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\newcommand{\ZZ}{\mathbb Z}
\newcommand{\NN}{\mathbb N}
\newcommand{\Prim}{\operatorname{Prim}}
\newcommand{\im}{\operatorname{im}}

\title{On Injectivity of Coalgebra Morphisms}
\author{GPT-5.6 Sol, with prompting by Darij Grinberg}
\date{\today}

\begin{document}

\maketitle

\begin{abstract}
\textbf{Abstract.}
We construct a surjective filtered morphism of connected graded coalgebras
over $\ZZ$ that is injective on primitive elements but is not injective.
As a consequence, we construct a nonzero coideal containing no nonzero
primitive element.  All tensor factors occurring in the coproduct
of the source are direct summands, so the construction does not rely on any
ambiguity in the notion of a subcoalgebra in the absence of flatness.
\medskip

\textbf{Manifest.} {This is a negative answer to
\url{https://mathoverflow.net/questions/140357}, generated by
GPT-5.6 Sol. -- DG}\footnote{This work is in the public domain.}
\end{abstract}

\section{Statement of the counterexample}

Let $R$ be a commutative ring, and let $C$ be a connected filtered $R$-coalgebra.
We recall that ``connected'' means that the $0$-th part of the filtration
on $C$ is isomorphic to $R$.
The unique preimage of $1_R \in R$ under this isomorphism shall be denoted by
$1_C$ or, for short, by $1$.
We write
\[
\Prim(C)
=
\{x\in C\mid \Delta(x)=1\otimes x+x\otimes 1\}
\]
for the $R$-module of primitive elements of $C$.

Recall that a \emph{coideal} of $C$ is an $R$-submodule $I\subseteq C$
satisfying
\[
\varepsilon(I)=0
\]
and
\[
\Delta(I)
\subseteq
\im(I\otimes_R C\longrightarrow C\otimes_R C)
+
\im(C\otimes_R I\longrightarrow C\otimes_R C).
\]
(The latter relation is often written as $\Delta(I) \subseteq
I \otimes_R C + C \otimes_R I$, understanding that the tensor
products on the right hand side actually refer to their images
in $C \otimes_R C$. We shall avoid this abuse of notation here.)
% The counital condition is necessary: without it, the assertion below is
% already false for the one-dimensional coalgebra $C=R$ and $I=C$.

We first recall a well-known result for connected \textbf{graded}
coalgebras (see, e.g., \cite[Exercise 1.4.35]{GR26}):

\begin{proposition}[The graded case]\label{prop.graded-case}
\ \ \phantomsection

\begin{enumerate}
\item[(a)]
Let $C$ be a connected graded $R$-coalgebra, and let $I$ be a graded
coideal of $C$.  If
\[
I\cap\Prim(C)=0,
\]
then $I=0$.

\item[(b)]
Let $C$ and $D$ be connected graded $R$-coalgebras, and
let $f:C\twoheadrightarrow D$ be a surjective graded coalgebra morphism.
If the restriction
\[
f\vert_{\Prim(C)}:\Prim(C)\longrightarrow\Prim(D)
\]
is injective, then $f$ is injective.

\end{enumerate}
\end{proposition}

\begin{proof}
(a) Assume, for contradiction,
that $I\neq0$.  Since $I$ is graded, there exists a smallest $n\geq0$
such that the degree-$n$ component $I_n$ is nonzero.  Consider this
$n$, and choose a nonzero element $x\in I_n$.
Since $\varepsilon(I)=0$ and $C_0=R1$, we have $I_0 = 0$, so that
$n>0$.

