The Stacks project

Proof. The composition is isomorphic to the projection map $\text{pr}_1 : R \times _{t, U, t} R \to R$ by Diagram (40.3.0.2). The projection is open by Morphisms, Lemma 29.23.4. $\square$

Proof. By Groupoids, Lemma 39.7.3 the stabilizer group scheme $G \to U$ is separated. By Groupoids, Lemma 39.22.2 the morphism $j = (t, s) : R \to U \times _ S U$ is separated. As $U$ is the spectrum of a field the scheme $U \times _ S U$ is affine (by the construction of fibre products in Schemes, Section 26.17). Hence $R$ is a separated scheme, see Schemes, Lemma 26.21.12. $\square$

Proof. This is a special case of Lemma 40.7.1 parts (1) and (2). $\square$

Proof. The morphism $U' \to U$ equals $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$, hence is affine, surjective and flat. The morphisms $s, t : R \to U$ and the morphism $U' \to U$ are universally open by Morphisms, Lemma 29.23.4. Since $R$ is not empty and $U$ is the spectrum of a field the morphisms $s, t : R \to U$ are surjective and flat. Then you conclude by using Morphisms, Lemmas 29.9.4, 29.9.2, 29.23.3, 29.11.8, 29.11.7, 29.25.8, and 29.25.6. $\square$

Proof. Translating the geometric statement into a statement on fields, this means that we can find a diagram

where $i : k \to k'$ is the embedding of $k$ into $k'$, the maps $s, t : k \to \kappa (r)$ are induced by $s, t : R \to U$, and the map $\tau : k' \to k'$ is an automorphism. To produce such a diagram we may proceed in the following way:

1. Pick $i : k \to k'$ a field map with $k'$ algebraically closed of very large transcendence degree over $k$.

2. Pick an embedding $\sigma : \kappa (r) \to k'$ such that $\sigma \circ s = i$. Such a $\sigma $ exists because we can just choose a transcendence basis $\{ x_\alpha \} _{\alpha \in A}$ of $\kappa (r)$ over $k$ and find $y_\alpha \in k'$, $\alpha \in A$ which are algebraically independent over $i(k)$, and map $s(k)(\{ x_\alpha \} )$ into $k'$ by the rules $s(\lambda ) \mapsto i(\lambda )$ for $\lambda \in k$ and $x_\alpha \mapsto y_\alpha $ for $\alpha \in A$. Then extend to $\tau : \kappa (\alpha ) \to k'$ using that $k'$ is algebraically closed.

3. Pick an automorphism $\tau : k' \to k'$ such that $\tau \circ i = \sigma \circ t$. To do this pick a transcendence basis $\{ x_\alpha \} _{\alpha \in A}$ of $k$ over its prime field. On the one hand, extend $\{ i(x_\alpha )\} $ to a transcendence basis of $k'$ by adding $\{ y_\beta \} _{\beta \in B}$ and extend $\{ \sigma (t(x_\alpha ))\} $ to a transcendence basis of $k'$ by adding $\{ z_\gamma \} _{\gamma \in C}$. As $k'$ is algebraically closed we can extend the isomorphism $\sigma \circ t \circ i^{-1} : i(k) \to \sigma (t(k))$ to an isomorphism $\tau ' : \overline{i(k)} \to \overline{\sigma (t(k))}$ of their algebraic closures in $k'$. As $k'$ has large transcendence degree we see that the sets $B$ and $C$ have the same cardinality. Thus we can use a bijection $B \to C$ to extend $\tau '$ to an isomorphism

   \[ \overline{i(k)}(\{ y_\beta \} ) \longrightarrow \overline{\sigma (t(k))}(\{ z_\gamma \} ) \]

   and then since $k'$ is the algebraic closure of both sides we see that this extends to an automorphism $\tau : k' \to k'$ as desired.

This proves the lemma. $\square$

Proof. This is completely obvious if you think about groupoids in a functorial way. But we will also spell it out completely. Denote $a : U \to R$ the morphism with image $r$ such that $s \circ a = \text{id}_ U$ which exists by the hypothesis that $s : k \to \kappa (r)$ is an isomorphism. Similarly, denote $b : U \to R$ the morphism with image $r$ such that $t \circ b = \text{id}_ U$. Note that $b = a \circ (t \circ a)^{-1}$, in particular $a \circ s \circ b = b$.

