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](https://mathoverflow.net/questions/46934/zariski-closed-subsemigroups-of-sl-nc-are-groups). Does anyone see a proof or a quick counterexample?



Is a biideal of a Noetherian Hopf algebra automatically Hopf?
The answer is yes.


In ["Quotients of Hopf Algebras"](https://doi.org/10.1080/00927877808822321 "zbMATH review at https://zbmath.org/0391.16006"), 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 cocommutative


What you want is a generalization of the above, see corollary 2.4 in ["The largest Hopf subalgebra of a bialgebra"](https://doi.org/10.1134/S0001434609110327 "Math. Notes 86, No. 6, 887-891 (2009); translation from Mat. Zametki 86, No. 6, 942-946 (2009). zbMATH review at https://zbmath.org/1184.16033") 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.


