not show how this is linked to n-firs here. We will show one proposition~(\ref{PRIDfir}), stating that a commutative fir is a principal ideal domain (PID). Note however, that every left or right fir is a semifir by the above remark. Let us now turn to the by far most important and perhaps nicest property of semifirs: We will show that in fact every semifir has a universal \foff. The next two sections will lead to this result. % -------------------------- FULLY INVERTING HOMOMOPRHISMS -------------------------- \subsection{Fully inverting homomorphisms}\label{Fullyinverting} The results of this and the following section was the aim of this project. We will characterize rings with a so called \emph{fully inverting} homomorphism to an \ERF. Let us define this in detail. In any $\R$-field the only matrices over $\R$ that can be inverted are the full matrices. \begin{defn} % Fully inverting A ring homomorphism is said to be \emph{fully inverting} if every full matrix has an image which is invertible. \end{defn} In particular, if the canonical map from a ring $\R$ to an \ERF\ \mbf{K} is fully inverting, then the singular kernel is the set of all non-full matrices over $\R$, and \mbf{K} must be the universal \foff\ of $\R$. Clearly every non-zero element is full as a $1\times 1$ matrix and so is inverted over \mbf{K}. % Bla bla The corresponding localization of rings with a fully inverting homomorphism to an $\R$-field is called a \emph{universal localization}. We will see that rings with a fully inverting homomorphism to a field are characterized as \emph{Sylvester domains}, which we define in the next section~\ref{Sylvesterdomains}. This will then lead to the wanted result, that every semifir has a universal \foff. \begin{defn} % honest A ring homomorphism $f:\R\rightarrow\mathbf{S}$ is called \emph{honest} if it maps full matrices to full matrices. \end{defn} Clearly a homomorphism to an $\R$-field is honest \IFF\ it is fully inverting. Also note, that every honest homomorphism is injective.\footnote{This follows once more from the fact that every non-zero element is full as a $1\times 1$ matrix and so will become invertible.} To describe fully inverting homomorphisms we first give conditions for an \ERF\ to actually coincide with its localization. \begin{lem}\label{Rp=K} % Rp=K ! Let\/ $\R$ be any ring, $\ca{P}$ a prime matrix ideal of\/ $\R$ with universal localization $\R_{\ca{P}}$ and residue-class field\/ \mbf{K}. Suppose $f:\R\rightarrow\mathbf{K}$ is fully inverting, i.e.~$\ca{P}$ consists of all non-full matrices. Then $\R_{\ca{P}}=\mathbf{K}$ and $\ca{P}$ is a minimal prime matrix ideal. \end{lem} \begin{proof} % Rp=K We know that $\R_{\ca{P}}$ is a local ring by theorem~\ref{Rsigmalocalringthm}. So we need to show that under these conditions every non-unit of $\R_{\ca{P}}$ is $0$. Then $\R_{\ca{P}}$ will coincide with the residue-class field. So let $p\in\R_{\ca{P}}$ and take an admissible system for $p$: $Au=0$. Since $p$ is a non-unit, its numerator $(A_0\ A_*)$ is not invertible over $\R_{\ca{P}}$, and since $f$ is fully inverting, $(A_0\ A_*)$ is non-full over $\R$. Thus we can write $(A_0\ A_*)=PQ$ over $\R_{\ca{P}}$, where $P$ is $n\times (n-1)$, $Q$ is $(n-1)\times n$. Hence $$A=(A_0\ A_*\ A_n)=(PQ\ A_n)=(P\ A_n)\MATRIX{Q}{0}{0}{1}.$$ So $(P\ A_n)$ is a factor of the denominator (see definition~\ref{denominatorr}) $(A_*\ A_n)$ over $\R_{\ca{P}}$. This latter is a unit and the local ring $\R_{\ca{P}}$ has UGN. Therefore $(P\ A_n)$ is also invertible over $\R_{\ca{P}}$. So in our equation $Au=0$ we can cancel this factor, to obtain: $$\MATRIX{Q}{0}{0}{1}u=0,$$ and this has the solution $p=0$. Hence $\R_{\ca{P}}$ is a field, so $\R_{\ca{P}}=\mathbf{K}$. Now $\ca{P}$ is minimal, because when $\ca{P}'\subseteq \ca{P}$, then by the proof of theorem~\ref{spezzz} there is a homomorphism $f:\R_{\ca{P}}\rightarrow\R_{\ca{P}'}$ which must be injective, because $\R_{\ca{P}}$ is a field. Thus $f$ is an isomorphism and $\ca{P}'= \ca{P}$. \end{proof} We can now give the characterization of fully inverting homomorphisms to $\R$-fields. This is a sufficient condition for a ring to have a universal \foff. \begin{thm}[Sufficient Conditions]\label{SufficientConditions} % SUFFICIENT CONDITIONS ! Let $\R$ be a non-zero ring. Then there is an \ERF\/ \mbf{K} with a fully inverting homomorphism to \mbf{K} \IFF\ the following two conditions are satisfied: \begin{numlist}{BBB} \item the diagonal sum of full matrices over $\R$ is full, \item the determinantal sum of non-full matrices over $\R$ (where defined) is non-full. \end{numlist} \mbf{K} is then the universal localization\/ $\R_{\mathbf{\Phi}}$, where $\mathbf{\Phi}$ is the set of all full matrices over $\R$. \end{thm} \begin{proof} % SUFFICIENT CONDITION \noindent ($\Leftarrow$) We need that the set of non-full matrices is the unique least prime matrix ideal. Suppose such an \ERF\ \mbf{K} exists. Then its singular kernel consists of all non-full matrices. So the complement of $\mathbf{\Phi}$ in $\MAT$ is the least matrix ideal $\ca{N}$. Thus $\ca{N}$ is a prime ideal and therefore (i) and (ii) hold. \noindent ($\Rightarrow$) Now assume that (i) and (ii) hold. Then the set $\ca{N}$ of all non-full matrices form a matrix ideal by (ii). This is necessarily the least, and it is prime by (i). %- \newpage Hence we have an honest homomorphism to an \ERF\ \mbf{K}, the residue-class field of $\R_{\mathbf{\Phi}}$, which is fully