Let $k$ be a commutative ring.

Let $C$ be a filtered $k$-coalgebra. This means a $k$-coalgebra equipped with an increasing $k$-module filtration $C^0 \subseteq C^1 \subseteq C^2 \subseteq ...$ satisfying $\bigcup\limits_{n\geq 0}C^n = C$ and $\Delta\left(C^n\right) \subseteq \sum\limits_{k=0}^n C^k \otimes C^{n-k}$ for all nonnegative integers $n$, where $C^k \otimes C^{n-k}$ really means the image of $C^k \otimes C^{n-k}$ inside $C \otimes C$ by abuse of notation (even if the map $C^k \otimes C^{n-k} \to C \otimes C$ fails to be injective).

Assume further that $C$ is connected, i. e., the counit $\varepsilon$ restricted to $C^0$ is an isomorphism to $k$. This allows us to define an element $1_C \in C^0$ (often just called $1$ when no confusion can arise) as the preimage of $1 \in k$ under this isomorphism (even if $C$ has no algebra structure given).

An element $x\in C$ is said to be primitive if $\Delta x = x \otimes 1_C + 1_C \otimes x$. The $k$-submodule of $C$ formed by all primitive elements is called $\mathrm{Prim} C$. Every primitive element $x\in C$ satisfies $\varepsilon x = 0$ (this is easy to check).

The following two conjectures are equivalent versions of each other (I have given them different numbers for easier reference):

Conjecture 1: If $D$ is a further $k$-coalgebra, and $f : C \to D$ is a surjective homomorphism of $k$-coalgebras such that $f\mid_{\mathrm{Prim}C}$ is injective, then $f$ is injective.

Conjecture 2: If $I$ is a coideal of $C$ such that $I \cap \mathrm{Prim}C = 0$, then $I = 0$. (Here, a coideal of $C$ means a $k$-submodule $J$ of $C$ such that $\Delta J \subseteq J \otimes C + C \otimes J$, where again an abuse of notation is made.)

Background: Conjectures 1 and 2 are known to be true if $k$ is a field, and the easy proof (by induction) generalizes to the case when "everything in sight is flat" (I don't want to go into the details of what this means, but I can give the proof if needed). Moreover, Conjecture 2 is also true when $C$ is a graded coalgebra and $I$ is homogeneous (more or less because direct sums are just as good as flatness when it comes to preserving injectivity under tensor products). I am unable to come up with a counterexample in any other case, but this is because I have no real idea how to construct non-flat counterexamples to anything (a good reference for non-flat perversion would be highly appreciated). It seems to me that the conjectures should be true, if only because they are highly useful.

Note that when $k$ is a field, Conjecture 1 holds even without assuming that $f$ be surjective. But this does not generalize to general commutative rings $k$. Here is a counterexample (a variation of the example on pp. 56-57 of Warren Nichols and Moss Sweedler, Hopf Algebras and Combinatorics, in Robert Morris (ed.), Umbral Calculus and Hopf Algebras): Let $k = \mathbb Z$. Let $C = \mathbb Z \oplus \mathbb Z/2 \oplus \mathbb Z/2$ as $k$-module, with basis vectors $e_0 = \left(1,0,0\right)$, $e_1 = \left(0,1,0\right)$ and $e_2 = \left(0,0,1\right)$. Make $C$ into a graded $k$-module by setting $\deg e_i = i$. Define a coproduct on $C$ by $\Delta e_0 = e_0 \otimes e_0$, $\Delta e_1 = e_0 \otimes e_1 + e_1 \otimes e_0$ and $\Delta e_2 = e_0 \otimes e_2 + e_1 \otimes e_1 + e_2 \otimes e_0$ (like in a divided-powers coalgebra). Define a counity on $C$ by $\varepsilon\left(e_i\right) = \delta_{i, 0}$. Thus, $C$ becomes a connected filtered (and even graded) $k$-coalgebra with $1_C = e_0$ and $\mathrm{Prim}C = \left\langle e_1 \right\rangle$. Now, let $D = \mathbb Z \oplus \mathbb Z/4$, with basis vectors $f_0 = \left(1,0\right)$ and $f_1 = \left(0,1\right)$. Grade $D$ by $\deg f_i = i$. Define a coproduct on $D$ by $\Delta f_0 = f_0 \otimes f_0$ and $\Delta f_1 = f_0 \otimes f_1 + f_1 \otimes f_0$. Define a counity on $D$ by $\varepsilon\left(f_i\right) = \delta_{i, 0}$. This makes $D$ into a connected filtered (and, again, graded) $k$-coalgebra with $1_D = f_0$. Let now $f : C \to D$ be the $k$-module morphism defined by $f\left(e_0\right) = f_0$, $f\left(e_1\right) = 2f_1$ and $f\left(e_2\right) = 0$. It is straightforward to check that $f$ is a coalgebra homomorphism, and $f$ is injective on $\mathrm{Prim}C = \left\langle e_1 \right\rangle$; but clearly, $f$ is not injective on the whole of $C$.