The Stacks project

Lemma 39.23.7. Let $S$ be a scheme. Let $f : (U', R', s', t') \to (U, R, s, t, c)$ be a morphism of groupoid schemes over $S$.

  1. $U$, $R$, $U'$, $R'$ are affine,

  2. $s, t, s', t'$ are finite locally free,

  3. the diagrams

    \[ \xymatrix{ R' \ar[d]_{s'} \ar[r]_ f & R \ar[d]^ s \\ U' \ar[r]^ f & U } \quad \quad \xymatrix{ R' \ar[d]_{t'} \ar[r]_ f & R \ar[d]^ t \\ U' \ar[r]^ f & U } \quad \quad \xymatrix{ G' \ar[d] \ar[r]_ f & G \ar[d] \\ U' \ar[r]^ f & U } \]

    are cartesian where $G$ and $G'$ are the stabilizer group schemes, and

  4. $f : U' \to U$ is étale.

Then the map $C \to C'$ from the $R$-invariant functions on $U$ to the $R'$-invariant functions on $U'$ is étale and $U' = \mathop{\mathrm{Spec}}(C') \times _{\mathop{\mathrm{Spec}}(C)} U$.

Proof. Set $M = \mathop{\mathrm{Spec}}(C)$ and $M' = \mathop{\mathrm{Spec}}(C')$. Write $U = \mathop{\mathrm{Spec}}(A)$, $U' = \mathop{\mathrm{Spec}}(A')$, $R = \mathop{\mathrm{Spec}}(B)$, and $R' = \mathop{\mathrm{Spec}}(B')$. We will use the results of Lemmas 39.23.4, 39.23.5, and 39.23.6 without further mention.

Assume $C$ is a strictly henselian local ring. Let $p \in M$ be the closed point and let $p' \in M'$ map to $p$. Claim: in this case there is a disjoint union decomposition $(U', R', s', t', c') = (U, R, s, t, c) \amalg (U'', R'', s'', t'', c'')$ over $(U, R, s, t, c)$ such that for the corresponding disjoint union decomposition $M' = M \amalg M''$ over $M$ the point $p'$ corresponds to $p \in M$.

The claim implies the lemma. Suppose that $M_1 \to M$ is a flat morphism of affine schemes. Then we can base change everything to $M_1$ without affecting the hypotheses (1) – (4). From Lemma 39.23.5 we see $M_1$, resp. $M_1'$ is the spectrum of the $R_1$-invariant functions on $U_1$, resp. the $R'_1$-invariant functions on $U'_1$. Suppose that $p' \in M'$ maps to $p \in M$. Let $M_1$ be the spectrum of the strict henselization of $\mathcal{O}_{M, p}$ with closed point $p_1 \in M_1$. Choose a point $p'_1 \in M'_1$ mapping to $p_1$ and $p'$. From the claim we get

\[ (U'_1, R'_1, s'_1, t'_1, c'_1) = (U_1, R_1, s_1, t_1, c_1) \amalg (U''_1, R''_1, s''_1, t''_1, c''_1) \]

and correspondingly $M'_1 = M_1 \amalg M''_1$ as a scheme over $M_1$. Write $M_1 = \mathop{\mathrm{Spec}}(C_1)$ and write $C_1 = \mathop{\mathrm{colim}}\nolimits C_ i$ as a filtered colimit of étale $C$-algebras. Set $M_ i = \mathop{\mathrm{Spec}}(C_ i)$. The $M_1 = \mathop{\mathrm{lim}}\nolimits M_ i$ and similarly for the other schemes. By Limits, Lemmas 32.4.11 and 32.8.11 we can find an $i$ such that

\[ (U'_ i, R'_ i, s'_ i, t'_ i, c'_ i) = (U_ i, R_ i, s_ i, t_ i, c_ i) \amalg (U''_ i, R''_ i, s''_ i, t''_ i, c''_ i) \]

We conclude that $M'_ i = M_ i \amalg M''_ i$. In particular $M' \to M$ becomes étale at a point over $p'$ after an étale base change. This implies that $M' \to M$ is étale at $p'$ (for example by Morphisms, Lemma 29.36.17). We will prove $U' \cong M' \times _ M U$ after we prove the claim.