Consider the morphism $\Psi : R \to R$ given on $T$-valued points by

To see this is defined we have to check that $s \circ a \circ t \circ f = t \circ f$ which is obvious as $s \circ a = 1$. Note that $\Phi (e) = a$, so that in order to prove the lemma it suffices to show that $\Phi $ is an automorphism of $R$. Let $\Phi : R \to R$ be the morphism given on $T$-valued points by

This is defined because $s \circ i \circ b \circ t \circ g = t \circ b \circ t \circ g = t \circ g$. We claim that $\Phi $ and $\Psi $ are inverse to each other. To see this we compute

where we have used the relation $a \circ s \circ b = b$ shown above. In the other direction we have

The lemma is proved. $\square$

Proof. Write $U = \mathop{\mathrm{Spec}}(k)$. Consider a field extension $k'/k$. Denote $U' = \mathop{\mathrm{Spec}}(k')$. Let $R'$ be the restriction of $R$ via $U' \to U$. Set $Z' = Z \times _ R R'$ and $W' = R' \times _ R W$. Consider a point $\xi = (z, w)$ of $Z \times _{s, U, t} W$. Let $r \in R$ be the image of $z$ under $Z \to R$. Pick $k' \supset k$ and $r' \in R'$ as in Lemma 40.10.5. We can choose $z' \in Z'$ mapping to $z$ and $r'$. Then we can find $\xi ' \in Z' \times _{s', U', t'} W'$ mapping to $z'$ and $\xi $. The open $c(r', W')$ (Lemma 40.10.6) is contained in the image of $Z' \times _{s', U', t'} W' \to R'$. Observe that $Z' \times _{s', U', t'} W' = (Z \times _{s, U, t} W) \times _{R \times _{s, U, t} R} (R' \times _{s', U', t'} R')$. Hence the image of $Z' \times _{s', U', t'} W' \to R' \to R$ is contained in the image of $Z \times _{s, U, t} W \to R$. As $R' \to R$ is open (Lemma 40.10.4) we conclude the image contains an open neighbourhood of the image of $\xi $ as desired. $\square$

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 $t$. $\square$

Proof. Let $r, r' \in R$. Then $\dim _ r(R) = \dim _{r'}(R)$ by Lemma 40.10.3 and Morphisms, Lemma 29.28.3. By Morphisms, Lemma 29.28.1 we have

On the other hand, the dimension of $R$ (or any open subset of $R$) is the supremum of the dimensions of the local rings of $R$, see Properties, Lemma 28.10.3. Clearly this is maximal for closed points $r$ in which case $\text{trdeg}_ k(\kappa (r)) = 0$ (by the Hilbert Nullstellensatz, see Morphisms, Section 29.16). Hence the lemma follows. $\square$

Proof. Let $Z \subset R$ be the irreducible component passing through $e$ (see Lemma 40.10.8) thought of as an integral closed subscheme of $R$. Let $k'_ s$, resp. $k'_ t$ be the integral closure of $s(k)$, resp. $t(k)$ in $\Gamma (Z, \mathcal{O}_ Z)$. Recall that $k'_ s$ and $k'_ t$ are fields, see Varieties, Lemma 33.28.4. By Varieties, Proposition 33.31.1 we have $k'_ s = k'_ t$ as subrings of $\Gamma (Z, \mathcal{O}_ Z)$. As $e$ factors through $Z$ we obtain a commutative diagram

This on the one hand shows that $k'_ s = s(k)$, $k'_ t = t(k)$, so $s(k) = t(k)$, which combined with the diagram above implies that $s = t$! In other words, we conclude that $Z$ is a closed subscheme of $G = R \times _{(t, s), U \times _ S U, \Delta } U$. The lemma follows as both $G$ and $R$ are equidimensional, see Lemma 40.10.9 and Groupoids, Lemma 39.8.1. $\square$

Proof. By Lemma 40.4.1 the sheaf of differentials of $R \to U$ is free. Hence smoothness follows from Varieties, Lemma 33.25.1. $\square$

Proof. By Lemma 40.4.1 the sheaf $\Omega _{R/U}$ is free. Hence the lemma follows from Varieties, Lemma 33.25.2. $\square$