Let us spell out the grading on the tensor product
$C \otimes_R C$.  Since
\[
C=\bigoplus_{k\geq0}C_k,
\]
the tensor product has the direct-sum decomposition
\begin{equation}\label{eq.tensor-bigrading}
C\otimes_R C
=
\bigoplus_{i,j\geq0} C_i\otimes_R C_j.
\end{equation}
An element of $C_i\otimes_R C_j$ is said to have \emph{bidegree}
$(i,j)$; its \emph{total degree} is $i+j$.  Thus, the degree-$n$
component of $C\otimes_R C$ for the total grading is
\[
(C\otimes_R C)_n
=
\bigoplus_{i+j=n} C_i\otimes_R C_j.
\]
By definition, the coproduct of a graded coalgebra is homogeneous of
degree $0$ with respect to this total grading.  In other words,
\[
\Delta(C_n)
\subseteq
(C\otimes_R C)_n
=
\bigoplus_{i+j=n} C_i\otimes_R C_j.
\]
Since $x\in C_n$, this gives
\begin{equation}\label{eq.graded-delta-x}
\Delta(x)\in\bigoplus_{i+j=n} C_i\otimes_R C_j.
\end{equation}
Thus, all $(i,j)$-bidegree components of $\Delta(x)$
are zero except for those with $i+j=n$.

On the other hand, $x$ belongs to the coideal $I$, so that
$\Delta(x)$ belongs to
\begin{equation}\label{eq.graded-coideal-images}
\im(I\otimes_R C\longrightarrow C\otimes_R C)
+
\im(C\otimes_R I\longrightarrow C\otimes_R C).
\end{equation}
Since $I$ is graded, we have
\[
I=\bigoplus_{k\geq0}I_k,
\qquad\text{where } I_k=I\cap C_k.
\]
Consequently, the two maps occurring in
\eqref{eq.graded-coideal-images} respect the bigrading
\eqref{eq.tensor-bigrading}: the first sends
$I_i\otimes_R C_j$ into $C_i\otimes_R C_j$, whereas the second sends
$C_i\otimes_R I_j$ into $C_i\otimes_R C_j$.  Hence, for any
$i,j\geq0$, the bidegree-$(i,j)$ component of the module in
\eqref{eq.graded-coideal-images} is contained in
\[
\im(I_i\otimes_R C_j\longrightarrow C_i\otimes_R C_j)
+
\im(C_i\otimes_R I_j\longrightarrow C_i\otimes_R C_j).
\]
If $i<n$ and $j<n$, then this component must therefore be zero
(since the minimality of $n$ yields $I_i=I_j=0$),
and therefore the bidegree-$(i,j)$ component of $\Delta(x)$
must be $0$ (since $\Delta(x)$ lies in this module).
But we have previously also shown (using \eqref{eq.graded-delta-x})
that all $(i,j)$-bidegree components of $\Delta(x)$
are zero except for those with $i+j=n$.
Combining these two observations, we conclude that the only
$(i,j)$-bidegree components of $\Delta(x)$ that can be
nonzero are those that satisfy $i+j=n$ but not $i<n$ and $j<n$.
In other words, they are the components with
bidegrees $(n,0)$ and $(0,n)$.  Therefore,
\[
\Delta(x)\in C_n\otimes_R C_0+C_0\otimes_R C_n.
\]
Since $C_0=R1$, this means that
there exist $y,z\in C_n$ such that
\begin{equation}
\Delta(x)=y\otimes1+1\otimes z.
\label{eq.xyz}
\end{equation}
Because $n>0$ and $y,z\in C_n$,
we have $\varepsilon(y)=\varepsilon(z)=0$.  Applying
$\operatorname{id}\otimes\varepsilon$ to the equality \eqref{eq.xyz},
we obtain $x=y\varepsilon(1)+1\varepsilon(z)=y$, so that $y=x$; similarly,
applying $\varepsilon\otimes\operatorname{id}$ to \eqref{eq.xyz},
we obtain $z=x$.  Hence \eqref{eq.xyz} rewrites as
\[
\Delta(x)=x\otimes1+1\otimes x,
\]
so $x\in\Prim(C)$.  This contradicts $I\cap\Prim(C)=0$.  Thus $I=0$.
\medskip

