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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}

\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 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 (3)). 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.

\begin{vershort}
\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.
\end{vershort}

\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 three statements:
\par
\begin{quote}
\textit{Statement 1:} The set $S_{\geq x}$ is a subsemigroup of $S$ with
identity $x$.
\par
\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.
\par
\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)  $.
\par
\textit{Statement 4:} Let $a\in S_{\geq x}$ and $b\in S_{\geq x}$. Then,
$ab=a$ holds if and only if $\left(  \operatorname*{supp}\mid_{S_{\geq x}%
}\right)  b\leq\left(  \operatorname*{supp}\mid_{S_{\geq x}}\right)  a$.
\end{quote}
\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 $a$. This finishes the proof of Statement 1.
\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.
\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.
\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.
\par
Now, all four Statements are proven, and the proof is complete.}

\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 maximal elements
($\neq\widehat{1}$)\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{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 $A\left(  \left(  \dfrac{1}%
{z}\right)  \right)  $ (that is, of the ring of Laurent series in $\dfrac
{1}{z}$ over $A$). Later it becomes clear that $g\left(  z\right)  $ is indeed
a rational function in $z$ over $A$; but before this is shown, it is important
to understand whether $g\left(  z\right)  $ is viewed as a Laurent series in
$\dfrac{1}{z}$ (correct) or as a Laurent series in $z$ (incorrect).

In my opinion, it is also useful to stress that you regard $k\left(  z\right)
\otimes A$ as a subring of $A\left(  \left(  \dfrac{1}{z}\right)  \right)  $,
since $k\left(  z\right)  $ is canonically a subring of $k\left(  \left(
\dfrac{1}{z}\right)  \right)  $. This explains how exactly you can make sense
of the statement that $g\left(  z\right)  $ is a rational function (namely,
this statement means that $g\left(  z\right)  $ lies in the subring $k\left(
z\right)  \otimes A$ of $A\left(  \left(  \dfrac{1}{z}\right)  \right)  $).

\item \textbf{Page 24, proof of Proposition 2:} I think this proof is missing
a few steps. You leave the following statements unproven:

\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.

\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).

(Statement 1 is needed to prove the first sentence of Proposition 2. Statement
2 is needed to prove the second sentence.)

\textit{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.

\textit{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). 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 $g\left(  z\right)  \in k\left(  z\right)  \otimes A$). Hence, for every
$\mu\in k$, an element $\operatorname*{Res}\nolimits_{\mu}g\left(  z\right)  $
of $A$ 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
$\operatorname*{Res}\nolimits_{\mu}g\left(  z\right)  \neq0$), and the
elements $e_{i}$ of $A$ 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.

\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 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 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\ (twice).

\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 not for $X\in L$ but for $X\in\overline{L}$, so the correct
version of the claim \textquotedblleft$d\left(  \left[  \widehat{X},1\right]
\right)  =m_{X}$\textquotedblright\ should be \textquotedblleft$d\left(
\left[  \widehat{X},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}
$ (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 the first sight they would seem to be the numbers more directly
connected to the $d\left(  \left[  \widehat{X},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:} You should say 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:} Replace \textquotedblleft a flag $X_{1}%
<X_{2}<\cdots<X_{l}$\textquotedblright\ by \textquotedblleft a flag
$\widehat{0}<X_{1}<X_{2}<\cdots<X_{l}<\widehat{1}$\textquotedblright.

\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.
\end{itemize}


\end{document}