Lemma 10.138.14. Let $R \to S$ be a smooth ring map. Then there exists a subring $R_0 \subset R$ of finite type over $\mathbf{Z}$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$.
Proof. We are going to use that smooth is equivalent to finite presentation and formally smooth, see Proposition 10.138.13. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and denote $I = (f_1, \ldots , f_ m)$. Choose a right inverse $\sigma : S \to R[x_1, \ldots , x_ n]/I^2$ to the projection to $S$ as in Lemma 10.138.5. Choose $h_ i \in R[x_1, \ldots , x_ n]$ such that $\sigma (x_ i \bmod I) = h_ i \bmod I^2$. The fact that $\sigma $ is an $R$-algebra homomorphism $R[x_1, \ldots , x_ n]/I \to R[x_1, \ldots , x_ n]/I^2$ is equivalent to the condition that
\[ f_ j(h_1, \ldots , h_ n) = \sum \nolimits _{j_1 j_2} a_{j_1 j_2} f_{j_1} f_{j_2} \]
for certain $a_{kl} \in R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be the subring generated over $\mathbf{Z}$ by all the coefficients of the polynomials $f_ j, h_ i, a_{kl}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, with $I_0 = (f_1, \ldots , f_ m)$. Let $\sigma _0 : S_0 \to R_0[x_1, \ldots , x_ n]/I_0^2$ defined by the rule $x_ i \mapsto h_ i \bmod I_0^2$; this works since the $a_{lk}$ are defined over $R_0$ and satisfy the same relations. Thus by Lemma 10.138.5 the ring $S_0$ is formally smooth over $R_0$. $\square$
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