\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{framed} \usepackage{amsmath} \usepackage{comment} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Thursday, June 25, 2026 18:58:56} %TCIDATA{SuppressPackageManagement} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}[1]{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \theoremstyle{definition} \newtheorem{theo}{Theorem}[section] \newenvironment{theorem}[1][] {\begin{theo}[#1]\begin{leftbar}} {\end{leftbar}\end{theo}} \newtheorem{lem}[theo]{Lemma} \newenvironment{lemma}[1][] {\begin{lem}[#1]\begin{leftbar}} {\end{leftbar}\end{lem}} \newtheorem{prop}[theo]{Proposition} \newenvironment{proposition}[1][] {\begin{prop}[#1]\begin{leftbar}} {\end{leftbar}\end{prop}} \newtheorem{defi}[theo]{Definition} \newenvironment{definition}[1][] {\begin{defi}[#1]\begin{leftbar}} {\end{leftbar}\end{defi}} \newtheorem{remk}[theo]{Remark} \newenvironment{remark}[1][] {\begin{remk}[#1]\begin{leftbar}} {\end{leftbar}\end{remk}} \newtheorem{coro}[theo]{Corollary} \newenvironment{corollary}[1][] {\begin{coro}[#1]\begin{leftbar}} {\end{leftbar}\end{coro}} \newtheorem{conv}[theo]{Convention} \newenvironment{condition}[1][] {\begin{conv}[#1]\begin{leftbar}} {\end{leftbar}\end{conv}} \newtheorem{quest}[theo]{Question} \newenvironment{algorithm}[1][] {\begin{quest}[#1]\begin{leftbar}} {\end{leftbar}\end{quest}} \newtheorem{warn}[theo]{Warning} \newenvironment{conclusion}[1][] {\begin{warn}[#1]\begin{leftbar}} {\end{leftbar}\end{warn}} \newtheorem{conj}[theo]{Conjecture} \newenvironment{conjecture}[1][] {\begin{conj}[#1]\begin{leftbar}} {\end{leftbar}\end{conj}} \newtheorem{exmp}[theo]{Example} \newenvironment{example}[1][] {\begin{exmp}[#1]\begin{leftbar}} {\end{leftbar}\end{exmp}} \newenvironment{statement}{\begin{quote}}{\end{quote}} \iffalse \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \newenvironment{statement}[1][Statement]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \fi \newenvironment{noncompile}{}{} \excludecomment{noncompile} \newcommand{\kk}{\mathbf{k}} \newcommand{\id}{\operatorname{id}} \newcommand{\ev}{\operatorname{ev}} \newcommand{\bk}{\mathbf{k}} \newcommand{\Nplus}{\mathbb{N}_{+}} \newcommand{\NN}{\mathbb{N}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to ``Semigroups, rings, ...''} \ohead{\today} \begin{document} \begin{center} \textbf{Semigroups, rings, and Markov chains} \textit{Kenneth S. Brown} \texttt{\url{http://arxiv.org/abs/math/0006145v1}} version of 20 June 2000 (arXiv preprint arXiv:math/0006145v1). \textbf{Errata and addenda by Darij Grinberg} \setcounter{equation}{200} \bigskip \end{center} I will refer to the results appearing in the paper ``Semigroups, rings, and Markov chains'' by the numbers under which they appear in this paper (specifically, in its version of 20 June 2000, which appears on arXiv as preprint arXiv:math/0006145v1). \setcounter{section}{30} \section*{Errata} \begin{itemize} \item \textbf{Page 1, \S 1:} \textquotedblleft explcitly\textquotedblright% \ $\rightarrow$ \textquotedblleft explicitly\textquotedblright. \item \textbf{Page 3, \S 1.4:} Replace \textquotedblleft$q$% -elements\textquotedblright\ by \textquotedblleft$q$ elements\textquotedblright. (The hyphen should not be there.) \item \textbf{Page 4, Theorem 0:} \textquotedblleft maximal elements (co-atoms)\textquotedblright\ should just say \textquotedblleft co-atoms\textquotedblright. And the word \textquotedblleft co-atom\textquotedblright\ should be defined: A \emph{co-atom} of a lattice $L$ is a maximal element of the poset $L\setminus\left\{ \widehat{1}\right\} $. (In contrast, the only maximal element of $L$ -- in the literal sense of this word -- is $\widehat{1}$. This is not what you want the sum to range over.) \item \textbf{Page 4, \S 1.6:} \textquotedblleft the partial sums are eventually constant\textquotedblright\ $\rightarrow$ \textquotedblleft the partial products are eventually constant\textquotedblright. \item \textbf{Page 4, \S 1.7:} \textquotedblleft restarting\textquotedblright% \ might be meant to be \textquotedblleft restating\textquotedblright... though \textquotedblleft restarting\textquotedblright\ also makes sense. \item \textbf{Page 6, \S 2.1:} At the end of the first paragraph on this page (i.e., right after the sentence \textquotedblleft Thus $S$ has the deletion property (D) stated in Section 1.1\textquotedblright), I suggest adding something like the following definitions: \textquotedblleft When $S$ is a LRB, the lattice $L$ and the map $\operatorname*{supp}$ in the above definition are determined uniquely up to isomorphism (this follows