## 10.128 Openness of the flat locus

Lemma 10.128.1. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $f_1, \ldots , f_ i$ be elements of $S$. Assume that $S$ is Cohen-Macaulay and equidimensional of dimension $d$, and that $\dim V(f_1, \ldots , f_ i) \leq d - i$. Then equality holds and $f_1, \ldots , f_ i$ forms a regular sequence in $S_{\mathfrak q}$ for every prime $\mathfrak q$ of $V(f_1, \ldots , f_ i)$.

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 10.113.7. By Proposition 10.102.4 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 $d - i$. $\square$

Lemma 10.128.2. Suppose that $R \to S$ is a ring map which is finite type, flat. Let $d$ be an integer such that all fibres $S \otimes _ R \kappa (\mathfrak p)$ are Cohen-Macaulay and equidimensional of dimension $d$. Let $f_1, \ldots , f_ i$ be elements of $S$. The set

$\{ \mathfrak q \in V(f_1, \ldots , f_ i) \mid f_1, \ldots , f_ i \text{ are a regular sequence in } S_{\mathfrak q}/\mathfrak p S_{\mathfrak q} \text{ where }\mathfrak p = R \cap \mathfrak q \}$

is open in $V(f_1, \ldots , f_ i)$.

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 10.124.1. 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 10.115.3. Since $f_1, \ldots , f_ i$ form a regular sequence in the Noetherian local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ general dimension theory tells us that $\dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) - i$. By the same Lemma 10.115.3 we then 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 10.124.6 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 10.128.1. $\square$

Lemma 10.128.3. Let $R \to S$ is a ring map. Consider a finite homological complex of finite free $S$-modules:

$F_{\bullet } : 0 \to S^{n_ e} \xrightarrow {\varphi _ e} S^{n_{e-1}} \xrightarrow {\varphi _{e-1}} \ldots \xrightarrow {\varphi _{i + 1}} S^{n_ i} \xrightarrow {\varphi _ i} S^{n_{i-1}} \xrightarrow {\varphi _{i-1}} \ldots \xrightarrow {\varphi _1} S^{n_0}$

For every prime $\mathfrak q$ of $S$ consider the complex $\overline{F}_{\bullet , \mathfrak q} = F_{\bullet , \mathfrak q} \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p$ is inverse image of $\mathfrak q$ in $R$. Assume there exists an integer $d$ such that $R \to S$ is finite type, flat with fibres $S \otimes _ R \kappa (\mathfrak p)$ Cohen-Macaulay of dimension $d$. The set

$\{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid \overline{F}_{\bullet , \mathfrak q}\text{ is exact}\}$

is open in $\mathop{\mathrm{Spec}}(S)$.

Proof. Let $\mathfrak q$ be an element of the set defined in the lemma. We are going to use Proposition 10.101.9 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 10.101.9 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 10.98.5 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 10.101.9 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 10.101.9 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 10.101.9 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 10.128.2. $\square$

Theorem 10.128.4. Let $R$ be a ring. Let $R \to S$ be a ring map of finite presentation. Let $M$ be a finitely presented $S$-module. The set

$\{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid M_{\mathfrak q}\text{ is flat over }R\}$

is open in $\mathop{\mathrm{Spec}}(S)$.

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 10.126.18. 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

$((R_\lambda )_{\mathfrak p_\lambda }, (S_\lambda )_{\mathfrak q_\lambda }, (M_\lambda )_{\mathfrak q_{\lambda }})$

is a system as in Lemma 10.126.13. Hence by Lemma 10.127.3 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 10.126.18 we have $M = M_\lambda \otimes _{R_\lambda } R$ and hence also $M_ g = (M_\lambda )_ g \otimes _{R_\lambda } R$. Thus by Lemma 10.38.7 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 10.70.1. 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 10.38.13. In addition, the sequence

$0 \to K_ n/\mathfrak p K_ n \to F_{n-1}/ \mathfrak p F_{n-1} \to \ldots \to F_0 / \mathfrak p F_0 \to M/\mathfrak p M \to 0$

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 10.98.4 $(K_ n)_{\mathfrak q}$ 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 10.128.3 there exists a further localization $S_ g$ such that the complex

$0 \to K_ n \to F_{n-1} \to \ldots \to F_0$

is exact on all fibres of $R \to S$. By Lemma 10.98.5 this implies that the cokernel of $F_1 \to F_0$ is flat. This proves the theorem in the Noetherian case. $\square$

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