(b) Set $I=\ker f$.  Since $f$ is graded, this submodule
$I$ is graded.  Since $f$ is surjective,
right exactness of tensor products shows that
\[
\ker(f\otimes f) = \im(I \otimes C \longrightarrow C \otimes C)
+ \im(C \otimes I \longrightarrow C \otimes C) .
\]
But $f$ is a coalgebra morphism, so that
$(f\otimes f)\circ\Delta = \Delta\circ f$, and thus
$(f\otimes f)(\Delta(I)) = \Delta(f(I)) = \Delta(0) = 0$,
so that
\[
\Delta(I) \subseteq \ker(f\otimes f)
= \im(I \otimes C \longrightarrow C \otimes C)
+ \im(C \otimes I \longrightarrow C \otimes C) .
\]
It is also easy to see that $\varepsilon(I) = 0$ (since
$\varepsilon_C = \varepsilon_D\circ f$). Hence,
$I$ is a coideal.  The injectivity of $f\vert_{\Prim(C)}$ yields
$I\cap\Prim(C)=0$.  Part (a) therefore gives $I=0$, so $f$
is injective.
\end{proof}

It is natural to wonder whether the gradedness of $f$ in the
former fact can be replaced by a weaker property, such as
filteredness (i.e., the image of the $n$-th degree component
of $C$ must be contained in the sum of the $0$-th, $1$-st, \ldots,
$n$-th degree components of $D$).
In this note, we shall give a counterexample showing that it
cannot:

\begin{theorem}\label{thm.main}
There exist connected graded $\ZZ$-coalgebras $C$ and $D$ and
a surjective filtered\footnote{A linear map
$f : C \to D$ between two graded modules $C$ and $D$
is said to be \emph{filtered} if and only if
each $n \geq 0$ satisfies
$f\left(C_0 + C_1 + \cdots + C_n\right) \subseteq
D_0 + D_1 + \cdots + D_n$, where $M_k$ denotes the $k$-th
degree component of a graded module $M$.} coalgebra morphism
\[
f:C\twoheadrightarrow D
\]
such that the restriction
\[
f\vert_{\Prim(C)}:\Prim(C)\longrightarrow \Prim(D)
\]
is injective, whereas $f$ itself is not injective.

In particular, there exists a connected graded $\ZZ$-coalgebra
$C$ and a nonzero coideal $I\subseteq C$ such that
\[
I\cap\Prim(C)=0.
\]
\end{theorem}

We shall construct these $C$, $D$, $f$ and $I$ in what follows.
The construction rests on a simple module-theoretic phenomenon: an
injective map can have zero tensor square when the modules
are not flat.

\section{The module-theoretic ingredient}

Fix a prime number $p$, and set
\[
a_n=2^{n-1}
\qquad\text{for every }n\geq 1.
\]
Thus
\begin{equation}\label{eq.an-recursion}
a_{n+1}=2a_n
\qquad\text{for every }n\geq 1.
\end{equation}
Define the two (isomorphic) $\ZZ$-modules
\[
L=\bigoplus_{n\geq 1}(\ZZ/p^{a_n}\ZZ)e_n
\qquad\text{and}\qquad
N=\bigoplus_{n\geq 1}(\ZZ/p^{a_n}\ZZ)u_n.
\]

\begin{lemma}\label{lem.module-data}
There exist homomorphisms
\[
\alpha:L\longrightarrow N
\qquad\text{and}\qquad
\beta:L\otimes_{\ZZ}L\longrightarrow N
\]
with the following properties:
\begin{enumerate}
\item the map $\alpha$ is injective;
\item the map $\alpha\otimes\alpha:L\otimes L\to N\otimes N$ is zero;
\item the map $\beta$ is surjective.
\end{enumerate}
\end{lemma}

\begin{proof}
Define $\alpha$ on the distinguished generators by
\begin{equation}\label{eq.alpha-definition}
\alpha(e_n)=p^{a_n}u_{n+1}
\qquad\text{for every }n\geq 1.
\end{equation}
The element $u_{n+1}$ has order $p^{a_{n+1}}=p^{2a_n}$, by
\eqref{eq.an-recursion}.  Hence $p^{a_n}u_{n+1}$ has order $p^{a_n}$,
just like $e_n$ does.
Thus the restriction of $\alpha$ to every direct summand
$(\ZZ/p^{a_n}\ZZ)e_n$ is injective.  Since the images of distinct summands
belong to distinct direct summands of $N$, the map $\alpha$ is injective.