easily from the surjectivity of $\operatorname*{supp}$ and the equivalence \begin{equation} \left( xy=x\right) \ \Longleftrightarrow\ \left( \operatorname*{supp}% y\leq\operatorname*{supp}x\right) , \label{equiv3}% \end{equation} which holds for all $x,y\in S$\ \ \ \ \footnote{\textit{Proof of (\ref{equiv3}):} Let $x,y\in S$. We must prove the equivalence $\left( xy=x\right) \ \Longleftrightarrow\ \left( \operatorname*{supp}% y\leq\operatorname*{supp}x\right) $. The \textquotedblleft$\Longleftarrow $\textquotedblright\ direction follows immediately from (3). The \textquotedblleft$\Longrightarrow$\textquotedblright\ direction can be proved as follows: From (2), we know that $\operatorname*{supp}\left( xy\right) =\operatorname*{supp}x\vee\operatorname*{supp}y\geq\operatorname*{supp}y$. Hence, $\operatorname*{supp}y\leq\operatorname*{supp}\left( xy\right) =\operatorname*{supp}x$ if $xy=x$. This proves the \textquotedblleft% $\Longrightarrow$\textquotedblright\ direction of (\ref{equiv3}).}). The lattice $L$ is called the \textit{lattice of supports} (or the \textit{support lattice}) of $S$. The surjection $\operatorname*{supp}:S\rightarrow L$ is called the \textit{support map} of $S$. For any $s\in S$, the element $\operatorname*{supp}s$ of $L$ is called the \textit{support} of $s$.\textquotedblright (Of course, the purpose of these sentences is to introduce some notations that you use several times; you'll probably find a more succinct way to introduce them.) \item \textbf{Page 6, \S 2.1:} The last comma in \textquotedblleft Sections 4, 5, and 6,\textquotedblright\ probably should be a period. \item \textbf{Page 6, \S 2.2:} You write that $S_{\geq x}$ \textquotedblleft is a LRB in its own right, the associated lattice being the interval $\left[ X,\widehat{1}\right] $ in $L$, where $X=\operatorname*{supp}x$% \textquotedblright. This is correct, but (in my opinion) not obvious enough to be left to the reader.\footnote{Here is a proof, for the sake of completeness: \par Let $x\in S$. Let $X=\operatorname*{supp}x$. The interval $\left[ X,\widehat{1}\right] $ of the poset $L$ is a lattice (since $L$ is a lattice). Its join operation is the restriction of the join operation $\vee$ of the lattice $L$; therefore, we shall denote it by $\vee$ as well. \par It remains to prove that $S_{\geq x}$ is a LRB with support lattice $\left[ X,\widehat{1}\right] $ and support map $\operatorname*{supp}\mid_{S_{\geq x}% }$. In order to do so, we need to prove the following four statements: \par \begin{statement} \textit{Statement 1:} The set $S_{\geq x}$ is a subsemigroup of $S$ with identity $x$. \end{statement} \par \begin{statement} \textit{Statement 2:} The map $\operatorname*{supp}\mid_{S_{\geq x}}:S_{\geq x}\rightarrow\left[ X,\widehat{1}\right] $ is well-defined and surjective. \end{statement} \par \begin{statement} \textit{Statement 3:} For any $a\in S_{\geq x}$ and $b\in S_{\geq x}$, we have $\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) \left( ab\right) =\left( \left( \operatorname*{supp}\mid_{S_{\geq x}}\right) a\right) \vee\left( \left( \operatorname*{supp}\mid_{S_{\geq x}}\right) b\right) $. \end{statement} \par \begin{statement} \textit{Statement 4:} Let $a\in S_{\geq x}$ and $b\in S_{\geq x}$. Then, $ab=a$ if $\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) b\leq\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) a$. \end{statement} \par (These four Statements are precisely the requirements in the definition of a LRB, except that we have renamed $x$ and $y$ as $a$ and $b$ since the letter $x$ is already taken for something else.) \par \textit{Proof of Statement 1:} Let $a\in S_{\geq x}$ and $b\in S_{\geq x}$. We have $a\geq x$ (since $a\in S_{\geq x}$). In other words, $x\leq a$. But (5) (applied to $y=a$) shows that $x\leq a\Longleftrightarrow xa=a$. Thus, $xa=a$ (since $x\leq a$). But (5) (applied to $y=ab$) shows that $x\leq ab\Longleftrightarrow xab=ab$. Hence, $x\leq ab$ (since $\underbrace{xa}% _{=a}b=ab$). In other words, $ab\geq x$, so that $ab\in S_{\geq x}$. \par Now, let us forget that we fixed $a$ and $b$. We thus have shown that $ab\in S_{\geq x}$ for every $a\in S_{\geq x}$ and $b\in S_{\geq x}$. In other words, $S_{\geq x}$ is a subsemigroup of $S$, although we do not yet know whether it has an identity. \par Now, let $a\in S_{\geq x}$. Then, $a\geq x$ (since $a\in S_{\geq x}$). In other words, $x\leq a$. But (5) (applied to $y=a$) shows that $x\leq a\Longleftrightarrow xa=a$. Thus, $xa=a$ (since $x\leq a$). Also, $\underbrace{a}_{=xa}x=xax=xa$ (by (4), applied to $y=a$), so that $ax=xa=a$. \par Now, let us forget