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.
$\square$

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