The Stacks project

Proof. If $S$ is Cohen-Macaulay and equidimensional of dimension $d$, then we have $\dim (S_{\mathfrak m}) = d$ for all maximal ideals $\mathfrak m$ of $S$, see Lemma 10.114.7. By Proposition 10.103.4 we see that for all maximal ideals $\mathfrak m \in V(f_1, \ldots , f_ i)$ the sequence is a regular sequence in $S_{\mathfrak m}$ and the local ring $S_{\mathfrak m}/(f_1, \ldots , f_ i)$ is Cohen-Macaulay of dimension $d - i$. This actually means that $S/(f_1, \ldots , f_ i)$ is Cohen-Macaulay and equidimensional of dimension $d - i$. $\square$

Proof. Write $\overline{S} = S/(f_1, \ldots , f_ i)$. Suppose $\mathfrak q$ is an element of the set defined in the lemma, and $\mathfrak p$ is the corresponding prime of $R$. We will use relative dimension as defined in Definition 10.125.1. First, note that $d = \dim _{\mathfrak q}(S/R) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{trdeg}_{\kappa (\mathfrak p)}\ \kappa (\mathfrak q)$ by Lemma 10.116.3. Since $f_1, \ldots , f_ i$ form a regular sequence in the Noetherian local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ Lemma 10.60.13 tells us that $\dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) - i$. We conclude that $\dim _{\mathfrak q}(\overline{S}/R) = \dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) + \text{trdeg}_{\kappa (\mathfrak p)}\ \kappa (\mathfrak q) = d - i$ by Lemma 10.116.3. By Lemma 10.125.6 we have $\dim _{\mathfrak q'}(\overline{S}/R) \leq d - i$ for all $\mathfrak q' \in V(f_1, \ldots , f_ i) = \mathop{\mathrm{Spec}}(\overline{S})$ in a neighbourhood of $\mathfrak q$. Thus after replacing $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ we may assume that the inequality holds for all $\mathfrak q'$. The result follows from Lemma 10.129.1. $\square$

Proof. Let $\mathfrak q$ be an element of the set defined in the lemma. We are going to use Proposition 10.102.9 to show there exists a $g \in S$, $g \not\in \mathfrak q$ such that $D(g)$ is contained in the set defined in the lemma. In other words, we are going to show that after replacing $S$ by $S_ g$, the set of the lemma is all of $\mathop{\mathrm{Spec}}(S)$. Thus during the proof we will, finitely often, replace $S$ by such a localization. Recall that Proposition 10.102.9 characterizes exactness of complexes in terms of ranks of the maps $\varphi _ i$ and the ideals $I(\varphi _ i)$, in case the ring is local. We first address the rank condition. Set $r_ i = n_ i - n_{i + 1} + \ldots + (-1)^{e - i} n_ e$. Note that $r_ i + r_{i + 1} = n_ i$ and note that $r_ i$ is the expected rank of $\varphi _ i$ (in the exact case).

By Lemma 10.99.5 we see that if $\overline{F}_{\bullet , \mathfrak q}$ is exact, then the localization $F_{\bullet , \mathfrak q}$ is exact. In particular the complex $F_\bullet$ becomes exact after localizing by an element $g \in S$, $g \not\in \mathfrak q$. In this case Proposition 10.102.9 applied to all localizations of $S$ at prime ideals implies that all $(r_ i + 1) \times (r_ i + 1)$-minors of $\varphi _ i$ are zero. Thus we see that the rank of $\varphi _ i$ is at most $r_ i$.

Let $I_ i \subset S$ denote the ideal generated by the $r_ i \times r_ i$-minors of the matrix of $\varphi _ i$. By Proposition 10.102.9 the complex $\overline{F}_{\bullet , \mathfrak q}$ is exact if and only if for every $1 \leq i \leq e$ we have either $(I_ i)_{\mathfrak q} = S_{\mathfrak q}$ or $(I_ i)_{\mathfrak q}$ contains a $S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$-regular sequence of length $i$. Namely, by our choice of $r_ i$ above and by the bound on the ranks of the $\varphi _ i$ this is the only way the conditions of Proposition 10.102.9 can be satisfied.

