The Stacks project

Proof. We are going to prove this by reducing to the Noetherian case. By openness of flatness (see Theorem 37.15.1) we may assume, after replacing $X$ by an open neighbourhood of $x$, that $X \to S$ is flat. We may also assume that $X$ and $S$ are affine. After possible shrinking $X$ a bit we may assume that there exists an $h \in \Gamma (X, \mathcal{O}_ X)$ which maps to our given $h$.

We may write $S = \mathop{\mathrm{Spec}}(A)$ and we may write $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ as a directed colimit of finite type $\mathbf{Z}$ algebras. Then by Algebra, Lemma 10.168.1 or Limits, Lemmas 32.10.1, 32.8.2, and 32.10.1 we can find a cartesian diagram

with $f_0$ flat and of finite presentation, $X_0$ affine, and $S_0$ affine and Noetherian. Let $x_0 \in X_0$, resp. $s_0 \in S_0$ be the image of $x$, resp. $s$. We may also assume there exists an element $h_0 \in \Gamma (X_0, \mathcal{O}_{X_0})$ which restricts to $h$ on $X$. (If you used the algebra reference above then this is clear; if you used the references to the chapter on limits then this follows from Limits, Lemma 32.10.1 by thinking of $h$ as a morphism $X \to \mathbf{A}^1_ S$.) Note that $\mathcal{O}_{X_ s, x}$ is a localization of $\mathcal{O}_{(X_0)_{s_0}, x_0} \otimes _{\kappa (s_0)} \kappa (s)$, so that $\mathcal{O}_{(X_0)_{s_0}, x_0} \to \mathcal{O}_{X_ s, x}$ is a flat local ring map, in particular faithfully flat. Hence the image $\overline{h}_0 \in \mathcal{O}_{(X_0)_{s_0}, x_0}$ is contained in $\mathfrak m_{(X_0)_{s_0}, x_0}$ and is a nonzerodivisor. We claim that after replacing $X_0$ by a principal open neighbourhood of $x_0$ the element $h_0$ is a nonzerodivisor in $B_0 = \Gamma (X_0, \mathcal{O}_{X_0})$ such that $B_0/h_0B_0$ is flat over $A_0 = \Gamma (S_0, \mathcal{O}_{S_0})$. If so then

is a short exact sequence of flat $A_0$-modules. Hence this remains exact on tensoring with $A$ (by Algebra, Lemma 10.39.12) and the lemma follows.

It remains to prove the claim above. The corresponding algebra statement is the following (we drop the subscript ${}_0$ here): Let $A \to B$ be a flat, finite type ring map of Noetherian rings. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. Assume $h \in \mathfrak q$ maps to a nonzerodivisor in $B_{\mathfrak q}/\mathfrak p B_{\mathfrak q}$. Goal: show that after possible replacing $B$ by $B_ g$ for some $g \in B$, $g \not\in \mathfrak q$ the element $h$ becomes a nonzerodivisor and $B/hB$ becomes flat over $A$. By Algebra, Lemma 10.99.2 we see that $h$ is a nonzerodivisor in $B_{\mathfrak q}$ and that $B_{\mathfrak q}/hB_{\mathfrak q}$ is flat over $A$. By openness of flatness, see Algebra, Theorem 10.129.4 or Theorem 37.15.1 we see that $B/hB$ is flat over $A$ after replacing $B$ by $B_ g$ for some $g \in B$, $g \not\in \mathfrak q$. Finally, let $I = \{ b \in B \mid hb = 0\} $ be the annihilator of $h$. Then $IB_{\mathfrak q} = 0$ as $h$ is a nonzerodivisor in $B_{\mathfrak q}$. Also $I$ is finitely generated as $B$ is Noetherian. Hence there exists a $g \in B$, $g \not\in \mathfrak q$ such that $IB_ g = 0$. After replacing $B$ by $B_ g$ we see that $h$ is a nonzerodivisor. $\square$

Proof. (Our conventions on regular sequences imply that $h_ i \in \mathfrak m_ x$ for each $i$.) The case $r = 1$ follows from Lemma 37.23.1 combined with Divisors, Lemma 31.18.1 to see that $V(h_1)$ remains an effective Cartier divisor after base change. The case $r > 1$ follows from a straightforward induction on $r$ (applying the result for $r = 1$ exactly $r$ times; details omitted).

