The Stacks project

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

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

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

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

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

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. $\square$