that we fixed $a$. We thus have shown that $ax=xa=a$ for every $a\in S_{\geq x}$. In other words, the subsemigroup $S_{\geq x}$ of $S$ has identity $x$. This finishes the proof of Statement 1. $\blacksquare$ \par \textit{Proof of Statement 2:} Let $y\in S_{\geq x}$. Thus, $y\geq x$, so that $x\leq y$. Hence, $xy=y$ (by (5)). Thus, $\operatorname*{supp}\underbrace{y}% _{=xy}=\operatorname*{supp}\left( xy\right) =\operatorname*{supp}% x\vee\operatorname*{supp}y$ (by (2)), so that $\operatorname*{supp}% y=\operatorname*{supp}x\vee\operatorname*{supp}y\geq\operatorname*{supp}x=X$. In other words, $\operatorname*{supp}y\in\left[ X,\widehat{1}\right] $. \par Let us now forget that we fixed $y$. We thus have shown that $\operatorname*{supp}y\in\left[ X,\widehat{1}\right] $ for every $y\in S_{\geq x}$. Hence, $\operatorname*{supp}\left( S_{\geq x}\right) \subseteq\left[ X,\widehat{1}\right] $. Therefore, the map $\operatorname*{supp}\mid_{S_{\geq x}}:S_{\geq x}\rightarrow\left[ X,\widehat{1}\right] $ is well-defined. \par We shall now show that this map is surjective. Indeed, let $Y\in\left[ X,\widehat{1}\right] $. Thus, $Y\in L$ and $Y\geq X$. Since the map $\operatorname*{supp}:S\rightarrow L$ is surjective, we thus see that there exists some $y\in S$ satisfying $Y=\operatorname*{supp}y$. Consider such a $y$. We have $\operatorname*{supp}x=X\leq Y=\operatorname*{supp}y$. Hence, (3) (applied to $y$ and $x$ instead of $x$ and $y$) shows that $yx=y$. \par But (5) (applied to $xy$ instead of $y$) shows that $x\leq xy\Longleftrightarrow xxy=xy$. Hence, $x\leq xy$ (since $\underbrace{xx}% _{=x}y=xy$). In other words, $xy\geq x$, so that $xy\in S_{\geq x}$. Now, (2) shows that $\operatorname*{supp}\left( xy\right) =\underbrace{\operatorname*{supp}x}_{=X}\vee\underbrace{\operatorname*{supp}% y}_{=Y}=X\vee Y=Y$ (since $X\leq Y$). Hence, $Y=\operatorname*{supp}\left( \underbrace{xy}_{\in S_{\geq x}}\right) \in\operatorname*{supp}\left( S_{\geq x}\right) =\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) \left( S_{\geq x}\right) $. \par Now, let us forget that we fixed $Y$. We thus have shown that $Y\in\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) \left( S_{\geq x}\right) $ for every $Y\in\left[ X,\widehat{1}\right] $. In other words, $\left[ X,\widehat{1}\right] \subseteq\left( \operatorname*{supp}\mid_{S_{\geq x}% }\right) \left( S_{\geq x}\right) $. In other words, the map $\operatorname*{supp}\mid_{S_{\geq x}}:S_{\geq x}\rightarrow\left[ X,\widehat{1}\right] $ is surjective. This finishes the proof of Statement 2. $\blacksquare$ \par \textit{Proof of Statement 3:} Let $a\in S_{\geq x}$ and $b\in S_{\geq x}$. Since $S_{\geq x}$ is a subsemigroup of $S$ (by Statement 1), this shows that $ab\in S_{\geq x}$. Now, (2) (applied to $a$ and $b$ instead of $x$ and $y$) shows that $\operatorname*{supp}\left( ab\right) =\operatorname*{supp}% a\vee\operatorname*{supp}b$. This rewrites as $\left( \operatorname*{supp}% \mid_{S_{\geq x}}\right) \left( ab\right) =\left( \left( \operatorname*{supp}\mid_{S_{\geq x}}\right) a\right) \vee\left( \left( \operatorname*{supp}\mid_{S_{\geq x}}\right) b\right) $ (since $\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) \left( ab\right) =\operatorname*{supp}\left( ab\right) $, $\left( \operatorname*{supp}% \mid_{S_{\geq x}}\right) a=\operatorname*{supp}a$ and $\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) b=\operatorname*{supp}b$). This proves Statement 3. $\blacksquare$ \par \textit{Proof of Statement 4:} Assume that $\left( \operatorname*{supp}% \mid_{S_{\geq x}}\right) b\leq\left( \operatorname*{supp}\mid_{S_{\geq x}% }\right) a$. This rewrites as $\operatorname*{supp}b\leq\operatorname*{supp}% a$ (since $\left( \operatorname*{supp}\mid_{S_{\geq x}}\right) a=\operatorname*{supp}a$ and $\left( \operatorname*{supp}\mid_{S_{\geq x}% }\right) b=\operatorname*{supp}b$). Hence, (3) (applied to $a$ and $b$ instead of $x$ and $y$) yields $ab=a$. This proves Statement 4. $\blacksquare$ \par Now, all four Statements are proven, and the proof is complete. $\blacksquare $} \item \textbf{Page 9, \S 4.2:} In \textquotedblleft Let $\mathcal{G}% _{0}=\left\{ G\in\mathcal{G}:\sigma_{i}\left( G\right) =+\right\} $\textquotedblright, replace \textquotedblleft$\sigma_{i}\left( G\right) =+$\textquotedblright\ by \textquotedblleft$\sigma_{i}\left( G\right) =+$ for all $i\in J$\textquotedblright. \item \textbf{Page 9, \S 4.2:} \textquotedblleft faces in $G$% \textquotedblright\ should be \textquotedblleft faces in $\mathcal{G}% $\textquotedblright. \item \textbf{Page 13, \S 5.1, Remark:} \textquotedblleft maximal