inverting, so $\mathbf{K}=\R_{\mathbf{\Phi}}$ by lemma~\ref{Rp=K}. \end{proof} This is what we wanted. We next try to find a class of rings to which this result can be applied. As mentioned before, these turn out to be precisely \emph{Sylvester domains}. % -------------------------- SYLVESTER DOMAINS -------------------------- \subsection{Sylvester domains}\label{Sylvesterdomains} Sylvester domains are n-firs for all n, which satisfy a special \emph{nullity condition}. They are interesting, because they have a universal field of fractions and this is in fact the localization at the set of all full matrices. To define them in detail we need the following property on the inner rank\footnote{We defined the inner rank in section~\ref{Innerrank}} in $n$-firs: \begin{prop}\label{rPrQcondition} % rP+rQ leq 0 If\/ $\R$ is an $n$-fir and $P$ is an $\,r\times n$, $Q$ an $\,n\times s$ matrix over $\R$, such that $$PQ=0, \mbox{ then }\ \rho(P) + \rho(Q) \leq n.$$ \end{prop} \noindent To prove this proposition we shall make use of a technical lemma: \begin{lem} % AB=0 trivializable. Let\/ $\R$ be an $n$-fir. Let $A$ be an $m\times r$ and $B$ be an $r\times s$ matrix, such that $AB=0$, and $r\leq n$. Then there is an invertible $r\times r$ matrix $P$, such that for each $i=1,\ldots,r$, either the $i$th column of $AP^{-1}$ or the $ith$ row of $PB$ is zero. \end{lem} \begin{proof} % AB=0 trivializable. \noindent Right multiplication by the matrix $B$ defines a linear map $f:\R^{r}\rightarrow\R^{s}$, where $im(f)$ is an $r$-generated submodule of $\R^{s}$. Since $r\leq n$, $im(f)$ is free of unique rank, as $\R$ is an $n$-fir. Thus it has a basis. Also $ker(f)$ is a submodule of $\R^{r}$ and hence is free of unique rank. So we can change the basis in $\R^{r}$ to one adapted to the decomposition $\R^{r}\cong ker(f)\oplus im(f)$, where $dim(im(f))$ is $t$, say. This change of basis can be described by an invertible $r\times r$ matrix $P$ over $\R$. Let $A'=AP^{-1},\ B'=PB$, then since $AP^{-1}PB=AB=0$, the columns of $A'$ lie in $ker(f)$. Thus the columns of $A'$ after the first $t$ are zero, while the first $t$ rows of $B'$ are zero. \end{proof} \noindent Now we can easily prove proposition~\ref{rPrQcondition}:\\ \begin{proof} % rP+rQ leq 0 By the above lemma, there is an invertible matrix $U$, such that $$PU=(P'\ 0),\ U^{-1}Q=\left(\begin{array}{c} 0 \\ Q' \end{array}\right),$$ where $P'$ is $r\times n_{1}$, $Q'$ is $n_{2}\times s$, such that $n_{1}+n_{2}=n$. So we have $$ \rho(P)\leq n_{1},\ \rho(Q)\leq n_{2},\ \Rightarrow \rho(P) + \rho(Q) \leq n_{1}+n_{2}=n.$$ \end{proof} \noindent Using this we can define Sylvester domains as follows: \begin{defn} % Sylvester domains A non-trivial ring $\R$ is called a \emph{Sylvester domain} if for any matrices $A,B$, such that the number of columns of $A$ equals the number of rows of $B$, equals to $n$ say, the following \emph{nullity condition} holds: \begin{equation}\label{nullitycondition} AB=0\ \Rightarrow \rho(A)+\rho(B)\leq n. \end{equation} \end{defn} \rems % Sylvetser domain \begin{enumerate} \item By the above proposition~(\ref{rPrQcondition}), every semifir is a Sylvester domain. \item Let us apply the nullity condition~(\ref{nullitycondition}) to elements $a,b \in \R$, that is consider $1\times 1$ matrices (i.e.~$n=1$). Since $0$ has rank $\rho(0)=0$, and a non-zero element of $\R$ has rank 1, we get: $$ ab=0\ \Rightarrow \rho(a)+\rho(b)\leq 1,$$ i.e.~$ab=0\, \Rightarrow\, a=0 \mbox{ or } b=0$, which is nothing else than the condition to be an \ID. Hence Sylvester domains are indeed \ID s which, as we know, is a necessary condition to have a \foff. \item More generally we see that over Sylvester domains every full matrix is regular. \end{enumerate} We will now prove that in Sylvester domains the following two properties of full matrices hold: The first is that the diagonal sum of full matrices is again full. The second is the fact that the determinantal sum, when defined, of non-full matrices remains non-full. As we have seen before, these are the key properties for having a fully-inverting homomorphism to a field, and applying the ``Sufficient Conditions'' theorem~\ref{SufficientConditions}, we then have shown that Sylvester domains have a universal \foff. Moreover, since every semifir is a Sylvester domain by the above remark, we finally obtain that every semifir can be embedded in a field. \begin{prop}\label{diagsum} % A+B is full Let\/ $\R$ be a Sylvester domain, $A,B$ full matrices over\/ $\R$. Then $A\oplus B$ is a full matrix over\/ $\R$. \end{prop} \begin{proof} % A+B is full Since $A,B$ are full, their inner rank is equal to the number of rows (by definition~\ref{fullrowrank}). It suffices then to show that $\rho(A\oplus B)=\rho(A)+\rho(B)$. Clearly $\rho(A\oplus B)\leq \rho(A) + \rho(B)$: If we take rank factorizations of $A$ and $B$ such that $A=PQ,\ B=P'Q'$ (i.e.