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\ihead{Errata to ``Semigroups, rings, ...''}
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\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 $X_{1}%
<X_{2}<\cdots<X_{l}$ is the set $\left\{  \operatorname*{rank}X_{i}\right\}
_{1\leq i\leq l}$\textquotedblright, the word \textquotedblleft
flag\textquotedblright\ should probably be replaced by \textquotedblleft
chain\textquotedblright\ both times. (At least, the word \textquotedblleft
chain\textquotedblright\ is more standard for this usage.)

\item \textbf{Page 46, Proposition 11:} Again, (50) holds only for $n>0$.

\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}