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\begin{document}

\title{Cohn's theorem on the kernel of the Dynkin operator}
\author{GPT-5.5, edited by Darij Grinberg}
\date{\today}
\maketitle

\begin{abstract}
\textbf{Abstract.} In his 1951 thesis, P. M. Cohn described the kernel of the
Dynkin operator on the tensor algebra. This note gives a modern exposition of
this result with its proof, both in the case of a free module and in the case
of an arbitrary module over a $\mathbb{Q}$-algebra. (In the fully general
case, a counterexample is given.) No novelty is claimed. \medskip

\textbf{Manifest.} This note has been written by GPT-5.5 based on an excerpt
of Cohn's thesis and some of my own work; it has then been substantially
edited by myself. I thank Christophe Reutenauer for sending me the relevant
part of Cohn's thesis. -- DG\footnote{This work is in the public domain.}

\end{abstract}

Let $\mathbf{k}$ be a commutative ring, and let $V$ be a $\mathbf{k}$-module.
The tensor algebra of $V$ is
\[
T\left(  V\right)  :=\bigoplus_{n\in\mathbb{N}}V^{\otimes n},
\]
and its positive part is its ideal%
\[
T\left(  V\right)  ^{+}:=\bigoplus_{n\geq1}V^{\otimes n}.
\]
We write multiplication in $T\left(  V\right)  $ by juxtaposition, and we use
the notation%
\[
\left[  a,b\right]  :=ab-ba
\]
for its commutator bracket.

The \emph{Dynkin operator} is the $\mathbf{k}$-linear map
\begin{align*}
D:T\left(  V\right)  ^{+}  &  \longrightarrow T\left(  V\right)  ^{+},\\
v_{1}v_{2}\cdots v_{n}  &  \longmapsto\left[  \left[  \cdots\left[  \left[
v_{1},v_{2}\right]  ,v_{3}\right]  ,\ldots\right]  ,v_{n}\right]
\ \ \ \ \ \ \ \ \ \ \text{for }v_{1},v_{2},\ldots,v_{n}\in V.
\end{align*}
This is recursively defined by%
\begin{align}
D(v)  &  =v\ \ \ \ \ \ \ \ \ \ \text{for all }v\in
V,\ \ \ \ \ \ \ \ \ \ \text{and}\label{eq.D.base}\\
D\left(  av\right)   &  =\left[  D\left(  a\right)  ,v\right]
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in T\left(  V\right)  ^{+}\text{ and
}v\in V. \label{eq.D.rec}%
\end{align}


Let $Z$ be the right ideal of $T\left(  V\right)  $ generated by the elements
$xD\left(  x\right)  $ with $x\in T\left(  V\right)  ^{+}$. In other words,
\begin{equation}
Z:=\sum_{x\in T\left(  V\right)  ^{+}}xD\left(  x\right)  T\left(  V\right)  .
\label{eq.intro.Z}%
\end{equation}
The main result is the following theorem of P.~M.~Cohn. It appears, in a
somewhat compressed form, as Theorem 10.1 of his Ph.D. thesis
\cite[pp.~88--89]{Cohn51}.

\begin{theorem}
[Cohn's kernel theorem]\label{thm.main} We always have
\begin{equation}
Z\subseteq\operatorname{Ker} D. \label{eq.main.easy-inclusion}%
\end{equation}
Moreover, equality holds in each of the following two cases:

\begin{enumerate}
\item The $\mathbf{k}$-module $V$ is free.

\item The ring $\mathbf{k}$ is a $\mathbb{Q}$-algebra.
\end{enumerate}
\end{theorem}

Here and below, $\operatorname{Ker}D$ means the kernel of the map $D$; thus,
it is a submodule of $T\left(  V\right)  ^{+}$.

The proof below follows Cohn's three auxiliary maps, but replaces the last
induction in his proof by a short ideal argument. We shall also prove two
additional facts. First, Cohn's proposed homogeneous spanning description of
$Z$ is valid without any hypothesis on $\mathbf{k}$ or $V$. Second, the
equality $Z=\operatorname{Ker} D$ is false for arbitrary modules; the
obstruction is already visible in degree $2$.

\section{Two identities for the Dynkin operator}

We begin with an identity that Cohn attributes to Baker. We include a proof,
since this identity drives the whole argument.

\begin{lemma}
[Baker's identity]\label{lem.Baker} Let $u,v\in T\left(  V\right)  ^{+}$.
Then,
\begin{equation}
D\left(  uD\left(  v\right)  \right)  =\left[  D\left(  u\right)  ,D\left(
v\right)  \right]  . \label{eq.Baker}%
\end{equation}

\end{lemma}

\begin{proof}
By $\mathbf{k}$-linearity, it is enough to prove \eqref{eq.Baker} when $u$ and
$v$ are pure tensors. We induct on the degree of $v$.

If $v\in V$, then $D\left(  v\right)  =v$, and the recursion (\ref{eq.D.rec})
yields $D\left(  uv\right)  =\left[  D\left(  u\right)  ,v\right]  $. In view
of $D\left(  v\right)  =v$, we can rewrite this as $D\left(  uD\left(
v\right)  \right)  =\left[  D\left(  u\right)  ,D\left(  v\right)  \right]  $.
Thus, \eqref{eq.Baker} is proved in the base case $v\in V$.

Now, assume that $v=v^{\prime}a$ for some pure tensor $v^{\prime}\in T\left(
V\right)  ^{+}$ and some $a\in V$. Then, $v^{\prime}$ has lower degree than
$v$, so our induction hypothesis yields%
\[
D\left(  uD\left(  v^{\prime}\right)  \right)  =\left[  D\left(  u\right)
,D\left(  v^{\prime}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ D\left(  uaD\left(  v^{\prime}\right)  \right)  =\left[
D\left(  ua\right)  ,D\left(  v^{\prime}\right)  \right]  .
\]
But $v=v^{\prime}a$, so that
\begin{align*}
D\left(  v\right)   &  =D\left(  v^{\prime}a\right)  =\left[  D\left(
v^{\prime}\right)  ,a\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.D.rec})}\right) \\
&  =D\left(  v^{\prime}\right)  a-aD\left(  v^{\prime}\right)  .
\end{align*}
Hence,
\begin{align*}
D\left(  uD\left(  v\right)  \right)   &  =D\left(  u\left(  D\left(
v^{\prime}\right)  a-aD\left(  v^{\prime}\right)  \right)  \right)
=\underbrace{D\left(  uD\left(  v^{\prime}\right)  a\right)  }%
_{\substack{=\left[  D\left(  uD\left(  v^{\prime}\right)  \right)  ,a\right]
\\\text{(by (\ref{eq.D.rec}))}}}-\underbrace{D\left(  uaD\left(  v^{\prime
}\right)  \right)  }_{\substack{=\left[  D\left(  ua\right)  ,D\left(
v^{\prime}\right)  \right]  }}\\
&  =\left[  \underbrace{D\left(  uD\left(  v^{\prime}\right)  \right)
}_{\substack{=\left[  D\left(  u\right)  ,D\left(  v^{\prime}\right)  \right]
}},a\right]  -\left[  \underbrace{D\left(  ua\right)  }_{\substack{=\left[
D\left(  u\right)  ,a\right]  \\\text{(by (\ref{eq.D.rec}))}}},D\left(
v^{\prime}\right)  \right] \\
&  =\left[  \left[  D\left(  u\right)  ,D\left(  v^{\prime}\right)  \right]
,a\right]  -\left[  \left[  D\left(  u\right)  ,a\right]  ,D\left(  v^{\prime
}\right)  \right]  =\left[  D\left(  u\right)  ,\underbrace{\left[  D\left(
v^{\prime}\right)  ,a\right]  }_{=D\left(  v\right)  }\right] \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the Jacobi identity for commutators in the}\\
\text{form }\left[  \left[  x,y\right]  ,z\right]  -\left[  \left[
x,z\right]  ,y\right]  =\left[  x,\left[  y,z\right]  \right]
\end{array}
\right) \\
&  =\left[  D\left(  u\right)  ,D\left(  v\right)  \right]  .
\end{align*}
This completes the induction.
\end{proof}

\begin{corollary}
\label{cor.ker-right-ideal} The submodule $\operatorname{Ker} D$ is a right
ideal of $T\left(  V\right)  $. Moreover, $Z\subseteq\operatorname{Ker} D$.
\end{corollary}

\begin{proof}
Let $g\in\operatorname{Ker}D$, and let $a\in V$. Then, (\ref{eq.D.rec}) yields%
\[
D\left(  ga\right)  =\left[  D\left(  g\right)  ,a\right]
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }D\left(  g\right)  =0\right)  .
\]
An induction on the degree of a pure tensor $w\in T\left(  V\right)  $ now
shows that $D\left(  gw\right)  =0$ for each pure tensor $w\in T\left(
V\right)  $. By linearity, this must therefore hold for every $w\in T\left(
V\right)  $. Hence, $\operatorname{Ker}D$ is a right ideal.