~such that $P,P'$ have least possible inner rank), then $$\MATRIX{A}{0}{0}{B} = \MATRIX{P}{0}{0}{P'}\MATRIX{Q}{0}{0}{Q'}$$ and therefore $\rho(A\oplus B)\leq \rho(A) + \rho(B)$. But we also get $\rho(A\oplus B)\geq \rho(A) + \rho(B)$: Rank factorize $\rho(A\oplus B)$ as $$\MATRIX{A}{0}{0}{B}= \left(\begin{array}{c} P \\ P' \end{array}\right) (Q\quad Q')$$ where $P$ and $Q$ are square and the number of columns of $P$ is $\rho(A\oplus B)=n$. We see that $PQ'=0$, and since $\R$ is a Sylvester domain, we have \begin{eqnarray*} n & \geq & \rho(P)+\rho(Q')\qquad \mbox{and thus} \\ \rho(A\oplus B) & \geq & \rho(P)+\rho(Q') \\ & \geq & \rho(PQ)+\rho(P'Q') = \rho(A)+\rho(B). \end{eqnarray*} Combining the two inequalities we get $\rho(A\oplus B)=\rho(A)+\rho(B)$, and thus $A\oplus B$ is full, as its inner rank equals the number of rows. \end{proof} This fulfills the first condition. For several proofs which follow, it is useful to take notice of the following:\\ \begin{rem} % AB=CD as MATRIX Recall (see section~\ref{Innerrank}) that a matrix relation $PQ=0$ is called \emph{full} if $P$ is left full and $Q$ is right full. This notion can also be applied to relations of the form $AB=CD$. This is because we can write equivalently \begin{equation}\label{equivalently} (A\ C) \left(\begin{array}{c} -B \\ D \end{array}\right) = 0. \end{equation} This is often useful to apply the nullity condition~(\ref{nullitycondition}).\\ \end{rem} \noindent We are left to show the following: \begin{prop}\label{detsum} % A det B is non-full Let\/ $\R$ be a Sylvester domain, $A,B$ non-full matrices over\/ $\R$. Then $A\nabla B$ is a non-full matrix over\/ $\R$, whenever the determinantal sum is defined with respect to some column. \end{prop} \begin{proof} % A det B is non-full We can take $A=(A'\ a)$, $B=(A'\ b)$ non-full $n\times n$ matrices, which differ in the last columns $a,b$ for simplicity. Let $C=(A'\ a+b)=A\nabla B$, then $C$ is again $n\times n$ and $A'$ is a $n\times (n-1)$ matrix. Now two cases can happen: \noindent If $A'$ is full, i.e.~$\rho(A')=n-1$, then $n-1=\rho(A')=\rho(A)=\rho(B)$, since $A,B$ are non-full. To show that the determinantal sum $C=(A'\ a+b)$ is non-full, we will show that $\rho(C)=\rho(A')=n-1$. \noindent For this we shall need the following lemma, which we will prove for general matrices $A,B,C$ over $\R$: \begin{lem} % LEMMA A B C Let $A,B,C$ be matrices with the same number of rows over a Sylvester domain\/ $\R$. Then \begin{equation}\label{ABCeqn} \rho(A\ B)=\rho(A\ C)=\rho(A)\ \Rightarrow\ \rho(A\ B\ C)=\rho(A). \end{equation} \end{lem} \begin{proof} % LEMMA A B C \noindent Take suitably partitioned rank factorizations $(A\ B)=D(E\ E'),$ $(A\ C)=F(G\ G')$. Since with our choices of matrices $A,B,C$ we have $\rho(A\ B)=\rho(A\ C)=\rho(A)$, then $A=DE=FG$ are also rank factorizations of $A$. Therefore \begin{equation}\label{rhorholeq} \rho(A)\leq \rho(D)\ \mbox{ and }\ \rho(A)\leq \rho(G). \end{equation} Thus the number of columns of $(D\ F)$ is $2\rho(A)$ and $DE=FG$, so using~(\ref{equivalently}): $$(D\ F)\left(\begin{array}{c} E \\ -G \end{array}\right)=0.$$ Since $\R$ is a Sylvester domain, we can apply the nullity condition~(\ref{nullitycondition}) to get: \begin{eqnarray*} 2\rho(A) & \geq & \rho(D\ F)+\rho\left(\begin{array}{c} E \\ -G \end{array}\right) \\ & \geq&\rho(D) + \rho(G) \\ &\geq & \rho(DE) + \rho(FG) \\ &=&\rho(A)+\rho(A) \end{eqnarray*} Hence by ``sandwich'' we have equality throughout. Because of~(\ref{rhorholeq}) we now get $\rho(A)=\rho(D)=\rho(G)$. Further, we also have the rank equality $\rho(A)=\rho(D\ F)$ (since $\rho(A)\leq\rho(D\ F)$), so there is a rank factorization $(D\ F)=H(J\ K)$, where the number of columns of $H$ is $\rho(A)=\rho(D\ F).$ So: \begin{eqnarray*} (A\ B\ C)&=&(DE\ \, DE'\ \, FG') \\ & = & (HJE\ \, HJE'\ \, HKG') \\ & = & H(JE\ \, JE'\ \, KG') \end{eqnarray*} and therefore $\rho(A\ B\ C)\leq\rho(H)\leq\rho(A)$, which proves~(\ref{ABCeqn}) and the lemma. \end{proof} \noindent Recall that our objective was to show that $\rho(C)=\rho(A')=n-1$, i.e.~$C$ is non-full. Since we had $\rho(A'\ a)=\rho(A'\ b)=\rho(A')=n-1$, we can apply above lemma to our matrices $A,B,C$ and get $\rho(C)=\rho(A')=n-1$, and thus $C$ is non-full. This proves the first case. The second case is trivial, as if we assume that $A'$ is non-full, then clearly $\rho(A')< n-1$ and in that case $\rho(C) det sum non-full. Let\/ $\R$ be any ring. If the product of two full matrices, when defined, is again full, then the determinantal sum of two non-full matrices, when defined, is non-full. \end{prop} \begin{proof} % Product full -> det sum non-full. Let $A,B$ be square non-full matrices, such that $C=A\nabla B$. Write $A=(A_{1},A_{2},\ldots,A_{n}),\ B=(B_{1},A_{2},\ldots,A_{n})$. Since $B$ is non-full, we can write $B=PQ$, where $P$ is a $n\times (n-1)$ and $Q$ a $(n-1)\times n$ matrix. Now also write $Q=(Q_{1}\ Q')$, where $Q'$ is a square $(n-1)\times(n-1)$ matrix. Then \begin{equation}\label{eq1} A=(A_{1}\ PQ')=(A_{1}\ P)\MATRIX{1}{0}{0}{Q'},\ \mbox{ and } \end{equation} $$B=PQ=(PQ_{1}\ PQ')=(PQ_{1}\ P)\MATRIX{1}{0}{0}{Q'},\ \mbox{ and hence }$$ $$C=(A_{1}+PQ_{1}\ \ PQ')=(A_{1}+PQ_{1}\ \ P)\MATRIX{1}{0}{0}{Q'},$$ \begin{equation}\label{eq4} \mbox{which is }\ C=(A_{1}\ P)\MATRIX{1}{0}{Q'}{I}\MATRIX{1}{0}{0}{Q'} \end{equation} Assume C is full. Then each product in the matrix factorization must be full. So comparing~(\ref{eq1}) with~(\ref{eq4}), and assuming the product of full matrices is full, then so must be $A$. This contradicts the fact that $A$ is non-full. Therefore $C$ is non-full, which is the desired result. \end{proof} \noindent This means proposition~\ref{fullfull} implies proposition~\ref{detsum} in a Sylvester domain.\footnote{To be a Sylvester domain is needed in the first proposition~\ref{fullfull}.} Let us now show that the necessary condition to have UGN\footnote{UGN stands for unbounded generating number as defined in chapter~\ref{UGN}.