Prove that $\Delta(x) = x \otimes1 + 1 \otimes x$ mod $I \otimes I $ for all $x \in I$. I got stuck on this problem, so if anyone can give me a hint on this, I really appreciate.
Let $I$ be the augmentation ideal in Hopf algebra $A$. Prove that $\Delta(x) = x \otimes1 + 1 \otimes x$
mod $I \otimes I $ for all $x \in I$.
Because $I$ is a Hopf ideal, so $\Delta(I) \subset I \otimes A + A \otimes I$. So I can write $\Delta(x) = \sum a_i \otimes b_i + \sum c_i \otimes d_i$, here $a_i, d_i \in I$ and $b_i, c_i \in A$, and the sums are finite. I applied $\epsilon \otimes id$ to have a presentation $x = \sum \epsilon(c_i) \Delta(d_i)$. And I got stuck here. Thanks in advance for your help.
 A: Since the discussion in the chat is getting too long, let's write down our final conclusions in an answer. We first observe the following:

*

*Since $(A,\Delta,\epsilon)$ is a coalgebra, the composites
\begin{aligned}
A &\xrightarrow{\Delta} A \otimes A \xrightarrow{\epsilon \otimes \operatorname{id}} k \otimes A \cong A \\
A &\xrightarrow{\Delta} A \otimes A \xrightarrow{\operatorname{id}\otimes \epsilon} A \otimes k \cong A
\end{aligned}
are the identity of $A$, i.e. we have
\begin{aligned}
(\epsilon \otimes \operatorname{id})(\Delta(x)) = 1 \otimes x \hspace{20pt} \text{and} \hspace{20pt} (\operatorname{id} \otimes \epsilon)(\Delta(x)) = x \otimes 1.
\end{aligned}

*Since $\epsilon: A \to k$ is an algebra homomorphism, we have $\epsilon(1) = 1$.

It thus follows that
\begin{aligned}
(\epsilon \otimes \operatorname{id})(\Delta(x) - x \otimes 1 - 1 \otimes x) &= 
1 \otimes x - 0 \otimes 1 - 1 \otimes x = 0 \\
(\operatorname{id} \otimes \epsilon)(\Delta(x) - x \otimes 1 - 1 \otimes x) &= 
x \otimes 1 - 1 \otimes 0 - x \otimes 1 = 0
\end{aligned}
and thus $\Delta(x) - x \otimes 1 - 1 \otimes x \in \ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon)$. It thus remains to prove that
\begin{aligned}
\ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon) = I \otimes I.
\end{aligned}
Edit (updated after it turned out $k$ is not assumed to be a field.)
We claim that the short exact sequence
\begin{aligned}
0 \rightarrow I \rightarrow A \xrightarrow{\epsilon} k \rightarrow 0
\end{aligned}
splits. Indeed, the condition $\epsilon(1) = 1$ precisely tells us that the composition
\begin{aligned}
k \to A \xrightarrow{\epsilon} k
\end{aligned}
is the identity. Since split short exact sequences are closed under tensoring, we can tensor the sequence with $A$ to get a short exact sequence
\begin{aligned}
0 \rightarrow I \otimes A \rightarrow A \otimes A \xrightarrow{\epsilon \otimes \operatorname{id}} k \otimes A \rightarrow 0,
\end{aligned}
proving that $\ker(\epsilon \otimes \operatorname{id}) = I \otimes A$. Similarly we can tensor the above sequence on the left with $I$, giving a short exact sequence
\begin{aligned}
0 \rightarrow I \otimes I \rightarrow I \otimes A \xrightarrow{\operatorname{id} \otimes \epsilon} I \otimes k \rightarrow 0,
\end{aligned}
which shows that
\begin{aligned}
\ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon) = I \otimes I.
\end{aligned}
