Is the kernel of an action of a Hopf algebra on an algebra a biideal? I think, this must be simple, but I am not a specialist in this field, so excuse me. I asked this a week ago at MSE, but without success.
S.Dascalescu, C.Nastasescu and S.Raianu define the action of a Hopf algebra $H$ on an (associative) algebra $A$ as a map $H\times A\owns (h,a)\mapsto h\cdot a\in A$ which

*

*is an action of $H$ on $A$ as an algebra on a vector space, and

*satisfies two supplementary conditions:
$$
h\cdot(a\cdot b)=\sum(h_1\cdot a)\cdot (h_2\cdot b),\qquad 
h\cdot 1_A=\varepsilon(h)\cdot 1_A.
$$
(where $\sum h_1\otimes h_2=\Delta(h)$ is the comultiplication, and $\varepsilon$, the counit).

I think that the kernel of such an action, i.e. the set
$$
I=\{h\in H:\quad \forall a\in A\quad h\cdot a=0\},
$$
must be (not only a two-sided ideal in $H$ as in an algebra, but also) a biideal in $H$ as in a bialgebra, i.e.
$$
\Delta(I)\subseteq I\otimes H+H\otimes I.
$$
Is this true?
 A: No, it is not true (in general).
For a specific counterexample, let me first recall that what you call an
algebra $A$ equipped with an action of $H$ is the same as what is classically
called a (left) $H$-module algebra. When $H$ is the group algebra $k\left[
G\right]  $ of a finite group (where $k$ is the base field), this is the same
as a $k$-algebra on which the group $G$ acts by $k$-algebra automorphisms.
(Indeed, your two "supplementary conditions" boil down to $g\cdot\left(
a\cdot b\right)  =\left(  g\cdot a\right)  \cdot\left(  g\cdot b\right)  $ and
$g\cdot1_{A}=1_{A}$ for all $g\in G$ in this case.) My counterexample will use
a group algebra, so we can forget about Hopf algebras and just think about
groups acting on algebras by automorphisms instead.
Consider the symmetric group $S_{3}$ (of size $6$) acting on the polynomial
ring $k\left[  x,y,z\right]  $ by permuting the three variables. Let $A$ be
the quotient ring of $k\left[  x,y,z\right]  $ by the ideal generated by all
degree-$2$ monomials. Thus, $A$ is a commutative $k$-algebra; as a $k$-vector
space, it has a basis $\left(  \overline{1},\overline{x},\overline
{y},\overline{z}\right)  $, with multiplication given by $\overline{x}
^{2}=\overline{x}\cdot\overline{y}=\overline{x}\cdot\overline{z}=\overline
{y}^{2}=\overline{y}\cdot\overline{z}=\overline{z}^{2}=0$.
Let $\alpha\in k\left[  S_{3}\right]  $ be the element $\sum_{\sigma\in S_{3}
}\left(  -1\right)  ^{\sigma}\sigma$ (where $\left(  -1\right)  ^{\sigma}$
denotes the sign of a permutation $\sigma$). We call this element $\alpha$ the
antisymmetrizer. Let $\left\langle \alpha\right\rangle $ denote the
$k$-vector subspace of $k\left[  S_{3}\right]  $ spanned by $\alpha$. This
subspace $\left\langle \alpha\right\rangle $ is actually an ideal of the
$k$-algebra $k\left[  S_{3}\right]  $, since each $\sigma\in S_{3}$ satisfies
$\alpha\sigma=\sigma\alpha=\left(  -1\right)  ^{\sigma}\alpha$.
However, this subspace $\left\langle \alpha\right\rangle $ is not a coideal of
the Hopf algebra $k\left[  S_{3}\right]  $. The easiest way I know to check
this is to show that the orthogonal space $\left\langle \alpha\right\rangle
^{\perp}$ in the dual algebra $\left(  k\left[  S_{3}\right]  \right)  ^{\ast
}\cong k^{S_{3}}$ (which is a Cartesian product of $6$ copies of $k$, indexed
by the permutations $\sigma\in S_{3}$) is not a subalgebra of $k^{S_{3}}$ (but
the orthogonal space of a coideal of a coalgebra in the dual algebra is always
a subalgebra of this dual algebra). The latter fact is pretty easy to see,
since
\begin{align*}