} is satisfied in a Sylvester domain too: \begin{prop} % Sylvester domain satisfies UGN A Sylvester domain satisfies UGN. \end{prop} \begin{proof} % Sylvester domain satisfies UGN Recall from equation~\ref{matUGN}, chapter~\ref{UGN}, that UGN holds if whenever $AB=I_{n}, \mbox{ then } m \geq n$, for matrices $A,B$ of the form $n\times m,\ m\times n$ respectively. We must show that in a Sylvester domain $\R$, the matrix $I_{n}$ is full for all $n$. But since $\R\neq 0$ and $\rho(1)=1$ is full, then by induction using the fact that in $\R$ the diagonal sum of full matrices is full, we get $\rho(I_{n})=n$, which is what we wanted. \end{proof} A final remark on the names of these rings: They are called Sylvester domains, because they satisfy \emph{Sylvester's Law of Nullity}\footnote{The Law of Nullity was first mentioned by J.J.~Sylvester in 1884 in the case of fields. Sylvester domains however were first introduced by W.~Dicks and E.D.~Sontag in 1978.} for the inner rank: If $\R$ is a Sylvester domain, and $A$ an $r\times n$, $B$ an $n\times s$ matrix. Then $$ \rho(A)+\rho(B)\leq n+\rho(AB).$$ The proof of this follows from proposition~\ref{rPrQcondition}. Let us finally mention that, of course, any field satisfies it. % -------------------------- EXAMPLES -------------------------- \section{Examples}\label{Examples} %\remafresh In the last chapter we saw that every semifir is a Sylvester domain and thus has a universal \foff. The converse is not true: There are Sylvester domains that are not semifirs. An example is the ring $\mathbf{A}\langle X \rangle$, where \mbf{A} is a commutative principal ideal domain, and $X$ any set. It can be proven that this is a Sylvester domain, which is not a semifir.\footnote{See Cohn~\cite{CohnFRTR}, theorem~5.5.12, p.~260.} % -------------------------- NOETHERIAN and Bezout DOMAINS -------------------------- \subsection{Noetherian and Bezout domains}\label{NoetherianBezout} Goldie observed in 1958 that the Ore condition is a consequence of the Noetherian condition: \begin{thm}\label{noetore} % IDs are always Ore or contain free ideals Let\/ $\R$ be an \ID. Then either\/ $\R$ is a right Ore domain or it contains a right ideal which is free of infinite rank as\/ $\R$-module. In particular, every right Noetherian domain is right Ore.\footnote{This is easily remembered by the following anagram: ``noether'' $\rightarrow$ ``then Ore''. Note however, that it is essential to speak of \emph{domains}, as in general a right Noetherian ring need not be a right Ore ring. For a counter-example see Lam~\cite{Lam}, p.~354.} \end{thm} \begin{proof} % IDs are always Ore or contain free ideals Suppose $\R$ is an \ID\ which is not right Ore. Then there exist $a, b \in \R\setminus\{0\}$, such that $a\R\cap b\mathbf{R}=0$. Claim: the elements $b,ab,a^{2}b,\ldots$ are right linearly independent over $\R$, so the right ideal generated by them is free, of infinite rank. If not, then there would be a relation $\sum a^{i}bc_{i}=0$, where the $c_{i}$ are not all $0$. Let $c_{r}$ be the first non-zero coefficient. Then we obtain the relation $$ bc_{r}+abc_{r+1}+\ldots+a^{n-r}bc_{n}=0$$ by canceling by $a^{r}$. Hence $$ a(bc_{r+1}+\ldots+a^{n-r-1}bc_{n})=-bc_{r}\neq 0.$$ This is clearly a contradiction to our assumptions on $a,b$. %- \newpage Note that a free right ideal of infinite rank cannot be finitely generated. Therefore every right Noetherian domain has to be right Ore. \end{proof} \begin{cor} % NOETHERIAN Any right Noetherian domain can be embedded in a skew field. In particular, any principal right ideal domain (PRID) can be embedded in a skew field. \end{cor} This is clear from the fact that a PRID is right Noetherian and applying above theorem. This means that a PRID has a \foff, since it is right Ore. \noindent Another interesting result is the following: \begin{prop}\label{PRIDfir} % PRID <-> right fir + Ore A ring is a PRID \IFF\ it is a right fir and satisfies the right Ore condition. \end{prop} \begin{proof} % PRID <-> right fir + Ore \noindent ($\Leftarrow$) In a right fir $\R$ every right ideal is free and if $\R$ is also right Ore, then any two elements of $\R$ are right linearly dependant. Hence no right ideal needs more than one generator and so $\R$ is a right principal ideal. \noindent ($\Rightarrow$) Since a PRID is right Noetherian, it is also right Ore by theorem~\ref{noetore}. Thus all right ideals are free and we have a right fir. \end{proof} \noindent This shows that a commutative fir is a PID. % -------------------------- BEZOUT DOMAINS -------------------------- \subsubsection*{Bezout domains}\label{Bezoutdomains} \begin{defn} % Bezout domain A \emph{Bezout domain} is an \ID\ in which every\\ finitely generated right ideal is principal. \end{defn} \begin{prop} % Bezout -> Ore Every right Bezout domain is a right Ore domain. \end{prop} \begin{proof} % Bezout -> Ore Suppose not, that is a Bezout domain $\R$ does not satisfy the right Ore condition for domains (see equation~(\ref{oredomain}), section~\ref{Orecase}), i.e.~$a\R\cap b\R=0$, for $a, b \in \R\setminus\{0\}$. Choose $c\in\R$, such that $a\R \oplus b\R =c\R$ and let $b=cd$. $$a\R= {{a\R}\over{a\R \cap b\R}},\ \mbox{ since }\ a\R\cap b\R=0.