## 40.5 Local structure

Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. Let $u \in U$ be a point. In this section we explain what kind of structure we obtain on the local rings

$A = \mathcal{O}_{U, u} \quad \text{and}\quad B = \mathcal{O}_{R, e(u)}$

The convention we will use is to denote the local ring homomorphisms induced by the morphisms $s, t, c, e, i$ by the corresponding letters. In particular we have a commutative diagram

$\xymatrix{ A \ar[rd]_ t \ar[rrd]^1 \\ & B \ar[r]^ e & A \\ A \ar[ru]^ s \ar[rru]_1 }$

of local rings. Thus if $I \subset B$ denotes the kernel of $e : B \to A$, then $B = s(A) \oplus I = t(A) \oplus I$. Let us denote

$C = \mathcal{O}_{R \times _{s, U, t} R, (e(u), e(u))}$

Then we have

$C = (B \otimes _{s, A, t} B)_{\mathfrak m_ B \otimes B + B \otimes \mathfrak m_ B}$

Let $J \subset C$ be the ideal of $C$ generated by $I \otimes B + B \otimes I$. Then $J$ is also the kernel of the local ring homomorphism

$(e, e) : C \longrightarrow A$

The composition law $c : R \times _{s, U, t} R \to R$ corresponds to a ring map

$c : B \longrightarrow C$

sending $I$ into $J$.

Lemma 40.5.1. The map $I/I^2 \to J/J^2$ induced by $c$ is the composition

$I/I^2 \xrightarrow {(1, 1)} I/I^2 \oplus I/I^2 \to J/J^2$

where the second arrow comes from the equality $J = (I \otimes B + B \otimes I)C$. The map $i : B \to B$ induces the map $-1 : I/I^2 \to I/I^2$.

Proof. To describe a local homomorphism from $C$ to another local ring it is enough to say what happens to elements of the form $b_1 \otimes b_2$. Keeping this in mind we have the two canonical maps

$e_2 : C \to B,\ b_1 \otimes b_2 \mapsto b_1s(e(b_2)),\quad e_1 : C \to B,\ b_1 \otimes b_2 \mapsto t(e(b_1))b_2$

corresponding to the embeddings $R \to R \times _{s, U, t} R$ given by $r \mapsto (r, e(s(r)))$ and $r \mapsto (e(t(r)), r)$. These maps define maps $J/J^2 \to I/I^2$ which jointly give an inverse to the map $I/I^2 \oplus I/I^2 \to J/J^2$ of the lemma. Thus to prove statement we only have to show that $e_1 \circ c : B \to B$ and $e_2 \circ c : B \to B$ are the identity maps. This follows from the fact that both compositions $R \to R \times _{s, U, t} R \to R$ are identities.

The statement on $i$ follows from the statement on $c$ and the fact that $c \circ (1, i) = e \circ t$. Some details omitted. $\square$

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