Lemma 40.10.8. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(k)$ with $k$ a field. By abuse of notation denote $e \in R$ the image of the identity morphism $e : U \to R$. Then

every local ring $\mathcal{O}_{R, r}$ of $R$ has a unique minimal prime ideal,

there is exactly one irreducible component $Z$ of $R$ passing through $e$, and

$Z$ is geometrically irreducible over $k$ via either $s$ or $t$.

**Proof.**
Let $r \in R$ be a point. In this proof we will use the correspondence between irreducible components of $R$ passing through a point $r$ and minimal primes of the local ring $\mathcal{O}_{R, r}$ without further mention. Choose $k \subset k'$ and $r' \in R'$ as in Lemma 40.10.5. Note that $\mathcal{O}_{R, r} \to \mathcal{O}_{R', r'}$ is faithfully flat and local, see Lemma 40.10.4. Hence the result for $r' \in R'$ implies the result for $r \in R$. In other words we may assume that $s, t : k \to \kappa (r)$ are isomorphisms. By Lemma 40.10.6 there exists an automorphism moving $e$ to $r$. Hence we may assume $r = e$, i.e., part (1) follows from part (2).

We first prove (2) in case $k$ is separably algebraically closed. Namely, let $X, Y \subset R$ be irreducible components passing through $e$. Then by Varieties, Lemma 33.8.4 and 33.8.3 the scheme $X \times _{s, U, t} Y$ is irreducible as well. Hence $c(X \times _{s, U, t} Y) \subset R$ is an irreducible subset. We claim it contains both $X$ and $Y$ (as subsets of $R$). Namely, let $T$ be the spectrum of a field. If $x : T \to X$ is a $T$-valued point of $X$, then $c(x, e \circ s \circ x) = x$ and $e \circ s \circ x$ factors through $Y$ as $e \in Y$. Similarly for points of $Y$. This clearly implies that $X = Y$, i.e., there is a unique irreducible component of $R$ passing through $e$.

Proof of (2) and (3) in general. Let $k \subset k'$ be a separable algebraic closure, and let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ via $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. By the previous paragraph there is exactly one irreducible component $Z'$ of $R'$ passing through $e'$. Denote $e'' \in R \times _{s, U} U'$ the base change of $e$. As $R' \to R \times _{s, U} U'$ is faithfully flat, see Lemma 40.10.4, and $e' \mapsto e''$ we see that there is exactly one irreducible component $Z''$ of $R \times _{s, k} k'$ passing through $e''$. This implies, as $R \times _ k k' \to R$ is faithfully flat, that there is exactly one irreducible component $Z$ of $R$ passing through $e$. This proves (2).

To prove (3) let $Z''' \subset R \times _ k k'$ be an arbitrary irreducible component of $Z \times _ k k'$. By Varieties, Lemma 33.8.13 we see that $Z''' = \sigma (Z'')$ for some $\sigma \in \text{Gal}(k'/k)$. Since $\sigma (e'') = e''$ we see that $e'' \in Z'''$ and hence $Z''' = Z''$. This means that $Z$ is geometrically irreducible over $\mathop{\mathrm{Spec}}(k)$ via the morphism $s$. The same argument implies that $Z$ is geometrically irreducible over $\mathop{\mathrm{Spec}}(k)$ via the morphism $t$.
$\square$

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