$$ Then by the second isomorphism theorem we get: $${{a\R}\over{a\R \cap b\R}} \cong {{a\R \oplus b\R}\over{b\R}}$$ But since $a\R \oplus b\R =c\R$ $$ {{a\R \oplus b\R}\over{b\R}} = {{c\R}\over{b\R}} \cong {{\R}\over{d\R}}, $$ as we chose $b=cd$. This is then a contradiction as $b,d\neq 0$. \end{proof} \rems % BEZOUT \begin{enumerate} \item This means a right Bezout domain is embeddable in a field, since it is right Ore. \item A commutative \ID\ is a 2-fir precisely if every 2-generator ideal is principal. Thus a commutative 2-fir is a Bezout domain. \item Every commutative 2-fir, and hence every Bezout domain is a semifir. Also, a commutative semifir is just a Bezout domain. \end{enumerate} % -------------------------- FREE ALGEBRA -------------------------- \subsection{The free algebra $\mathbf{k}\langle X\rangle$} % k The free algebra $\mathbf{k}\langle X \rangle$ (where \mbf{k} is a commutative field, $X$ a set) may be regarded as a generalization of the polynomial ring $\mathbf{k}[x]$, to which it reduces when $X$ consists of a single element $x$. As is well-known, the polynomial ring $\mathbf{k}[x]$ is a principal ideal domain (PID). For the non-commutative polynomial ring $\mathbf{k}\langle X \rangle$ the PID property can be generalized by free ideal rings as defined in~\ref{semifir}.\\ \rems % k is a fir \begin{enumerate} \item The fact that $\mathbf{k}\langle X\rangle$ is a fir can be proved by means of the weak algorithm, a generalization of the Euclidean algorithm, to which it reduces in the commutative case. Details can be found in chapter 2 of P.M.~Cohn's Free Rings and their Relations~\cite{CohnFRTR}. \item The free algebra $\mathbf{k}\langle X \rangle$ (where $|X|\geq 2$) has many non-isomorphic \foff. \item To show that not every ring embeddable in a field needs to be Ore, take for example $|X|=2$. Then $\mathbf{k}\langle X \rangle=\mathbf{k}\langle x,y \rangle$, where the elements $x,y$ do not commute in $\mathbf{k}\langle x,y \rangle$. Hence $\mathbf{k}\langle x,y \rangle$ is not Ore, because $x\mathbf{k}\cap y\mathbf{k}=0$. However $\mathbf{k}\langle x,y \rangle$ is a semifir and therefore, as we have seen, embeddable in a field. \item One of the characteristics of the free algebra $\mathbf{k}\langle X\rangle$ is the normal form property for its elements, which can easily be described. Consider the free monoid \mbf{G} on a set $X$, whose elements can be expressed as products $x_i=x_{i_1}x_{i_2}\ldots x_{i_n}$, where $x\in X$ and $n\geq0$. If $n=0$, the empty product represents $1$. The free algebra $\mathbf{k}\langle X\rangle$ on $X$ over $k$ may be described as the monoid algebra of \mbf{G}. Its elements are uniquely expressible in the form \begin{equation}\label{monomial} f=\sum_i \alpha_i x_i\ (\alpha_i\in\mathbf{k},\, \mbox{almost all 0}). \end{equation} Let us define a \emph{monomial} of $\mathbf{k}\langle X\rangle$, which we will need in the proof of~\ref{MALCEVVV}, as the product $x_1x_2\ldots x_n$, where we denote its \emph{length} by $n=l(x_1x_2\ldots x_n)$. For a general element~(\ref{monomial}) of $\mathbf{k}\langle X\rangle$ we define its \emph{degree} as $deg(f)=max\{l(x_i):\alpha_i\neq0\}$. \item In the commutative case the rational function field $\mathbf{k}(x_{1},\ldots,x_{d})$ may be regarded to be ``universal'' in the sense that all other fields generated by $d$ elements over $\mathbf{k}$ can be obtained from it by specialization. Moreover, it is the \foff\ of the polynomial ring $\mathbf{k}[x_{1},\ldots,x_{d}]$ and as such it is uniquely determined. By contrast, and as we have remarked above, the free algebra $\mathbf{k}\langle x_1,\ldots,x_d \rangle$ has a universal \foff, but several different fields of fractions. \end{enumerate} % -------------------------- MALCEV -------------------------- \subsection{Malcev and more}\label{Malmore} % Malcev and more %\remafresh % ------------------ Counter Example ------------------ \subsubsection*{Malcev's counter-example}\label{MalcevCounter} A. Malcev was the first to give a negative answer to the question whether any (commutative or not) \ID\ can be embedded in a skew field. More generally, Malcev was working on the problem of embedding a semigroup \mbf{H} into a group \mbf{G}. He published a famous paper in 1937 with an example of a cancellation\footnote{A cancellation semigroup is a semigroup in which both cancellation laws hold: If $am=an$ or $mb=nb$ then $m=n$.} semigroup\footnote{Semigroups containing $1$ are usually called \emph{monoids} and are just groups that do not admit inverses.} \mbf{H} that cannot be embedded into a group. He showed further that the semigroup ring \mbf{Q[H]} is an \ID. Finally he proved that this \ID\ cannot be embedded in a skew field. Thus he was the first to show that in general for a ring to be an \ID\ was not sufficient to have a \foff. Malcev showed however that it was a necessary condition and, coming back to his example, he showed that if his \ID\ had a \foff, then \mbf{H} would be embeddable in the group of units of the \foff. We will state here all theorems used to construct this counter-example, but will not give proofs. These can be found in~\cite{Lam}, chapter 4. However we will be using Malcev's construction later on. \begin{thm}[Malcev's semigroup]\label{MalcevsSemigroup} % Malcev's semigroup $\ $ There exists a cancellation semigroup \mbf{H} with $a,b,c,d,x,y,u,v\in\mathbf{H}$, such that $ax=by,\ cx=dy,\ au=bv$, but $cu\neq dv$. In particular, \mbf{H} cannot be embedded in any group\/ \mbf{G}. \end{thm} \begin{rem} % Malcev's semigroup Note that commutative semigroups are embeddable in groups \IFF\ they satisfy the cancellation law(s)\footnote{The actual embedding is constructed in a similar way to the embedding of a commutative \ID\ as done in chapter~\ref{comidFOFF}. Here, of course, we only need the multiplicative operation $(a,b)(c,d)=(ac,bd)$ and identity $(1,1)$.