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\begin{document}

\begin{center}
\textbf{Algebras, Rings and Modules: Lie Algebras and Hopf Algebras}

\textit{Michiel Hazewinkel, Nadiya Gubareni, V. V. Kirichenko}

[\texttt{hazewinkel gubareni kirichenko - algebras rings and modules 3.pdf}]

\textbf{Darij's list of errata and comments}

\bigskip
\end{center}

\begin{itemize}
\item \textbf{Page 232:} \textquotedblleft counably\textquotedblright%
\ $\rightarrow$ \textquotedblleft countably\textquotedblright.

\item \textbf{Page 232:} \textquotedblleft Frobenious\textquotedblright%
\ $\rightarrow$ \textquotedblleft Frobenius\textquotedblright.

\item \textbf{Page 234, (6.1.10):} Replace \textquotedblleft$Q_{n-2}%
$\textquotedblright\ by \textquotedblleft$Q_{n-2}\left(  Z\right)
$\textquotedblright.

\item \textbf{Page 235:} In the second displayed equality on this page,
replace \textquotedblleft$\sum_{r+s=m}Z_{r}\otimes Z_{r}$\textquotedblright%
\ by \textquotedblleft$\sum_{r+s=m}Z_{r}\otimes Z_{s}$\textquotedblright.

\item \textbf{Page 237, (6.3.9):} Add an equality sign after \textquotedblleft%
$\beta\times_{osh}\gamma$\textquotedblright.

\item \textbf{Page 237:} It would be better to replace \textquotedblleft the
RHS of (6.3.8) has a contribution $1$ for each pair $\left(  \alpha
,\beta\otimes\gamma\right)  $ such that $\alpha$ is a column sum of a matrix
in $\mathcal{M}_{\beta,\gamma}$\textquotedblright\ by \textquotedblleft the
RHS of (6.3.8) has a contribution $1$ from each matrix in $\mathcal{M}%
_{\beta,\gamma}$ having column sum $\alpha$\textquotedblright.

\item \textbf{Page 238:} On the first line of this page, replace
\textquotedblleft$\left[  r_{1},...r_{m}\right]  $\textquotedblright\ by
\textquotedblleft$\left[  r_{1},...,r_{m}\right]  $\textquotedblright.
Similarly, on the second line of this page, replace \textquotedblleft$\left[
s_{1},...s_{m}\right]  $\textquotedblright\ by \textquotedblleft$\left[
s_{1},...,s_{m}\right]  $\textquotedblright.

\item \textbf{Page 241, Definition 6.5.3:} Replace \textquotedblleft%
$\alpha^{\prime\prime}=\left[  a_{i},a_{i+l},...a_{m}\right]  $%
\textquotedblright\ by \textquotedblleft$\alpha^{\prime\prime}=\left[
a_{i},a_{i+1},...,a_{m}\right]  $\textquotedblright.

\item \textbf{Page 241, Definition 6.5.3:} Replace \textquotedblleft A word is
\textbf{Lyndon} iff it is\textquotedblright\ by \textquotedblleft A word is
\textbf{Lyndon} if it is nonempty and\textquotedblright. (Otherwise, the empty
word would be Lyndon, which would break uniqueness of CFL factorization.)

\item \textbf{Page 242, Theorem 6.5.5:} The word \textquotedblleft
nonincreasing\textquotedblright\ is very confusing here: it suggests (falsely)
that each of the $\lambda_{1},\lambda_{2},...,\lambda_{s}$ itself is
nonincreasing, whereas it is supposed to mean that the sequence $\left(
\lambda_{1},\lambda_{2},...,\lambda_{s}\right)  $ is nonincreasing.

\item \textbf{Page 242, proof of Theorem 6.5.5:} Replace \textquotedblleft for
some prefix $\beta$\textquotedblright\ by \textquotedblleft for some nonempty
prefix $\beta$\textquotedblright. (Otherwise, $\lambda_{1}%
<_{\operatorname*{lex}}\beta$ would be false.)

\item \textbf{Page 243, footnote }$^{13}$\textbf{:} You don't prove the claim
of this footnote, but you later use it in the proof of Lemma 6.7.15.

