If $R$ is a commutative ring, is the kernel of any coalgebra homomorphism $f:C\to D$ a (two sided) coideal of $C$?
For $R$ a field this is the case, since we have $(f\otimes f)\circ\Delta_C=\Delta_D\circ f=0$ so that $\Delta_C(\text{ker}(f))\subseteq \text{ker}(f\otimes f)=\text{ker}(f)\otimes C+C\otimes\text{ker}(f)$. But for general rings $R$ this argument does not work, since we need not have $\text{ker}(f\otimes f)=\text{ker}(f)\otimes C+C\otimes\text{ker}(f)$.
It depends on what you mean by coideal. In Brzezinski, Wisbauer, Corings and Comodules, a coideal in a coalgebra $C$ over a commutative ring $R$ is exactly the kernel of a surjective coalgebra morphism (see chap 1 §2.3). Since the image of any coalgebra morphism is always a subcoalgebra of the codomain (see chap 1 §2.8), every kernel of a coalgebra morphism is a coideal.
However, the condition to check that an $R$-submodule $K$ of $C$ is a coideal now states that (see chap 1 §2.4) $$K \text{ coideal} \iff \Delta(K)\subseteq\mathsf{ker}(p\otimes p)\text{ and }\varepsilon(K)=0,$$ where $p:C\to C/K$ is the canonical projection.
The analogue of the "field-like" condition holds only under some additional assumptions (see §2.4 again).
