The Stacks project

Proof. Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is a relative global complete intersection. By Lemma 10.136.11 there exists a finite type $\mathbf{Z}$-algebra $R$, a ring map $R \to A \otimes _{A_0} B_0$, a relative global complete intersection $R \to S$, and an isomorphism

Because $R$ is of finite type (and hence finite presentation) over $\mathbf{Z}$, there exists an $i$ and a map $R \to A_ i \otimes _{A_0} B_0$ lifting the map $R \to A \otimes _{A_0} B_0$, see Lemma 10.127.3. Using the same lemma, there exists an $i' \geq i$ such that $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A \otimes _{A_0} C_0$ comes from a map $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A_{i'} \otimes _{A_0} C_0$. Thus we may assume, after replacing $i$ by $i'$, that the displayed map comes from an $A_ i \otimes _{A_0} B_0$-algebra map

By Lemma 10.168.6 after increasing $i$ this map is an isomorphism. This finishes the proof in this case because the base change of a relative global complete intersection is a relative global complete intersection by Lemma 10.136.9.

Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is syntomic. Then there exist elements $g_1, \ldots , g_ m$ in $A \otimes _{A_0} C_0$ generating the unit ideal such that $A \otimes _{A_0} B_0 \to (A \otimes _{A_0} C_0)_{g_ j}$ is a relative global complete intersection, see Lemma 10.136.15. We can find an $i$ and elements $g_{i, j} \in A_ i \otimes _{A_0} C_0$ mapping to $g_ j$. After increasing $i$ we may assume $g_{i, 1}, \ldots , g_{i, m}$ generate the unit ideal of $A_ i \otimes _{A_0} C_0$. The result of the previous paragraph implies that, after increasing $i$, we may assume the maps $A_ i \otimes _{A_0} B_0 \to (A_ i \otimes _{A_0} C_0)_{g_{i, j}}$ are relative global complete intersections. Then $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is syntomic by Lemma 10.136.4 (and the already used Lemma 10.136.15). $\square$