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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