The Stacks project

Proof. By Lemma 10.69.2 part (3) follows from part (1).

Assume $R$ is Noetherian. Let $\mathfrak p = R \cap \mathfrak q'$. By Lemma 10.135.4 for example we see that $f_1, \ldots , f_ c$ form a regular sequence in the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'} \otimes _ R \kappa (\mathfrak p)$. Moreover, the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'}$ is flat over $R_{\mathfrak p}$. Since $R$, and hence $R[x_1, \ldots , x_ n]_{\mathfrak q'}$ is Noetherian we see from Lemma 10.99.3 that (1) and (2) hold.

Let $R$ be general. Write $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda$ as the filtered colimit of finite type $\mathbf{Z}$-subalgebras (compare with Section 10.127). We may assume that $f_1, \ldots , f_ c \in R_\lambda [x_1, \ldots , x_ n]$ for all $\lambda$. Let $R_0 \subset R$ be as in Lemma 10.136.11. Then we may assume $R_0 \subset R_\lambda$ for all $\lambda$. It follows that $S_\lambda = R_\lambda [x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection (as base change of $S_0$ via $R_0 \to R_\lambda$, see Lemma 10.136.9). Denote $\mathfrak p_\lambda$, $\mathfrak q_\lambda$, $\mathfrak q'_\lambda$ the prime of $R_\lambda$, $S_\lambda$, $R_\lambda [x_1, \ldots , x_ n]$ induced by $\mathfrak p$, $\mathfrak q$, $\mathfrak q'$. With this notation, we have (1) and (2) for each $\lambda$. Since

we deduce flatness in (2) over $R$ from Lemma 10.39.6. Since we have

and since filtered colimits are exact (Lemma 10.8.8) we conclude that we have (1). $\square$