}. However cancellation is a necessary but not sufficient condition for the non-commutative case. Another necessary condition needs to be satisfied, which Malcev denoted by ``Condition Z'', and which is automatic in the commutative case: \end{rem} \begin{lem}[Condition Z]\label{ConditionZ} % Condition Z Let\/ \mbf{H} be a semigroup and $a,b,c,d,x,y,u,\quad $ \mbox{$v\in\mathbf{H}$}. If\/ \mbf{H} is embeddable in a group\/ \mbf{G}, then \begin{equation}\label{LemmaMalcev} ax=by,\ cx=dy,\ au=bv\ \Rightarrow cu=dv\ \mbox{ in } \mathbf{H}.\end{equation} \end{lem} \begin{rem} % Condition Z The implication~(\ref{LemmaMalcev}) may be helpfully rewritten in matrix form as follows: The three relations on the LHS may be expressed formally by the matrix equation: \begin{equation}\label{mateqmal} \MATRIX{a}{b}{c}{d} \MATRIX{x}{u}{-y}{-v} = \MATRIX{0}{0}{0}{cu-dv} \end{equation} So we can say, that when the implication~(\ref{LemmaMalcev}) holds, then the RHS of the above equation must in fact be the zero matrix since $cu=dv$ holds, i.e.~$cu-dv=0$.\\ \end{rem} \noindent Let us finally state Malcev's famous counter-example: \begin{thm}[Non-embeddable \ID]\label{nonembed} % Non-embeddable ID Let\/ \mbf{H} be Malcev's semigroup and let\/ $\R=\mathbf{K[H]}$ be the semigroup algebra of\/ \mbf{H}, where\/ \mbf{K} is any \ID. Then\/ $\R$ is an \ID, which cannot be embedded into any skew field. \end{thm} % ------------------ Rs=0 ------------------ \subsubsection*{$\R_{\mathbf{S}}$ may be zero}\label{Rszero} In chapter~\ref{Localization} on the general process of Localization we mentioned in a remark that the localization $\Rs$ may be the zero ring even though $\R\neq 0$ and $0\not\in\mathbf{S}$. Let us first construct an example for this happening:\\ \ex % Rs=0 LAM EXAMPLE 1 Let $\R=\mathcal{M}_{n}(\mathbf{K})$, where \mbf{K} is a non-zero ring. Let \mbf{S} be the multiplicative subset $\{1,E_{11}\}$, where $E_{11}$ is the matrix unit with a $1$ in the first position. Now look at the kernel of the \mbf{S}-inverting homomorphism $\lambda:\R \rightarrow \Rs$. We will show that it is the whole of \mbf{R}, i.e.~$ker(\lambda)=\mathbf{R}$. This means that everything gets mapped to $0$ and $\mathbf{R_{S}}=0$. \noindent Since $ker(\lambda)$ is an ideal of $\R$, it must have the form $\mathcal{M}_{n}(\ca{I})$, where $\ca{I}$ is an ideal of \mbf{K}. Recall the equation for the kernel: $$ker(\lambda)=\{ A\in \R : AB=0 \mbox{ for some } B\in\mathbf{S} \}$$ But now for example $E_{22}E_{11}=0$, which implies that $E_{22}\in ker(\lambda)$. This means that the element $1\in\ca{I}$, but being an ideal, it follows that $\ca{I}=\mathbf{K}$. Thus $ker(\lambda)=\mathcal{M}_{n}(\mathbf{K})=\R$, and therefore $\lambda$ is the zero map and $\Rs=0$ as requested.\\ Note that in this example, the matrix ring is of course not even an \ID. More interesting is the following example based on Malcev's construction, in which $\Rs=0$, yet $\R$ is an \ID: \begin{prop}\label{MALCEVVV} % Rs=0 LAM EXAMPLE Let\/ $\R=\mathbf{K[H]}$ as in theorem~\ref{nonembed}. Let\/ $\mathbf{D}=\R/\ca{I}$, where\/ $\ca{I}$ is the ideal of\/ $\R$ generated by $(cu-dv)+1$. Then, for any multiplicative subset\/ $\mathbf{S}\subseteq \mathbf{D}$ containing $b,d,x,u$ (which are also elements in \mbf{H}), $\mathbf{D_{S}}=0$. \end{prop} \begin{proof} % Rs=0 LAM EXAMPLE There are two parts we need to prove: First we will show that the localization under these conditions is zero. Secondly we need to show that \mbf{D} is indeed an \ID. \mbf{D} is the algebra with the following generators and relations:\\ $\mathbf{K}\langle a,b,c,d,x,y,u,v\, :\,ax=by,\,cx=dy,\,au=bv,cu=dv-1 \rangle$. To show that the localization is zero, we assume to get a contradiction, that $\mathbf{D_{S}}\neq0$, whenever we are given $b,d,x,u\in\mathbf{S}$. Then $\mathbf{D_{S}}$ is constructed as in chapter~\ref{Localization} by adjoining elements, in particular $b^{-1},d^{-1},x^{-1},u^{-1}$, satisfying the relations $bb^{-1}=1=b^{-1}b,\,dd^{-1}=1=d^{-1}d,\ xx^{-1}=1=x^{-1}x,\ uu^{-1}=1=u^{-1}u$. But we also have the relations $ax=by,\, cx=dy,\, au=bv$ and \begin{equation}\label{malrel} (cu-dv)+1=0 \end{equation} holding in $\mathbf{D_{S}}$. This will lead to a contradiction. Manipulate the given relations as follows: \begin{eqnarray*} ax=by\ \Longrightarrow & b^{-1}axx^{-1}=b^{-1}byx^{-1}&\Longrightarrow\ b^{-1}a=yx^{-1} \\ cx=dy\ \Longrightarrow & d^{-1}cxx^{-1}=d^{-1}yx^{-1}&\Longrightarrow\ d^{-1}c=yx^{-1} \\ au=bv\ \Longrightarrow & b^{-1}auu^{-1}=b^{-1}bvu^{-1}&\Longrightarrow\ b^{-1}a=vu^{-1} \end{eqnarray*} Then we have $vu^{-1}=b^{-1}a=yx^{-1}=dc^{-1}$, i.e. \begin{eqnarray*} vu^{-1} = d^{-1}c\ \Longrightarrow\ dv=cu\ \Longrightarrow\ cu-dv=0. \end{eqnarray*} This contradicts equation~(\ref{malrel}), since $cu-dv+1=0$ in \mbf{D}. Thus we must have that whenever $b,d,x,u\in\mathbf{S}$, the localization $\mathbf{D_{S}}=0$.