For each $x\in T\left(  V\right)  ^{+}$, Lemma \ref{lem.Baker} gives
\[
D\left(  xD\left(  x\right)  \right)  =\left[  D\left(  x\right)  ,D\left(
x\right)  \right]  =0.
\]
Thus, every generator $xD\left(  x\right)  $ of $Z$ belongs to
$\operatorname{Ker} D$. Since $\operatorname{Ker} D$ is a right ideal, we
obtain $Z\subseteq\operatorname{Ker} D$.
\end{proof}

We shall also need the usual Dynkin--Specht--Wever identity.

\begin{lemma}
[Dynkin--Specht--Wever identity]\label{lem.DSW} Let $n\geq1$. Then,%
\begin{equation}
D^{2}=nD\ \ \ \ \ \ \ \ \ \ \text{on }V^{\otimes n}. \label{eq.DSW}%
\end{equation}

\end{lemma}

\begin{proof}
We induct on $n$. The claim is clear for $n=1$. Let $n>1$, and let $u\in
V^{\otimes\left(  n-1\right)  }$ and $a\in V$. Then, $D^{2}\left(  u\right)
=\left(  n-1\right)  D\left(  u\right)  $ by the induction hypothesis. But
(\ref{eq.D.rec}) yields $D\left(  ua\right)  =\left[  D\left(  u\right)
,a\right]  =D\left(  u\right)  a-aD\left(  u\right)  $. Applying $D$ to this
equality, we obtain%
\begin{align*}
D^{2}\left(  ua\right)   &  =D\left(  D\left(  u\right)  a-aD\left(  u\right)
\right)  =\underbrace{D\left(  D\left(  u\right)  a\right)  }%
_{\substack{=\left[  D\left(  D\left(  u\right)  \right)  ,a\right]
\\\text{(by (\ref{eq.D.rec}))}}}-\underbrace{D\left(  aD\left(  u\right)
\right)  }_{\substack{=\left[  D\left(  a\right)  ,D\left(  u\right)  \right]
\\\text{(by Lemma \ref{lem.Baker})}}}\\
&  =\left[  \underbrace{D\left(  D\left(  u\right)  \right)  }_{=D^{2}\left(
u\right)  =\left(  n-1\right)  D\left(  u\right)  },a\right]  -\left[
\underbrace{D\left(  a\right)  }_{=a},D\left(  u\right)  \right]
=\underbrace{\left[  \left(  n-1\right)  D\left(  u\right)  ,a\right]
}_{=\left(  n-1\right)  \left[  D\left(  u\right)  ,a\right]  }%
-\underbrace{\left[  a,D\left(  u\right)  \right]  }_{=-\left[  D\left(
u\right)  ,a\right]  }\\
&  =\left(  n-1\right)  \left[  D\left(  u\right)  ,a\right]  +\left[
D\left(  u\right)  ,a\right]  =n\underbrace{\left[  D\left(  u\right)
,a\right]  }_{=D\left(  ua\right)  }=nD\left(  ua\right)  .
\end{align*}
This proves that $D^{2}=nD$ holds on each pure tensor $ua\in V^{\otimes n}$
with $u\in V^{\otimes\left(  n-1\right)  }$ and $a\in V$. By linearity, it
follows that $D^{2}=nD$ on all of $V^{\otimes n}$. Thus the induction is complete.
\end{proof}

\begin{remark}
\label{rem.Q-projection}Let $n\geq1$. On $V^{\otimes n}$, the map $\frac{1}%
{n}D$ is an idempotent whenever $n$ is invertible in $\mathbf{k}$. Indeed,
Lemma \ref{lem.DSW} gives
\[
\left(  \frac{1}{n}D\right)  ^{2}=\frac{1}{n}D\ \ \ \ \ \ \ \ \ \ \text{on
}V^{\otimes n}.
\]
When $\mathbf{k}$ is a $\mathbb{Q}$-algebra, the endomorphism of $T\left(
V\right)  ^{+}$ that is given by $\frac{1}{n}D$ on each $V^{\otimes n}$ is
known as the \emph{Dynkin idempotent}. (Usually it is extended to $T\left(
V\right)  $ by letting it act as $0$ on the $0$-th degree component.)
\end{remark}

\section{The free-module case}

Throughout this section, assume that $V$ is free as a $\mathbf{k}$-module. Fix
a basis $X$ of $V$. Let $\mathcal{M}$ be the free magmatic $\mathbf{k}%
$-algebra\footnote{A \emph{magmatic }$\mathbf{k}$\emph{-algebra} means a
$\mathbf{k}$-module $A$ equipped with a $\mathbf{k}$-bilinear operation
$A\times A\rightarrow A$ that is written as multiplication, but is not
required to satisfy any further axioms; i.e., it is a nonunital nonassociative
$\mathbf{k}$-algebra.} on symbols $\overline{x}$ indexed by $x\in X$. Thus,
$\mathcal{M}$ is the free $\mathbf{k}$-module on all nonassociative (fully
parenthesized) words in the symbols $\overline{x}$. Combinatorialists will
recognize the latter words as plane binary trees with their leaves colored by
the elements of $X$. For instance, if $x,y,z$ are three elements of $X$, then
$\left(  \overline{x}\left(  \overline{y}\overline{x}\right)  \right)  \left(
\overline{x}\overline{z}\right)  $ is such a word, corresponding to the binary
tree
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We use the following standard description of the free Lie algebra. This is the
only structural result about free Lie algebras that will enter our proof.

\begin{lemma}
[The defining relations of a free Lie algebra]\label{lem.free-Lie-relations}
Define a $\mathbf{k}$-linear map
\[
\lambda:\mathcal{M}\longrightarrow T\left(  V\right)  ^{+}%
\]
recursively by%
\begin{align*}
\lambda\left(  \overline{x}\right)   &  :=x\ \ \ \ \ \ \ \ \ \ \text{for each
}x\in X,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\lambda\left(  pq\right)   &  :=\left[  \lambda\left(  p\right)
,\lambda\left(  q\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for all
nonassociative words }p,q.
\end{align*}
(Thus, $\lambda$ transforms products in $\mathcal{M}$ into commutators in
$T\left(  V\right)  ^{+}$. For instance, $\lambda\left(  \overline{x}\left(
\overline{y}\overline{z}\right)  \right)  =\left[  x,\left[  y,z\right]
\right]  $ for any $x,y,z\in X$.) Then, $\operatorname{Ker}\lambda$ is the
two-sided ideal of the nonassociative algebra $\mathcal{M}$ generated by the
elements
\begin{equation}
p^{2}\qquad\text{and}\qquad\left(  pq\right)  r+\left(  qr\right)  p+\left(
rp\right)  q \label{eq.free.Lie-relators}%
\end{equation}
with $p,q,r\in\mathcal{M}$.
\end{lemma}

\begin{proof}
The quotient of $\mathcal{M}$ by the ideal generated by
\eqref{eq.free.Lie-relators} is the free Lie $\mathbf{k}$-algebra on $X$ (by
abstract nonsense: quotienting out \eqref{eq.free.Lie-relators} imposes
precisely the axioms of a Lie algebra on the quotient). Therefore, the
standard embedding theorem for a free Lie algebra (see, e.g., \cite[Theorem
0.5]{Reutenauer93}, \cite[\S II.3.1, Theorem 1 (a)]{Bou89}, \cite[Chapter IV,
Theorem 4.2 part 2]{Serre06}, \cite[Hauptsatz 4.10 (4)]{Laue13} or
\cite[Remark 6.1.42]{GR26}) identifies this quotient with the Lie subalgebra
of $T\left(  V\right)  $ generated by $V$. Thus, the projection of
$\mathcal{M}$ onto this quotient can be identified with the map $\lambda
:\mathcal{M}\rightarrow T\left(  V\right)  ^{+}$. Consequently, the ideal
generated by \eqref{eq.free.Lie-relators} (being the kernel of this
projection) is precisely $\operatorname{Ker}\lambda$. See, for example,
\cite[\S \S 9--10]{Cohn51} or \cite[Chapter~0]{Reutenauer93}.
\end{proof}

