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

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

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

3. 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. 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 10.136.12. 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 10.136.10). 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

$R[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ i) = \mathop{\mathrm{colim}}\nolimits R_\lambda [x_1, \ldots , x_ n]_{\mathfrak q_\lambda '}/(f_1, \ldots , f_ i)$

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

\begin{align*} R[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ i) \xrightarrow {f_{i + 1}} R[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ i) \\ = \mathop{\mathrm{colim}}\nolimits \left( R_\lambda [x_1, \ldots , x_ n]_{\mathfrak q_\lambda '}/(f_1, \ldots , f_ i) \xrightarrow {f_{i + 1}} R_\lambda [x_1, \ldots , x_ n]_{\mathfrak q_\lambda '}/(f_1, \ldots , f_ i) \right) \end{align*}

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

Comment #3254 by Dario Weißmann on

We could define $\mathfrak{p}=R\cap \mathfrak{q}$. It is clear from the context, but still.

Comment #4721 by comment_bot on

A pedantic comment: I suggest adding "such that every nonempty $R$-fiber of $S$ has dimension $n - c$" to the second sentence of the statement. This would make the statement less ambiguous: theoretically, in that sentence it is not clear whether we are choosing a presentation as in the definition of a "global complete intersection" or whether we are choosing an arbitrary presentation for the $R$-algebra $S$ (knowing that $S$ happens to be a global complete intersection, as witnessed by some other presentation).

The same pedantic comment applies to other statements in this section.

Comment #4811 by on

Tried to improve the wording of these lemmas. See changes here.

Comment #6606 by WhatJiaranEatsTonight on

(2) is equivalent to that $R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$ is flat over $R$. And since flatness is preserved under base change, we can reduce (2) to Noetherian case.

But I don't know how to reduce the case to Noetherian for (1).

Comment #6607 by on

The point is that $R$ will be the filtered union of Noetherian rings for which the result is true. Then you use that if $R = colim_{i \in I} R_i$ is a filtered colimit and if $f_1, \ldots, f_r \in R_0$ for some $0 \in I$ form a regular sequence in each $R_i$ for $i \geq 0$, then $f_1, \ldots, f_r$ form a regular sequence in $R$.

Comment #6851 by on

@#6606: Thanks and I have now added the extra arguments here.

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