The Stacks project

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 $\sigma (x_ i \bmod I) = h_ i \bmod I^2$. Since $x_ i - h_ i \in I$, there exist $b_{ij} \in R[x_1, \ldots , x_ n]$ such that

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

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 $f_ j, h_ i, a_{kl}, b_{ij}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, with $I_0 = (f_1, \ldots , f_ m)$. Since the second displayed equation holds in $R_0[x_1, \ldots , x_ n]$ we can let $\sigma _0 : S_0 \to R_0[x_1, \ldots , x_ n]/I_0^2$ be the $R_0$-algebra map defined by the rule $x_ i \mapsto h_ i \bmod I_0^2$. Since the first displayed equation holds in $R_0[x_1, \ldots , x_ n]$ we see that $\sigma _0$ is a right inverse to the projection $R_0[x_1, \ldots , x_ n] / I_0^2 \to R_0[x_1, \ldots , x_ n] / I_0 = S_0$. Thus by Lemma 10.138.5 the ring $S_0$ is formally smooth over $R_0$. $\square$