Proposition 39.23.9. Let S be a scheme. Let (U, R, s, t, c) be a groupoid scheme over S. Assume
U = \mathop{\mathrm{Spec}}(A), and R = \mathop{\mathrm{Spec}}(B) are affine,
s, t : R \to U finite locally free, and
j = (t, s) is an equivalence.
In this case, let C \subset A be as in (39.23.0.1). Then U \to M = \mathop{\mathrm{Spec}}(C) is finite locally free and R = U \times _ M U. Moreover, M represents the quotient sheaf U/R in the fppf topology (see Definition 39.20.1).
Proof.
During this proof we use the notation s, t : A \to B instead of the notation s^\sharp , t^\sharp . By Lemma 39.20.3 it suffices to show that C \to A is finite locally free and that the map
t \otimes s : A \otimes _ C A \longrightarrow B
is an isomorphism. First, note that j is a monomorphism, and also finite (since already s and t are finite). Hence we see that j is a closed immersion by Morphisms, Lemma 29.44.15. Hence A \otimes _ C A \to B is surjective.
We will perform base change by flat ring maps C \to C' as in Lemma 39.23.5, and we will use that formation of invariants commutes with flat base change, see part (3) of the lemma cited. We will show below that for every prime \mathfrak p \subset C, there exists a local flat ring map C_{\mathfrak p} \to C_{\mathfrak p}' such that the result holds after a base change to C_{\mathfrak p}'. This implies immediately that A \otimes _ C A \to B is injective (use Algebra, Lemma 10.23.1). It also implies that C \to A is flat, by combining Algebra, Lemmas 10.39.17, 10.39.18, and 10.39.8. Then since U \to \mathop{\mathrm{Spec}}(C) is surjective also (Lemma 39.23.6) we conclude that C \to A is faithfully flat. Then the isomorphism B \cong A \otimes _ C A implies that A is a finitely presented C-module, see Algebra, Lemma 10.83.2. Hence A is finite locally free over C, see Algebra, Lemma 10.78.2.
By Lemma 39.23.3 we know that A is a finite product of rings A_ r and B is a finite product of rings B_ r such that the groupoid scheme decomposes accordingly (see the proof of Lemma 39.23.4). Then also C is a product of rings C_ r and correspondingly C' decomposes as a product. Hence we may and do assume that the ring maps s, t : A \to B are finite locally free of a fixed rank r.
The local ring maps C_{\mathfrak p} \to C_{\mathfrak p}' we are going to use are any local flat ring maps such that the residue field of C_{\mathfrak p}' is infinite. By Algebra, Lemma 10.159.1 such local ring maps exist.
Assume C is a local ring with maximal ideal \mathfrak m and infinite residue field, and assume that s, t : A \to B is finite locally free of constant rank r > 0. Since C \subset A is integral (Lemma 39.23.4) all primes lying over \mathfrak m are maximal, and all maximal ideals of A lie over \mathfrak m. Similarly for C \subset B. Pick a maximal ideal \mathfrak m' of A lying over \mathfrak m (exists by Lemma 39.23.6). Since t : A \to B is finite locally free there exist at most finitely many maximal ideals of B lying over \mathfrak m'. Hence we conclude (by Lemma 39.23.6 again) that A has finitely many maximal ideals, i.e., A is semi-local. This in turn implies that B is semi-local as well. OK, and now, because t \otimes s : A \otimes _ C A \to B is surjective, we can apply Algebra, Lemma 10.78.8 to the ring map C \to A, the A-module M = B (seen as an A-module via t) and the C-submodule s(A) \subset B. This lemma implies that there exist x_1, \ldots , x_ r \in A such that M is free over A on the basis s(x_1), \ldots , s(x_ r). Hence we conclude that C \to A is finite free and B \cong A \otimes _ C A by applying Lemma 39.23.8.
\square
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