Let $G$ be a linear algebraic group over a field $k$, i. e. a functor from the category of $k$-algebras to the category of groups. Let $A = k[G]$ be the coordinate ring of $G$.
The definition of the derived group $\mathcal D(G)$ of $G$ goes as follows: We write $[a,b] = aba^{-1}b^{-1}$ for elements $a,b$ in some group. For all $n\in\mathbb N$ we have natural transformations $\psi_n\colon G^{2n}\rightarrow G$ given by $$ \psi_n(R)\colon G^{2n}(R) \rightarrow G(R),\quad (g_1,h_1,g_2,h_2,\dotsc,g_n,h_n)\mapsto [g_1,h_1][g_2,h_2] \dotsm [g_n,h_n] $$ for each $k$-algebra $R$. This induces by the Yoneda Lemma the pullback maps $\psi_n^*\colon A\rightarrow A^{\otimes 2n}$. Let $I_n := \ker(\psi_n^*)$. Then we have a descending chain of ideals $$ I_1\supseteq I_2\supseteq\dotsb \supseteq I_n\supseteq\dotsb $$ Finally, put $I:= \bigcap_{n\in\mathbb N}I_n$. I would like to understand, why this is a Hopf ideal.
I am well aware that none of the $I_n$’s need to be a Hopf ideal, but at least the counit and antipode of $A$ restrict to each of them and in particular to $I$. So my problem is to understand, why we have $\Delta(I) \subseteq A\otimes_kI + I\otimes_kA$ (as a subset of $A\otimes_kA$), where $\Delta\colon A\rightarrow A\otimes_kA$ is the comultiplication of $A$.
If $\mu\colon G^2\rightarrow G$ denotes the multiplication of $G$, then we have $\mu\circ(\psi_n\times \psi_n) = \psi_{2n}$ by definition and hence $\psi_{2n}^* = (\psi_n^*\otimes \psi_n^*) \circ\Delta$ by taking pullbacks. This explains why $$ \Delta(I_{2n}) \subseteq \ker(\psi_n^*\otimes \psi_n^*) = A\otimes_kI_n + I_n\otimes_kA\quad \text{for all $n\in\mathbb N$} $$ and hence $\Delta(I) \subseteq \bigcap_{n\in\mathbb N} (A\otimes_kI_n + I_n\otimes_kA)$. My question therefore boils down to: Why do we have $\bigcap_{n\in\mathbb N}(A\otimes_kI_n + I_n\otimes_kA) \subseteq A\otimes_kI + I\otimes_kA$?
I have tried looking it up in Milne’s Algebraic Groups, Lie Groups, and their Arithmetic Subgroups, but on page 188 he only explains that we have morphisms $\Delta\colon A/I_{2n}\rightarrow A/I_n \otimes_k A/I_n$ and hence(?!) $\Delta\colon A\rightarrow A/I \otimes_kA/I$ factors through $A/I$.