\item \textbf{Page 243, proof of Theorem 6.5.8, Case A:} Replace
\textquotedblleft The element $a_{i,1}$\textquotedblright\ by
\textquotedblleft The element $a_{1,1}$\textquotedblright.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case A2:} I think the
condition for this case should be \textquotedblleft There exists some
$i\in\left\{  1,2,...,\min\left\{  n_{1},n_{2}\right\}  \right\}  $ such that
$a_{1,i}>a_{2,i}$ and $\left(  a_{1,j}=a_{2,j}\text{ for all }%
j=1,2,...,i-1\right)  $\textquotedblright\ rather than \textquotedblleft%
$\left[  a_{1,1},...,a_{1,n_{1}}\right]  >_{\operatorname*{lex}}\left[
a_{2,1},...,a_{2,n_{2}}\right]  $ and $n_{1}\leq n_{2}$\textquotedblright.
Otherwise, for example, the situation when $\alpha_{1}=132$ and $\alpha
_{2}=12$ does not fit in any of the Cases A1, A2 and A3.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case A3:} Replace
\textquotedblleft the equal prefixes $\left[  a_{1,1},...,,a_{1,n}\right]
=\left[  a_{2,1},...,,a_{2,n_{2}}\right]  $\textquotedblright\ by
\textquotedblleft the equal prefixes $\left[  a_{1,1},...,a_{1,n_{2}}\right]
=\left[  a_{2,1},...,a_{2,n_{2}}\right]  $\textquotedblright.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case A3:} In \textquotedblleft%
$\left[  b_{1},...,b_{s}\right]  >_{\operatorname*{lex}}\left[  a_{1,1}%
,...,a_{1,n}\right]  >_{\operatorname*{lex}}\left[  a_{2,1},...,a_{2,n_{2}%
}\right]  $\textquotedblright, replace \textquotedblleft$a_{1,n}%
$\textquotedblright\ by \textquotedblleft$a_{1,n_{1}}$\textquotedblright. But
you might want to remove this relation altogether, since I don't think you
ever use $\left[  b_{1},...,b_{s}\right]  >_{\operatorname*{lex}}\left[
a_{2,1},...,a_{2,n_{2}}\right]  $.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case A3:} Replace
\textquotedblleft the CFL factorization of $\left[  \beta_{1},...,\beta
_{s}\right]  $\textquotedblright\ by \textquotedblleft the CFL factorization
of $\left[  b_{1},...,b_{s}\right]  $\textquotedblright.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case A3:} I would replace
\textquotedblleft$\alpha>_{\operatorname*{lex}}\beta$\textquotedblright\ by
\textquotedblleft$\alpha\geq_{\operatorname*{lex}}\beta$\textquotedblright%
\ since the latter is easier to prove and just as fine.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case B:} In the second
displayed equation of Case B, replace \textquotedblleft$i_{2}$%
\textquotedblright\ by \textquotedblleft$i_{1}$\textquotedblright.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case B1:} I would replace
\textquotedblleft$\alpha>_{\operatorname*{lex}}\beta$\textquotedblright\ by
\textquotedblleft$\alpha\geq_{\operatorname*{lex}}\beta$\textquotedblright%
\ since the latter is easier to prove and just as fine.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case B3:} I would replace
\textquotedblleft$\alpha>_{\operatorname*{lex}}\beta$\textquotedblright\ by
\textquotedblleft$\alpha\geq_{\operatorname*{lex}}\beta$\textquotedblright%
\ since the latter is easier to prove and just as fine.

\item \textbf{Page 244, proof of Theorem 6.5.8, Case B3:} I have my doubts
about this argument. You write that \textquotedblleft relabel the latter
$a_{1,j}$ with the equal letters $a_{2,j}$, $j=1,...,i_{1}$\textquotedblright.
But can't it happen that, after the relabelling, the letters of $\alpha_{2}$
will be out of order in $\beta^{\prime\prime}$, which means $\beta
^{\prime\prime}$ might no longer be a shuffle of $\beta_{1},\beta
_{2},...,\beta_{t},\alpha_{2},...,\alpha_{m}$ ? For an example, let
$\alpha=12141213$, so that $m=2$, $\alpha_{1}=1214$ and $\alpha_{2}=1213$. Let
$\beta$ be the shuffle $a_{2,1}a_{2,2}a_{2,3}a_{1,1}a_{2,4}a_{1,2}%
a_{1,3}a_{1,4}=12113214$ of $\alpha_{1}$ and $\alpha_{2}$. Then, we are in
Case B3, with $i_{1}=3$ and $\left[  a_{1,1},a_{1,2},a_{1,3}\right]  =\left[
a_{2,1},a_{2,2},a_{2,3}\right]  =121$. But removing this prefix $\left[
a_{2,1},a_{2,2},a_{2,3}\right]  $ from $\beta$ leaves us with $a_{1,1}%
a_{2,4}a_{1,2}a_{1,3}a_{1,4}=13214$, which cannot be obtained as a shuffle of
$\left[  a_{1,4}\right]  $ with $\alpha_{2}$.

