# Is a coalgebra comodule cosemisimple if and only if every subcomodule is a direct summand?

It is well known that if $$R$$ is a ring, then every $$R$$-module $$M$$ is semisimple (that is, $$M$$ is the direct sum of simple $$R$$-modules) if and only if every submodule of $$M$$ is a direct summand. Is it true that if $$C$$ is a coalgebra over a field $$k$$, then the analogous result holds: a coalgebra comodule $$N$$ is cosemisimple if and only if every subcomodule is a direct summand?

Yes, if your coalgebra is flat over the base ring $$\Bbbk$$ (for example when $$\Bbbk$$ is a field). You can find it in Brzezinski-Wisbauer, Corings and Comodules, §19.13 and results around.
Broadly speaking, if $$P$$ is a simple subcomodule of $$N$$ and $$M$$ is any subcomodule of $$N$$ then $$P\cap M$$ is still a subcomodule and it can only be $$0$$ or $$P$$. Therefore, if $$N$$ is cosemisimple then $$M$$ is the direct sum of the simple subcomodules it contains and hence is a direct summand of $$N$$ and conversely, if every subcomodule is a direct summand, then you check iteratively that $$N$$ is the direct sum of its simple subcomodules.
I would dare to say that the only difference is that in this case you don't know that $$N$$ is the direct sum of a finite number of summands, as it happens for modules over a semisimple ring. However, I am not completely sure of this claim at the present moment so, please, check and forgive me if I am wrong.