We now introduce Cohn's other two maps. Define a $\mathbf{k}$-linear map
\begin{align*}
\delta:T\left(  V\right)  ^{+}  &  \longrightarrow\mathcal{M},\\
x_{1}x_{2}\cdots x_{n}  &  \longmapsto\left(  \cdots\left(  \left(
\overline{x_{1}}\overline{x_{2}}\right)  \overline{x_{3}}\right)
\cdots\overline{x_{n}}\right)  \ \ \ \ \ \ \ \ \ \ \text{for }x_{1}%
,x_{2},\ldots,x_{n}\in X.
\end{align*}
Thus, $\delta$ puts a basis word into completely left-associated form. (Of
course, this is again shorthand for a recursive definition: $\delta\left(
x\right)  =\overline{x}$ for any single letter $x\in X$, and $\delta\left(
ax\right)  =\delta\left(  a\right)  \overline{x}$ for any $a\in T\left(
V\right)  ^{+}$ and $x\in X$.)

Define another $\mathbf{k}$-linear map
\begin{align*}
\gamma:\mathcal{M}  &  \longrightarrow T\left(  V\right)  ^{+}%
\ \ \ \ \ \ \ \ \ \ \text{by}\\
\gamma\left(  \overline{x}\right)   &  :=x\ \ \ \ \ \ \ \ \ \ \text{for all
}x\in X,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\gamma\left(  pq\right)   &  :=\gamma\left(  p\right)  \lambda\left(
q\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }p,q\in\mathcal{M}%
\end{align*}
(where the product in $\gamma\left(  p\right)  \lambda\left(  q\right)  $ is
the associative product in $T\left(  V\right)  $). (Strictly speaking, the
equality $\gamma\left(  pq\right)  =\gamma\left(  p\right)  \lambda\left(
q\right)  $ is imposed only on nonassociative words $p$ and $q$, since each
nonassociative word of length $>1$ can be uniquely factored as $pq$ for two
nonassociative words $p$ and $q$. But then, by linearity, the same equality
holds for all $p,q\in\mathcal{M}$.) For instance, for any $x,y,z\in X$, we
have%
\begin{align*}
\gamma\left(  \overline{x}\overline{y}\right)   &  =\underbrace{\gamma\left(
\overline{x}\right)  }_{=x}\underbrace{\lambda\left(  \overline{y}\right)
}_{=y}=xy;\\
\gamma\left(  \left(  \overline{x}\overline{y}\right)  \overline{z}\right)
&  =\underbrace{\gamma\left(  \overline{x}\overline{y}\right)  }%
_{=xy}\underbrace{\lambda\left(  \overline{z}\right)  }_{=z}=xyz;\\
\gamma\left(  \overline{x}\left(  \overline{y}\overline{z}\right)  \right)
&  =\underbrace{\gamma\left(  \overline{x}\right)  }_{=x}\underbrace{\lambda
\left(  \overline{y}\overline{z}\right)  }_{=\left[  y,z\right]  }=x\left[
y,z\right]  .
\end{align*}


\begin{lemma}
\label{lem.three-maps} The maps $\delta$, $\lambda$, and $\gamma$ satisfy
\begin{align}
\gamma\circ\delta &  =\operatorname{id}_{T\left(  V\right)  ^{+}%
},\label{eq.three.gamma-delta}\\
\lambda\circ\delta &  =D,\label{eq.three.lambda-delta}\\
D\circ\gamma &  =\lambda. \label{eq.three.D-gamma}%
\end{align}

\end{lemma}

\begin{proof}
Equation \eqref{eq.three.gamma-delta} follows immediately by induction on the
length of a basis word. Equation \eqref{eq.three.lambda-delta} is simply the
definition of $D$.

It remains to prove \eqref{eq.three.D-gamma}. By linearity, it suffices to
prove that $D\left(  \gamma\left(  w\right)  \right)  =\lambda\left(
w\right)  $ for any nonassociative word $w$ in $\mathcal{M}$. We induct on the
length (or the structure) of $w$. The claim is clear if $w$ is a generator
$\overline{x}$. Let $p$ and $q$ be nonassociative words, and assume that the
claim holds for $w=p$ and $w=q$. We must prove that it also holds for $w=pq$.
The induction hypothesis yields $D\left(  \gamma\left(  p\right)  \right)
=\lambda\left(  p\right)  $ and $D\left(  \gamma\left(  q\right)  \right)
=\lambda\left(  q\right)  $. Now,%
\begin{align*}
D\left(  \gamma\left(  pq\right)  \right)   &  =D\left(  \gamma\left(
p\right)  \lambda\left(  q\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the recursive definition of }\gamma\right) \\
&  =D\left(  \gamma\left(  p\right)  D\left(  \gamma\left(  q\right)  \right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda\left(  q\right)
=D\left(  \gamma\left(  q\right)  \right)  \right) \\
&  =\left[  D\left(  \gamma\left(  p\right)  \right)  ,D\left(  \gamma\left(
q\right)  \right)  \right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.Baker}}\right) \\
&  =\left[  \lambda\left(  p\right)  ,\lambda\left(  q\right)  \right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }D\left(  \gamma\left(  p\right)
\right)  =\lambda\left(  p\right)  \text{ and }D\left(  \gamma\left(
q\right)  \right)  =\lambda\left(  q\right)  \right) \\
&  =\lambda\left(  pq\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
recursive definition of }\lambda\right)  .
\end{align*}
This completes the induction, and thus proves \eqref{eq.three.D-gamma}.
\end{proof}

For $a,b\in T\left(  V\right)  ^{+}$, set
\begin{equation}
P\left(  a,b\right)  :=aD\left(  b\right)  +bD\left(  a\right)  .
\label{eq.free.P}%
\end{equation}
Linearity of $D$ yields%
\begin{align*}
\left(  a+b\right)  D\left(  a+b\right)   &  =\left(  a+b\right)  \left(
D\left(  a\right)  +D\left(  b\right)  \right)  =aD\left(  a\right)
+\underbrace{aD\left(  b\right)  +bD\left(  a\right)  }_{=P\left(  a,b\right)
}+\,bD\left(  b\right) \\
&  =aD\left(  a\right)  +P\left(  a,b\right)  +bD\left(  b\right)  ,
\end{align*}
so that
\begin{equation}
P\left(  a,b\right)  =\underbrace{\left(  a+b\right)  D\left(  a+b\right)
}_{\in Z}-\underbrace{aD\left(  a\right)  }_{\in Z}-\underbrace{bD\left(
b\right)  }_{\in Z}\in Z. \label{eq.free.P-in-Z}%
\end{equation}
This simple computation (which did not use the freeness of $V$) will be used
several times in what follows.

\begin{lemma}
\label{lem.gamma-kernel-in-Z} We have
\begin{equation}
\gamma\left(  \operatorname{Ker}\lambda\right)  \subseteq Z.
\label{eq.gamma-kernel-in-Z}%
\end{equation}

\end{lemma}

\begin{proof}
Let
\begin{equation}
\mathcal{J}:=\left\{  p\in\mathcal{M}\mid\gamma\left(  p\right)  \in
Z\right\}  . \label{eq.free.J}%
\end{equation}
Clearly, $\mathcal{J}$ is a $\mathbf{k}$-submodule of $\mathcal{M}$ (since
$\gamma$ is $\mathbf{k}$-linear, and $Z$ is a $\mathbf{k}$-module). We shall
now show that $\mathcal{J}$ is a two-sided ideal of $\mathcal{M}$.