\item \textbf{Page 245, \S 6.5.17:} Replace \textquotedblleft$Shuffle\otimes
_{Z}\mathbf{Z}/\left(  2\right)  $\textquotedblright\ by \textquotedblleft%
$Shuffle\otimes_{\mathbf{Z}}\mathbf{Z}/\left(  2\right)  $\textquotedblright.

\item \textbf{Page 247, proof of Lemma 6.5.32:} Typo: \textquotedblleft the
the\textquotedblright.

\item \textbf{Page 248, proof of Construction and Lemma 6.5.33:} I don't see
how you conclude that $\beta>_{\operatorname*{lex}}\alpha^{\prime}$. In my
opinion, some more arguments are needed here.

\item \textbf{Page 249:} Remove the opening parenthesis at the beginning of
\textquotedblleft(from the left $\mathbf{Symm}$ of (6.6.1) to the right
$\mathbf{Symm}$\textquotedblright\ (since no matching closing parenthesis exists).

\item \textbf{Page 249, footnote }$^{19}$\textbf{:} Replace \textquotedblleft%
$h_{l}$\textquotedblright\ by \textquotedblleft$h_{1}$\textquotedblright.

\item \textbf{Page 250:} Replace \textquotedblleft a word as in
(6.7.2)\textquotedblright\ by \textquotedblleft a word as in
(6.7.3)\textquotedblright.

\item \textbf{Page 251, between (6.7.7) and (6.7.8):} Replace
\textquotedblleft$\Psi^{\prime}$\textquotedblright\ by \textquotedblleft%
$\Psi^{i}$\textquotedblright.

\item \textbf{Page 251, one line below (6.7.8):} Replace \textquotedblleft%
([\textbf{23}], p 49ff), and section 4.9 above.\textquotedblright\ by
\textquotedblleft([\textbf{23}], p 49ff, and section 4.9
above).\textquotedblright.

\item \textbf{Page 252, proof of Lemma 6.7.15:} Replace \textquotedblleft%
$\lambda_{n}\left(  \alpha\right)  $\textquotedblright\ by \textquotedblleft%
$\lambda^{n}\left(  \alpha\right)  $\textquotedblright\ (throughout the proof,
for different values of $n$).

\item \textbf{Page 253, proof of Theorem 6.7.5:} Replace \textquotedblleft%
$\beta=\left[  b_{1},b_{2},...,b_{n}\right]  $\textquotedblright\ by
\textquotedblleft$\beta=\left[  b_{1},b_{2},...,b_{m}\right]  $%
\textquotedblright\ (since the letter $n$ will soon be used for something
different). In the same vein, replace \textquotedblleft$\alpha=\beta
_{red}=\left[  g\left(  \beta\right)  ^{-1}b_{1},g\left(  \beta\right)
^{-1}b_{2},...,g\left(  \beta\right)  ^{-1}b_{n}\right]  $\textquotedblright%
\ by \textquotedblleft$\alpha=\beta_{\operatorname*{red}}=\left[  g\left(
\beta\right)  ^{-1}b_{1},g\left(  \beta\right)  ^{-1}b_{2},...,g\left(
\beta\right)  ^{-1}b_{m}\right]  $\textquotedblright.

\item \textbf{Page 253, proof of Theorem 6.7.5:} It would clarify things if
you replace \textquotedblleft and formula (6.7.18)\textquotedblright\ by
\textquotedblleft and the first equation of (6.7.18) (applied to $r_{1}$
instead of $n$) we have\textquotedblright.

\item \textbf{Page 255:} In the first displayed formula on page 255, I believe
the summation index \textquotedblleft$\operatorname*{wt}\left(  \alpha\right)
$\textquotedblright\ should be \textquotedblleft$\operatorname*{wt}\left(
\alpha\right)  =j$\textquotedblright.