\left\langle \alpha\right\rangle ^{\perp}  & =\left\{  \left(  p_{\sigma
}\right)  _{\sigma\in S_{3}}\ \mid\ \sum_{\sigma\in S_{3}}\left(  -1\right)
^{\sigma}p_{\sigma}=0\right\}  \\
& =\left\{  \left(  p_{1},p_{2},p_{3},p_{4},p_{5},p_{6}\right)  \in
k^{6}\ \mid\ p_{1}+p_{3}+p_{5}=p_{2}+p_{4}+p_{6}\right\}
\end{align*}
(where we have numbered the elements of $S_{3}$ by $1,2,3,4,5,6$ in such a way
that even permutations correspond to even numbers) is not closed under
(entrywise) multiplication.
However, it is easy to see that the set $I$ you defined is precisely
$\left\langle \alpha\right\rangle $. (Indeed, if $\sum_{\sigma\in S_{3}
}\left(  -1\right)  ^{\sigma}p_{\sigma}\sigma$ is an element of $I$ with
$p_{\sigma}\in k$, then the definition of $I$ yields $\sum_{\sigma\in S_{3}
}\left(  -1\right)  ^{\sigma}p_{\sigma}\sigma\cdot\overline{x}=0$ and
$\sum_{\sigma\in S_{3}}\left(  -1\right)  ^{\sigma}p_{\sigma}\sigma
\cdot\overline{y}=0$ and $\sum_{\sigma\in S_{3}}\left(  -1\right)  ^{\sigma
}p_{\sigma}\sigma\cdot\overline{z}=0$; but these easily entail $p_{\sigma
}=p_{\sigma s_{1}}=p_{\sigma s_{2}}$ for all $\sigma\in S_{3}$, and therefore
the coefficients $p_{\sigma}$ are all equal, which shows that $\sum_{\sigma\in
S_{3}}\left(  -1\right)  ^{\sigma}p_{\sigma}\sigma\in\left\langle
\alpha\right\rangle $. Conversely, it is easy to see that $\left\langle
\alpha\right\rangle \subseteq I$.)
Now, we know that $I=\left\langle \alpha\right\rangle $ is not a coideal of
$k\left[  S_{3}\right]  $, and hence not a biideal either.
That said, it is closed under the antipode. I'm wondering if this generalizes?
EDIT: Here is an easier way to generate counterexamples:
Let $H$ be any Hopf algebra, and let $V$ be any left $H$-module. Let $A$ be the commutative algebra $k \times V$ (where $k$ is the base field), with addition being defined entrywise and with multiplication given by $\left(\lambda, v\right)\left(\mu, w\right) = \left(\lambda\mu, \lambda w + \mu v\right)$. (This is the famous Dorroh extension, or square-zero extension, of $V$ over the base ring $k$. Note that the product of any two elements of $V$ in $A$ is $0$.) Then, $A$ becomes a left $H$-module algebra (by setting $h \cdot \left(\lambda, v\right) = \left(\varepsilon\left(h\right)\lambda, h\cdot v\right)$ for any $\left(\lambda, v\right) \in k \times V = A$). Your ideal $I$ is then the intersection of the annihilator of $V$ with the augmentation ideal of $H$. But any ideal of $H$ can be written as the annihilator of an appropriate left $H$-module. Thus, any ideal of $H$ that is contained in the augmentation of ideal of $H$ can be written as $I$ for an appropriate choice of $A$. Now it is easy to construct an example of an ideal of $H$ that is contained in the augmentation ideal but is neither a coideal nor closed under the antipode. Here is an explicit such example: Let $k = \mathbb{C}$, let $C_3 = \left\{1,u,u^2\right\}$ be the cyclic group with generator $u$, let $H = k\left[C_3\right]$ be its group algebra over $k = \mathbb{C}$, and let $I$ be the ideal of $H$ generated (and also spanned) by the single element $1 + \zeta u + \zeta^2 u^2$, where $\zeta = e^{2\pi i/3}$ is a primitive $3$-rd root of uniy. This ideal $I$ is contained in the augmentation ideal but is neither a coideal nor closed under the antipode. (Fun and helpful fact: If $I$ is a biideal of a Hopf algebra $H$ over a field $k$, and if the $k$-vector space $H/I$ is finite-dimensional, then $I$ is a Hopf ideal of $H$. Thus, showing that the $I$ in the above example is not closed under the antipode automatically reveals that it is not a coideal.)