Let $p,q\in\mathcal{M}$.

If $p\in\mathcal{J}$, then $\gamma\left(  p\right)  \in Z$ and
\[
\gamma\left(  pq\right)  =\gamma\left(  p\right)  \lambda\left(  q\right)  \in
Z,
\]
since $Z$ is a right ideal. Thus, $pq\in\mathcal{J}$.

If $q\in\mathcal{J}$, then $\gamma\left(  q\right)  \in Z\subseteq
\operatorname{Ker}D$ by Corollary \ref{cor.ker-right-ideal}. Hence, $D\left(
\gamma\left(  q\right)  \right)  =0$, so that \eqref{eq.three.D-gamma} gives
$\lambda\left(  q\right)  =D\left(  \gamma\left(  q\right)  \right)  =0$, and
therefore
\[
\gamma\left(  pq\right)  =\gamma\left(  p\right)  \lambda\left(  q\right)
=0\in Z.
\]
Thus, $pq\in\mathcal{J}$ again. So we have proved that $pq\in\mathcal{J}$
holds both if $p\in\mathcal{J}$ and if $q\in\mathcal{J}$. This proves that
$\mathcal{J}$ is a two-sided ideal.

We next show that $\mathcal{J}$ contains the generators in
\eqref{eq.free.Lie-relators}. For $p\in\mathcal{M}$, the recursive definition
of $\gamma$ and the equation \eqref{eq.three.D-gamma} give%
\[
\gamma\left(  p^{2}\right)  =\gamma\left(  p\right)  \lambda\left(  p\right)
=\gamma\left(  p\right)  D\left(  \gamma\left(  p\right)  \right)  \in Z.
\]
Hence, $p^{2}\in\mathcal{J}$.

Now, let $p,q,r\in\mathcal{M}$, and abbreviate
\[
a:=\gamma\left(  p\right)  ,\qquad b:=\gamma\left(  q\right)  ,\qquad
c:=\gamma\left(  r\right)  .
\]
Then, $D\left(  a\right)  =D\left(  \gamma\left(  p\right)  \right)
=\lambda\left(  p\right)  $ by \eqref{eq.three.D-gamma}, and similarly
$D\left(  b\right)  =\lambda\left(  q\right)  $ and $D\left(  c\right)
=\lambda\left(  r\right)  $. Using the recursive definition of $\gamma$ and
the equation \eqref{eq.three.D-gamma}, we now have%
\[
\gamma\left(  \left(  pq\right)  r\right)  =\underbrace{\gamma\left(
pq\right)  }_{=\gamma\left(  p\right)  \lambda\left(  q\right)  }%
\lambda\left(  r\right)  =\underbrace{\gamma\left(  p\right)  }_{=a}%
\underbrace{\lambda\left(  q\right)  }_{=D\left(  b\right)  }%
\underbrace{\lambda\left(  r\right)  }_{=D\left(  c\right)  }=aD\left(
b\right)  D\left(  c\right)
\]
and similarly%
\begin{align*}
\gamma\left(  \left(  qr\right)  p\right)   &  =bD\left(  c\right)  D\left(
a\right)  \qquad\text{and}\\
\gamma\left(  \left(  rp\right)  q\right)   &  =cD\left(  a\right)  D\left(
b\right)  .
\end{align*}
Adding these three equalities together, we obtain
\begin{align}
&  \gamma\left(  \left(  pq\right)  r+\left(  qr\right)  p+\left(  rp\right)
q\right) \nonumber\\
&  =aD\left(  b\right)  D\left(  c\right)  +bD\left(  c\right)  D\left(
a\right)  +cD\left(  a\right)  D\left(  b\right)  .
\label{eq.free.gamma-Jacobi-1}%
\end{align}
On the other hand, Lemma \ref{lem.Baker} gives
\begin{equation}
D\left(  cD\left(  a\right)  \right)  =\left[  D\left(  c\right)  ,D\left(
a\right)  \right]  . \label{eq.free.Baker-ca}%
\end{equation}
Consequently, \eqref{eq.free.gamma-Jacobi-1} becomes%
\begin{align*}
&  \gamma\left(  \left(  pq\right)  r+\left(  qr\right)  p+\left(  rp\right)
q\right) \\
&  =\underbrace{aD\left(  b\right)  }_{\substack{=P\left(  a,b\right)
-bD\left(  a\right)  \\\text{(since }P\left(  a,b\right)  =aD\left(  b\right)
+bD\left(  a\right)  \text{)}}}D\left(  c\right)  +bD\left(  c\right)
D\left(  a\right)  +\underbrace{cD\left(  a\right)  D\left(  b\right)
}_{\substack{=P\left(  b,cD\left(  a\right)  \right)  -bD\left(  cD\left(
a\right)  \right)  \\\text{(since }P\left(  b,cD\left(  a\right)  \right)
=bD\left(  cD\left(  a\right)  \right)  +cD\left(  a\right)  D\left(
b\right)  \text{)}}}\\
&  =\left(  P\left(  a,b\right)  -bD\left(  a\right)  \right)  D\left(
c\right)  +bD\left(  c\right)  D\left(  a\right)  +P\left(  b,cD\left(
a\right)  \right)  -bD\left(  cD\left(  a\right)  \right) \\
&  =P\left(  a,b\right)  D\left(  c\right)  +\underbrace{bD\left(  c\right)
D\left(  a\right)  -bD\left(  a\right)  D\left(  c\right)  }_{=b\left[
D\left(  c\right)  ,D\left(  a\right)  \right]  }+\,P\left(  b,cD\left(
a\right)  \right)  -b\underbrace{D\left(  cD\left(  a\right)  \right)
}_{\substack{=\left[  D\left(  c\right)  ,D\left(  a\right)  \right]
\\\text{(by (\ref{eq.free.Baker-ca}))}}}\\
&  =P\left(  a,b\right)  D\left(  c\right)  +b\left[  D\left(  c\right)
,D\left(  a\right)  \right]  +P\left(  b,cD\left(  a\right)  \right)
-b\left[  D\left(  c\right)  ,D\left(  a\right)  \right] \\
&  =\underbrace{P\left(  a,b\right)  D\left(  c\right)  }_{\substack{\in
Z\\\text{(since (\ref{eq.free.P-in-Z}) yields }P\left(  a,b\right)  \in
Z\text{,}\\\text{and since }Z\text{ is a right ideal)}}}+\underbrace{P\left(
b,cD\left(  a\right)  \right)  }_{\substack{\in Z\\\text{(by
(\ref{eq.free.P-in-Z}))}}}\\
&  \in Z.
\end{align*}
Thus, $\left(  pq\right)  r+\left(  qr\right)  p+\left(  rp\right)  q$ belongs
to $\mathcal{J}$.

