Understanding a map in "Algebraic Group Schemes in characteristic 0 are reduced" Let $G$ be a group scheme of finite type over a field $k$.
In his paper Algebraic Group Schemes in characteristic 0 are reduced, Oort says that since $G$ is not affine a homomorphism $\mathscr{O}_{G}\to \mathscr{O}_G\otimes \mathscr{O}_G$ does not exist in general. But for $e\in G$, the identity element of $G$ and $\mathfrak{m}\subseteq \mathscr{O}_{G,e}$ the maximal ideal, the multpilication map $G\times G\to G$, induces a ring homomorphism
$\mathscr{O}_{G,e}\to \mathscr{O}_{G,e}/\mathfrak{m}^q\otimes \mathscr{O}_{G,e}/\mathfrak{m}^q$ (for any integer $q$).
My question: How is this map defined?
 A: Comment: If $e∈U:=Spec(A)⊆G$ is open affine, let $p⊆A⊗_k A$ be the maximal ideal of $(e,e)$ and let $q⊆A$ be the ideal of $e$. There is a canonical map $A_q→(A⊗_k A)_p$. But there appears to be no natural map $(A⊗_k A)_p→A_q⊗_k A_q$. If there was such a map you could compose with the map
$$A_q⊗_k A_q→A_q/q^lA_q⊗_k A_q/q^lA_q$$
and arrive at a definition of your map. There is a map in the other direction
$$A_q⊗_k A_q→(A⊗_k A)_p.$$
Your comment: I totally agree, this is the reason for my question. In general $(A\otimes_k A)_p$ is a localization of $A_q⊗_k A_q$. So what is the reason for this? A somewhat similar argument appears in Mumford's "Lectures on Curves on an Algebraic Surface" Lecture 25.
Comment: Look for an example where the map $(A⊗_k A)_p→A_q⊗_k A_q$ does not exist. If you find an example, you have a counterexample.
There is a canonical map
$$\phi:A\otimes_k A \rightarrow A_q \otimes_k A_q$$
and if $S:=A\otimes_k A-p$ has the property that $\phi(S)$ consists of units, there is an induced map $(A\otimes_k A)_p \rightarrow A_q \otimes_k A_q$.
