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](https://math.stackexchange.com/questions/4064375/is-the-kernel-of-an-action-of-a-hopf-algebra-on-an-algebra-a-biideal?noredirect=1#comment8397383_4064375), but without success.


[S.Dascalescu, C.Nastasescu and S.Raianu](https://rads.stackoverflow.com/amzn/click/com/0824704819) 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](https://ncatlab.org/nlab/show/biideal) in $H$ as in a bialgebra, i.e.
$$
\Delta(I)\subseteq I\otimes H+H\otimes I.
$$
Is this true?



Is the kernel of an action of a Hopf algebra on an algebra a biideal?
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.)