We next prove that $\alpha\otimes\alpha=0$.  It is enough to evaluate this
map on the tensors $e_m\otimes e_n$.  Assume without loss of generality that
$m\leq n$.  Then
\begin{equation}\label{eq.tensor-alpha}
(\alpha\otimes\alpha)(e_m\otimes e_n)
=
p^{a_m+a_n}u_{m+1}\otimes u_{n+1}.
\end{equation}
But it is well-known that
$(\ZZ/u\ZZ) \otimes_{\ZZ} (\ZZ/v\ZZ)
\cong \ZZ / \gcd\left(u,v\right)\ZZ$ for any integers $u$ and $v$.
Thus, the tensor $u_{m+1}\otimes u_{n+1}$ has order
$\gcd\left(p^{a_{m+1}},p^{a_{n+1}}\right)
=
p^{\min(a_{m+1},a_{n+1})}
=
p^{a_{m+1}}
=
p^{2a_m}$.
Since $a_n\geq a_m$, we have
$a_m+a_n\geq 2a_m$.
Therefore the right hand side of \eqref{eq.tensor-alpha} is zero.

Finally, define the map $\beta:L\otimes_{\ZZ}L\longrightarrow N$
by
\begin{equation}\label{eq.beta-definition}
\beta(e_m\otimes e_n)
=
\begin{cases}
 u_n,&\text{if }m=n,\\
 0,&\text{if }m\neq n.
\end{cases}
\end{equation}
This is well-defined.  Indeed, the $(m,n)$-summand of $L\otimes L$ is
isomorphic to
\[
(\ZZ/p^{a_m}\ZZ)\otimes(\ZZ/p^{a_n}\ZZ)
\cong
\ZZ/p^{\min(a_m,a_n)}\ZZ,
\]
and when $m=n$, the prescribed image $u_n$ has exactly order $p^{a_n}$.
The map $\beta$ is surjective because $u_n=\beta(e_n\otimes e_n)$ for every
$n\geq 1$.
\end{proof}

\section{The coalgebras}

We now use the maps $\alpha$ and $\beta$ from Lemma~\ref{lem.module-data}.
As a $\ZZ$-module, let
\begin{equation}\label{eq.C-definition}
C=\ZZ 1\oplus L\oplus(L\otimes L).
\end{equation}
To avoid confusing the last direct summand of $C$ with the tensor product
$C\otimes C$, we denote by
\[
[x\mid y]\in C
\]
the element of the last summand corresponding to $x\otimes y\in L\otimes L$.

Define a coproduct on $C$ by
\begin{align}
\Delta_C(1)
&=1\otimes 1,\label{eq.delta-one}\\
\Delta_C(x)
&=1\otimes x+x\otimes 1
&&\text{for every }x\in L,\label{eq.delta-degree-one}\\
\Delta_C([x\mid y])
&=1\otimes[x\mid y]+x\otimes y+[x\mid y]\otimes 1
&&\text{for every }x,y\in L.\label{eq.delta-degree-two}
\end{align}
Let the counit $\varepsilon_C:C\to\ZZ$ be the projection onto the first
summand in \eqref{eq.C-definition}.

The formulas above make $C$ into a (coassociative, counital) coalgebra.
Indeed, $\Delta_C$ is the well-known deconcatenation coproduct of
the shuffle bialgebra on $L$, truncated after degree $2$.
Alternatively, coassociativity in degree $2$ follows by expanding both
$(\Delta_C\otimes\operatorname{id})\Delta_C([x\mid y])$ and
$(\operatorname{id}\otimes\Delta_C)\Delta_C([x\mid y])$; both expansions are
\[
1\otimes1\otimes[x\mid y]
+1\otimes x\otimes y
+x\otimes 1\otimes y
+x\otimes y\otimes 1
+1\otimes[x\mid y]\otimes 1
+[x\mid y]\otimes1\otimes1.
\]

Give $C$ the grading
\[
C_0=\ZZ 1,
\qquad
C_1=L,
\qquad
C_2=L\otimes L
\]
and $C_k = 0$ for all $k>2$.
Thus $C$ is a connected graded coalgebra.
  % We also use the induced
% filtration
% \[
% C^0=\ZZ1,
% \qquad
% C^1=\ZZ1\oplus L,
% \qquad
% C^2=C.
% \]

