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
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