Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

The answer is yes.

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.

    share|improve this answer

    Your Answer

     
    discard

    By posting your answer, you agree to the privacy policy and terms of service.

    Not the answer you're looking for? Browse other questions tagged or ask your own question.