\\ To show that \mbf{D} is an \ID, we proceed as follows: Let $f=\sum_{i}\alpha_{i}s_{i},\ g=\sum_{j}\beta_{j}r_{j} \in \mathbf{D}$, where $\alpha_{i},\beta_{j}\in\mathbf{K}$. We may assume $s_i,\,r_j$ to be distinct monomials with no $ax,cx,au,cu$ present as subwords. That is, if these subwords occur, we can transform or ``reduce'' them by replacing each occurrence of them by $by,dy,bv,dv-1$ respectively. This gives us a normal form for the elements in \mbf{D}, as the elements can be written in terms of unique reduced monomials. Now suppose \mbf{D} is not an \ID. Then there is an equation \begin{equation}\label{countercounter} fg=\left( \sum_{i}\alpha_{i}s_{i}\right) \left( \sum_{j}\beta_{j}r_{j} \right)=\sum_{i,j}\alpha_{i}\beta_{j}s_{i}r_{j}=0\ \in \mathbf{D}, \end{equation} where $\alpha_i\neq0,\ \beta_j\neq0$. In the above equation~(\ref{countercounter}) the terms with the highest degrees cannot cancel out in combination with lower degree terms, so it will be sufficient to look at such terms of highest degrees only. In order to cancel out $s_1r_1$, we must have $s_1r_1=s_ir_j$ for some $i\neq1,\,j\neq 1$. Since $s_{1}\neq s_{i},\,r_{1}\neq r_{j}$, one possible way to obtain $s_1r_1=s_ir_j$ is the following case (e.g.~for $ax=by$), in which $k,l,m,n$ are just taken to be other elements not interfering with the replacement: $$s_1=k\ldots la,\quad r_1=xmn\ldots,\quad \mbox{ so }\ s_1r_1=k\ldots laxmn\ldots $$ $$s_i=k\ldots lb,\quad r_j=ymn\ldots,\quad \mbox{ so }\ s_ir_j=k\ldots lbymn\ldots $$ But then in equation~(\ref{countercounter}) we have a term $\alpha_1\beta_j s_1 r_j$ corresponding to the reduced monomial $k\ldots laymn\ldots$, which \emph{cannot} be cancelled out by any other term. This leads to a contradiction, since this means the product of sums in~(\ref{countercounter}) does not equal zero. Similarly for the other three cases of replacement, and even the last transformation does not give a possibility, which will not lead to a contradiction. Hence \mbf{D} is an \ID. There is another more rigorous way to show that \mbf{D} is an \ID. This needs the so-called $n$-term weak algorithm\footnote{For a complete reference see Cohn~\cite{CohnFRTR}, chapter 2 and in particular 2.11.}: Recall that if a ring $\R$ satisfies the $n$-term weak algorithm, then $\R$ is an $n$-fir. For our case let $n=1$. Then it will be sufficient to show that \mbf{D} satisfies the $1$-term weak algorithm. Then we have a $1$-fir, which is nothing else than an \ID, as seen in chapter~\ref{semifir}. Recall that \mbf{D} is the \mbf{K}-algebra with $\langle a,b,c,d,x,y,u,v\, :\,ax=by,\,cx=dy,\,au=bv,cu=dv-1 \rangle$ as generators and relations. We can rewrite this using equation~(\ref{mateqmal}) as: $$\MATRIX{a}{b}{c}{d} \MATRIX{x}{u}{-y}{-v} = \MATRIX{0}{0}{0}{-1}$$ Now we need to show that these relation form an \ID, i.e.~a $1$-fir. For this we will be using P.M.~Cohn's theorem~2.11.1 in~\cite{CohnFRTR}, and show that all conditions are satisfied. We begin by remarking that each element $f\in\mathbf{D}$ can be expressed as a linear combination of products of the generators. Any such $f$ can be expressed in reduced form by writing down an expression for $f$ and then applying the transformations $ax=by,\,cx=dy,\,au=bv,cu=dv-1$, for all the defining relations, wherever possible. We will eventually reach an expression with no $ax,cx,au,cu$ present (called $0$-reduced). This final $0$-reduced expression is then unique and thus the \emph{normal form} of $f$. Now consider the given relations: $$ax-by=0,\ cx-dy=0,\ au-bv=0,\ cu-dv=-1.$$ We can form the sets $X=\{a,b,c,d\},\ Y=\{x,-y,u,-v\}$, in which the elements of $X$, respectively $Y$, are distinct. What we are left to show is that whenever $sr\in\mathbf{D}$ is $0$ and $1$-reduced, then $s[rt]_\nu$ is in normal form for $\nu=0,1$. Applying Cohn's theorem~2.11.1, this implies that $\R$ is a $1$-fir. We will show the case when $\nu=0$, since the other case is treated similarly: Assume, to get a contradiction, that $s[rt]_\nu$ is not $0$-reduced, i.e.~not in normal form. Then $s[rt]_\nu$ contains one of the following terms: $a,c,x$ or $u$. Thus $s$ must then contain either $a$ or $c$, and then $r$ is either $x$ or $u$. But that contradicts $sr$ being $0$-reduced. Thus $s[rt]_\nu$ must be $0$-reduced too in \mbf{D}. We conclude that \mbf{D} satisfies the $1$-term weak algorithm and is thus a $1$-fir, i.e.~an \ID. \end{proof} % ------------------ Klein's Nilpotence Condition ------------------ \subsubsection*{And more...}\label{more} As we have seen, the example of an \ID\ $\R$ not embeddable in a skew field given by Malcev, is such that the semigroup $\R\setminus\{0\}$ is already not embeddable in a group. This led Malcev to ask if there also exist \ID s $\R$ for which $\R\setminus\{0\}$ is embeddable in a group, yet $\R$ is still not embeddable in a skew field. It took nearly thirty years before A.~Klein found such an example in 1967.\footnote{See Journal of Algebra 7 (1967), p.~100-125 on ``Rings non-embeddable in fields with multiplicative semigroups embeddable in groups''.} Malcev however solved the general problem of finding necessary and sufficient conditions for a semigroup to be embeddable in a group. He published these in another paper in 1939, and the conditions are in the form of an infinite sequence of implication statements of the form: $$A_1,A_2,\ldots,A_n\ \Longrightarrow\ A,$$ where $A$ and the $A_i$'s are equations in the semigroup. Malcev called these statements \emph{quasi-identities}.\footnote{For more detail see A.~I.~Malcev ``Ueber die Einbettung von assoziativen Systemen in Gruppen I \& II'', Mat.~Sbornik NS~6(48) 1939, p.~331-336 \& 8(50) 1940, p.~251-264.} As we have seen, the simplest quasi-identities required for embeddability are the cancellation laws and Condition Z. For the embeddability of an \ID\ into a skew field, in addition to these semigroup quasi-identities, there are also other, more ring-theoretic\footnote{i.e.