Proof of the claim. Observe that $U_ p$ and $U'_{p'}$ have finitely many points. For $u \in U_ p$ we have $\kappa (u)/\kappa (p)$ is algebraic, hence $\kappa (u)$ is separably closed. As $U' \to U$ is étale, we conclude the morphism $U'_{p'} \to U_ p$ induces isomorphisms on residue field extensions. Let $u' \in U'_{p'}$ with image $u \in U_ p$. By assumption (3) the morphism of scheme theoretic fibres $(s')^{-1}(u') \to s^{-1}(u)$, $(t')^{-1}(u') \to t^{-1}(u)$, and $G'_{u'} \to G_ u$ are isomorphisms. Observing that $U_ p = t(s^{-1}(u))$ (set theoretically) we conclude that the points of $U'_{p'}$ surject onto the points of $U_ p$. Suppose that $u'_1$ and $u'_2$ are points of $U'_{p'}$ mapping to the same point $u$ of $U_ p$. Then there exists a point $r' \in R'_{p'}$ with $s'(r') = u'_1$ and $t'(r') = u'_2$. Consider the two towers of fields

\[ \kappa (r')/\kappa (u'_1)/\kappa (u)/\kappa (p) \quad \kappa (r')/\kappa (u'_2)/\kappa (u)/\kappa (p) \]

whose “ends” are the same as the two “ends” of the two towers

\[ \kappa (r')/\kappa (u'_1)/\kappa (p')/\kappa (p) \quad \kappa (r')/\kappa (u'_2)/\kappa (p')/\kappa (p) \]

These two induce the same maps $\kappa (p') \to \kappa (r')$ as $(U'_{p'}, R'_{p'}, s', t', c')$ is a groupoid over $p'$. Since $\kappa (u)/\kappa (p)$ is purely inseparable, we conclude that the two induced maps $\kappa (u) \to \kappa (r')$ are the same. Therefore $r'$ maps to a point of the fibre $G_ u$. By assumption (3) we conclude that $r' \in (G')_{u'_1}$. Namely, we may think of $G$ as a closed subscheme of $R$ viewed as a scheme over $U$ via $s$ and use that the base change to $U'$ gives $G' \subset R'$. In particular we have $u'_1 = u'_2$. We conclude that $U'_{p'} \to U_ p$ is a bijective map on points inducing isomorphisms on residue fields. It follows that $U'_{p'}$ is a finite set of closed points (Algebra, Lemma 10.35.9) and hence $U'_{p'}$ is closed in $U'$. Let $J' \subset A'$ be the radical ideal cutting out $U'_{p'}$ set theoretically.

Second part proof of the claim. Let $\mathfrak m \subset C$ be the maximal ideal. Observe that $(A, \mathfrak m A)$ is a henselian pair by More on Algebra, Lemma 15.11.8. Let $J = \sqrt{\mathfrak m A}$. Then $(A, J)$ is a henselian pair (More on Algebra, Lemma 15.11.7) and the étale ring map $A \to A'$ induces an isomorphism $A/J \to A'/J'$ by our deliberations above. We conclude that $A' = A \times A''$ by More on Algebra, Lemma 15.11.6. Consider the corresponding disjoint union decomposition $U' = U \amalg U''$. The open $(s')^{-1}(U)$ is the set of points of $R'$ specializing to a point of $R'_{p'}$. Similarly for $(t')^{-1}(U)$. Similarly we have $(s')^{-1}(U'') = (t')^{-1}(U'')$ as this is the set of points which do not specialize to $R'_{p'}$. Hence we obtain a disjoint union decomposition

\[ (U', R', s', t', c') = (U, R, s, t, c) \amalg (U'', R'', s'', t'', c'') \]

This immediately gives $M' = M \amalg M''$ and the proof of the claim is complete.

We still have to prove that the canonical map $U' \to M' \times _ M U$ is an isomorphism. It is an étale morphism (Morphisms, Lemma 29.36.18). On the other hand, by base changing to strictly henselian local rings (as in the third paragraph of the proof) and using the bijectivity $U'_{p'} \to U_ p$ established in the course of the proof of the claim, we see that $U' \to M' \times _ M U$ is universally bijective (some details omitted). However, a universally bijective étale morphism is an isomorphism (Descent, Lemma 35.25.2) and the proof is complete. $\square$

Comments (0)

There are also:

  • 6 comment(s) on Section 39.23: Finite flat groupoids, affine case

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DT9. Beware of the difference between the letter 'O' and the digit '0'.