The Stacks project

Proof. By Lemma 10.138.14 we can extend the diagram to a commutative diagram

with $R_0 \to S_0$ smooth, $R_0$ of finite type over $\mathbf{Z}$, and $S = S_0 \otimes _{R_0} R$. Let $x_1, \ldots , x_ n \in S_0$ be generators of $S_0$ over $R_0$. Let $a_1, \ldots , a_ n$ be elements of $A$ which map to the same elements in $A/I$ as the elements $x_1, \ldots , x_ n$. Denote $A_0 \subset A$ the subring generated by the image of $R_0$ and the elements $a_1, \ldots , a_ n$. Set $I_0 = A_0 \cap I$. Then $A_0/I_0 \subset A/I$ and $S_0 \to A/I$ maps into $A_0/I_0$. Thus it suffices to find the dotted arrow in the diagram

The ring $A_0$ is of finite type over $\mathbf{Z}$ by construction. Hence $A_0$ is Noetherian, whence $I_0$ is nilpotent, see Lemma 10.32.5. Say $I_0^ n = 0$. By Proposition 10.138.13 we can successively lift the $R_0$-algebra map $S_0 \to A_0/I_0$ to $S_0 \to A_0/I_0^2$, $S_0 \to A_0/I_0^3$, $\ldots$, and finally $S_0 \to A_0/I_0^ n = A_0$. $\square$