elements\textquotedblright\ $\rightarrow$ \textquotedblleft co-atoms\textquotedblright. (See the correction to Theorem 0 for a precise definition of \textquotedblleft co-atom\textquotedblright.) \item \textbf{Page 14:} In \textquotedblleft where $c_{X}$ is the number of maximal chains in the interval $\left[ X,V\right] $ in $L$\textquotedblright% , the lattice $L$ should be defined. (It is not the lattice of supports of $\overline{F}_{n,q}$, but rather the lattice of all subspaces of $V$, including those of dimension $n-1$.) \item \textbf{Page 14:} \textquotedblleft get a line $l_{1}$\textquotedblright% \ $\rightarrow$ \textquotedblleft get a line $\ell_{1}$\textquotedblright. \item \textbf{Page 15, \S 6.1:} \textquotedblleft the rank of any maximal\textquotedblright\ $\rightarrow$ \textquotedblleft the size of any maximal\textquotedblright. \item \textbf{Page 16:} \textquotedblleft any fixed basis of $X$ to a basis\textquotedblright\ $\rightarrow$ \textquotedblleft any fixed ordered basis of $X$ to an ordered basis\textquotedblright. \item \textbf{Page 18:} \textquotedblleft the first $e_{i}$ such that $\left\{ e,e_{1},\ldots,e_{i}\right\} $ contains a cycle\textquotedblright% \ $\rightarrow$ \textquotedblleft the first $e_{i}$ such that the list $\left( e,e_{1},\ldots,e_{i}\right) $ contains a cycle or two equal elements\textquotedblright. \item \textbf{Pages 24--25, \S 8.2:} Any appearance of \textquotedblleft% $A$\textquotedblright\ in \S 8.2 should be replaced by an \textquotedblleft% $R$\textquotedblright. (There are three appearances on page 24, and six appearances on page 25, not counting the two \textquotedblleft$A^{\prime}% $\textquotedblright s.) \item \textbf{Page 24, \S 8.2:} It would be useful to point out that $g\left( z\right) $ is understood to be an element of $R\left( \left( \dfrac{1}% {z}\right) \right) $ (that is, of the ring of Laurent series in $\dfrac {1}{z}$ over $R$). It can indeed be shown that $g\left( z\right) $ is a rational function in $z$ over $R$, in an appropriate sense\footnote{Namely, $g\left( z\right) $ can be written as $\dfrac{1}{p\left( z\right) }q\left( z\right) $ for some nonzero polynomial $p\left( z\right) \in k\left[ z\right] $ and some polynomial $q\left( z\right) \in R\left[ z\right] $. The easiest way to see this is as follows: \par The powers $a^{0},a^{1},a^{2},\ldots$ of $a$ are $k$-linearly dependent (since they lie in the finite-dimensional $k$-algebra $R$). That is, some nontrivial $k$-linear combination of these powers is $0$. In other words, there exists a nonzero polynomial $p\left( z\right) \in k\left[ z\right] $ such that $p\left( a\right) =0$. Consider this $p$. Thus, $a$ is a root of $p$ in the commutative ring $R$. Hence, in the polynomial ring $R\left[ z\right] $, we have $z-a\mid p\left( z\right) $. Thus, $p\left( z\right) $ can be written as $\left( z-a\right) \cdot q\left( z\right) $ for some $q\left( z\right) \in R\left[ z\right] $. Consider this $q\left( z\right) $. Now, from $p\left( z\right) =\left( z-a\right) \cdot q\left( z\right) $, we obtain $\dfrac{1}{z-a}=\dfrac{1}{p\left( z\right) }q\left( z\right) $. That is, $g\left( z\right) =\dfrac{1}{p\left( z\right) }q\left( z\right) $ (since $g\left( z\right) =\dfrac{1}{z-a}$), qed.}; but before this is shown, it is important to know whether $g\left( z\right) $ is viewed as a Laurent series in $\dfrac{1}{z}$ (correct) or as a Laurent series in $z$ (incorrect). \item \textbf{Page 24, proof of Proposition 2:} I think this proof is missing a few steps. You leave the following statements unproven: \begin{statement} \textit{Statement 1:} Suppose that $R$ is split semisimple with primitive idempotents $\left( e_{i}\right) _{i\in I}$, and write $a=\sum_{i}% \lambda_{i}e_{i}$ with $\lambda_{i}\in k$. Then, the $\lambda_{i}$ (for $i\in I$) are distinct. \end{statement} \begin{statement} \textit{Statement 2:} Up to relabelling, there is only one choice of a set $I$ and two families $\left( e_{i}\right) _{i\in I}\in R^{I}$ and $\left( \lambda_{i}\right) _{i\in I}\in k^{I}$ satisfying (18) (with the $\lambda _{i}$ being distinct elements of $k$ and the $e_{i}$ being nonzero elements of $R$). \end{statement} (Statement 1 is needed to prove the first sentence of Proposition 2. Statement 2 is needed to prove the second sentence.) \begin{proof} [Proof of Statement 1.]Let $j$ and $j^{\prime}$ be two distinct elements of $I$. Let $G$ be the subset $\left\{ \sum_{i}\mu_{i}e_{i}\ \mid\ \left( \mu _{i}\right) _{i\in I}\in k^{I}\text{ and }\mu_{j}=\mu_{j^{\prime}}\right\} $ of $R$. Then, $G$ is a $k$-subalgebra of $R$ (this is easy to see) and