Another way to prove the lemma is using the material from Divisors, Section 31.22. Namely, first by openness of flatness (see Theorem 37.15.1) we may assume, after replacing $X$ by an open neighbourhood of $x$, that $X \to S$ is flat. We may also assume that $X$ and $S$ are affine. After possible shrinking $X$ a bit we may assume that we have $h_1, \ldots , h_ r \in \Gamma (X, \mathcal{O}_ X)$. Set $Z = V(h_1, \ldots , h_ r)$. Note that $X_ s$ is a Noetherian scheme (because it is an algebraic $\kappa (s)$-scheme, see Varieties, Section 33.20) and that the topology on $X_ s$ is induced from the topology on $X$ (see Schemes, Lemma 26.18.5). Hence after shrinking $X$ a bit more we may assume that $Z_ s \subset X_ s$ is a regular immersion cut out by the $r$ elements $h_ i|_{X_ s}$, see Divisors, Lemma 31.20.8 and its proof. It is also clear that $r = \dim _ x(X_ s) - \dim _ x(Z_ s)$ because

the first two equalities by Algebra, Lemma 10.116.3 and the second by $r$ times applying Algebra, Lemma 10.60.13. Hence Divisors, Lemma 31.22.7 part (3) applies to show that (after Zariski shrinking $X$) the morphism $Z \to X$ is a regular immersion to which Divisors, Lemma 31.22.4 applies (which gives the flatness and the statement on base change). $\square$

Proof. Pick any $h \in \mathfrak m_ x \subset \mathcal{O}_{X, x}$ which maps to a nonzerodivisor in $\mathcal{O}_{X_ s, x}$ and apply Lemma 37.23.1. $\square$

Proof. We may and do replace $S$ by an affine open neighbourhood of $s$. We will prove the lemma for affine $S$ by induction on $d = \dim _ x(X_ s)$.

The case $d = 0$. In this case we show that we may take $Z$ to be an open neighbourhood of $x$. (Note that an open immersion is a regular immersion.) Namely, if $d = 0$, then $X \to S$ is quasi-finite at $x$, see Morphisms, Lemma 29.29.5. Hence there exists an affine open neighbourhood $U \subset X$ such that $U \to S$ is quasi-finite, see Morphisms, Lemma 29.56.2. Thus after replacing $X$ by $U$ we see that the fibre $X_ s$ is a finite discrete set. Hence after replacing $X$ by a further affine open neighbourhood of $X$ we see that $f^{-1}(\{ s\} ) = \{ x\} $ (because the topology on $X_ s$ is induced from the topology on $X$, see Schemes, Lemma 26.18.5). This proves the lemma in this case.

Next, assume $d > 0$. Note that because $x$ is a closed point of its fibre the extension $\kappa (x)/\kappa (s)$ is finite (by the Hilbert Nullstellensatz, see Morphisms, Lemma 29.20.3). Thus we see

the first equality as $\mathcal{O}_{X_ s, x}$ is Cohen-Macaulay and the second by Morphisms, Lemma 29.28.1. Thus we may apply Lemma 37.23.3 to find a diagram

with $x \in D$. Note that $\mathcal{O}_{D_ s, x} = \mathcal{O}_{X_ s, x}/(\overline{h})$ for some nonzerodivisor $\overline{h}$, see Divisors, Lemma 31.18.1. Hence $\mathcal{O}_{D_ s, x}$ is Cohen-Macaulay of dimension one less than the dimension of $\mathcal{O}_{X_ s, x}$, see Algebra, Lemma 10.104.2 for example. Thus the morphism $D \to S$ is flat, locally of finite presentation, and Cohen-Macaulay at $x$ with $\dim _ x(D_ s) = \dim _ x(X_ s) - 1 = d - 1$. By induction hypothesis we can find a regular immersion $Z \to D$ having properties (a), (b), (c). As $Z \to D \to U$ are both regular immersions, we see that also $Z \to U$ is a regular immersion by Divisors, Lemma 31.21.7. This finishes the proof. $\square$

Proof. The fibre $X_ s$ is not empty by assumption. Hence there exists a closed point $x \in X_ s$ where $f$ is Cohen-Macaulay, see Lemma 37.22.7. Apply Lemma 37.23.4 and set $S' = S$. $\square$

Proof. For every $s \in S$ there exists an $i \in I$ such that $s$ is in the image of $S_ i \to S$. By Lemma 37.23.5 we can find a morphism $g_ s : T_ s \to S$ such that $s \in g_ s(T_ s)$ which is flat, locally of finite presentation and locally quasi-finite and such that $g_ s$ factors through $S_ i \to S$. Hence $\{ T_ s \to S\} $ is the desired covering of $S$ that refines $\mathcal{U}$. $\square$