\item \textbf{Page 255, Theorem 6.9.4 (ii):} Replace \textquotedblleft%
$\varphi_{\ast}\left(  h\left(  t\right)  \right)  $\textquotedblright\ by
\textquotedblleft$\varphi_{\ast}\left(  Z\left(  t\right)  \right)
$\textquotedblright.

\item \textbf{Page 256, proof of Theorem 6.9.4:} Remove the period after
\textquotedblleft$\mathbf{NSymm}$\textquotedblright.

\item \textbf{Page 257, (6.9.12):} Replace \textquotedblleft$\mathbf{v}%
_{n}\left(  ?\mathbf{Symm}\right)  $\textquotedblright\ by \textquotedblleft%
$\mathbf{v}_{n}\left(  ?\mathbf{Symm}_{k}\right)  $\textquotedblright.

\item \textbf{Page 258:} Remove the word \textquotedblleft a\textquotedblright%
\ in \textquotedblleft Now there exist a natural lifts\textquotedblright.

\item \textbf{Page 258, (6.9.15):} Replace \textquotedblleft$\mathbf{f}%
_{n}^{\mathbf{QSymm}}$\textquotedblright\ by \textquotedblleft$\mathbf{f}%
_{p}^{\mathbf{QSymm}}$\textquotedblright.

\item \textbf{Page 258, (6.9.15):} Replace the equality sign by an
\textquotedblleft$\equiv$\textquotedblright\ sign.

\item \textbf{Page 259, (6.9.23):} Replace the \textquotedblleft$\mathbf{Z}%
$\textquotedblright\ by a \textquotedblleft$\mathbf{Q}$\textquotedblright.

\item \textbf{Page 259, footnote }$^{26}$\textbf{:} Here you conjecture that
\textquotedblleft The corresponding ideals in $\mathbf{NSymm}$ are most likely
the iterated commutator ideals\textquotedblright. If by \textquotedblleft the
corresponding ideals\textquotedblright\ you mean the orthogonal spaces
$\left(  F_{i}\left(  \mathbf{QSymm}\right)  \right)  ^{\perp}$ of
$\mathbf{QSymm}$ with respect to the bilinear form, then I think the
conjecture is false. It is true that $\left(  \underbrace{F_{1}\left(
\mathbf{QSymm}\right)  }_{=\mathbf{Symm}}\right)  ^{\perp}=\mathbf{Symm}%
^{\perp}=\left[  \mathbf{NSymm},\mathbf{NSymm}\right]  $, but it is not true
that $\left(  F_{2}\left(  \mathbf{QSymm}\right)  \right)  ^{\perp}=\left[
\mathbf{NSymm},\left[  \mathbf{NSymm},\mathbf{NSymm}\right]  \right]  $. For
example, $M_{12}^{2}=e_{1}\left(  M_{12}\right)  \cdot e_{1}\left(
M_{12}\right)  $ (I write $M_{\alpha}$ for the monomial quasisymmetric
function you call $\alpha$) has scalar product $2$ with $\left[  H_{2},\left[
H_{3},H_{1}\right]  \right]  $, as the following sage code shows:

\texttt{sage: QSym = QuasiSymmetricFunctions(QQ)}

\texttt{sage: M = QSym.M()}

\texttt{sage: NSym = NonCommutativeSymmetricFunctions(QQ)}

\texttt{sage: S = NSym.S()}

\texttt{sage: print
(M[1,2]**2).duality\_pairing(S[2,3,1]-S[2,1,3]-S[3,1,2]+S[1,3,2])}

\texttt{2}

\item \textbf{Page 260, Theorem 6.9.27 (i):} Replace \textquotedblleft%
$\mathbf{v}$\textquotedblright\ by \textquotedblleft$\mathbf{v}_{n}%
$\textquotedblright.

\item \textbf{Page 260, Theorem 6.9.27 (i):} Replace \textquotedblleft$\left[
a_{1},...a_{m}\right]  $\textquotedblright\ by \textquotedblleft$\left[
a_{1},...,a_{m}\right]  $\textquotedblright. Similarly, replace
\textquotedblleft$\left[  n^{-1}a_{1},...n^{-1}a_{m}\right]  $%
\textquotedblright\ by \textquotedblleft$\left[  n^{-1}a_{1},...,n^{-1}%
a_{m}\right]  $\textquotedblright.

\item \textbf{Page 260, Theorem 6.9.27 (iv):} Replace the equality sign by an
\textquotedblleft$\equiv$\textquotedblright\ sign.
\end{itemize}


\end{document}