Lemma 79.9.8. In Situation 79.9.2 the algebraic space R is geometrically unibranch. In Situation 79.9.1 the algebraic space G is geometrically unibranch.
Proof. Let r \in |R|. We have to show that R is geometrically unibranch at r. Combining Lemma 79.9.5 with Descent on Spaces, Lemma 74.9.1 we see that it suffices to prove this in case k is algebraically closed and r comes from a morphism r : \mathop{\mathrm{Spec}}(k) \to R such that s \circ r and t \circ r are automorphisms of \mathop{\mathrm{Spec}}(k). By Lemma 79.9.7 we reduce to the case that r = e is the identity of R and k is algebraically closed.
Assume r = e and k is algebraically closed. Let A = \mathcal{O}_{R, e} be the étale local ring of R at e and let C = \mathcal{O}_{R \times _{s, U, t} R, (e, e)} be the étale local ring of R \times _{s, U, t} R at (e, e). By More on Algebra, Lemma 15.107.4 the minimal prime ideals \mathfrak q of C correspond 1-to-1 to pairs of minimal primes \mathfrak p, \mathfrak p' \subset A. On the other hand, the composition law induces a flat ring map
\xymatrix{ A \ar[r]_{c^\sharp } & C & \mathfrak q \\ & A \otimes _{s^\sharp , k, t^\sharp } A \ar[u] & \mathfrak p \otimes A + A \otimes \mathfrak p' \ar@{|}[u] }
Note that (c^\sharp )^{-1}(\mathfrak q) contains both \mathfrak p and \mathfrak p' as the diagrams
\xymatrix{ A \ar[r]_{c^\sharp } & C \\ A \otimes _{s^\sharp , k} k \ar[u] & A \otimes _{s^\sharp , k, t^\sharp } A \ar[l]_{1 \otimes e^\sharp } \ar[u] } \quad \quad \xymatrix{ A \ar[r]_{c^\sharp } & C \\ k \otimes _{k, t^\sharp } A \ar[u] & A \otimes _{s^\sharp , k, t^\sharp } A \ar[l]_{e^\sharp \otimes 1} \ar[u] }
commute by (79.3.0.1). Since c^\sharp is flat (as c is a flat morphism by Lemma 79.9.3), we see that (c^\sharp )^{-1}(\mathfrak q) is a minimal prime of A. Hence \mathfrak p = (c^\sharp )^{-1}(\mathfrak q) = \mathfrak p'. \square
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