Lemma 10.136.13. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection (Definition 10.136.5). For every prime $\mathfrak q$ of $S$, let $\mathfrak q'$ denote the corresponding prime of $R[x_1, \ldots , x_ n]$. Then

$f_1, \ldots , f_ c$ is a regular sequence in the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'}$,

each of the rings $R[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ i)$ is flat over $R$, and

the $S$-module $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free with basis given by the elements $f_ i \bmod (f_1, \ldots , f_ c)^2$.

**Proof.**
First, by Lemma 10.69.2, part (3) follows from part (1). Parts (1) and (2) immediately reduce to the Noetherian case by Lemma 10.136.12 (some minor details omitted). 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 may apply Lemma 10.99.3 to conclude.
$\square$

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