If $(I_ i)_{\mathfrak q} = S_{\mathfrak q}$, then after localizing $S$ at some element $g \not\in \mathfrak q$ we may assume that $I_ i = S$. Clearly, this is an open condition.

If $(I_ i)_{\mathfrak q} \not= S_{\mathfrak q}$, then we have a sequence $f_1, \ldots , f_ i \in (I_ i)_{\mathfrak q}$ which form a regular sequence in $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$. Note that for any prime $\mathfrak q' \subset S$ such that $(f_1, \ldots , f_ i) \not\subset \mathfrak q'$ we have $(I_ i)_{\mathfrak q'} = S_{\mathfrak q'}$. Thus the result follows from Lemma 10.129.2. $\square$

Proof. Let $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$ be a prime. Let $\mathfrak p \subset R$ be the inverse image of $\mathfrak q$ in $R$. Note that $M_{\mathfrak q}$ is flat over $R$ if and only if it is flat over $R_{\mathfrak p}$. Let us assume that $M_{\mathfrak q}$ is flat over $R$. We claim that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $M_ g$ is flat over $R$.

We first reduce to the case where $R$ and $S$ are of finite type over $\mathbf{Z}$. Choose a directed set $\Lambda$ and a system $(R_\lambda \to S_\lambda , M_\lambda )$ as in Lemma 10.127.18. Set $\mathfrak p_\lambda$ equal to the inverse image of $\mathfrak p$ in $R_\lambda$. Set $\mathfrak q_\lambda$ equal to the inverse image of $\mathfrak q$ in $S_\lambda$. Then the system

is a system as in Lemma 10.127.13. Hence by Lemma 10.128.3 we see that for some $\lambda$ the module $M_\lambda$ is flat over $R_\lambda$ at the prime $\mathfrak q_{\lambda }$. Suppose we can prove our claim for the system $(R_\lambda \to S_\lambda , M_\lambda , \mathfrak q_{\lambda })$. In other words, suppose that we can find a $g \in S_\lambda$, $g \not\in \mathfrak q_\lambda$ such that $(M_\lambda )_ g$ is flat over $R_\lambda$. By Lemma 10.127.18 we have $M = M_\lambda \otimes _{R_\lambda } R$ and hence also $M_ g = (M_\lambda )_ g \otimes _{R_\lambda } R$. Thus by Lemma 10.39.7 we deduce the claim for the system $(R \to S, M, \mathfrak q)$.

At this point we may assume that $R$ and $S$ are of finite type over $\mathbf{Z}$. We may write $S$ as a quotient of a polynomial ring $R[x_1, \ldots , x_ n]$. Of course, we may replace $S$ by $R[x_1, \ldots , x_ n]$ and assume that $S$ is a polynomial ring over $R$. In particular we see that $R \to S$ is flat and all fibres rings $S \otimes _ R \kappa (\mathfrak p)$ have global dimension $n$.

Choose a resolution $F_\bullet$ of $M$ over $S$ with each $F_ i$ finite free, see Lemma 10.71.1. Let $K_ n = \mathop{\mathrm{Ker}}(F_{n-1} \to F_{n-2})$. Note that $(K_ n)_{\mathfrak q}$ is flat over $R$, since each $F_ i$ is flat over $R$ and by assumption on $M$, see Lemma 10.39.13. In addition, the sequence

is exact upon localizing at $\mathfrak q$, because of vanishing of $\text{Tor}_ i^{R_\mathfrak p}(\kappa (\mathfrak p), M_{\mathfrak q})$. Since the global dimension of $S_\mathfrak q/\mathfrak p S_{\mathfrak q}$ is $n$ we conclude that $K_ n / \mathfrak p K_ n$ localized at $\mathfrak q$ is a finite free module over $S_\mathfrak q/\mathfrak p S_{\mathfrak q}$. By Lemma 10.99.4 $(K_ n)_{\mathfrak q}$ is free over $S_{\mathfrak q}$. In particular, there exists a $g \in S$, $g \not\in \mathfrak q$ such that $(K_ n)_ g$ is finite free over $S_ g$.

By Lemma 10.129.3 there exists a further localization $S_ g$ such that the complex

is exact on all fibres of $R \to S$. By Lemma 10.99.5 this implies that the cokernel of $F_1 \to F_0$ is flat. This proves the theorem in the Noetherian case. $\square$