satisfies $G\neq R$ (since $e_{j}\notin G$). But every $k$-subalgebra of $R$ which contains $a$ must be $R$ itself (since $a$ generates the $k$-algebra $R$). Hence, if we had $a\in G$, then we would have $G=R$ (since $G$ would be a $k$-subalgebra of $R$ which contains $a$), which would contradict $G\neq R$. Thus, we cannot have $a\in G$. In other words, we have $a\notin G$. In other words, $\lambda_{j}\neq\lambda_{j^{\prime}}$. Now, let us forget that we fixed $j$ and $j^{\prime}$. We thus have shown that $\lambda_{j}\neq\lambda_{j^{\prime}}$ for any two distinct elements $j$ and $j^{\prime}$ of $I$. In other words, the $\lambda_{i}$ (for $i\in I$) are distinct. This proves Statement 1. \end{proof} \begin{proof} [First proof of Statement 2.]Consider a set $I$ and two families $\left( e_{i}\right) _{i\in I}\in R^{I}$ and $\left( \lambda_{i}\right) _{i\in I}\in k^{I}$ satisfying (18) (with the $\lambda_{i}$ being distinct elements of $k$ and the $e_{i}$ being nonzero elements of $R$). We shall show that the $\lambda_{i}$ and the corresponding $e_{i}$ can be reconstructed from $g\left( z\right) $ (up to labelling). This will clearly prove Statement 2. The function $g\left( z\right) $ is a rational function in $z$ (in the sense that $g\left( z\right) $ can be written as $\dfrac{1}{p\left( z\right) }q\left( z\right) $ for some nonzero polynomial $p\left( z\right) \in k\left[ z\right] $ and some polynomial $q\left( z\right) \in R\left[ z\right] $), and thus\footnote{Here we are using the fact that a rational function in $z$ can be expanded as a Laurent series in $z-\mu$ for each $\mu\in k$. The way to do this is straightforward: Write the rational function as $\dfrac{1}{p\left( z\right) }q\left( z\right) $ for some nonzero polynomial $p\left( z\right) \in k\left[ z\right] $ and some polynomial $q\left( z\right) \in R\left[ z\right] $; then compute $\dfrac{1}{p\left( w+\mu\right) }q\left( w+\mu\right) $ in the Laurent series ring $R\left( \left( w\right) \right) $; this will be the desired expansion in $z-\mu$ (of course, $w$ stands for $z-\mu$). A quick verification is required to show that this result is well-defined (the polynomial $p\left( w+\mu\right) $ is nonzero since $p\left( z\right) $ is nonzero) and does not depend on the exact choice of $p$ and $q$ (because if $\dfrac{1}{p\left( z\right) }q\left( z\right) =\dfrac{1}{\widetilde{p}\left( z\right) }\widetilde{q}% \left( z\right) $, then $\widetilde{p}\left( z\right) q\left( z\right) =p\left( z\right) \widetilde{q}\left( z\right) $ in $R\left[ z\right] $, and thus $\widetilde{p}\left( w+\mu\right) q\left( w+\mu\right) =p\left( w+\mu\right) \widetilde{q}\left( w+\mu\right) $, whence $\dfrac{1}{p\left( w+\mu\right) }q\left( w+\mu\right) =\dfrac{1}{\widetilde{p}\left( w+\mu\right) }\widetilde{q}\left( w+\mu\right) $).} can be expanded as a Laurent series in $z-\mu$ for each $\mu\in k$. Hence, for every $\mu\in k$, an element $\operatorname*{Res}\nolimits_{\mu}g\left( z\right) $ of $R$ is well-defined (namely, it is defined as the coefficient of $\left( z-\mu\right) ^{-1}$ when $g\left( z\right) $ is expanded as a Laurent series in $z-\mu$). Now, (18) shows that the elements $\lambda_{i}$ of $k$ are the poles of $g\left( z\right) $ (that is, the elements $\mu\in k$ for which $\operatorname*{Res}\nolimits_{\mu}g\left( z\right) \neq0$% \ \ \ \ \footnote{This is not the general definition of a pole, but in our case it does its job just fine, since all our poles are simple. (Non-simple poles could have residue equal to $0$.)}), and the elements $e_{i}$ of $R$ are their corresponding residues (i.e., we have $e_{i}=\operatorname*{Res}% \nolimits_{\lambda_{i}}g\left( z\right) $ for every $i\in I$). Thus, the $\lambda_{i}$ and the corresponding $e_{i}$ can be reconstructed from $g\left( z\right) $ (up to labelling). This proves Statement 2. \end{proof} \begin{proof} [Second proof of Statement 2.]Here is a more pedestrian proof of Statement 2: We WLOG assume (for the sake of simplicity) that the family $\left( \lambda_{i}\right) _{i\in I}$ is self-indexing, i.e., that each $\lambda_{i}$ equals the corresponding index $i$. This assumption can be made, since the $\lambda_{i}$ are required to be distinct. So we must prove the following claim: There is only one choice of a finite subset $I$ of $k$ and a family $\left( e_{\lambda}\right) _{\lambda\in I}\in R^{I}$ of nonzero elements $e_{\lambda}\in R$ satisfying $g\left( z\right) =\sum_{\lambda\in I}% \dfrac{e_{\lambda}}{z-\lambda}$. In order to prove this, we must show that any such two choices of $I$ and $\left( e_{\lambda}\right) _{\lambda\in I}$ are equal. So let us consider two such choices: $I$ and $\left( e_{\lambda}\right) _{\lambda\in I}$ on one side, and $J$ and $\left( f_{\lambda}\right) _{\lambda\in J}$ on another. So $I$ and $J$ are two finite subsets of $k$, whereas $\left( e_{\lambda}\right) _{\lambda\in I}$ and $\left( f_{\lambda }\right) _{\lambda\in J}$ are two families of nonzero elements of $R$ satisfying $g\left( z\right) =\sum_{\lambda\in I}\dfrac{e_{\lambda}% }{z-\lambda}$ and $g\left( z\right) =\sum_{\lambda\in J}\dfrac{f_{\lambda}% }{z-\lambda}$. We must show that $I=J$ and $\left( e_{\lambda}\right) _{\lambda\in I}=\left( f_{\lambda}\right) _{\lambda\in J}$. Set $U:=I\cup J$; this is again a finite subset of $k$. Extend the family $\left( e_{\lambda}\right) _{\lambda\in I}\in R^{I}$ to a family $\left( e_{\lambda}\right) _{\lambda\in U}\in R^{U}$ by setting $e_{\lambda}:=0$ for each $\lambda\notin I$. Likewise, extend the family $\left( f_{\lambda }\right) _{\lambda\in J}\in R^{J}$ to a family $\left( f_{\lambda}\right) _{\lambda\in U}\in R^{U}$ by setting $f_{\lambda}:=0$ for each $\lambda\notin J$. Note that% \begin{equation} I=\left\{ \lambda\in U\ \mid\ e_{\lambda}\neq0\right\} \label{st2.I=}% \end{equation} (since the original $e_{\lambda}$'s for $\lambda\in I$ were nonzero by assumption, whereas the new $e_{\lambda}$'s for $\lambda\notin I$ are zero by definition) and% \begin{equation} J=\left\{ \lambda\in U\ \mid\ f_{\lambda}\neq0\right\} \label{st2.J=}% \end{equation} (similarly). The sum $\sum_{\lambda\in U}\dfrac{e_{\lambda}}{z-\lambda}$ equals the sum $\sum_{\lambda\in I}\dfrac{e_{\lambda}}{z-\lambda}$ (since the former sum differs from the latter only in the presence of some addends with $\lambda\notin I$, but all these addends vanish because of $e_{\lambda}=0$). Thus,% \[ \sum_{\lambda\in U}\dfrac{e_{\lambda}}{z-\lambda}=\sum_{\lambda\in I}% \dfrac{e_{\lambda}}{z-\lambda}=g\left( z\right) . \] Similarly, $\sum_{\lambda\in U}\dfrac{f_{\lambda}}{z-\lambda}=g\left( z\right) $. Comparing these two equalities,% \[ \sum_{\lambda\in U}\dfrac{e_{\lambda}}{z-\lambda}=\sum_{\lambda\in U}% \dfrac{f_{\lambda}}{z-\lambda}. \] Subtracting the left-hand side from the right-hand side, we obtain% \[ 0=\sum_{\lambda\in U}\dfrac{f_{\lambda}}{z-\lambda}-\sum_{\lambda\in U}% \dfrac{e_{\lambda}}{z-\lambda}=\sum_{\lambda\in U}\dfrac{f_{\lambda }-e_{\lambda}}{z-\lambda}. \] Multiplying this equality by $\prod_{\mu\in U}\left( z-\mu\right) $, we find% \begin{equation} 0=\sum_{\lambda\in U}\left( f_{\lambda}-e_{\lambda}\right) \prod_{\mu\in U\setminus\left\{ \lambda\right\} }\left( z-\mu\right) .\label{st2.5}% \end{equation} This is an equality in the polynomial ring $R\left[ z\right] $ (since $R\left[ z\right] $ is a subring of $R\left( \left( \dfrac{1}{z}\right) \right) $). Hence, for each $\nu\in U$, we can substitute $z=\nu$ in (\ref{st2.5}), and obtain% \begin{align*} 0 & =\sum_{\lambda\in U}\left( f_{\lambda}-e_{\lambda}\right) \underbrace{\prod_{\mu\in U\setminus\left\{ \lambda\right\} }\left( \nu -\mu\right) }_{\substack{=0\text{ whenever }\lambda\neq\nu\\\text{(because if }\lambda\neq\nu\text{, then }\nu\in U\setminus\left\{ \lambda\right\} \text{, and thus}\\\text{this product contains the factor }\nu-\nu=0\text{)}% }}\\ & =\sum_{\substack{\lambda\in U;\\\lambda=\nu}}\left( f_{\lambda}% -e_{\lambda}\right) \prod_{\mu\in U\setminus\left\{ \lambda\right\} }\left( \nu-\mu\right) \\ & =\left( f_{\nu}-e_{\nu}\right) \prod_{\mu\in U\setminus\left\{ \nu\right\} }\left( \nu-\mu\right) . \end{align*} Dividing this equality by the nonzero scalar $\prod_{\mu\in U\setminus\left\{ \nu\right\} }\left( \nu-\mu\right) \in k$ (which is nonzero because it is a product of the nonzero scalars $\nu-\mu\in k$), we obtain $0=f_{\nu}-e_{\nu}$, so that $e_{\nu}=f_{\nu}$. Thus, we have shown that \begin{equation} e_{\nu}=f_{\nu}\ \ \ \ \ \ \ \ \ \ \text{for each }\nu\in U. \label{st2.at}% \end{equation} Thus, the right-hand sides of the equalities (\ref{st2.I=}) and (\ref{st2.J=}) agree. Hence, the left-hand sides agree, too. That is, $I=J$. Therefore, (\ref{st2.at}) yields $\left( e_{\lambda}\right) _{\lambda\in I}=\left( f_{\lambda}\right) _{\lambda\in J}$, and thus our proof is complete. \end{proof} \item \textbf{Page 24, proof of Proposition 2:} Here is an alternative proof of the \textquotedblleft if\textquotedblright\ direction of Proposition 2: Assume that $g\left( z\right) $ has the form (18) where the $\lambda_{i}$ are distinct elements of $k$ and where the $e_{i}$ are nonzero elements of $R$. We must show that $R$ is split semisimple. From $g\left( z\right) =\left( 1/z\right) f\left( 