The Stacks project

Proof. By assumption we may choose an isomorphism $\Phi : (R[x_1, \ldots , x_ n]/I)_{\mathfrak q} \to S$ where $I \subset R[x_1, \ldots , x_ n]$ is a finitely generated ideal, and $\mathfrak q \subset R[x_1, \ldots , x_ n]/I$ is a prime. (Note that $R \cap \mathfrak q$ is equal to the maximal ideal $\mathfrak m$ of $R$.) We also choose generators $f_1, \ldots , f_ m \in I$ for the ideal $I$. Write $R$ in any way as a colimit $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ over a directed set $(\Lambda , \leq )$, with each $R_\lambda $ local and essentially of finite type over $\mathbf{Z}$. There exists some $\lambda _0 \in \Lambda $ such that $f_ j$ is the image of some $f_{j, \lambda _0} \in R_{\lambda _0}[x_1, \ldots , x_ n]$. For all $\lambda \geq \lambda _0$ denote $f_{j, \lambda } \in R_{\lambda }[x_1, \ldots , x_ n]$ the image of $f_{j, \lambda _0}$. Thus we obtain a system of ring maps

Set $\mathfrak q_\lambda $ the inverse image of $\mathfrak q$. Set $S_\lambda = (R_\lambda [x_1, \ldots , x_ n]/ (f_{1, \lambda }, \ldots , f_{m, \lambda }))_{\mathfrak q_\lambda }$. We leave it to the reader to see that this works. $\square$