~a condition is expressed not only in terms of multiplication, but in terms of both addition and multiplication.} necessary conditions. We have seen for example that when an \ID\ $\R$ embeds in a field, then $\R$ needs to satisfy UGN (else the unit matrix is not full). Since we had UGN $\Rightarrow$ IBN (see chapter~\ref{UGN}), $\R$ satisfies IBN too. Another famous necessary condition is \emph{Klein's Nilpotence Condition}, which says the following: $$\mbox{For any nilpotent matrix $A\in\MATn$, we have $A^n=0$.}\footnote{Note, the same statement holds (by linear algebra) for nilpotent matrices over a field.}$$ Klein even showed that his condition implies that the semigroup $\R\setminus\{0\}$ embeds into a group. Then again Bergman produced a counter-example in 1974\footnote{See ``Coproducts and some universal ring constructions'', Trans.~Amer.~Math.~Soc.~200 (1974), p.~33-88.} showing that Klein's Nilpotence Condition was still not sufficient for the embeddability of an \ID\ into a skew field. %- \newpage % -------------------------- CONCLUSION -------------------------- \section{Conclusion} This project investigated the question what rings can be embedded in a field. Moreover we looked at sufficient conditions for a universal \foff\ to exist. It is impossible to list all the rings with an embedding into a field, but we have seen necessary and sufficient conditions for a \foff\ to exist in chapter~\ref{MatrixIdeals}. By studying special cases, we have shown that the following rings have an embeddeding into a field: \begin{enumerate} \item Commutative \ID s (chapter~\ref{comidFOFF}). \item Ore domains (chapter~\ref{Orecase}). These include Noetherian domains, Bezout domains and principal ideal domains (PIDs, PRIDs and PLIDs), because they are Noetherian (chapter~\ref{NoetherianBezout}). \item Sylvester domains (chapter~\ref{Sylvesterdomains}). These include semifirs and free ideal rings (firs)~(chapter~\ref{semifir}). The implications are as follows: $$\mbox{fir $\Rightarrow$ semifir $\Rightarrow$ Sylvester domain $\Rightarrow$ \ID.}$$ \end{enumerate} All of the above rings are \ID s. However, we have seen, being a non-commutative \ID\ is not a sufficient condition to be embeddable in a field. We outlined the important counter-example by Malcev and showed a case of a ring with zero localization, although the ring was an \ID. We also remarked that there are rings, such as free algebras ($\mathbf{k}\langle X \rangle$, where $|X|\geq 2$) having a universal \foff, but several non-isomorphic fields of fractions. Clearly, in the commutative case once we have an $\R \setminus \{0\}$-inverting homomorphism to a non-zero ring, we obtained the embedding into a field. This is actually the universal localization. Thus a commutative integral domain has a universal \foff, which is unique up to isomorphism. Note that any commutative ring is Ore. Another aim, which was the characterization of rings with a fully inverting homomorphism to a field was achieved, and they turned out to be precisely Sylvester domains. By using the so-called Sufficient Conditions Theorem~\ref{SufficientConditions}, we proved that every Sylvester domain (and hence every semifir) has a universal \foff. For Ore domains we proved uniqueness of the \foff\ in chapter~\ref{Orecase}. It can be shown that an Ore domain is a Sylvester domain \IFF\ every full matrix is regular. This links the two seemingly separated classes of rings together, and it is not an easy task to draw a picture of how all the above rings are related through implication. % --------------------- APPENDIX, BIBLIOGRAPHY & END --------------------- \begin{appendix} % APPENDIX \section{Bibliography} % REFERENCES \begin{thebibliography}{[0w]} \bibitem[1]{MacDonald} Atiyah,~M~F \& MacDonald,~I~G: \emph{Introduction to Commutative Algebra}, Addison-Wesley, London, 1969. \bibitem[2]{Beachy} Beachy,~J~A \& Blair, B: \emph{Abstract Algebra, Second Edition}, Waveland Press, 1995. \bibitem[3]{BeachyII} Beachy,~J~A: \emph{Abstract Algebra II}, Northern Illionis University, 1996. \bibitem[4]{BeachyCam} Beachy,~J~A: \emph{Introductory Lectures on Rings and Modules}, CUP, 1999. \bibitem[5]{Cohn2} Cohn,~P~M: \emph{Algebra Volume 2 (Second Edition)}, John Wiley \& Sons, Chichester, 1989. \bibitem[6]{Cohn3} Cohn,~P~M: \emph{Algebra Volume 3 (Second Edition)}, John Wiley \& Sons, Chichester, 1991. \bibitem[7]{CohnFRTR} \mbox{Cohn,~P~M}: \emph{Free Rings and their Relations (Second Edition)}, Academic Press, London, 1985. \bibitem[8]{CohnSkew} Cohn,~P~M: \emph{Skew Fields: Theory of General Division Rings}, Encyclopedia of Mathematics (Volume 57), Cambridge University Press, 1995. \bibitem[9]{Keating} Keating, M E: \emph{A First Course in Module Theory}, Imperial College Press, London, 1998. \bibitem[10]{Lam} Lam, T Y: \emph{Lectures on Modules and Rings}, Springer, New York, 1999. \bibitem[11]{Malcev} Malcev, A: \emph{On the Immersion of an Algebraic Ring into a Field}, Mathematische Annalen 113 p.686-691, 1937. \bibitem[12]{Stewart} Stewart, I: \emph{Galois Theory (Second Edition)}, Chapman \& Hall/CRC, London, 1989. \end{thebibliography} \subsection*{Acknowledgment} % THANKS I would like to express deep gratitude to my supervisor Dr.~Mark Roberts whose guidance and support were crucial for the successful completion of this project. \end{appendix} \end{document} % End of MSci-Project