1/z\right) $, we obtain $zg\left( z\right) =f\left( 1/z\right) $. Substituting $z=1/t$ in this, we obtain $\left( 1/t\right) g\left( 1/t\right) =f\left( t\right) $. But substituting $z=1/t$ in (18) and multiplying by $1/t$, we obtain% \[ \left( 1/t\right) g\left( 1/t\right) =\left( 1/t\right) \sum_{i\in I}\dfrac{e_{i}}{1/t-\lambda_{i}}=\sum_{i\in I}\dfrac{e_{i}}{1-\lambda_{i}% t}=\sum_{i\in I}\ \ \sum_{m\geq0}e_{i}\left( \lambda_{i}t\right) ^{m}% \] in the ring $R\left[ \left[ t\right] \right] $. Comparing this with $\left( 1/t\right) g\left( 1/t\right) =f\left( t\right) =\dfrac{1}% {1-at}=\sum_{m\geq0}\left( at\right) ^{m}$, we obtain% \[ \sum_{m\geq0}\left( at\right) ^{m}=\sum_{i\in I}\ \ \sum_{m\geq0}% e_{i}\left( \lambda_{i}t\right) ^{m}. \] Comparing the $t^{m}$-coefficients on both sides of this equality, we find% \begin{equation} a^{m}=\sum_{i\in I}e_{i}\lambda_{i}^{m}\ \ \ \ \ \ \ \ \ \ \text{for each }m\geq0. \label{p24.4}% \end{equation} Hence, for each polynomial $h\left( t\right) \in k\left[ t\right] $, we have% \begin{equation} h\left( a\right) =\sum_{i\in I}e_{i}h\left( \lambda_{i}\right) \label{p24.5}% \end{equation} (because (\ref{p24.4}) shows that this is true whenever $h\left( t\right) $ is a monomial $t^{m}$, and thus, by linearity, this is also true for all polynomials $h$). Now, let $h\left( t\right) \in k\left[ t\right] $ be the polynomial $\prod_{i\in I}\left( t-\lambda_{i}\right) $. Then, $h\left( t\right) $ is a product of distinct monic linear factors. Its roots are precisely the $\lambda_{i}$. In particular, $h\left( \lambda_{i}\right) =0$ for all $i\in I$. Hence, (\ref{p24.5}) simplifies to $h\left( a\right) =0$. In other words, $a$ is a root of $h$. Now, let $m\left( t\right) \in k\left[ t\right] $ be the minimal polynomial of $a$ over $k$. Then, $m\left( t\right) $ divides any polynomial in $k\left[ t\right] $ that has $a$ as a root. In particular, $m\left( t\right) $ divides $h\left( t\right) $ (since $h$ has $a$ as a root). Hence, $m\left( t\right) \mid h\left( t\right) =\prod_{i\in I}\left( t-\lambda_{i}\right) $. Therefore,% \[ m\left( t\right) =\prod_{i\in J}\left( t-\lambda_{i}\right) \ \ \ \ \ \ \ \ \ \ \text{for some subset }J\text{ of }I \] (since $m\left( t\right) $ is monic). We have $m\left( a\right) =0$ (since $m\left( t\right) $ is the minimal polynomial of $a$). Hence, the canonical $k$-algebra homomorphism% \begin{align*} k\left[ t\right] & \rightarrow R,\\ t & \mapsto a \end{align*} has $m\left( t\right) $ in its kernel, and thus factors via the quotient ring $k\left[ t\right] /\left( m\left( t\right) \right) $. We thus obtain a $k$-algebra homomorphism% \begin{align*} k\left[ t\right] /\left( m\left( t\right) \right) & \rightarrow R,\\ \overline{t} & \mapsto a \end{align*} (which sends each $\overline{p\left( t\right) }\in k\left[ t\right] /\left( m\left( t\right) \right) $ to $p\left( a\right) \in R$). This homomorphism is surjective (since $R$ is generated by $a$) and injective (since $m\left( t\right) $ is the minimal polynomial of $a$, so that every nonzero polynomial of smaller degree than $m\left( t\right) $ will take a nonzero value at $a$); thus, it is invertible, i.e., a $k$-algebra isomorphism. Therefore, as a $k$-algebra, we have% \begin{align*} R & \cong k\left[ t\right] /\left( m\left( t\right) \right) =k\left[ t\right] /\left( \prod_{i\in J}\left( t-\lambda_{i}\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }m\left( t\right) =\prod_{i\in J}\left( t-\lambda_{i}\right) \right) \\ & \cong\prod_{i\in J}\underbrace{\left( k\left[ t\right] /\left( t-\lambda_{i}\right) \right) }_{\cong k}\ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{by the Chinese Remainder Theorem,}\\ \text{since the }\lambda_{i}\text{ are distinct}% \end{array} \right) \\ & \cong\prod_{i\in J}k=k^{J}. \end{align*} This shows that $R$ is split semisimple. This proof also shows that Proposition 2 holds for any field $k$, not just for subfields of $\mathbb{C}$. (The original proof might also show this, but it would take more thought.) \item \textbf{Page 26:} After \textquotedblleft in terms of the function $g\left( z\right) $ of Section 8.2\textquotedblright, add \textquotedblleft% (for $R=\mathbb{R}\left[ w\right] $ and $a=w$)\textquotedblright. \item \textbf{Page 27, \S 8.4:} After \textquotedblleft We have therefore proved\textquotedblright, add \textquotedblleft(using Proposition 2)\textquotedblright. \item \textbf{Page 28, \S 8.6:} Replace \textquotedblleft(Section 2.3)\textquotedblright\ by \textquotedblleft(Section 5.1)\textquotedblright. \item \textbf{Page 32, Proposition 4:} Why does \textquotedblleft the smallest face $F\leq C^{\prime}$ such that $FC=C^{\prime}$\textquotedblright\ exist? \item \textbf{Page 36, \S 9.6:} Replace \textquotedblleft$k\left[ w\right] $\textquotedblright\ by \textquotedblleft$k\left[ p\right] $% \textquotedblright. Also, this claim about constructing primitive idempotents is tacitly understood to cover the case when $k$ is a field of characteristic $0$, since otherwise it is not clear what \textquotedblleft generic\textquotedblright\ means. \item \textbf{Page 36, Theorem 8:} \textquotedblleft the isomorphism of Theorem 7\textquotedblright\ should be \textquotedblleft the anti-isomorphism of Theorem 7\textquotedblright. \item \textbf{Page 37, \S 9.7:} Remove the comma at the end of \textquotedblleft Then $E_{0},\ldots,E_{n-2},E_{n},$\textquotedblright. \item \textbf{Page 37, \S A.1:} Replace \textquotedblleft and is denoted $\sigma\left( F\right) $\textquotedblright\ by \textquotedblleft and is denoted $\sigma\left( F\right) =\left( \sigma_{i}\left( F\right) \right) _{i\in I}$\textquotedblright. (This way, the notation $\sigma_{i}\left( F\right) $ is also defined.) \item \textbf{Page 37, \S A.1:} Somewhere here you should probably define $\mathcal{C}$ as the set of all chambers of $\mathcal{A}$. (You use this notation in \S A.7, and maybe earlier.) \item \textbf{Page 37, \S A.2:} The notation \textquotedblleft$F$ is a \textit{face} of $G$\textquotedblright\ is not defined. (It means \textquotedblleft$F\leq G$\textquotedblright.) \item \textbf{Page 40:} \textquotedblleft stated as in exercise\textquotedblright\ should be \textquotedblleft stated as an exercise\textquotedblright. \item \textbf{Page 41, \S B.2:} Replace \textquotedblleft If $S$ has an identity e\textquotedblright\ by \textquotedblleft If $S$ has an identity $e$\textquotedblright\ (the \textquotedblleft$e$\textquotedblright\ should be in mathmode). \item \textbf{Page 41, \S B.2:} Replace \textquotedblleft it is is a lattice\textquotedblright\ by \textquotedblleft it is a lattice\textquotedblright. \item \textbf{Page 41, proof of Proposition 9:} Replace \textquotedblleft% $cx=x$\textquotedblright\ by \textquotedblleft$cx=c$\textquotedblright\ in the second sentence of this proof (but not in the third sentence). \item \textbf{Page 42, \S C.1:} The paragraph starting with \textquotedblleft In case $L$ is a geometric lattice\textquotedblright\ has confused me for a while until I resolved the ambiguities. The problem is that the $m_{X}$ in (14) are defined only for $X\in\overline{L}$, not for all $X\in L$, so the correct version of the claim \textquotedblleft$d\left( \left[ X,\widehat{1}% \right] \right) =m_{X}$\textquotedblright\ should be \textquotedblleft% $d\left( \left[ X,\widehat{1}\right] \right) =% \begin{cases} m_{X}, & \text{if }\operatorname*{rank}\left( X\right) \neq n-1;\\ 0, & \text{if }\operatorname*{rank}\left( X\right) =n-1 \end{cases} $ for all $X\in L$ (where the $m_{X}$ are those of (14), not those of (13))\textquotedblright. In particular, I find it important to point out that the $m_{X}$ are not those of (13); at first sight they would seem to be the numbers more directly connected to the $d\left( \left[ X,\widehat{1}\right] \right) $. You might also want to replace \textquotedblleft random walk\textquotedblright% \ by \textquotedblleft random walk on the chambers of $\overline{S}% $\textquotedblright\ to get this point across again. \item \textbf{Page 43, Proposition 10:} Since the audience of this work is unlikely to consist entirely of combinatorialists, it is worth explaining that a \textquotedblleft cover of $X$\textquotedblright\ means an element $Y\in L$ such that $Y$ covers $X$. \item \textbf{Page 45, \S C.3:} The equality (46) holds only for $n>0$. \item \textbf{Page 45, \S C.3:} In the sentence \textquotedblleft For $J\subseteq\left[ n-1\right] $, let $f_{J}\left( L\right) $ be the number of flags in $L$ of type $J$, where the \textit{type} of a flag $X_{1}% 0$. \item \textbf{Page 46, proof of Proposition 11:} You say that $\left( -1\right) ^{n}h_{\left[ n-1\right] }\left( L\right) =\mu_{L}\left( \widehat{0},\widehat{1}\right) $ holds \textquotedblleft by (49)\textquotedblright. I do not know enough topology to understand this, but it might be worth pointing out that $\left( -1\right) ^{n}h_{\left[ n-1\right] }\left( L\right) =\mu_{L}\left( \widehat{0},\widehat{1}\right) $ also follows from (47) using Hall's formula for the M\"{o}bius function. \item \textbf{Page 49, reference [38]:} The reference needs volume and issue: Amer. J. Math. \textbf{57} (1935), no. 3, 509--533. \end{itemize} \end{document}