Lemma 10.129.3. Let $R \to S$ be 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 $R$ is Noetherian and 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 10.102.9 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. $\square$

Comment #6259 by Laurent Moret-Bailly on

Typo in statement: "Let $R\to S$ is a ring map". Also, shouldn't there be a noetherian hypothesis? There is one, at least, in Lemma 00MI which is used in the proof.

Comment #6261 by on

Say you have a ring $S$, an ideal $I$ and $f_1, \ldots, f_r \in I$. Suppose you have a subset $E \subset \text{Spec}(S)$. If $E$ contains every $\mathfrak q \subset S$ with $I_\mathfrak q = S_\mathfrak q$ and $V(f_1, \ldots, f_r) \cap E$ is open in $V(f_1, \ldots, f_r)$ then $E$ is open. Namely, write $V(f_1, \ldots, f_r) \cap E = V(f_1, \ldots, f_r) \cap U$ for some open $U \subset S$. Then $E = (\text{Spec}(S) \setminus V(I)) \cup U$ is the union of two opens and hence open. Does this help?

Comment #6262 by on

@#6259. Argh, yes! I have fixed this here. Thanks very much. (If you replace finite type by finite presentation in the statement, then you can remove the Noetherian assumption. But that would have to go later in the Stacks project and does not seem very useful.)

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