Let
\begin{equation}\label{eq.D-definition}
D=\ZZ1\oplus N,
\end{equation}
and make every element of $N$ primitive:
\begin{align*}
\Delta_D(1)&=1\otimes1,\\
\Delta_D(v)&=1\otimes v+v\otimes1
&&\text{for every }v\in N.
\end{align*}
The counit $\varepsilon_D$ is the projection onto $\ZZ1$.
We grade $D$ by
\[
D_0=\ZZ1
\qquad\text{and}\qquad
D_1 = N
\]
and $D_k = 0$ for all $k>1$.
Thus, $D$ is a connected graded coalgebra.

Define a $\ZZ$-linear map $f:C\to D$ by
\begin{align}
f(1)&=1,\label{eq.f-one}\\
f(x)&=\alpha(x)
&&\text{for every }x\in L,\label{eq.f-degree-one}\\
f([x\mid y])&=\beta(x\otimes y)
&&\text{for every }x,y\in L.\label{eq.f-degree-two}
\end{align}

\begin{proposition}\label{prop.f-coalgebra}
The map $f:C\to D$ is a surjective filtered coalgebra morphism.
\end{proposition}

\begin{proof}
The map preserves counits by construction.  It also preserves coproducts on
$1$ and on $L$, since every element of $\alpha(L)$ is primitive in $D$.
For $x,y\in L$, formulas \eqref{eq.delta-degree-two} and
\eqref{eq.f-degree-two} give
\begin{align*}
(f\otimes f)\Delta_C([x\mid y])
&=
1\otimes\beta(x\otimes y)
+\alpha(x)\otimes\alpha(y)
+\beta(x\otimes y)\otimes1\\
&=
1\otimes\beta(x\otimes y)
+\beta(x\otimes y)\otimes1
\qquad \left(\text{since } \alpha\otimes\alpha=0 \right) \\
&= \Delta_D(\beta(x\otimes y)) \\
&= \Delta_D(f([x\mid y])).
\end{align*}
Hence $f$ is a coalgebra morphism.

The map $f$ is filtered since $f(C_0) \subseteq D_0$,
$f(C_1) \subseteq D_1$ and $f(C_2) \subseteq D_1$.
Finally, $f$ is
surjective because its restriction to the summand $L\otimes L$ is the
surjection $\beta:L\otimes L\twoheadrightarrow N$.
\end{proof}

\section{Primitive elements and failure of injectivity}

% Define the reduced coproduct $\overline{\Delta}$ on the kernel of
% the counit by
% \[
% \overline{\Delta}(c)
% =
% \Delta(c)-1\otimes c-c\otimes1.
% \]

\begin{proposition}\label{prop.primitives}
We have
\[
\Prim(C)=L.
\]
Consequently, the restriction
\[
f\vert_{\Prim(C)}:\Prim(C)\longrightarrow\Prim(D)
\]
is injective.
\end{proposition}

\begin{proof}
Every element of $L$ is primitive by \eqref{eq.delta-degree-one}, so
$L\subseteq\Prim(C)$.

Conversely, let $c\in\Prim(C)$.  Since primitive elements belong to the
kernel of the counit, we may write uniquely
\[
c=x+z
\qquad\text{with }x\in L\text{ and }z\in L\otimes L.
\]
By \eqref{eq.delta-degree-one} and \eqref{eq.delta-degree-two}, the
difference $\Delta(c)-1\otimes c-c\otimes1$ is the image of $z$ under
the natural map
\begin{equation}\label{eq.split-inclusion}
L\otimes L\longrightarrow C\otimes C
\end{equation}
that places the first factor in the degree-$1$ summand of the first copy of
$C$ and the second factor in the degree-$1$ summand of the second copy.
The map \eqref{eq.split-inclusion} is injective: it is split by applying the
projection $C\to L$ to both tensor factors.  But
$\Delta(c)-1\otimes c-c\otimes1 = 0$, since $c$ is primitive.
Hence $z=0$, and therefore $c=x\in L$.
Thus $\Prim(C)=L$.