We have shown that the two-sided ideal $\mathcal{J}$ contains all the elements
$p^{2}$ and $\left(  pq\right)  r+\left(  qr\right)  p+\left(  rp\right)  q$
from \eqref{eq.free.Lie-relators}. Since these elements generate the two-sided
ideal $\operatorname{Ker}\lambda$ (by Lemma \ref{lem.free-Lie-relations}), we
thus conclude that $\operatorname{Ker}\lambda\subseteq\mathcal{J}$. But this
is precisely \eqref{eq.gamma-kernel-in-Z}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.main} when $V$ is free.]Let $f\in\operatorname{Ker}%
D$. By \eqref{eq.three.lambda-delta}, we then have%
\[
\lambda\left(  \delta\left(  f\right)  \right)  =D\left(  f\right)  =0.
\]
Thus, $\delta\left(  f\right)  \in\operatorname{Ker}\lambda$, so that
$\gamma\left(  \delta\left(  f\right)  \right)  \in\gamma\left(
\operatorname*{Ker}\lambda\right)  \subseteq Z$ by Lemma
\ref{lem.gamma-kernel-in-Z}. Now, \eqref{eq.three.gamma-delta} yields%
\[
f=\gamma\left(  \delta\left(  f\right)  \right)  \in Z.
\]
Since we have shown this for each $f\in\operatorname*{Ker}D$, we thus obtain
$\operatorname{Ker}D\subseteq Z$. The reverse inclusion is Corollary
\ref{cor.ker-right-ideal}.
\end{proof}

The same argument immediately yields a slight extension.

\begin{corollary}
\label{cor.projective} If the $\mathbf{k}$-module $V$ is projective, then
$Z=\operatorname{Ker} D$.
\end{corollary}

\begin{proof}
We denote the Dynkin operator $D$ and the right ideal $Z$ by $D_{V}$ and
$Z_{V}$, respectively, in order to contrast them for different choices of $V$.
Note that the Dynkin operator is natural in $V$. That is, if $M$ and $N$ are
two $\mathbf{k}$-modules, and if $f:M\rightarrow N$ is a $\mathbf{k}$-linear
map, then the canonically induced $\mathbf{k}$-algebra morphism $T\left(
f\right)  :T\left(  M\right)  \rightarrow T\left(  N\right)  $ commutes with
the respective Dynkin operators $D$, in the sense that the diagram%
\begin{equation}
\xymatrix{ T\left(M\right)^+ \ar[r]^{D_M} \ar[d]_{T\left(f\right)} & T\left(M\right)^+ \ar[d]^{T\left(f\right)} \\ T\left(N\right)^+ \ar[r]_{D_N} & T\left(N\right)^+ }
\label{eq.D-nat}%
\end{equation}
is commutative. Similarly, the right ideal $Z_{V}$ is natural in $V$, meaning
that any $\mathbf{k}$-linear map $f:M\rightarrow N$ satisfies
\begin{equation}
T\left(  f\right)  \left(  Z_{M}\right)  \subseteq Z_{N} \label{eq.Z-nat}%
\end{equation}
(since $T\left(  f\right)  $ sends the generators $xD_{M}\left(  x\right)  $
of $Z_{M}$ to generators $T\left(  f\right)  \left(  x\right)  \cdot
D_{N}\left(  T\left(  f\right)  \left(  x\right)  \right)  $ of $Z_{N}$).

Now, assume that $V$ is projective. Thus, $V$ is a direct addend of a free
$\mathbf{k}$-module $F$. Let $i:V\rightarrow F$ and $r:F\rightarrow V$ be the
corresponding injection and projection, so that $r\circ i=\operatorname{id}%
_{V}$. The induced tensor algebra maps $T\left(  i\right)  :T\left(  V\right)
\rightarrow T\left(  F\right)  $ and $T\left(  r\right)  :T\left(  F\right)
\rightarrow T\left(  V\right)  $ commute with the Dynkin operators $D_{V}$ and
$D_{F}$ and send the corresponding right ideals $Z_{V}$ and $Z_{F}$ into one
another, as we saw in the preceding paragraph.

Hence, $T\left(  r\right)  \left(  Z_{F}\right)  \subseteq Z_{V}$ and
$T\left(  i\right)  \left(  \operatorname*{Ker}D_{V}\right)  \subseteq
\operatorname*{Ker}D_{F}$. But $F$ is a free $\mathbf{k}$-module, so we have
$\operatorname{Ker}D_{F}=Z_{F}$ by Theorem \ref{thm.main}. Hence, $T\left(
i\right)  \left(  \operatorname*{Ker}D_{V}\right)  \subseteq
\operatorname*{Ker}D_{F}=Z_{F}$. Applying $T\left(  r\right)  $ to this
relation, we obtain
\[
\left(  T\left(  r\right)  \circ T\left(  i\right)  \right)  \left(
\operatorname*{Ker}D_{V}\right)  \subseteq T\left(  r\right)  \left(
Z_{F}\right)  \subseteq Z_{V}.
\]
Since $T\left(  r\right)  \circ T\left(  i\right)  =\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$ (because $r\circ i=\operatorname*{id}%
\nolimits_{V}$), this rewrites as $\operatorname*{Ker}D_{V}\subseteq Z_{V}$.
The reverse inclusion again follows from Corollary \ref{cor.ker-right-ideal}.
\end{proof}

\section{The case of a $\mathbb{Q}$-algebra}

Now we shall no longer assume that $V$ is free. We work towards the proof of
Theorem \ref{thm.main} in the case when $\mathbf{k}$ is a $\mathbb{Q}%
$-algebra. A weaker statement holds in full generality:

\begin{theorem}
[torsion version of Cohn's kernel theorem]\label{thm.torsion} Let $n\geq1$.
Then,%
\[
n\left(  V^{\otimes n}\cap\operatorname{Ker}D\right)  \subseteq V^{\otimes
n}\cap Z\subseteq V^{\otimes n}\cap\operatorname{Ker}D.
\]

\end{theorem}

\begin{proof}
[First proof of Theorem \ref{thm.torsion}.]Corollary \ref{cor.ker-right-ideal}
yields $Z\subseteq\operatorname*{Ker}D$, from which $V^{\otimes n}\cap
Z\subseteq V^{\otimes n}\cap\operatorname{Ker}D$ immediately follows. It
remains to show that $n\left(  V^{\otimes n}\cap\operatorname{Ker}D\right)
\subseteq V^{\otimes n}\cap Z$.

Let%
\[
z\in V^{\otimes n}\cap\operatorname{Ker}D.
\]
We must prove that $nz\in V^{\otimes n}\cap Z$.

Choose a free $\mathbf{k}$-module $F$ and a surjective $\mathbf{k}$-linear map
$\pi:F\rightarrow V$. The induced map
\[
\pi_{n}:=\pi^{\otimes n}:F^{\otimes n}\longrightarrow V^{\otimes n}%
\]
is surjective. Choose $f\in F^{\otimes n}$ satisfying $\pi_{n}\left(
f\right)  =z$.

The Dynkin operator is natural in the module (see the proof of Corollary
\ref{cor.projective}), so we have
\begin{equation}
\pi_{n}\left(  D_{F}\left(  f\right)  \right)  =D_{V}\left(  \pi_{n}\left(
f\right)  \right)  =D_{V}\left(  z\right)  =0 \label{eq.Q-lift-Dzero}%
\end{equation}
(since $z\in\operatorname*{Ker}D=\operatorname*{Ker}D_{V}$). Set%
\begin{equation}
f_{0}:=nf-D_{F}\left(  f\right)  . \label{eq.Q.f0}%
\end{equation}
Thus,%
\[
D_{F}\left(  f_{0}\right)  =nD_{F}\left(  f\right)  -D_{F}^{2}\left(
f\right)  =0\ \ \ \ \ \ \ \ \ \ \text{by Lemma \ref{lem.DSW}}.
\]
Thus, $f_{0}\in\operatorname*{Ker}D_{F}=Z_{F}$ (by the already-proved
free-module case of Theorem \ref{thm.main}, since $F$ is a free $\mathbf{k}$-module).

But \eqref{eq.Q.f0} gives $\pi_{n}\left(  f_{0}\right)  =n\underbrace{\pi
_{n}\left(  f\right)  }_{=z}-\underbrace{\pi_{n}\left(  D_{F}\left(  f\right)
\right)  }_{\substack{=0\\\text{(by \eqref{eq.Q-lift-Dzero})}}}=nz$. Thus,
$nz=\pi_{n}\left(  f_{0}\right)  $. In terms of the tensor algebra map
$T\left(  \pi\right)  :T\left(  F\right)  \rightarrow T\left(  V\right)  $, we
can rewrite this as $nz=T\left(  \pi\right)  \left(  f_{0}\right)  $ (since
$\pi_{n}=T\left(  \pi\right)  \mid_{F^{\otimes n}}$).

