# Irreducibility of product bicomodules

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?

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.
• Sorry but I don't see why what I am asking for is equivalent to $(C^*_M)^{op} \otimes C^*_N$ being simple. (Also, I guess $C^*_M$ denotes the linear dual of $C^*_M$ endowed with convolution multiplication?) – Jake Wetlock Apr 14 '20 at 11:29
• You are asking for $M\otimes N$ to be a simple $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$-module. And you know that $M\otimes N$ is a faithful finite-dimensional $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$-module. Now simplicity of the algebra $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$ implies simplicity of the module $M\otimes N$. In the opposite direction, hit it with Jacobson Density Theorem. – Bugs Bunny Apr 14 '20 at 11:59
• Yes or no, since you have a typo: $C^{\ast}_M$ is the linear dual of coalgebra $C_M$. – Bugs Bunny Apr 14 '20 at 12:01
• Yes, it should read "$C^*_M$ denotes the linear dual of $C_M$". – Jake Wetlock Apr 14 '20 at 13:04