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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?

hm2020
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user4231
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  • note that the tensor product of two local rings is not a local ring: $k[x]/(x^2)\otimes_k k[x]/(x^2)$ is not a local ring. – hm2020 Aug 27 '21 at 08:14
  • There is for any group scheme $G$ an isomorphism of tangent space $T_{(e,e)}(G\times G) \cong T_e(G)\oplus T_e(G)$. There is a similar isomorphism for cotangent-spaces: https://math.stackexchange.com/questions/3541848/tangent-space-of-a-product-of-algebraic-group/3987723#3987723 – hm2020 Aug 27 '21 at 09:59
  • @hm2020 - can you elaborate on the connection to the question? – user4231 Aug 27 '21 at 10:38
  • for any ideal $I\subseteq A$ where $A$ is a commutative $k$-algebra, there are several maps $A \rightarrow A/I\otimes_k A/I$: $p(a):=a\otimes 1, q(a):=1\otimes a$. You should explain the construction in Oorts paper - how does Oort arrive at the specified map? – hm2020 Aug 27 '21 at 14:06
  • He says that this map is induced by the multiplication map $G\times G\to G$. – user4231 Aug 27 '21 at 14:22
  • If you assume that $e$ is a $k$-rational point it follows $\mathcal{P}^l_{G/k}\otimes_{\mathcal{O}{G,e}} \kappa(e) \cong \mathcal{O}{G,e}/\mathfrak{m}e^{l+1}$, where $\mathcal{P}^l{G/k}$is the l'th jet bundle of $G$ (= a higher order module of Kahler differentials) - maybe this can be used. – hm2020 Aug 27 '21 at 14:43
  • For the module of Kahler differentials there is always an map induced by the multiplication: $m^*\Omega^1_{G/k} \rightarrow \Omega^1_{G\times G/k}$. There are similar maps for higher order jet bundles. – hm2020 Aug 27 '21 at 14:44
  • A problem is that the cotangent space is "additive" in the following sense: There is an isomorphism $\Omega^1_{G\times G/k}\otimes \kappa(e,e) \cong \Omega^1_{G/k}\otimes \kappa(e) \oplus \Omega^1_{G/k}\otimes \kappa(e)$ (see the above post). – hm2020 Aug 27 '21 at 14:46
  • The cotangent space of the product is the direct sum of the cotangent space of the factors. There is an isomorphism of $\kappa(e)$-vector spaces $\Omega^1_{G/k}\otimes \kappa(e) \cong \mathfrak{m}_e/\mathfrak{m}_e^2$. – hm2020 Aug 27 '21 at 14:46
  • If $e\in U:=Spec(A)\subseteq G$ is open affine, let $\mathfrak{p} \subseteq A\otimes_k A$ be the maximal ideal of $(e,e)$ and let $\mathfrak{q} \subseteq A$ be the ideal of $e$. There is a canonical map $A_{\mathfrak{q}} \rightarrow (A\otimes_k A){\mathfrak{p}}$. But there appears to be no natural map $(A\otimes_k A){\mathfrak{p}} \rightarrow A_{\mathfrak{q}}\otimes_k A_{\mathfrak{q}}$. If there was such a map you could compose with the map – hm2020 Aug 27 '21 at 15:01
  • $A_{\mathfrak{q}}\otimes_k A_{\mathfrak{q}} \rightarrow A_{\mathfrak{q}}/ \mathfrak{q}^lA_{\mathfrak{q}} \otimes_k A_{\mathfrak{q}}/\mathfrak{q}^lA_{\mathfrak{q}} $ and arrive at a definition of your map. There is a map in the other direction $A_{\mathfrak{q}}\otimes_k A_{\mathfrak{q}} \rightarrow (A\otimes_k A)_{\mathfrak{p}}$. – hm2020 Aug 27 '21 at 15:03
  • I totally agree, this is the reason for my question. In general $(A\times_k A)_{\mathfrak{p}}$ is a localization of $A_q\otimes A_q$. So what is the reason for this? A somewhat similar argument appears in Mumford's "Lectures on Curves on an Algebraic Surface" Lecture 25. – user4231 Aug 27 '21 at 15:26
  • Look for an example where the map $(A\otimes_k A){\mathfrak{p}} \rightarrow A{\mathfrak{q}}\otimes_k A_{\mathfrak{q}}$ does not exist. If you find an example, you have a counterexample. – hm2020 Aug 28 '21 at 10:30

1 Answers1

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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 $(A\otimes_k A)_p \rightarrow A_q \otimes_k A_q$.

hm2020
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