But as we saw in the above proof of Corollary \ref{cor.projective}, the tensor
algebra map $T\left(  \pi\right)  :T\left(  F\right)  \rightarrow T\left(
V\right)  $ satisfies $T\left(  \pi\right)  \left(  Z_{F}\right)  \subseteq
Z_{V}$. We conclude that $nz=T\left(  \pi\right)  \underbrace{\left(
f_{0}\right)  }_{\in Z_{F}}\in T\left(  \pi\right)  \left(  Z_{F}\right)
\subseteq Z_{V}=Z$. Hence, $nz\in V^{\otimes n}\cap Z$ (since $nz\in
V^{\otimes n}$).

Since we have chosen $z$ to be an arbitrary element of $V^{\otimes n}%
\cap\operatorname{Ker}D$, we thus have proved that $n\left(  V^{\otimes n}%
\cap\operatorname{Ker}D\right)  \subseteq V^{\otimes n}\cap Z$. This completes
our proof.
\end{proof}

\begin{proof}
[Second proof of Theorem \ref{thm.torsion}.]We give a proof based on cyclic
rotation, avoiding the use of the free-module case of Theorem \ref{thm.main} entirely.

As in the first proof, it suffices to show that
\begin{equation}
n\left(  V^{\otimes n}\cap\operatorname{Ker}D\right)  \subseteq V^{\otimes
n}\cap Z. \label{eq.Q-cyclic.goal}%
\end{equation}
The case $n=1$ is clear, since $D$ is the identity on $V^{\otimes1}=V$. Thus,
fix $n\geq2$.

Let
\[
c_{n}:V^{\otimes n}\longrightarrow V^{\otimes n}%
\]
be the cyclic rotation defined by
\begin{equation}
c_{n}\left(  v_{1}v_{2}\cdots v_{n}\right)  :=v_{n}v_{1}v_{2}\cdots v_{n-1}
\label{eq.Q-cyclic.cn}%
\end{equation}
for $v_{1},v_{2},\ldots,v_{n}\in V$. Also, define the $\mathbf{k}$-linear map
\[
\rho_{n}:V^{\otimes n}\longrightarrow V^{\otimes n}%
\]
by
\begin{equation}
\rho_{n}\left(  uv\right)  :=D\left(  u\right)  v \label{eq.Q-cyclic.rho}%
\end{equation}
for $u\in V^{\otimes\left(  n-1\right)  }$ and $v\in V$. Thus, $\rho_{n}$ is
the map $D\mid_{V^{\otimes\left(  n-1\right)  }}\otimes\operatorname*{id}_{V}$
under the canonical identification $V^{\otimes n}=V^{\otimes\left(
n-1\right)  }\otimes V$. Since $\left(  D\mid_{V^{\otimes\left(  n-1\right)
}}\right)  ^{2}=\left(  n-1\right)  \left(  D\mid_{V^{\otimes\left(
n-1\right)  }}\right)  $ (by Lemma \ref{lem.DSW}, applied to $n-1$ instead of
$n$), we thus obtain%
\begin{equation}
\rho_{n}^{2}=\left(  n-1\right)  \rho_{n}. \label{eq.Q-cyclic.rho2}%
\end{equation}


However, using the recursion \eqref{eq.D.rec}, we can see that%
\begin{equation}
D=\left(  \operatorname*{id}-c_{n}\right)  \circ\rho_{n}\qquad\text{on
}V^{\otimes n}. \label{eq.Q-cyclic.D-factor}%
\end{equation}
Indeed, any $u\in V^{\otimes\left(  n-1\right)  }$ and $v\in V$ satisfy
\begin{align*}
\left(  \left(  \operatorname*{id}-c_{n}\right)  \circ\rho_{n}\right)  \left(
uv\right)   &  =\left(  \operatorname*{id}-c_{n}\right)  \underbrace{\left(
\rho_{n}\left(  uv\right)  \right)  }_{=D\left(  u\right)  v}=\left(
\operatorname*{id}-c_{n}\right)  \left(  D\left(  u\right)  v\right) \\
&  =D\left(  u\right)  v-\underbrace{c_{n}\left(  D\left(  u\right)  v\right)
}_{=vD\left(  u\right)  }=D\left(  u\right)  v-vD\left(  u\right)  =\left[
D\left(  u\right)  ,v\right] \\
&  =D\left(  uv\right)  \qquad\left(  \text{by \eqref{eq.D.rec}}\right)  ,
\end{align*}
so that \eqref{eq.Q-cyclic.D-factor} follows by linearity.

Now, let
\[
z\in V^{\otimes n}\cap\operatorname*{Ker}D.
\]
We must prove that $nz\in V^{\otimes n}\cap Z$. Let us set
\begin{equation}
q:=\rho_{n}\left(  z\right)  . \label{eq.Q-cyclic.q}%
\end{equation}
Since $z\in\operatorname*{Ker}D$, we have%
\begin{align*}
0  &  =D\left(  z\right)  =\left(  \operatorname*{id}-c_{n}\right)
\underbrace{\left(  \rho_{n}\left(  z\right)  \right)  }_{=q}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq.Q-cyclic.D-factor}}\right) \\
&  =\left(  \operatorname*{id}-c_{n}\right)  \left(  q\right)  =q-c_{n}\left(
q\right)  ,
\end{align*}
and therefore
\begin{equation}
c_{n}\left(  q\right)  =q. \label{eq.Q-cyclic.q-invariant}%
\end{equation}


Write $z$ as a finite sum
\[
z=\sum_{i}u_{i}v_{i}%
\]
with $u_{i}\in V^{\otimes\left(  n-1\right)  }$ and $v_{i}\in V$. Applying the
map $\rho_{n}$ to this equality (and recalling that $\rho_{n}\left(  z\right)
=q$), we find%
\begin{equation}
q=\sum_{i}\underbrace{\rho_{n}\left(  u_{i}v_{i}\right)  }%
_{\substack{=D\left(  u_{i}\right)  v_{i}\\\text{(by \eqref{eq.Q-cyclic.rho})}%
}}=\sum_{i}D\left(  u_{i}\right)  v_{i}. \label{eq.Q-cyclic.q-=3}%
\end{equation}
Applying the map $c_{n}$ to this equality, we find%
\[
c_{n}\left(  q\right)  =\sum_{i}\underbrace{c_{n}\left(  D\left(
u_{i}\right)  v_{i}\right)  }_{=v_{i}D\left(  u_{i}\right)  }=\sum_{i}%
v_{i}D\left(  u_{i}\right)  .
\]
Compared with \eqref{eq.Q-cyclic.q-invariant}, this becomes%
\begin{equation}
q=\sum_{i}v_{i}D\left(  u_{i}\right)  . \label{eq.Q-cyclic.q-=5}%
\end{equation}


For any $a,b\in T\left(  V\right)  ^{+}$, we have $aD\left(  b\right)
+bD\left(  a\right)  =P\left(  a,b\right)  \in Z$ (by \eqref{eq.free.P-in-Z}).
Applying this with $a=u_{i}$ and $b=v_{i}$, and then summing over $i$, we
obtain%
\[
\sum_{i}\left(  u_{i}D\left(  v_{i}\right)  +v_{i}D\left(  u_{i}\right)
\right)  \in Z.
\]
Since%
\begin{align*}
&  \sum_{i}\left(  u_{i}D\left(  v_{i}\right)  +v_{i}D\left(  u_{i}\right)
\right) \\
&  =\sum_{i}u_{i}\underbrace{D\left(  v_{i}\right)  }_{\substack{=v_{i}%
\\\text{(since }v_{i}\in V\text{)}}}+\underbrace{\sum_{i}v_{i}D\left(
u_{i}\right)  }_{\substack{=q\\\text{(by \eqref{eq.Q-cyclic.q-=5})}}}\\
&  =\underbrace{\sum_{i}u_{i}v_{i}}_{=z}+\,q=z+q,
\end{align*}
this rewrites as%
\begin{equation}
z+q\in Z. \label{eq.Q-cyclic.z-plus-q}%
\end{equation}
Since $q=\rho_{n}\left(  z\right)  $, we can rewrite this as
\begin{equation}
z+\rho_{n}\left(  z\right)  \in Z. \label{eq.Q-cyclic.z-plus-rhoz}%
\end{equation}


