Let $H$ be a Hopf algebra, and $V$ and $W$ a left, and a right, 
$H$-comodule respectively. The tensor product  

$$
V \otimes W
$$
has an obvious $H$-$H$-bicomodule structure.
If $V$ and $W$ are irreducible as left and right comodules, then  

is $V \otimes W$ irreducible as a $H$-$H$-bicomodule?



Irreducibility of product bicomodules
No, off course, not. You need a slightly stronger condition. One of your comodules need to be absolutely irreducible.


To prove it, apply the fundamental theorem of coalgebra. The comodules give you simple subcoalgebras $C\_M$ and $C\_N$ in the coradical of $H$. You are asking for the algebra $(C^{\ast}\_M)^{op}\otimes C^{\ast}\_N$ to be simple.


For an elementary example let ${\mathbb R}$ be the ground field, $H$ the free cocommutative Hop algebra generated by the trigonometric coalgebra ${\mathbb C}^{\ast}$. Let $M=N={\mathbb C}$ be the $\mathbb C$-module, hence, ${\mathbb C}^\ast$-comodule, hence $H$-comodule. The $H$-bicomodule structure on $M\otimes N$ is essentially its module structure over ${\mathbb C}\otimes\_{\mathbb R}{\mathbb C}\cong {\mathbb C}\oplus{\mathbb C}$. This is not irreducible because $dim(M\otimes N)=4$ but its simple modules are 2-dimensional.


