## Is a biideal of a Noetherian Hopf algebra automatically Hopf?

Let $H$ be a Hopf algebra over a field $k$, and $I$ be a biideal of $H$. I am looking for conditions that guarantee that $I$ is a Hopf ideal (that means $S\left(I\right)\subseteq I$).

One condition that definitely works is that $\dim\left(H / I\right) < \infty$ (where $\dim$ means dimension as a $k$-vector space). This is a well-known consequence of the criterion that a bialgebra $A$ is a Hopf algebra if and only if the $k$-linear map $A\otimes A\to A\otimes A,\ x\otimes y\mapsto xy_{(1)}\otimes y_{(2)}$ is bijective. (This criterion must be applied to $H$ and $H / I$.)

I suspect that some kind of Noetherianness of $H$ or $H / I$ (note that I am not specifying left or right or bi, because I have no idea) would make another criterion. My suspicion is based on the commutative $H$ case. Does anyone see a proof or a quick counterexample?

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In "Quotients of Hopf Algebras", Warren D. Nichols, Comm. Algebra 6(1978), 1789-1800, proves that if $I$ is a bi-ideal then it will be a Hopf ideal under any of the following conditions:

• $H/I$ is finite dimensional
• $H/I$ is commutative
• $H$ is pointed
• $H$ is cocomutative
• What you want is a generalization of the above, see corollary 2.4 in "The largest Hopf subalgebra of a bialgebra" by M. S. Eryashkin and S. M. Skryabin. They prove that a bi-ideal $I$ is a Hopf ideal provided that $H/I$ is weakly finite. In particular every bi-ideal is a Hopf ideal when $H$ is left or right Noetherian, or when $H$ satisfies a polynomial identity.

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