We next apply the same observation to $q$. First, \eqref{eq.Q-cyclic.rho2}
shows that $\rho_{n}$ acts on the image of $\rho_{n}$ as multiplication by
$n-1$. Since $q=\rho_{n}\left(  z\right)  $ belongs to the image of $\rho_{n}%
$, we thus obtain%
\begin{equation}
\rho_{n}\left(  q\right)  =\left(  n-1\right)  q. \label{eq.Q-cyclic.rhoq}%
\end{equation}
Hence, \eqref{eq.Q-cyclic.D-factor} yields%
\begin{align*}
D\left(  q\right)   &  =\left(  \operatorname*{id}-c_{n}\right)
\underbrace{\left(  \rho_{n}\left(  q\right)  \right)  }_{=\left(  n-1\right)
q}=\left(  n-1\right)  \left(  \left(  \operatorname*{id}-c_{n}\right)
\left(  q\right)  \right) \\
&  =\left(  n-1\right)  \underbrace{\left(  q-c_{n}\left(  q\right)  \right)
}_{\substack{=0\\\text{(by \eqref{eq.Q-cyclic.q-invariant})}}}=0.
\end{align*}
Thus, $q\in\operatorname*{Ker}D$, so that $q\in V^{\otimes n}\cap
\operatorname*{Ker}D$.

The argument that led from $z\in V^{\otimes n}\cap\operatorname*{Ker}D$ to
\eqref{eq.Q-cyclic.z-plus-rhoz} can therefore be applied with $q$ in place of
$z$. It gives
\[
q+\rho_{n}\left(  q\right)  \in Z.
\]
Using \eqref{eq.Q-cyclic.rhoq}, we can rewrite this as $q+\left(  n-1\right)
q\in Z$. In other words,
\begin{equation}
nq\in Z. \label{eq.Q-cyclic.nq}%
\end{equation}
On the other hand, \eqref{eq.Q-cyclic.z-plus-q} gives
\[
n\left(  z+q\right)  \in Z.
\]
Subtracting \eqref{eq.Q-cyclic.nq}, we obtain $nz\in Z$. Hence, $nz\in
V^{\otimes n}\cap Z$ (since $nz\in V^{\otimes n}$ is obvious). Since we have
proved this for every $z\in V^{\otimes n}\cap\operatorname*{Ker}D$, we thus
have shown that $n\left(  V^{\otimes n}\cap\operatorname*{Ker}D\right)
\subseteq V^{\otimes n}\cap Z$. This completes the second proof.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.main} when $\mathbf{k}$ is a $\mathbb{Q}$%
-algebra.]Assume that $\mathbf{k}$ is a $\mathbb{Q}$-algebra. We must show
that $\operatorname*{Ker}D\subseteq Z$ (since $Z\subseteq\operatorname*{Ker}D$
was already shown in Corollary \ref{cor.ker-right-ideal}). Since the map $D$
is graded (i.e., preserves degree), it suffices to prove the claim in each
positive degree, i.e., to show that
\[
V^{\otimes n}\cap\operatorname{Ker}D\subseteq Z\ \ \ \ \ \ \ \ \ \ \text{for
each }n\geq1.
\]
But this follows from Theorem \ref{thm.torsion}: Since $n$ is invertible in
$\mathbf{k}$, we have%
\begin{align*}
V^{\otimes n}\cap\operatorname*{Ker}D  &  =n\left(  V^{\otimes n}%
\cap\operatorname*{Ker}D\right)  \subseteq V^{\otimes n}\cap
Z\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.torsion}}\right) \\
&  \subseteq Z.\qedhere
\end{align*}

\end{proof}

\section{A homogeneous spanning set for $Z$}

The following proposition is a nicer version of Cohn's \cite[Theorem
10.1]{Cohn51}. It does not require either of the hypotheses in Theorem
\ref{thm.main}.

\begin{proposition}
\label{prop.homogeneous-span} For every commutative ring $\mathbf{k}$ and
every $\mathbf{k}$-module $V$, the $\mathbf{k}$-module $Z$ is spanned by
elements of the following two forms:

\begin{itemize}
\item elements of the form
\begin{equation}
xD\left(  x\right)  , \label{eq.homogeneous.type1}%
\end{equation}
where $x\in T\left(  V\right)  ^{+}$ is homogeneous, and

\item elements of the form
\begin{equation}
\left(  yD\left(  z\right)  +zD\left(  y\right)  \right)  w,
\label{eq.homogeneous.type2}%
\end{equation}
where $y,z\in T\left(  V\right)  ^{+}$ and $w\in T\left(  V\right)  $ are homogeneous.
\end{itemize}
\end{proposition}

\begin{proof}
Every element of the form \eqref{eq.homogeneous.type1} belongs to $Z$ by
definition. Every element of the form \eqref{eq.homogeneous.type2} belongs to
$Z$ by \eqref{eq.free.P-in-Z} (indeed, $yD\left(  z\right)  +zD\left(
y\right)  =P\left(  y,z\right)  \in Z$ by \eqref{eq.free.P-in-Z}, and
therefore $\left(  yD\left(  z\right)  +zD\left(  y\right)  \right)  w\in Z$
since $Z$ is a right ideal). Thus, the span of these elements is contained in
$Z$. It remains to prove the reverse inclusion.

By its definition, $Z$ is spanned by the elements
\begin{equation}
xD\left(  x\right)  w\qquad\text{with }x\in T\left(  V\right)  ^{+}\text{ and
}w\in T\left(  V\right)  . \label{eq.homogeneous.general-generator}%
\end{equation}
It thus suffices to show that every such element lies in the span of the
elements of the forms \eqref{eq.homogeneous.type1} and \eqref{eq.homogeneous.type2}.

Since \eqref{eq.homogeneous.general-generator} depends linearly on $w$, we
WLOG assume that $w$ is homogeneous. We cannot assume the same for $x$, but we
can still break $x$ into its homogeneous components:

Write $x$ as $x=\sum_{n\geq1}x_{n}$, where each $x_{n}\in V^{\otimes n}$.
Then,%
\begin{align*}
xD\left(  x\right)   &  =\left(  \sum_{n\geq1}x_{n}\right)  D\left(
\sum_{n\geq1}x_{n}\right)  =\left(  \sum_{n\geq1}x_{n}\right)  \sum_{n\geq
1}D\left(  x_{n}\right) \\
&  =\sum_{n\geq1}x_{n}D\left(  x_{n}\right)  +\sum_{1\leq m<n}\left(
x_{m}D\left(  x_{n}\right)  +x_{n}D\left(  x_{m}\right)  \right)  .
\end{align*}
Upon multiplication by $w$, this becomes%
\begin{equation}
xD\left(  x\right)  w=\sum_{n\geq1}x_{n}D\left(  x_{n}\right)  w+\sum_{1\leq
m<n}\left(  x_{m}D\left(  x_{n}\right)  +x_{n}D\left(  x_{m}\right)  \right)
w. \label{pf.prop.homogeneous-span.5}%
\end{equation}
Every addend in the second sum on the right-hand side already has the form
\eqref{eq.homogeneous.type2}. It remains to show that the addends
$x_{n}D\left(  x_{n}\right)  w$ of the first sum also lie in the span of the
elements of the forms \eqref{eq.homogeneous.type1} and \eqref{eq.homogeneous.type2}.