Under this identification, the restriction of $f$ to $\Prim(C)$ is exactly
$\alpha:L\to N$, which is injective by Lemma~\ref{lem.module-data}.
\end{proof}

\begin{proposition}\label{prop.not-injective}
The map $f$ is not injective.
\end{proposition}

\begin{proof}
For every $n\geq1$, define
\begin{equation}\label{eq.wn-definition}
w_n
=
e_n-p^{a_n}[e_{n+1}\mid e_{n+1}]
\in C.
\end{equation}
Using \eqref{eq.alpha-definition} and \eqref{eq.beta-definition}, we obtain
\begin{align*}
f(w_n)
&=
\alpha(e_n)-p^{a_n}\beta(e_{n+1}\otimes e_{n+1})
=
p^{a_n}u_{n+1}-p^{a_n}u_{n+1}
=0.
\end{align*}
On the other hand, $w_n\neq0$, because its component in the direct summand
$L$ is $e_n\neq0$.  Thus $f$ is not injective.
% Notice also that $w_n$ is not primitive.  Indeed,
% \[
% \overline{\Delta}_C(w_n)
% =
% -p^{a_n}e_{n+1}\otimes e_{n+1}.
% \]
% The element $e_{n+1}\otimes e_{n+1}$ has order
% $p^{a_{n+1}}=p^{2a_n}$, so its multiple by $p^{a_n}$ is nonzero.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm.main}]
Propositions~\ref{prop.f-coalgebra}, \ref{prop.primitives}, and
\ref{prop.not-injective} prove the first assertion.

Let $I=\ker f$.  Since $f$ is a surjective coalgebra morphism, $I$ is a
coideal.  More explicitly, identifying $D$ with $C/I$, right
exactness of tensor products gives
\[
\ker(f\otimes f)
=
\im(I\otimes C\longrightarrow C\otimes C)
+
\im(C\otimes I\longrightarrow C\otimes C).
\]
For $x\in I$, we have
\[
(f\otimes f)\Delta_C(x)
=
\Delta_D(f(x))
=0,
\]
so $\Delta_C(x)$ belongs to the displayed sum.  Also
$\varepsilon_C(x)=\varepsilon_D(f(x))=0$.
Thus $I$ is a coideal.

The coideal $I$ is nonzero by Proposition~\ref{prop.not-injective}.  By
Proposition~\ref{prop.primitives}, the restriction of $f$ to $\Prim(C)$ is
injective, so
\[
I\cap\Prim(C)=0.
\]
Finally, the connected gradings of $C$ and $D$ were exhibited above.
\end{proof}

\section{Remarks}

% \begin{remark}[Where non-flatness enters]
% The decisive phenomenon is that the injective map $\alpha:L\to N$ has zero
% tensor square.  Thus injectivity on the degree-$1$ part does not prevent a
% degree-$2$ reduced coproduct from disappearing after applying
% $f\otimes f$.  This is exactly the tensor-injectivity step that is automatic
% over a field and can fail over a general commutative ring.
% \end{remark}

\begin{remark}[No ambiguity about subcoalgebras]
The source is defined as the direct sum
\[
C=\ZZ1\oplus L\oplus(L\otimes L).
\]
In particular, the map $L\otimes L\to C\otimes C$ used in the reduced
coproduct is split injective.  Hence the construction does not exploit any
ambiguity caused by the possible failure of $S\otimes S\to C\otimes C$ to
be injective for an arbitrary submodule $S\subseteq C$.
\end{remark}

% \begin{remark}[Why the positive homogeneous case is not contradicted]
% The coalgebra $C$ is graded, but $I=\ker f$ is not homogeneous.  Indeed,
% each element $w_n$ in \eqref{eq.wn-definition} is a sum of a degree-$1$
% term and a degree-$2$ term.  Thus the example is compatible with the usual
% minimal-degree argument for homogeneous coideals in graded coalgebras.
% \end{remark}

\begin{thebibliography}{99999}                                                                                            %
\bibitem[GR26]{GR26}\href{https://arxiv.org/abs/1409.8356v7}{Darij Grinberg,
Victor Reiner, \textit{Hopf algebras in combinatorics}, arXiv:1409.8356v7.}

\end{thebibliography}


\end{document}