# Coalgebra requirement?

If $({C},\Delta,\epsilon)$ is a K-Coalgebra, does K need to be a field? Is it okay for it to be a commutative ring??? Since it is okay to define an algebra A over a commutative ring, is okay to do the same for a coalgebra?

1. Yes, it is okay to define a coalgebra over a commutative ring (with 1), but prepare for nasty surprises when you start working with subcoalgebras, coideals and quotient coalgebras. The problem is that if $D$ is a $K$-submodule of a coalgebra $C$ satisfying $\Delta (D) \subseteq D \otimes D$ (where $D\otimes D$ is understood as the $K$-submodule of $C$ generated by all $d\otimes e$ with $d,e\in D$), then this does not make $D$ into a coalgebra… because the tensor product of injective $K$-linear maps needs no longer be $K$-injective. See page 56 of Robert Morris (ed.), “Umbral calculus and Hopf algebras”, Contemporary Mathematics vol. 6, AMS (easily found online if you know where to look) for a particular counterexample. The basic results of coalgebra theory can still be salvaged, though, if one restricts oneself to flat modules, or projective modules, or free modules, and since most coalgebras that appear in combinatorics are free, this is more than enough (but one should keep in mind that submodules of free modules needn’t be free…).
2. $K$-submodule of $C$” meant “$K$-submodule of $C\otimes C$“, sorry.