The degree-$0$ part of $w$ merely gives a scalar multiple of an element of the
form \eqref{eq.homogeneous.type1}. Hence, it suffices to handle the
positive-degree part of $w$. This part is a sum of elements of the form $vu$
with $v\in V$ and $u\in T\left(  V\right)  $. Thus, it is enough to show that
each of these elements $x_{n}D\left(  x_{n}\right)  vu$ has the form
\eqref{eq.homogeneous.type2}. Since (\ref{eq.D.rec}) yields
\[
D\left(  x_{n}v\right)  =\left[  D\left(  x_{n}\right)  ,v\right]  =D\left(
x_{n}\right)  v-vD\left(  x_{n}\right)  ,
\]
we have%
\[
D\left(  x_{n}\right)  v=D\left(  x_{n}v\right)  +vD\left(  x_{n}\right)
\]
and therefore
\[
x_{n}\underbrace{D\left(  x_{n}\right)  v}_{=D\left(  x_{n}v\right)
+vD\left(  x_{n}\right)  }u=\underbrace{\left(  x_{n}D\left(  x_{n}v\right)
+x_{n}vD\left(  x_{n}\right)  \right)  u}_{\text{an element of the form
\eqref{eq.homogeneous.type2}}}.
\]
This is precisely what we needed to show; thus,
\eqref{pf.prop.homogeneous-span.5} expresses $xD\left(  x\right)  w$ as a
linear combination of elements of the forms \eqref{eq.homogeneous.type1} and
\eqref{eq.homogeneous.type2}. This proves that the elements of the latter
forms span $Z$.
\end{proof}

As we said, Proposition \ref{prop.homogeneous-span} is a variant of
\cite[Theorem 10.1]{Cohn51}. In the latter theorem, Cohn uses $yD\left(
y\right)  w$ instead of $\left(  yD\left(  z\right)  +zD\left(  y\right)
\right)  w$ in (\ref{eq.homogeneous.type2}), and does not impose any
homogeneity conditions on $x,y,z,w$. It is easy to see that his generating set
spans ours (thanks to the polarization identity $yD\left(  z\right)
+zD\left(  y\right)  =\left(  y+z\right)  D\left(  y+z\right)  -yD\left(
y\right)  -zD\left(  z\right)  $), and thus \cite[Theorem 10.1]{Cohn51}
follows from our Proposition \ref{prop.homogeneous-span}.

\section{The unrestricted statement is false}

The question whether $Z=\operatorname{Ker}D$ holds for every module $V$ (with
no further restrictions) has a negative answer. The obstruction is already
present in degree $2$.

Let
\begin{align*}
\tau:V\otimes V  &  \longrightarrow V\otimes V,\\
v\otimes w  &  \longmapsto w\otimes v
\end{align*}
be the switch map. On $V^{\otimes2}$, the Dynkin operator $D$ acts as
$\operatorname{id}-\tau$. Thus,
\begin{equation}
\operatorname{Ker}\left(  D|_{V^{\otimes2}}\right)  =\left(  V\otimes
V\right)  ^{\langle\tau\rangle}, \label{eq.degree2.symmetric-tensors}%
\end{equation}
the module of symmetric tensors. On the other hand, the degree-$2$ part of $Z$
is
\begin{equation}
Z\cap V^{\otimes2}=\operatorname{span}_{\mathbf{k}}\left\{  v\otimes v\mid
v\in V\right\}  \label{eq.degree2.Z}%
\end{equation}
(this is particularly easy to see from Proposition \ref{prop.homogeneous-span}%
). The right-hand side of \eqref{eq.degree2.Z} is the image of the canonical
map
\begin{equation}
\Gamma_{\mathbf{k}}^{2}\left(  V\right)  \longrightarrow\operatorname{TS}%
_{\mathbf{k}}^{2}\left(  V\right)  \label{eq.degree2.divided-to-symmetric}%
\end{equation}
from the second divided-powers module of $V$ to the symmetric tensors. This
map is an isomorphism when $V$ is flat, and it is an isomorphism when $2$ is
invertible in $\mathbf{k}$; however, it need not be surjective for an
arbitrary module. Explicit counterexamples are given in \cite[Example~4.6]%
{Lundkvist08}. For any such counterexample, a symmetric tensor outside the
image of \eqref{eq.degree2.divided-to-symmetric} belongs to
$\operatorname{Ker}D$ but not to $Z$. Hence, $Z=\operatorname{Ker}D$ fails in general.

\begin{noncompile}
\begin{remark}
\label{rem.counterexample-explicit} Lundkvist's example can be summarized as
follows. Let $\mathbb{f}$ be a field of characteristic $2$, let
\[
A=\mathbb{f}\left[  x_{1},x_{2},x_{3},y_{1},y_{2},y_{3},z_{1},z_{2}%
,z_{3}\right]  ,
\]
and quotient $A$ by the ideal described in \cite[Example~4.6]{Lundkvist08}.
Over the resulting ring $A^{\prime}$, let
\[
V=A^{\prime3}/A^{\prime}\left(  z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3}\right)  ,
\]
and let $m_{i}$ be the image of $e_{i}$. The tensor
\[
\left(  x_{1}z_{2}+y_{2}z_{1}\right)  m_{1}\otimes m_{2}
\]
is symmetric, but is not in the span of the tensors $v\otimes v$. Therefore,
it lies in $\operatorname{Ker} D$ but not in $Z$.
\end{remark}
\end{noncompile}

\begin{noncompile}
Summary

Cohn's proof has the following structure. The free nonassociative algebra
$\mathcal{M}$ records arbitrary bracketings. The map $\lambda$ evaluates a
bracketing as a Lie commutator, while $\delta$ inserts a tensor word into
left-associated form and $\gamma$ removes the parentheses in a controlled way.
The identities
\[
\gamma\delta=\operatorname{id}, \qquad\lambda\delta=D, \qquad D\gamma=\lambda
\]
translate the kernel of $D$ into the defining alternating and Jacobi relations
of a free Lie algebra. The ideal argument in Lemma \ref{lem.gamma-kernel-in-Z}
then shows that both kinds of relations map into $Z$.

For a $\mathbb{Q}$-algebra, the identity $D^{2}=nD$ supplies the missing lift
from an arbitrary module to a free one. For arbitrary modules over arbitrary
rings, the conclusion is false: in degree $2$, it would assert that every
symmetric tensor is generated by diagonal tensors, which is not true in the
nonflat case.
\end{noncompile}

\begin{thebibliography}{99999}                                                                                            %


\bibitem[Bou89]{Bou89}N. Bourbaki, \textit{Lie groups and Lie algebras,
Chapters 1--3}, Elements of Mathematics, Springer, 2nd printing 1989.

\bibitem[Coh51]{Cohn51}P.~M.~Cohn, \textit{Integral modules, Lie rings and
free groups}, Ph.D. thesis, University of Cambridge, 1951, especially \S \S 9--10.

\bibitem[GR26]{GR26}\href{https://arxiv.org/abs/1409.8356v7}{Darij Grinberg,
Victor Reiner, \textit{Hopf algebras in combinatorics}, arXiv:1409.8356v7.}

\bibitem[Lau13]{Laue13}Hartmut Laue, \textit{Freie algebraische Strukturen},
16 September 2013.\newline%
\url{https://web.archive.org/web/20250526180315/https://www.math.uni-kiel.de/algebra/laue/vorlesungen/frei/freiealgstr.pdf}\newline
See \url{https://www.cip.ifi.lmu.de/~grinberg/algebra/laue-errata1.pdf} for
unofficial errata.

\bibitem[Lun08]{Lundkvist08}Christian Lundkvist, \textit{Counterexamples
regarding symmetric tensors and divided powers}, Journal of Pure and Applied
Algebra \textbf{212} (2008), no.~10, pp.~2236--2249. \newline\url{https://doi.org/10.1016/j.jpaa.2008.03.024}

\bibitem[Reu93]{Reutenauer93}Christophe Reutenauer, \textit{Free Lie
algebras}, London Mathematical Society Monographs, New Series, vol.~7, Oxford
University Press, New York, 1993.

\bibitem[Ser06]{Serre06}%
\href{https://doi.org/10.1007/978-3-540-70634-2}{Jean-Pierre Serre,
\textit{Lie Algebras and Lie Groups: 1964 Lectures given at Harvard
University}, Corrected 5th printing 2006.}
\end{thebibliography}


\end{document}