Prove that $\Delta(x) = x \otimes1 + 1 \otimes x$ mod $I \otimes I $ for all $x \in I$.

Question

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.

Answer 1

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}